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Cutnell & Johnson Physics

Eleventh Edition

DAVID YOUNG SHANE STADLER

Louisiana State University

VICE PRESIDENT AND DIRECTOR, SCIENCE Petra Recter

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iii

About the Authors

DAVID YOUNG received his Ph.D. in experimental condensed matter physics from Florida State University in 1998. He then held a

post-doc position in the Department of Chemistry and the Princeton

Materials Institute at Princeton University before joining the fac-

ulty in the Department of Physics and Astronomy at Louisiana State

University in 2000. His research focuses on the synthesis and char-

acterization of high-quality single crystals of novel electronic and

magnetic materials. The goal of his research group is to understand

the physics of electrons in materials under extreme conditions, i.e.,

at temperatures close to absolute zero, in high magnetic fi elds, and

under high pressure. He is the coauthor of over 200 research publica-

tions that have appeared in peer-reviewed journals, such as Physical Review B, Physical Review Letters, and Nature. Professor Young has taught introductory physics with the Cutnell & Johnson text since he

was a senior undergraduate over 20 years ago. He routinely lectures

to large sections, often in excess of 300 students. To engage such a

large number of students, he uses WileyPLUS, electronic response systems, tutorial-style recitation sessions, and in-class demonstra-

tions. Professor Young has received multiple awards for outstanding

teaching of undergraduates. David enjoys spending his free time with

his family, playing basketball, and working on his house.

I would like to thank my family for their continuous love and support.

—David Young

SH ANE STADLER Shane Stadler earned a Ph.D. in experi- mental condensed matter physics from Tulane University in 1998.

Afterwards, he accepted a National Research Council Postdoctoral

Fellowship with the Naval Research Laboratory in Washington, DC,

where he conducted research on artifi cially structured magnetic ma-

terials. Three years later, he joined the faculty in the Department of

Physics at Southern Illinois University (the home institution of John

Cutnell and Ken Johnson, the original authors of this textbook), be-

fore joining the Department of Physics and Astronomy at Louisiana

State University in 2008. His research group studies novel magnetic

materials for applications in the areas of spintronics and magnetic

cooling.

Over the past fi fteen years, Professor Stadler has taught the full

spectrum of physics courses, from physics for students outside the

sciences, to graduate-level physics courses, such as classical electro-

dynamics. He teaches classes that range from fewer than ten students

to those with enrollments of over 300. His educational interests are

focused on developing teaching tools and methods that apply to both

small and large classes, and which are applicable to emerging teach-

ing strategies, such as “fl ipping the classroom.”

In his spare time, Shane writes science fi ction/thriller novels.

I would like to thank my parents, George and Elissa, for their constant

support and encouragement. —Shane Stadler

C o u rt

es y D

av id

Y o u n g

C o u rt

es y S

h an

e S

ta d le

r

Dear Students and Inst ructors:

Welcome to college ph ysics! To the students:

We know there is a ne gative stigma associate

d with physics, and yo u yourself may har-

bor some trepidation a s you begin this course

. But fear not! We’re h ere to help. Whether y

ou’re worried about yo ur math profi ciency,

understanding the con cepts, or developing yo

ur problem-solving ski lls, the resources avail

able to you are designe d to address all of

these areas and more. Research has shown th

at learning styles vary greatly among student

s. Maybe some of you have a more visual

preference, or auditory preference, or some o

ther preferred learning modality. In any case,

the resources availabl e to you in this course

will satisfy all of these preferences and impro

ve your chance of succ ess. Take a moment to

explore below what th e textbook and

online course have to o ff er. We suspect that, a

s you continue to impr ove throughout the cou

rse, some of that initia l trepidation will be

replaced with exciteme nt.

To start, we have creat ed a new learning med

ium specifi c to this boo k in the form of a comp

rehensive set of LECTURE VIDE OS – one

for every section (259 in all). These animated

lectures (created and n arrated by the authors)

are 2–10 minutes in le ngth, and explain the

basic concepts and lear ning objectives of each

section. They are assig nable within WileyPLU

S and can be paired wi th follow-up ques-

tions that are gradable. In addition to supplem

enting traditional lectu ring, the videos can be

used in a variety of wa ys, including fl ipping

the classroom, a comp lete set of lectures for o

nline courses, and revi ewing for exams. Next

, we have enhanced “T he Physics of …”

examples by increasing the bio-inspired exam

ples by 40%. Although they are of general ins

tructional value, they a re also similar to what

premed students will e ncounter in the Chemical and Ph

ysical Foundations of Biological Systems Passages section of the

MCAT. Finally,

we have introduced new “team problems” in th

e end-of-chapter proble ms that are designed fo

r group problem-solvin g exercises. These

are context-rich proble ms of medium diffi cult

y designed for group c ooperation, but may al

so be tackled by the in dividual student.

One of the great streng ths of this text is the sy

nergistic relationship i t develops between pro

blem solving and conc eptual understand-

ing. For instance, avail able in WileyPLUS are animated

Chalkboard Videos, which cons ist of short (2–3 min) v

ideos demonstrat-

ing step-by-step practi cal solutions to typical

homework problems. Also available are num

erous Guided Online (GO) Tuto rials that

implement a step-by-s tep pedagogical approa

ch, which provides stu dents a low-stakes env

ironment for refi ning t heir problem solving

skills. One of the most important techniques

developed in the text f or solving problems in

volving multiple force s is the free-body

diagram (FBD). Many problems in the force-intensive

chapters, such as chap ters 4 and 18, take adv

antage of the new FBD capabilities

now available online in WileyPLUS, where students can

construct the FBD’s f or a select number of p

roblems and be graded on them.

Finally, ORION, an online adapt ive learning environme

nt, is seamlessly integr ated into WileyPLUS for Cutnell

& Johnson.

The content and functi onality of WileyPLUS, and the a

daptive learning enviro nment of ORION (see below), w

ill provide students wi th

all the resources they n eed to be successful in

the course.

• The Lecture Videos created b y the authors for each

section include questio ns with intelligent feed

back when a student en ters the

wrong answer.

• The multi-step GO Tutorial p roblems created in WileyPLUS a

re designed to provide targeted, intelligent fe

edback.

• The Free-body Diagram vecto r drawing tools provid

e students an easy way to enter answers requi

ring vector drawing, an d also

provide enhanced feed back.

• Chalkboard Video Solutions t ake the students step-b

y-step through the solu tion and the thought pr

ocess of the authors. P roblem-

solving strategies are d iscussed, and common

misconceptions and p otential pitfalls are add

ressed. The students ca n then apply these

techniques to solve sim ilar, but diff erent probl

ems.

All of these features ar e designed to encourag

e students to remain w ithin the WileyPLUS environme

nt, as opposed to pursu ing the

“pay-for solutions” we bsites that short circui

t the learning process. To the students – We s

trongly recommend th at you take this honest

approach to the course . Take full advantage o

f the many features an d learning resources th

at accompany the text and the online con-

tent. Be engaged with the material and push

yourself to work throu gh the exercises. Physi

cs may not be the easie st subject to under-

stand, but with the Wi ley resources at your d

isposal and your hard w ork, you CAN be succ

essful.

We are immensely gra teful to all of you who

have provided feedbac k as we’ve worked on

this new edition, and t o our students who

have taught us how to teach. Thank you for y

our guidance, and keep the feedback coming.

Best wishes for succes s in this course and

wherever your major m ay take you!

Sincerely,

David Young and Shan e Stadler, Louisiana St

ate University

email: dyoun14@gma il.com or stadler.ls

[email protected]

iv

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v

Brief Contents

1 Introduction and Mathematical Concepts 1

2 Kinematics in One Dimension 27

3 Kinematics in Two Dimensions 55

4 Forces and Newton’s Laws of Motion 80

5 Dynamics of Uniform Circular Motion 121

6 Work and Energy 144

7 Impulse and Momentum 175

8 Rotational Kinematics 200

9 Rotational Dynamics 223

10 Simple Harmonic Motion and Elasticity 257

11 Fluids 289

12 Temperature and Heat 326

13 The Transfer of Heat 360

14 The Ideal Gas Law and Kinetic Theory 380

15 Thermodynamics 401

16 Waves and Sound 433

17 The Principle of Linear Superposition and Interference Phenomena 465

18 Electric Forces and Electric Fields 489

19 Electric Potential Energy and the Electric Potential 523

20 Electric Circuits 551

21 Magnetic Forces and Magnetic Fields 590

22 Electromagnetic Induction 625

23 Alternating Current Circuits 661

24 Electromagnetic Waves 684

25 The Reflection of Light: Mirrors 711

26 The Refraction of Light: Lenses and Optical Instruments 733

27 Interference and the Wave Nature of Light 777

28 Special Relativity 808

29 Particles and Waves 832

30 The Nature of the Atom 853

31 Nuclear Physics and Radioactivity 885

32 Ionizing Radiation, Nuclear Energy, and Elementary Particles 911

Contents

1 Introduction and Mathematical Concepts 1

1.1 The Nature of Physics 1 1.2 Units 2 1.3 The Role of Units in Problem Solving 3 1.4 Trigonometry 6 1.5 Scalars and Vectors 8 1.6 Vector Addition and Subtraction 10 1.7 The Components of a Vector 12 1.8 Addition of Vectors by Means of Components 15 Concept Summary 19 Focus on Concepts 19 Problems 21 Additional Problems 24 Concepts and Calculations Problems 25 Team Problems 26

2 Kinematics in One Dimension 27 2.1 Displacement 27 2.2 Speed and Velocity 28 2.3 Acceleration 31 2.4 Equations of Kinematics for Constant

Acceleration 34 2.5 Applications of the Equations of Kinematics 37 2.6 Freely Falling Bodies 41 2.7 Graphical Analysis of Velocity and

Acceleration 45 Concept Summary 47 Focus on Concepts 48 Problems 49 Additional Problems 53 Concepts and Calculations Problems 54 Team Problems 54

3 Kinematics in Two Dimensions 55 3.1 Displacement, Velocity, and Acceleration 55 3.2 Equations of Kinematics in Two Dimensions 56 3.3 Projectile Motion 60 3.4 Relative Velocity 68 Concept Summary 72 Focus on Concepts 73 Problems 74 Additional Problems 77 Concepts and Calculations Problems 78 Team Problems 79

4 Forces and Newton’s Laws of Motion 80 4.1 The Concepts of Force and Mass 80 4.2 Newton’s First Law of Motion 81 4.3 Newton’s Second Law of Motion 83 4.4 The Vector Nature of Newton’s Second Law of Motion 85 4.5 Newton’s Third Law of Motion 86 4.6 Types of Forces: An Overview 88 4.7 The Gravitational Force 88 4.8 The Normal Force 92 4.9 Static and Kinetic Frictional Forces 95 4.10 The Tension Force 101 4.11 Equilibrium Applications of Newton’s Laws

of Motion 102 4.12 Nonequilibrium Applications of Newton’s Laws

of Motion 106 Concept Summary 111 Focus on Concepts 112 Problems 114 Additional Problems 118 Concepts and Calculations Problems 119 Team Problems 120

5 Dynamics of Uniform Circular Motion 121

5.1 Uniform Circular Motion 121 5.2 Centripetal Acceleration 122 5.3 Centripetal Force 125 5.4 Banked Curves 129 5.5 Satellites in Circular Orbits 130 5.6 Apparent Weightlessness and Artificial Gravity 133 5.7 *Vertical Circular Motion 136 Concept Summary 137 Focus on Concepts 138 Problems 139 Additional Problems 141 Concepts and Calculations Problems 142 Team Problems 143

6 Work and Energy 144 6.1 Work Done by a Constant Force 144 6.2 The Work–Energy Theorem and Kinetic Energy 147 6.3 Gravitational Potential Energy 153 6.4 Conservative Versus Nonconservative Forces 155 6.5 The Conservation of Mechanical Energy 157 6.6 Nonconservative Forces and the Work–Energy

Theorem 161

Contents vii

6.7 Power 162 6.8 Other Forms of Energy and the Conservation

of Energy 164 6.9 Work Done by a Variable Force 164 Concept Summary 166 Focus on Concepts 167 Problems 168 Additional Problems 172 Concepts and Calculations Problems 173 Team Problems 174

7 Impulse and Momentum 175 7.1 The Impulse–Momentum Theorem 175 7.2 The Principle of Conservation of Linear

Momentum 179 7.3 Collisions in One Dimension 184 7.4 Collisions in Two Dimensions 189 7.5 Center of Mass 189 Concept Summary 192 Focus on Concepts 193 Problems 194 Additional Problems 197 Concepts and Calculations Problems 198 Team Problems 199

8 Rotational Kinematics 200 8.1 Rotational Motion and Angular Displacement 200 8.2 Angular Velocity and Angular Acceleration 203 8.3 The Equations of Rotational Kinematics 205 8.4 Angular Variables and Tangential Variables 208 8.5 Centripetal Acceleration and Tangential

Acceleration 210 8.6 Rolling Motion 213 8.7 *The Vector Nature of Angular Variables 214 Concept Summary 215 Focus on Concepts 216 Problems 216 Additional Problems 220 Concepts and Calculations Problems 221 Team Problems 222

9 Rotational Dynamics 223 9.1 The Action of Forces and Torques on Rigid

Objects 223 9.2 Rigid Objects in Equilibrium 226 9.3 Center of Gravity 231 9.4 Newton’s Second Law for Rotational Motion About a

Fixed Axis 236 9.5 Rotational Work and Energy 241 9.6 Angular Momentum 244

Concept Summary 246 Focus on Concepts 247 Problems 248 Additional Problems 254 Concepts and Calculations Problems 255 Team Problems 256

10 Simple Harmonic Motion and Elasticity 257

10.1 The Ideal Spring and Simple Harmonic Motion 257 10.2 Simple Harmonic Motion and the Reference Circle 261 10.3 Energy and Simple Harmonic Motion 267 10.4 The Pendulum 270 10.5 Damped Harmonic Motion 273 10.6 Driven Harmonic Motion and Resonance 274 10.7 Elastic Deformation 275 10.8 Stress, Strain, and Hooke’s Law 279 Concept Summary 280 Focus on Concepts 281 Problems 282 Additional Problems 287 Concepts and Calculations Problems 288 Team Problems 288

11 Fluids 289 11.1 Mass Density 289 11.2 Pressure 291 11.3 Pressure and Depth in a Static Fluid 293 11.4 Pressure Gauges 297 11.5 Pascal’s Principle 298 11.6 Archimedes’ Principle 300 11.7 Fluids in Motion 305 11.8 The Equation of Continuity 307 11.9 Bernoulli’s Equation 309 11.10 Applications of Bernoulli’s Equation 311 11.11 *Viscous Flow 314 Concept Summary 317 Focus on Concepts 318 Problems 319 Additional Problems 323 Concepts and Calculations Problems 324 Team Problems 325

12 Temperature and Heat 326 12.1 Common Temperature Scales 326 12.2 The Kelvin Temperature Scale 328 12.3 Thermometers 329 12.4 Linear Thermal Expansion 330 12.5 Volume Thermal Expansion 337 12.6 Heat and Internal Energy 339

12.7 Heat and Temperature Change: Specific Heat Capacity 340

12.8 Heat and Phase Change: Latent Heat 343 12.9 *Equilibrium Between Phases of Matter 347 12.10 *Humidity 350 Concept Summary 352 Focus on Concepts 352 Problems 353 Additional Problems 358 Concepts and Calculations Problems 358 Team Problems 359

13 The Transfer of Heat 360 13.1 Convection 360 13.2 Conduction 363 13.3 Radiation 370 13.4 Applications 373 Concept Summary 375 Focus on Concepts 375 Problems 376 Additional Problems 378 Concepts and Calculations Problems 379 Team Problems 379

14 The Ideal Gas Law and Kinetic Theory 380

14.1 Molecular Mass, the Mole, and Avogadro’s Number 380 14.2 The Ideal Gas Law 383 14.3 Kinetic Theory of Gases 388 14.4 *Diff usion 392 Concept Summary 395 Focus on Concepts 396 Problems 397 Additional Problems 399 Concepts and Calculations Problems 400 Team Problems 400

15 Thermodynamics 401 15.1 Thermodynamic Systems and Their Surroundings 401 15.2 The Zeroth Law of Thermodynamics 402 15.3 The First Law of Thermodynamics 402 15.4 Thermal Processes 404 15.5 Thermal Processes Using an Ideal Gas 408 15.6 Specific Heat Capacities 411 15.7 The Second Law of Thermodynamics 412 15.8 Heat Engines 413 15.9 Carnot’s Principle and the Carnot Engine 414 15.10 Refrigerators, Air Conditioners, and Heat Pumps 417 15.11 Entropy 420 15.12 The Third Law of Thermodynamics 425

Concept Summary 425 Focus on Concepts 426 Problems 427 Additional Problems 431 Concepts and Calculations Problems 432 Team Problems 432

16 Waves and Sound 433 16.1 The Nature of Waves 433 16.2 Periodic Waves 435 16.3 The Speed of a Wave on a String 436 16.4 *The Mathematical Description of a Wave 439 16.5 The Nature of Sound 439 16.6 The Speed of Sound 442 16.7 Sound Intensity 446 16.8 Decibels 448 16.9 The Doppler Eff ect 450 16.10 Applications of Sound in Medicine 454 16.11 *The Sensitivity of the Human Ear 455 Concept Summary 456 Focus on Concepts 457 Problems 458 Additional Problems 463 Concepts and Calculations Problems 464 Team Problems 464

17 The Principle of Linear Superposition and Interference Phenomena 465

17.1 The Principle of Linear Superposition 465 17.2 Constructive and Destructive Interference of

Sound Waves 466 17.3 Diff raction 470 17.4 Beats 473 17.5 Transverse Standing Waves 474 17.6 Longitudinal Standing Waves 478 17.7 *Complex Sound Waves 481 Concept Summary 482 Focus on Concepts 483 Problems 484 Additional Problems 487 Concepts and Calculations Problems 488 Team Problems 488

18 Electric Forces and Electric Fields 489 18.1 The Origin of Electricity 489 18.2 Charged Objects and the Electric Force 490 18.3 Conductors and Insulators 493 18.4 Charging by Contact and by Induction 493 18.5 Coulomb’s Law 495 18.6 The Electric Field 500

viii Contents

Contents ix

18.7 Electric Field Lines 505 18.8 The Electric Field Inside a Conductor: Shielding 508 18.9 Gauss’ Law 510 18.10 *Copiers and Computer Printers 513 Concept Summary 516 Focus on Concepts 516 Problems 517 Additional Problems 521 Concepts and Calculations Problems 521 Team Problems 522

19 Electric Potential Energy and the Electric Potential 523

19.1 Potential Energy 523 19.2 The Electric Potential Diff erence 524 19.3 The Electric Potential Diff erence Created by Point

Charges 530 19.4 Equipotential Surfaces and Their Relation to the

Electric Field 534 19.5 Capacitors and Dielectrics 537 19.6 *Biomedical Applications of Electric Potential

Diff erences 541 Concept Summary 544 Focus on Concepts 544 Problems 546 Additional Problems 548 Concepts and Calculations Problems 549 Team Problems 550

20 Electric Circuits 551 20.1 Electromotive Force and Current 551 20.2 Ohm’s Law 553 20.3 Resistance and Resistivity 554 20.4 Electric Power 557 20.5 Alternating Current 559 20.6 Series Wiring 562 20.7 Parallel Wiring 565 20.8 Circuits Wired Partially in Series and Partially in

Parallel 569 20.9 Internal Resistance 570 20.10 Kirchhoff ’s Rules 571 20.11 The Measurement of Current and Voltage 574 20.12 Capacitors in Series and in Parallel 575 20.13 RC Circuits 577 20.14 Safety and the Physiological Eff ects of Current 579 Concept Summary 580 Focus on Concepts 581 Problems 582 Additional Problems 588 Concepts and Calculations Problems 589 Team Problems 589

21 Magnetic Forces and Magnetic Fields 590

21.1 Magnetic Fields 590 21.2 The Force That a Magnetic Field Exerts on a Moving

Charge 592 21.3 The Motion of a Charged Particle in a Magnetic Field 595 21.4 The Mass Spectrometer 599 21.5 The Force on a Current in a Magnetic Field 600 21.6 The Torque on a Current-Carrying Coil 602 21.7 Magnetic Fields Produced by Currents 605 21.8 Ampère’s Law 612 21.9 Magnetic Materials 613 Concept Summary 616 Focus on Concepts 617 Problems 618 Additional Problems 623 Concepts and Calculations Problems 624 Team Problems 624

22 Electromagnetic Induction 625 22.1 Induced Emf and Induced Current 625 22.2 Motional Emf 627 22.3 Magnetic Flux 631 22.4 Faraday’s Law of Electromagnetic Induction 634 22.5 Lenz’s Law 637 22.6 *Applications of Electromagnetic Induction to the

Reproduction of Sound 640 22.7 The Electric Generator 641 22.8 Mutual Inductance and Self-Inductance 646 22.9 Transformers 649 Concept Summary 652 Focus on Concepts 653 Problems 654 Additional Problems 659 Concepts and Calculations Problems 659 Team Problems 660

23 Alternating Current Circuits 661 23.1 Capacitors and Capacitive Reactance 661 23.2 Inductors and Inductive Reactance 664 23.3 Circuits Containing Resistance, Capacitance, and

Inductance 665 23.4 Resonance in Electric Circuits 670 23.5 Semiconductor Devices 672 Concept Summary 678 Focus on Concepts 679 Problems 680 Additional Problems 681 Concepts and Calculations Problems 682 Team Problems 683

24 Electromagnetic Waves 684 24.1 The Nature of Electromagnetic Waves 684 24.2 The Electromagnetic Spectrum 688 24.3 The Speed of Light 690 24.4 The Energy Carried by Electromagnetic Waves 692 24.5 The Doppler Eff ect and Electromagnetic Waves 695 24.6 Polarization 697 Concept Summary 704 Focus on Concepts 704 Problems 705 Additional Problems 708 Concepts and Calculations Problems 709 Team Problems 710

25 The Reflection of Light: Mirrors 711 25.1 Wave Fronts and Rays 711 25.2 The Reflection of Light 712 25.3 The Formation of Images by a Plane Mirror 713 25.4 Spherical Mirrors 716 25.5 The Formation of Images by Spherical Mirrors 718 25.6 The Mirror Equation and the Magnification

Equation 722 Concept Summary 728 Focus on Concepts 728 Problems 729 Additional Problems 731 Concepts and Calculations Problems 731 Team Problems 732

26 The Refraction of Light: Lenses and Optical Instruments 733

26.1 The Index of Refraction 733 26.2 Snell’s Law and the Refraction of Light 734 26.3 Total Internal Reflection 739 26.4 Polarization and the Reflection and Refraction of

Light 745 26.5 The Dispersion of Light: Prisms and Rainbows 746 26.6 Lenses 748 26.7 The Formation of Images by Lenses 749 26.8 The Thin-Lens Equation and the Magnification

Equation 752 26.9 Lenses in Combination 755 26.10 The Human Eye 756 26.11 Angular Magnification and the Magnifying Glass 761 26.12 The Compound Microscope 763 26.13 The Telescope 764 26.14 Lens Aberrations 765 Concept Summary 767 Focus on Concepts 768 Problems 769

Additional Problems 775 Concepts and Calculations Problems 775 Team Problems 776

27 Interference and the Wave Nature of Light 777

27.1 The Principle of Linear Superposition 777 27.2 Young’s Double-Slit Experiment 779 27.3 Thin-Film Interference 782 27.4 The Michelson Interferometer 786 27.5 Diff raction 787 27.6 Resolving Power 791 27.7 The Diff raction Grating 796 27.8 *Compact Discs, Digital Video Discs, and the Use of

Interference 798 27.9 X-Ray Diff raction 799 Concept Summary 801 Focus on Concepts 802 Problems 803 Additional Problems 805 Concepts and Calculations Problems 806 Team Problems 807

28 Special Relativity 808 28.1 Events and Inertial Reference Frames 808 28.2 The Postulates of Special Relativity 809 28.3 The Relativity of Time: Time Dilation 811 28.4 The Relativity of Length: Length Contraction 815 28.5 Relativistic Momentum 817 28.6 The Equivalence of Mass and Energy 819 28.7 The Relativistic Addition of Velocities 824 Concept Summary 827 Focus on Concepts 827 Problems 828 Additional Problems 830 Concepts and Calculations Problems 831 Team Problems 831

29 Particles and Waves 832 29.1 The Wave–Particle Duality 832 29.2 Blackbody Radiation and Planck’s Constant 833 29.3 Photons and the Photoelectric Eff ect 834 29.4 The Momentum of a Photon and the Compton

Eff ect 840 29.5 The De Broglie Wavelength and the Wave Nature

of Matter 843 29.6 The Heisenberg Uncertainty Principle 845 Concept Summary 849 Focus on Concepts 849 Problems 850

x Contents

Contents xi

Additional Problems 852 Concepts and Calculations Problems 852 Team Problems 852

30 The Nature of the Atom 853 30.1 Rutherford Scattering and the Nuclear Atom 853 30.2 Line Spectra 855 30.3 The Bohr Model of the Hydrogen Atom 857 30.4 De Broglie’s Explanation of Bohr’s Assumption About

Angular Momentum 861 30.5 The Quantum Mechanical Picture of the Hydrogen

Atom 862 30.6 The Pauli Exclusion Principle and the Periodic Table of

the Elements 866 30.7 X-Rays 868 30.8 The Laser 872 30.9 *Medical Applications of the Laser 874 30.10 *Holography 876 Concept Summary 878 Focus on Concepts 879 Problems 880 Additional Problems 883 Concepts and Calculations Problems 883 Team Problems 883

31 Nuclear Physics and Radioactivity 885 31.1 Nuclear Structure 885 31.2 The Strong Nuclear Force and the Stability of the

Nucleus 887 31.3 The Mass Defect of the Nucleus and Nuclear Binding

Energy 888 31.4 Radioactivity 890 31.5 The Neutrino 896 31.6 Radioactive Decay and Activity 897 31.7 Radioactive Dating 900 31.8 Radioactive Decay Series 903 31.9 Radiation Detectors 904

Concept Summary 906 Focus on Concepts 907 Problems 908 Additional Problems 910 Concepts and Calculations Problems 910 Team Problems 910

32 Ionizing Radiation, Nuclear Energy, and Elementary Particles 911

32.1 Biological Eff ects of Ionizing Radiation 911 32.2 Induced Nuclear Reactions 915 32.3 Nuclear Fission 916 32.4 Nuclear Reactors 919 32.5 Nuclear Fusion 920 32.6 Elementary Particles 922 32.7 Cosmology 928 Concept Summary 931 Focus on Concepts 932 Problems 932 Additional Problems 934 Concepts and Calculations Problems 935 Team Problems 935

Appendixes A-1

APPENDIX A Powers of Ten and Scientific Notation A-1 APPENDIX B Significant Figures A-1 APPENDIX C Algebra A-2 APPENDIX D Exponents and Logarithms A-3 APPENDIX E Geometry and Trigonometry A-4 APPENDIX F Selected Isotopes A-5

ANSWERS TO CHECK YOUR UNDERSTANDING A-10

ANSWERS TO ODD-NUMBERED PROBLEMS A-18

INDEX I -1

Note: Chapter sections marked with an asterisk (*) can be omitted with little impact to the overall development of the material.

Our Vision Our goal is to provide students with the skills they need to succeed in this course, and instructors with the tools they need to develop those skills.

Skills Development One of the great strengths of this text is the synergistic relationship

between conceptual understanding, problem solving, and establish-

ing relevance. We identify here some of the core features of the text

that support these synergies.

Conceptual Understanding Students often regard physics as a collection of equations that can be used blindly to solve problems. How-

ever, a good problem-solving technique does not begin with equations. It

starts with a fi rm grasp of physics concepts and how they fi t together to

provide a coherent description of natural phenomena. Helping students

develop a conceptual understanding of physics principles is a primary

goal of this text. The features in the text that work toward this goal are:

• Lecture Videos (one for each section of the text) • Conceptual Examples • Concepts & Calculations problems (now with video solutions) • Focus on Concepts homework material • Check Your Understanding questions • Concept Simulations (an online feature)

Problem Solving The ability to reason in an organized and mathematically correct manner is essential to solving problems, and

helping students to improve their reasoning skills is also one of our

primary goals. To this end, we have included the following features:

• Math Skills boxes for just-in-time delivery of math support • Explicit reasoning steps in all examples • Reasoning Strategies for solving certain classes of problems • Analyzing Multiple-Concept Problems • Video Support and Tutorials (in WileyPLUS)

Physics Demonstration Videos

Video Help

Concept Simulations

• Problem Solving Insights

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Chemical and Physical Foundations of Biological Systems Passages section of the MCAT. All biomedical examples and end-of-chapter

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EXAMPLE 7 BIO The Physics of Hearing Loss— Standing Waves in the Ear

Inner ear

Semicircular canals

Anvil

Hammer

Cochlea

Auditory nerve

Eustachian tubeOval

window Stirrup

Middle ear

Tympanic membrane

Outer ear

Auditory canal

Pinna

2.3 cm

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xvi Our V is ion and the Wi leyPLUS with ORION Advantage

The publishing world is changing rapidly! The digital age is here, and college

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Many of our physics colleagues and their students have generously

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About the cover: The cover image shows an artist’s rendition of a synaptic gap between an axon and a dendrite of a human nerve cell. Just like

the wires in the electrical system of your home, the nerve cells make connec-

tions in circuits called neural pathways. The transmission of chemical signals

between the axon and dendrite relies on the electrical potential diff erence

across the gap, which is a topic in Volume 2 of the text. Our hope is that this

book and its resources will help you develop some new neural pathways of

your own!

In spite of our best eff orts to produce an error-free book, errors no doubt remain. They are solely our responsibility, and we would appreciate hearing of any that

you fi nd. We hope that this text makes learning and teaching physics easier and more enjoyable, and we look forward to hearing about your experiences with

it. Please feel free to write us care of Physics Editor, Global Education, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, or contact the authors

at [email protected] or [email protected].

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

1.1 Describe the fundamental nature of physics.

1.2 Describe diff erent systems of units.

1.3 Solve unit conversion problems.

1.4 Solve trigonometry problems.

1.5 Distinguish between vectors and scalars.

1.6 Solve vector addition and subtraction problems by graphical methods.

1.7 Calculate vector components.

1.8 Solve vector addition and subtraction problems using components.

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P h o to

CHAPTER 1

Introduction and Mathematical Concepts

The animation techniques and special eff ects used in the fi lm The Avengers rely on computers and mathematical concepts such as trigonometry and vectors. Such mathematical concepts will be very useful

throughout this book in our discussion of physics.

1.1 The Nature of Physics Physics is the most basic of the sciences, and it is at the very root of subjects like chem-

istry, engineering, astronomy, and even biology. The discipline of physics has developed

over many centuries, and it continues to evolve. It is a mature science, and its laws en-

compass a wide scope of phenomena that range from the formation of galaxies to the in-

teractions of particles in the nuclei of atoms. Perhaps the most visible evidence of physics

in everyday life is the eruption of new applications that have improved our quality of life,

such as new medical devices, and advances in computers and high-tech communications.

The exciting feature of physics is its capacity for predicting how nature will be-

have in one situation on the basis of experimental data obtained in another situation.

Such predictions place physics at the heart of modern technology and, therefore, can

have a tremendous impact on our lives. Rocketry and the development of space travel

have their roots fi rmly planted in the physical laws of Galileo Galilei (1564–1642) and

Isaac Newton (1642–1727). The transportation industry relies heavily on physics in

the development of engines and the design of aerodynamic vehicles. Entire electronics

and computer industries owe their existence to the invention of the transistor, which

grew directly out of the laws of physics that describe the electrical behavior of solids.

The telecommunications industry depends extensively on electromagnetic waves, 1

2 CHAPTER 1 Introduction and Mathematical Concepts

whose existence was predicted by James Clerk Maxwell (1831–1879) in his theory of electricity and magnetism. The medical profession uses X-ray, ultrasonic, and magnetic resonance methods for obtaining images of the interior of the human body, and physics lies at the core of all these. Perhaps the most widespread impact in modern technology is that due to the laser. Fields ranging from space exploration to medicine benefi t from this incredible device, which is a direct applica- tion of the principles of atomic physics.

Because physics is so fundamental, it is a required course for students in a wide range of major areas. We welcome you to the study of this fascinating topic. You will learn how to see the world through the “eyes” of physics and to reason as a physicist does. In the process, you will learn how to apply physics principles to a wide range of problems. We hope that you will come to recognize that physics has important things to say about your environment.

1.2 Units Physics experiments involve the measurement of a variety of quantities, and a great deal of eff ort goes into making these measurements as accurate and reproducible as possible. The fi rst step toward ensuring accuracy and reproducibility is defi ning the units in which the measurements are made.

In this text, we emphasize the system of units known as SI units, which stands for the French phrase “Le Système International d’Unités.” By international agreement, this system employs the meter (m) as the unit of length, the kilogram (kg) as the unit of mass, and the second (s) as the unit of time. Two other systems of units are also in use, however. The CGS system utilizes the centimeter (cm), the gram (g), and the second for length, mass, and time, respectively, and the BE or British Engineering system (the gravitational version) uses the foot (ft), the slug (sl), and the second. Table 1.1 summarizes the units used for length, mass, and time in the three systems.

Originally, the meter was defi ned in terms of the distance measured along the earth’s surface between the north pole and the equator. Eventually, a more accurate measurement standard was needed, and by international agreement the meter became the distance between two marks on a bar of platinum–iridium alloy (see Figure 1.1) kept at a temperature of 0 °C. Today, to meet further demands for increased accuracy, the meter is defi ned as the distance that light travels in a vacuum in a time of 1/299 792 458 second. This defi nition arises because the speed of light is a universal constant that is defi ned to be 299 792 458 m/s.

The defi nition of a kilogram as a unit of mass has also undergone changes over the years. As Chapter 4 discusses, the mass of an object indicates the tendency of the object to continue in motion with a constant velocity. Originally, the kilogram was expressed in terms of a specifi c amount of water. Today, one kilogram is defi ned to be the mass of a standard cylinder of platinum– iridium alloy, like the one in Figure 1.2.

As with the units for length and mass, the present defi nition of the second as a unit of time is diff erent from the original defi nition. Originally, the second was defi ned according to the average time for the earth to rotate once about its axis, one day being set equal to 86 400 seconds. The earth’s rotational motion was chosen because it is naturally repetitive, occurring over and over again. Today, we still use a naturally occurring repetitive phenomenon to defi ne the second, but of a very diff erent kind. We use the electromagnetic waves emitted by cesium-133 atoms in an atomic clock like that in Figure 1.3. One second is defi ned as the time needed for 9 192 631 770 wave cycles to occur.*

The units for length, mass, and time, along with a few other units that will arise later, are regarded as base SI units. The word “base” refers to the fact that these units are used along with

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FIGURE 1.1 The standard platinum–iridium meter bar.

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FIGURE 1.2 The standard platinum–iridium kilogram is kept at the International Bureau of Weights and Measures in Sèvres, France. This copy of it was assigned to the United States in 1889 and is housed at the National Institute of Standards and Technology.

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FIGURE 1.3 This atomic clock, the NIST-F1, keeps time with an uncertainty of about one second in sixty million years. *See Chapter 16 for a discussion of waves in general and Chapter 24 for a discussion of electromagnetic waves in particular.

TABLE 1.1 Units of Measurement

System SI CGS BE

Length Meter (m) Centimeter (cm) Foot (ft)

Mass Kilogram (kg) Gram (g) Slug (sl)

Time Second (s) Second (s) Second (s)

1.3 The Role of Units in Problem Solving 3

various laws to defi ne additional units for other important physical quantities, such as force and energy. The units for such other physical quantities are referred to as derived units, since they are combinations of the base units. Derived units will be introduced from time to time, as they arise naturally along with the related physical laws.

The value of a quantity in terms of base or derived units is sometimes a very large or very small number. In such cases, it is convenient to introduce larger or smaller units that are related to the normal units by multiples of ten. Table 1.2 summarizes the prefi xes that are used to denote multiples of ten. For example, 1000 or 103 meters are referred to as 1 kilometer (km), and 0.001 or 10 −3 meter is called 1 millimeter (mm). Similarly, 1000 grams and 0.001 gram are referred to as 1 kilogram (kg) and 1 milligram (mg), respectively. Appendix A contains a discussion of scientifi c notation and powers of ten, such as 103 and 10 −3.

1.3 The Role of Units in Problem Solving The Conversion of Units Since any quantity, such as length, can be measured in several diff erent units, it is important to know how to convert from one unit to another. For instance, the foot can be used to express the distance between the two marks on the standard platinum–iridium meter bar. There are 3.281 feet in one meter, and this number can be used to convert from meters to feet, as the following example demonstrates.

TABLE 1.2 Standard Prefixes Used to Denote Multiples of Ten

Prefix Symbol Factora

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deka da 101

deci d 10−1

centi c 10−2

milli m 10−3

micro μ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

aAppendix A contains a discussion of powers of ten and

scientific notation.

EXAMPLE 1 The World’s Highest Waterfall

The highest waterfall in the world is Angel Falls in Venezuela, with a total drop of 979.0 m (see Figure 1.4). Express this drop in feet.

Reasoning When converting between units, we write down the units explicitly in the calculations and treat them like any algebraic quantity. In particular, we will take advantage of the following algebraic fact: Multiplying or dividing an equation by a factor of 1 does not alter an equation.

Solution Since 3.281 feet = 1 meter, it follows that (3.281 feet)/ (1 meter) = 1. Using this factor of 1 to multiply the equation “Length = 979.0 meters,” we fi nd that

Length = (979.0 m)(1) = (979.0 meters) (3.281 feet1 meter ) = 3212 feet The colored lines emphasize that the units of meters behave like any al- gebraic quantity and cancel when the multiplication is performed, leaving only the desired unit of feet to describe the answer. In this regard, note that 3.281 feet = 1 meter also implies that (1 meter)/(3.281 feet) = 1. However, we chose not to multiply by a factor of 1 in this form, because the units of meters would not have canceled.

A calculator gives the answer as 3212.099 feet. Standard proced- ures for signifi cant fi gures, however, indicate that the answer should be rounded off to four signifi cant fi gures, since the value of 979.0 meters is accurate to only four signifi cant fi gures. In this regard, the “1 meter” in the denominator does not limit the signifi cant fi gures of the answer, be- cause this number is precisely one meter by defi nition of the conversion factor. Appendix B contains a review of signifi cant fi gures. ©

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FIGURE 1.4 Angel Falls in Venezuela is the highest waterfall in the world.

Problem-Solving Insight In any conversion, if the units do not combine algebraically to give the desired result, the conversion has not been carried out properly.

4 CHAPTER 1 Introduction and Mathematical Concepts

EXAMPLE 2 Interstate Speed Limit

Express the speed limit of 65 miles/hour in terms of meters/second.

Reasoning As in Example 1, it is important to write down the units explicitly in the calculations and treat them like any algebraic quantity. Here, we take advantage of two well-known relationships—namely, 5280 feet = 1 mile and 3600 seconds = 1 hour. As a result, (5280 feet)/ (1 mile) = 1 and (3600 seconds)/(1 hour) = 1. In our solution we will use the fact that multiplying and dividing by these factors of unity does not alter an equation.

Solution Multiplying and dividing by factors of unity, we fi nd the speed limit in feet per second as shown below:

Speed = (65 mileshour ) (1)(1) =

(65 mileshour ) ( 5280 feet

1 mile ) ( 1 hour

3600 seconds) = 95 feet

second

To convert feet into meters, we use the fact that (1 meter)/(3.281 feet) = 1:

Speed = (95 feetsecond) (1) =

(95 feetsecond) ( 1 meter

3.281 feet) = 29 meters second

In addition to their role in guiding the use of conversion factors, units serve a useful purpose in solving problems. They can provide an internal check to eliminate errors, if they are carried along during each step of a calculation and treated like any algebraic factor.

Problem-Solving Insight In particular, remember that only quantities with the same units can be added or subtracted.

Thus, at one point in a calculation, if you fi nd yourself adding 12 miles to 32 kilometers, stop and reconsider. Either miles must be converted into kilometers or kilometers must be converted into miles before the addition can be carried out.

A collection of useful conversion factors is given on the page facing the inside of the front cover. The reasoning strategy that we have followed in Examples 1 and 2 for converting between units is outlined as follows:

REASONING STRATEGY Converting Between Units 1. In all calculations, write down the units explicitly. 2. Treat all units as algebraic quantities. In particular, when identical units are divided,

they are eliminated algebraically. 3. Use the conversion factors located on the page facing the inside of the front cover. Be

guided by the fact that multiplying or dividing an equation by a factor of 1 does not alter the equation. For instance, the conversion factor of 3.281 feet = 1 meter might be applied in the form (3.281 feet)/(1 meter) = 1. This factor of 1 would be used to multiply an equa- tion such as “Length = 5.00 meters” in order to convert meters to feet.

4. Check to see that your calculations are correct by verifying that the units combine algeb- raically to give the desired unit for the answer. Only quantities with the same units can be added or subtracted.

Sometimes an equation is expressed in a way that requires specifi c units to be used for the variables in the equation. In such cases it is important to understand why only certain units can be used in the equation, as the following example illustrates.

EXAMPLE 3 BIO The Physics of the Body Mass Index

The body mass index (BMI) takes into account your mass in kilograms (kg) and your height in meters (m) and is defi ned as follows:

BMI = Mass in kg

(Height in m)2

However, the BMI is often computed using the weight* of a person in pounds (lb) and his or her height in inches (in.). Thus, the expression for the BMI incorporates these quantities, rather than the mass in kilograms and the height in meters. Starting with the defi nition above, determine the expression for the BMI that uses pounds and inches.

*Weight and mass are different concepts, and the relationship between them will be discussed in Section 4.7.

With this in mind, the next example stresses the importance of writing down the units and illustrates a typical situation in which several conversions are required.

1.3 The Role of Units in Problem Solving 5

Dimensional Analysis We have seen that many quantities are denoted by specifying both a number and a unit. For

example, the distance to the nearest telephone may be 8 meters, or the speed of a car might be

25 meters/second. Each quantity, according to its physical nature, requires a certain type of unit. Distance must be measured in a length unit such as meters, feet, or miles, and a time unit will not

do. Likewise, the speed of an object must be specifi ed as a length unit divided by a time unit. In

physics, the term dimension is used to refer to the physical nature of a quantity and the type of unit used to specify it. Distance has the dimension of length, which is symbolized as [L], while

speed has the dimensions of length [L] divided by time [T], or [L/T]. Many physical quantities

can be expressed in terms of a combination of fundamental dimensions such as length [L], time

[T], and mass [M]. Later on, we will encounter certain other quantities, such as temperature,

which are also fundamental. A fundamental quantity like temperature cannot be expressed as a

combination of the dimensions of length, time, mass, or any other fundamental dimension.

Dimensional analysis is used to check mathematical relations for the consistency of their

dimensions. As an illustration, consider a car that starts from rest and accelerates to a speed υ in a time t. Suppose we wish to calculate the distance x traveled by the car but are not sure whether the correct relation is x = 12 𝜐t

2 or x = 12 𝜐t. We can decide by checking the quantities on both sides of the equals sign to see whether they have the same dimensions. If the dimensions are not the

same, the relation is incorrect. For x = 12 𝜐t 2, we use the dimensions for distance [L], time [T], and

speed [L/T] in the following way:

x = 12 𝜐t 2

Dimensions [L] ≟ [ LT ] [T]2 = [L][T] Dimensions cancel just like algebraic quantities, and pure numerical factors like

1

2 have no dimen-

sions, so they can be ignored. The dimension on the left of the equals sign does not match those

on the right, so the relation x = 12 𝜐t 2 cannot be correct. On the other hand, applying dimensional

analysis to x = 12 𝜐t, we fi nd that

x = 12 𝜐t 2

Dimensions [L] ≟ [ LT ] [T] = [L]

Reasoning We will begin with the BMI defi nition and work separately with the numerator and the denominator. We will determine the mass in

kilograms that appears in the numerator from the weight in pounds by using

the fact that 1 kg corresponds to 2.205 lb. Then, we will determine the height

in meters that appears in the denominator from the height in inches with the

aid of the facts that 1 m = 3.281 ft and 1 ft = 12 in. These conversion factors

are located on the page facing the inside of the front cover of the text.

Solution Since 1 kg corresponds to 2.205 lb, the mass in kilograms can be determined from the weight in pounds in the following way:

Mass in kg = (Weight in lb)( 1 kg

2.205 lb) Since 1 ft = 12 in. and 1 m = 3.281 ft, we have

Height in m = (Height in in.)( 1 ft12 in.)( 1 m

3.281 ft) Substituting these results into the numerator and denominator of the BMI

defi nition gives

BMI = Mass in kg

(Height in m)2 =

(Weight in lb)( 1 kg

2.205 lb) (Height in in.)2 ( 1 ft12 in.)

2

( 1 m3.281 ft) 2

= ( 1 kg

2.205 lb)( 12 in.

1 ft ) 2

(3.281 ft1 m ) 2 (Weight in lb)

(Height in in.)2

BMI = (703.0 kg · in.2

lb · m2 ) (Weight in lb)

(Height in in.)2

For example, if your weight and height are 180 lb and 71 in., your body

mass index is 25 kg/m2. The BMI can be used to assess approximately

whether your weight is normal for your height (see Table 1.3).

TABLE 1.3 The Body Mass Index

BMI (kg/m2) Evaluation Below 18.5 Underweight

18.5–24.9 Normal

25.0–29.9 Overweight

30.0–39.9 Obese

40 and above Morbidly obese

6 CHAPTER 1 Introduction and Mathematical Concepts

Problem-Solving Insight You can check for errors that may have arisen during algebraic manipulations by performing a dimensional analysis on the fi nal expression.

The dimension on the left of the equals sign matches that on the right, so this relation is dimen- sionally correct. If we know that one of our two choices is the right one, then x = 12 𝜐t is it. In the absence of such knowledge, however, dimensional analysis cannot identify the correct relation. It can only identify which choices may be correct, since it does not account for numerical factors like 12 or for the manner in which an equation was derived from physics principles.

Check Your Understanding

(The answers are given at the end of the book.) 1. (a) Is it possible for two quantities to have the same dimensions but diff erent units? (b) Is it possible for two quantities to have the same units but diff erent dimensions? 2. You can always add two numbers that have the same units (such as 6 meters + 3 meters). Can you always add

two numbers that have the same dimensions, such as two numbers that have the dimensions of length [L]?

3. The following table lists four variables, along with their units:

Variable Units x Meters (m) υ Meters per second (m/s) t Seconds (s) a Meters per second squared (m/s2)

These variables appear in the following equations, along with a few numbers that have no units. In which of the equations are the units on the left side of the equals sign consistent with the units on the right side?

(a) x = 𝜐t (d) 𝜐 = at + 12 at 3

(b) x = 𝜐t + 12 at 2 (e) 𝜐 3 = 2ax 2

(c) 𝜐 = at (f) t = √2xa 4. In the equation y = cnat2 you wish to determine the integer value (1, 2, etc.) of the exponent n. The di-

mensions of y, a, and t are known. It is also known that c has no dimensions. Can dimensional analysis be used to determine n?

1.4 Trigonometry Scientists use mathematics to help them describe how the physical universe works, and tri- gonometry is an important branch of mathematics. Three trigonometric functions are utilized throughout this text. They are the sine, the cosine, and the tangent of the angle θ (Greek theta), abbreviated as sin θ, cos θ, and tan θ, respectively. These functions are defi ned below in terms of the symbols given along with the right triangle in Interactive Figure 1.5.

DEFINITION OF SIN θ, COS θ, AND TAN θ

sin θ = ho h

(1.1)

cos θ = ha h

(1.2)

tan θ = ho ha

(1.3)

h = length of the hypotenuse of a right triangle ho = length of the side opposite the angle θ ha = length of the side adjacent to the angle θ

INTERACTIVE FIGURE 1.5 A right triangle.

90°

h = hypotenuse

θ

ho = length of side opposite the angle

ha = length of side adjacent to the angle θ

θ

1.4 Trigonometry 7

The sine, cosine, and tangent of an angle are numbers without units, because each is the ratio of the lengths of two sides of a right triangle. Example 4 illustrates a typical application of Equation 1.3.

EXAMPLE 4 Using Trigonometric Functions

On a sunny day, a tall building casts a shadow that is 67.2 m long. The angle between the sun’s rays and the ground is θ = 50.0°, as Figure 1.6 shows. Determine the height of the building.

Reasoning We want to fi nd the height of the building. Therefore, we begin with the colored right triangle in Figure 1.6 and identify the height as the length ho of the side opposite the angle θ. The length of the shadow is the length ha of the side that is adjacent to the angle θ. The ratio of the length of the opposite side to the length of the adjacent side is the tangent of the angle θ, which can be used to fi nd the height of the building.

Solution We use the tangent function in the following way, with θ = 50.0° and ha = 67.2 m:

tan θ = ho ha

(1.3)

ho = ha tan θ = (67.2 m)(tan 50.0°) = (67.2 m)(1.19) = 80.0 m

The value of tan 50.0° is found by using a calculator. FIGURE 1.6 From a value for the angle θ and the length ha of the

shadow, the height ho of the building can be found using trigonometry.

= 50.0°θ

= 67.2 mha

ho

The sine, cosine, or tangent may be used in calculations such as that in Example 4, depend- ing on which side of the triangle has a known value and which side is asked for.

Problem-Solving Insight However, the choice of which side of the triangle to label ho (opposite) and which to label ha (adjacent) can be made only after the angle θ is identifi ed.

Often the values for two sides of the right triangle in Interactive Figure 1.5 are available, and the value of the angle θ is unknown. The concept of inverse trigonometric functions plays an important role in such situations. Equations 1.4–1.6 give the inverse sine, inverse cosine, and inverse tangent in terms of the symbols used in the drawing. For instance, Equation 1.4 is read as “θ equals the angle whose sine is ho/h.”

θ = sin−1 ( ho h ) (1.4)

θ = cos−1 ( ha h ) (1.5)

θ = tan−1 ( ho ha) (1.6)

The use of −1 as an exponent in Equations 1.4–1.6 does not mean “take the reciprocal.” For instance, tan−1 (ho/ha) does not equal 1/tan (ho/ha). Another way to express the inverse trigono- metric functions is to use arc sin, arc cos, and arc tan instead of sin−1, cos−1, and tan−1. Example 5 illustrates the use of an inverse trigonometric function.

EXAMPLE 5 Using Inverse Trigonometric Functions

A lakefront drops off gradually at an angle θ, as Figure 1.7 indicates. For safety reasons, it is necessary to know how deep the lake is at vari- ous distances from the shore. To provide some information about the depth, a lifeguard rows straight out from the shore a distance of 14.0 m

and drops a weighted fi shing line. By measuring the length of the line, the lifeguard determines the depth to be 2.25 m. (a) What is the value of θ? (b) What would be the depth d of the lake at a distance of 22.0 m from the shore?

8 CHAPTER 1 Introduction and Mathematical Concepts

The right triangle in Interactive Figure 1.5 provides the basis for defi ning the various tri- gonometric functions according to Equations 1.1–1.3. These functions always involve an angle and two sides of the triangle. There is also a relationship among the lengths of the three sides of a right triangle. This relationship is known as the Pythagorean theorem and is used often in this text.

PYTHAGOREAN THEOREM The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides:

h2 = ho2 + ha2 (1.7)

1.5 Scalars and Vectors The volume of water in a swimming pool might be 50 cubic meters, or the winning time of a race could be 11.3 seconds. In cases like these, only the size of the numbers matters. In other words, how much volume or time is there? The 50 specifi es the amount of water in units of cubic meters, while the 11.3 specifi es the amount of time in seconds. Volume and time are examples of scalar quantities. A scalar quantity is one that can be described with a single number (including any units) giving its size or magnitude. Some other common scalars are temperature (e.g., 20 °C) and mass (e.g., 85 kg).

While many quantities in physics are scalars, there are also many that are not, and for these quantities the magnitude tells only part of the story. Consider Figure 1.8, which depicts a car that has moved 2 km along a straight line from start to fi nish. When describing the motion, it is incomplete to say that “the car moved a distance of 2 km.” This statement would indicate only that the car ends up somewhere on a circle whose center is at the starting point and whose radius

Reasoning Near the shore, the lengths of the opposite and adjacent sides of the right triangle in Figure 1.7 are ho = 2.25 m and ha = 14.0 m, relative to the angle θ. Having made this identifi cation, we can use the inverse tangent to fi nd the angle in part (a). For part (b) the opposite and adjacent sides farther from the shore become ho = d and ha = 22.0 m. With the value for θ obtained in part (a), the tangent function can be used to fi nd the unknown depth. Considering the way in which the lake bottom drops off in Figure 1.7, we expect the unknown depth to be greater than 2.25 m.

Solution (a) Using the inverse tangent given in Equation 1.6, we fi nd that

θ = tan−1 ( ho ha) = tan

−1 (2.25 m14.0 m) = 9.13° (b) With θ = 9.13°, the tangent function given in Equation 1.3 can be used to fi nd the unknown depth farther from the shore, where ho = d and ha = 22.0 m. Since tan θ = ho /ha, it follows that

ho = ha tan θ d = (22.0 m)(tan 9.13°) = 3.54 m

which is greater than 2.25 m, as expected.

FIGURE 1.7 If the distance from the shore and the depth of the water at any one point are known, the angle θ can be found with the aid of trigonometry. Knowing the value of θ is useful, because then the depth d at another point can be determined.

14.0 m

2.25 m d

22.0 m

θ

1.5 Scalars and Vectors 9

is 2 km. A complete description must include the direction along with the distance, as in the state- ment “the car moved a distance of 2 km in a direction 30° north of east.” A quantity that deals inherently with both magnitude and direction is called a vector quantity. Because direction is an important characteristic of vectors, arrows are used to represent them; the direction of the arrow gives the direction of the vector. The colored arrow in Figure 1.8, for example, is called the displacement vector, because it shows how the car is displaced from its starting point. Chapter 2 discusses this particular vector.

The length of the arrow in Figure 1.8 represents the magnitude of the displacement vector. If the car had moved 4 km instead of 2 km from the starting point, the arrow would have been drawn twice as long. By convention, the length of a vector arrow is proportional to the magnitude of the vector.

In physics there are many important kinds of vectors, and the practice of using the length of an arrow to represent the magnitude of a vector applies to each of them. All forces, for instance, are vectors. In common usage a force is a push or a pull, and the direction in which a force acts is just as important as the strength or magnitude of the force. The magnitude of a force is measured in SI units called newtons (N). An arrow representing a force of 20 newtons is drawn twice as long as one representing a force of 10 newtons.

The fundamental distinction between scalars and vectors is the characteristic of direction. Vectors have it, and scalars do not. Conceptual Example 6 helps to clarify this distinction and explains what is meant by the “direction” of a vector.

FIGURE 1.8 A vector quantity has a magnitude and a direction. The colored arrow in this drawing represents a displacement vector.

N

S

W E

30.0°

Start

Finish2 km

CONCEPTUAL EXAMPLE 6 Vectors, Scalars, and the Role of Plus and Minus Signs

There are places where the temperature is +20 °C at one time of the year and −20 °C at another time. Do the plus and minus signs that signify positive and negative temperatures imply that temperature is a vector quantity? (a) Yes (b) No

Reasoning A hallmark of a vector is that there is both a magnitude and a physical direction associated with it, such as 20 meters due east or 20 meters due west.

Answer (a) is incorrect. The plus and minus signs associated with +20 °C and −20 °C do not convey a physical direction, such as due east or due west. Therefore, temperature cannot be a vector quantity.

Answer (b) is correct. On a thermometer, the algebraic signs simply mean that the temperature is a number less than or greater than zero on the temperature scale being used and have nothing to do with east, west, or any other physical direction. Temperature, then, is not a vector. It is a scalar, and scalars can sometimes be negative.

*Vectors are also sometimes written in other texts as boldface symbols without arrows above them.

Often, for the sake of convenience, quantities such as volume, time, displacement, velo- city, and force are represented in physics by symbols. In this text, we write vectors in boldface symbols (this is boldface) with arrows above them* and write scalars in italic symbols (this is italic). Thus, a displacement vector is written as “A→ = 750 m, due east,” where the A→ is a boldface symbol. By itself, however, separated from the direction, the magnitude of this vector is a scalar quantity. Therefore, the magnitude is written as “A = 750 m,” where the A is an italic symbol without an arrow.

Check Your Understanding

(The answer is given at the end of the book.) 5. Which of the following statements, if any, involves a vector? (a) I walked 2 miles along the beach. (b) I

walked 2 miles due north along the beach. (c) I jumped off a cliff and hit the water traveling at 17 miles per hour. (d) I jumped off a cliff and hit the water traveling straight down at a speed of 17 miles per hour. (e) My bank account shows a negative balance of −25 dollars.

10 CHAPTER 1 Introduction and Mathematical Concepts

1.6 Vector Addition and Subtraction Addition Often it is necessary to add one vector to another, and the process of addition must take into ac- count both the magnitude and the direction of the vectors. The simplest situation occurs when the vectors point along the same direction—that is, when they are colinear, as in Figure 1.9. Here, a car fi rst moves along a straight line, with a displacement vector A→ of 275 m, due east. Then the car moves again in the same direction, with a displacement vector B→ of 125 m, due east. These two vectors add to give the total displacement vector R→ , which would apply if the car had moved from start to fi nish in one step. The symbol R→ is used because the total vector is often called the resultant vector. With the tail of the second arrow located at the head of the fi rst arrow, the two lengths simply add to give the length of the total displacement. This kind of vector addition is identical to the familiar addition of two scalar numbers (2 + 3 = 5) and can be carried out here only because the vectors point along the same direction. In such cases we add the individual magnitudes to get the magnitude of the total, knowing in advance what the direction must be. Formally, the addition is written as follows:

R→ = A→ + B→

R→ = 275 m, due east + 125 m, due east = 400 m, due east Perpendicular vectors are frequently encountered, and Figure 1.10 indicates how they can

be added. This fi gure applies to a car that fi rst travels with a displacement vector A→ of 275 m, due east, and then with a displacement vector B→ of 125 m, due north. The two vectors add to give a resultant displacement vector R→ . Once again, the vectors to be added are arranged in a tail-to- head fashion, and the resultant vector points from the tail of the fi rst to the head of the last vector added. The resultant displacement is given by the vector equation

R→ = A→ + B→

The addition in this equation cannot be carried out by writing R = 275 m + 125 m, because the vectors have diff erent directions. Instead, we take advantage of the fact that the triangle in Figure 1.10 is a right triangle and use the Pythagorean theorem (Equation 1.7). According to this theorem, the magnitude of R→ is

R = √(275 m)2 + (125 m)2 = 302 m

The angle θ in Figure 1.10 gives the direction of the resultant vector. Since the lengths of all three sides of the right triangle are now known, sin θ, cos θ, or tan θ can be used to determine θ. Noting that tan θ = B/A and using the inverse trigonometric function, we fi nd that:

θ = tan−1 (BA) = tan−1 ( 125 m 275 m) = 24.4°

Thus, the resultant displacement of the car has a magnitude of 302 m and points north of east at an angle of 24.4°. This displacement would bring the car from the start to the fi nish in Figure 1.10 in a single straight-line step.

When two vectors to be added are not perpendicular, the tail-to-head arrangement does not lead to a right triangle, and the Pythagorean theorem cannot be used. Figure 1.11a illustrates such a case for a car that moves with a displacement A→ of 275 m, due east, and then with a dis- placement B→ of 125 m, in a direction 55.0° north of west. As usual, the resultant displacement vector R→ is directed from the tail of the fi rst to the head of the last vector added. The vector addition is still given according to

R→ = A→ + B→

However, the magnitude of R→ is not R = A + B, because the vectors A→ and B→ do not have the same direction, and neither is it R = √A2 + B2, because the vectors are not perpendicular, so the Pythagorean theorem does not apply. Some other means must be used to fi nd the magnitude and direction of the resultant vector.

N

S

W E

Tail-to-head

Start Finish A

R

B

FIGURE 1.9 Two colinear displacement

vectors A→ and B→ add to give the resultant displacement vector R→ .

FIGURE 1.10 The addition of two perpendicular displacement vectors A→ and B→ gives the resultant vector R→ .

N

S

W E

Tail-to-head

Start

R B

Finish

90°θ

A

FIGURE 1.11 (a) The two displacement vectors A→ and B→ are neither colinear nor perpendicular, but even so they add to give the resultant vector R→. (b) In one method for adding them together, a graphical technique is used.

Finish

Start

Tail-to-head

55.0°

55.0°

( )a

( )b

22. 8 c

m

4 8

12

20

0

24 cm

16

28

N

S

W E

θ

θ

A

R B

A

R B

1.6 Vector Addition and Subtraction 11

One approach uses a graphical technique. In this method, a diagram is constructed in which the arrows are drawn tail to head. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor, perhaps). Then the length of the arrow representing the resultant vector is measured with a ruler. This length is converted to the mag- nitude of the resultant vector by using the scale factor with which the drawing is constructed. In Figure 1.11b, for example, a scale of one centimeter of arrow length for each 10.0 m of displacement is used, and it can be seen that the length of the arrow representing R→ is 22.8 cm. Since each centimeter corresponds to 10.0 m of displacement, the magnitude of R→ is 228 m. The angle θ, which gives the direction of R→ , can be measured with a protractor to be θ = 26.7° north of east.

Subtraction The subtraction of one vector from another is carried out in a way that depends on the follow- ing fact. When a vector is multiplied by −1, the magnitude of the vector remains the same, but the direction of the vector is reversed. Conceptual Example 7 illustrates the meaning of this statement.

Reasoning A displacement vector of −D→ is (−1) D→ . The presence of the (−1) factor reverses the direction of the vector, but does not change its magnitude. Similarly, a force vector of −F→ has the same magnitude as the vector F→ but has the opposite direction.

Answer (a) and (b) are incorrect. While scalars can sometimes be negative, magnitudes of vectors are never negative.

Answer (c) is correct. The vectors −D→ and −F→ have the same mag- nitudes as D→ and F→, but point in the opposite direction, as indicated in Figures 1.12b and 1.13b.

Related Homework: Problems 67

CONCEPTUAL EXAMPLE 7 Multiplying a Vector by −1

Consider two vectors described as follows:

1. A woman climbs 1.2 m up a ladder, so that her displacement vec- tor D→ is 1.2 m, upward along the ladder, as in Figure 1.12a.

2. A man is pushing with 450 N of force on his stalled car, trying to move it eastward. The force vector F→ that he applies to the car is 450 N, due east, as in Figure 1.13a.

What are the physical meanings of the vectors −D→ and −F→? (a) −D→ points upward along the ladder and has a magnitude of −1.2 m; −F→ points due east and has a magnitude of −450 N. (b) −D→ points downward along the ladder and has a magnitude of −1.2 m; −F→ points due west and has a magnitude of −450 N. (c) −D→ points downward along the ladder and has a magnitude of 1.2 m; −F→ points due west and has a magnitude of 450 N.

FIGURE 1.12 (a) The displacement vector for a woman climbing 1.2 m up a ladder is D→ . (b) The displacement vector for a woman climbing 1.2 m down a ladder is −D→ .

(a) (b)

–DD

FIGURE 1.13 (a) The force vector for a man pushing on a car with 450 N of force in a direction due east is F→ . (b) The force vector for a man pushing on a car with 450 N of force in a direction due west is −F→ .

(a)

(b)

F

–F

12 CHAPTER 1 Introduction and Mathematical Concepts

In practice, vector subtraction is carried out exactly like vector addition, except that one of the vectors added is multiplied by a scalar factor of −1. To see why, look at the two vectors A→ and B→ in Figure 1.14a. These vectors add together to give a third vector C→, according to C→ = A→ + B→. Therefore, we can calculate vector A→ as A→ = C→ − B→, which is an example of vector subtraction. However, we can also write this result as A→ = C→ + (−B→) and treat it as vector addi- tion. Figure 1.14b shows how to calculate vector A→ by adding the vectors C→ and −B→. Notice that vectors C→ and −B→ are arranged tail to head and that any suitable method of vector addition can be employed to determine A→.

Check Your Understanding

(The answers are given at the end of the book.) 6. Two vectors A→ and B→ are added together to give a resultant vector R→: R→ = A→ + B→. The magnitudes

of A→ and B→ are 3 m and 8 m, respectively, but the vectors can have any orientation. What are (a) the maximum possible value and (b) the minimum possible value for the magnitude of R→?

7. Can two nonzero perpendicular vectors be added together so their sum is zero? 8. Can three or more vectors with unequal magnitudes be added together so their sum is zero? 9. In preparation for this question, review Conceptual Example 7. Vectors A→ and B→ satisfy the vector

equation A→ + B→ = 0. (a) How does the magnitude of B→ compare with the magnitude of A→? (b) How does the direction of B→ compare with the direction of A→?

10. Vectors A→, B→, and C→ satisfy the vector equation A→ + B→ = C→, and their magnitudes are related by the scalar equation A2 + B2 = C2. How is vector A→ oriented with respect to vector B→?

11. Vectors A→, B→, and C→ satisfy the vector equation A→ + B→ = C→, and their magnitudes are related by the scalar equation A + B = C. How is vector A→ oriented with respect to vector B→?

1.7 The Components of a Vector Vector Components Suppose a car moves along a straight line from start to fi nish, as in Figure 1.15, the corres- ponding displacement vector being r→. The magnitude and direction of the vector r→ give the distance and direction traveled along the straight line. However, the car could also arrive at the fi nish point by fi rst moving due east, turning through 90°, and then moving due north. This alternative path is shown in the drawing and is associated with the two displacement vectors x→ and y→. The vectors x→ and y→ are called the x vector component and the y vector component of r→.

Vector components are very important in physics and have two basic features that are appar- ent in Figure 1.15. One is that the components add together to equal the original vector:

r→ = x→ + y→

The components x→ and y→, when added vectorially, convey exactly the same meaning as does the original vector r→: they indicate how the fi nish point is displaced relative to the starting point. The other feature of vector components that is apparent in Figure 1.15 is that x→ and y→ are not just any two vectors that add together to give the original vector r→: they are perpendicular vectors. This perpendicularity is a valuable characteristic, as we will soon see.

Any type of vector may be expressed in terms of its components, in a way similar to that illustrated for the displacement vector in Figure 1.15. Interactive Figure 1.16 shows an arbitrary vector A→ and its vector components A→ x and A

→ y. The components are drawn parallel

to convenient x and y axes and are perpendicular. They add vectorially to equal the original vector A→ :

A→ = A→ x + A →

y

FIGURE 1.15 The displacement vector r→ and its vector components x→ and y→.

FIGURE 1.14 (a) Vector addition according to C→ = A→ + B→ . (b) Vector subtraction accord- ing to A→ = C→ − B→ = C→ + (−B→ ).

A

B

C = A + B

( )a

B–

C

Tail-to-head

( )b

A = C – B

Start

Finish

90°

N

S

W E

x

yr

INTERACTIVE FIGURE 1.16 An arbitrary

vector A→ and its vector components A→ x and A →

y.

A

y+

y

x+ Ax

A

θ

1.7 The Components of a Vector 13

There are times when a drawing such as Interactive Figure 1.16 is not the most convenient way to represent vector components, and Figure 1.17 presents an alternative method. The disadvant- age of this alternative is that the tail-to-head arrangement of A→ x and A

→ y is missing, an arrange-

ment that is a nice reminder that A→ x and A →

y add together to equal A →

. The defi nition that follows summarizes the meaning of vector components:

DEFINITION OF VECTOR COMPONENTS In two dimensions, the vector components of a vector A→ are two perpendicular vectors A→ x and A

→ y that are parallel to the x and y axes, respectively, and add together vectorially

according to A→ = A→ x + A →

y.

Problem-Solving Insight In general, the components of any vector can be used in place of the vector itself in any calculation where it is convenient to do so.

The values calculated for vector components depend on the orientation of the vector relative to the axes used as a reference. Figure 1.18 illustrates this fact for a vector A→ by showing two sets of axes, one set being rotated clockwise relative to the other. With respect to the black axes, vector A→ has perpendicular vector components A→ x and A

→ y; with respect to the colored rotated axes,

vector A→ has diff erent vector components A→ x′ and A →

y′. The choice of which set of components to use is purely a matter of convenience.

Scalar Components It is often easier to work with the scalar components, Ax and Ay (note the italic symbols), rather than the vector components A→ x and A

→ y. Scalar components are positive or negative numbers (with

units) that are defi ned as follows: The scalar component Ax has a magnitude equal to that of A →

x and is given a positive sign if A→ x points along the +x axis and a negative sign if A

→ x points along

the −x axis. The scalar component Ay is defi ned in a similar manner. The following table shows an example of vector and scalar components:

In this text, when we use the term “component,” we will be referring to a scalar component, unless otherwise indicated.

Another method of expressing vector components is to use unit vectors. A unit vector is a vector that has a magnitude of 1, but no dimensions. We will use a caret (^) to distinguish it from other vectors. Thus,

x̂ is a dimensionless unit vector of length l that points in the positive x direction, and ŷ is a dimensionless unit vector of length l that points in the positive y direction.

These unit vectors are illustrated in Figure 1.19. With the aid of unit vectors, the vector com- ponents of an arbitrary vector A→ can be written as A→ x = Ax x̂ and A

→ y = Ay ŷ, where Ax and Ay are

its scalar components (see the drawing and the third column of the table above). The vector A→ is then written as A→ = Ax x̂ + Ay ŷ.

Resolving a Vector into Its Components If the magnitude and direction of a vector are known, it is possible to fi nd the components of the vector. The process of fi nding the components is called “resolving the vector into its components.” As Example 8 illustrates, this process can be carried out with the aid of trigo- nometry, because the two perpendicular vector components and the original vector form a right triangle.

Ay

+y

+x Ax

A

θ

FIGURE 1.17 This alternative way of

drawing the vector A→ and its vector components is completely equivalent to that shown in Interactive Figure 1.16.

+y

+x

+y´

+x´

Ax

Áy

Á x

A

Ay

FIGURE 1.18 The vector components of the vector depend on the orientation of the axes used as a reference.

x

y Axx

Ayy

+x

+y

FIGURE 1.19 The dimensionless unit vectors x̂ and ŷ have magnitudes equal to 1, and they point in the +x and +y directions, respectively. Expressed in terms of unit vectors, the vector components of the vector A→ are Ax x̂ and Ay ŷ.

Vector Components Scalar Components Unit Vectors

A→ x = 8 meters, directed along the +x axis Ax = +8 meters A →

x = (+8 meters) x̂

A→ y = 10 meters, directed along the –y axis Ay = −10 meters A →

y = (–10 meters) ŷ

14 CHAPTER 1 Introduction and Mathematical Concepts

EXAMPLE 8 Finding the Components of a Vector

A displacement vector r→ has a magnitude of r = 175 m and points at an angle of 50.0° relative to the x axis in Figure 1.20. Find the x and y com- ponents of this vector.

Reasoning We will base our solution on the fact that the triangle formed in Figure 1.20 by the vector r→ and its components x→ and y→ is a right triangle. This fact enables us to use the trigonometric sine and cosine functions, as defi ned in Equations 1.1 and 1.2.

Problem-Solving Insight You can check to see whether the components of a vector are correct by substituting them into the Pythagorean theorem in order to calculate the magnitude of the original vector.

Solution The y component can be obtained using the 50.0° angle and Equation 1.1, sin θ = y/r:

y = r sin θ = (175 m)(sin 50.0°) = 134 m

In a similar fashion, the x component can be obtained using the 50.0° angle and Equation 1.2, cos θ = x/r:

x = r cos θ = (175 m)(cos 50.0°) = 112 m

Math Skills Either acute angle of a right triangle can be used to determine the components of a vector. The choice of angle is a matter of convenience. For instance, instead of the 50.0° angle, it is also possible to use the angle α in Figure 1.20. Since α + 50.0° = 90.0°, it follows that α = 40.0°. The solution using α yields the same answers as the solution using the 50.0° angle:

cos α = y r

y = r cos α = (175 m)(cos 40.0°) = 134 m

sin α = x r

x = r sin α = (175 m)(sin 40.0°) = 112 m90.0°50.0°

x

yr

α

FIGURE 1.20 The x and y components of the displacement vector r→ can be found using trigonometry.

Since the vector components and the original vector form a right triangle, the Pythagorean theorem can be applied to check the validity of calculations such as those in Example 8. Thus, with the components obtained in Example 8, the theorem can be used to verify that the magnitude of the original vector is indeed 175 m, as given initially:

r = √(112 m) 2 + (134 m) 2 = 175 m

It is possible for one of the components of a vector to be zero. This does not mean that the vector itself is zero, however.

Problem-Solving Insight For a vector to be zero, every vector component must individually be zero.

Thus, in two dimensions, saying that A→ = 0 is equivalent to saying that A→ x = 0 and A →

y = 0. Or, stated in terms of scalar components, if A→ = 0, then Ax = 0 and Ay = 0.

Problem-Solving Insight Two vectors are equal if, and only if, they have the same magnitude and direction.

Thus, if one displacement vector points east and another points north, they are not equal, even if each has the same magnitude of 480 m. In terms of vector components, two vectors A→ and B →

are equal if, and only if, each vector component of one is equal to the corresponding vector component of the other. In two dimensions, if A→ = B

→ , then A→ x = B

→ x and A

→ y = A

→ y. Alternatively,

using scalar components, we write that Ax = Bx and Ay = By.

Check Your Understanding

(The answers are given at the end of the book.) 12. Which of the following displacement vectors

(if any) are equal?

Variable Magnitude Direction A→ 100 m 30° north of east

B→ 100 m 30° south of west

C→ 50 m 30° south of west

D→ 100 m 60° east of north

1.8 Addition of Vectors by Means of Components 15

13. Two vectors, A→ and B→, are shown in CYU Figure 1.1. (a) What are the signs (+ or −) of the scalar components, Ax and Ay, of the vector A→? (b) What are the signs of the scalar components, Bx and By, of the vector B

→ ? (c) What are the signs of the scalar

components, Rx and Ry, of the vector R →

, where R→ = A→ + B→? 14. Are two vectors with the same magnitude necessarily equal? 15. The magnitude of a vector has doubled, its direction remaining

the same. Can you conclude that the magnitude of each com- ponent of the vector has doubled?

16. The tail of a vector is fi xed to the origin of an x, y axis sys- tem. Originally the vector points along the +x axis and has a magnitude of 12 units. As time passes, the vector rotates coun- terclockwise. What are the sizes of the x and y components of the vector for the following rotational angles? (a) 90° (b) 180° (c) 270° (d) 360°

17. A vector has a component of zero along the x axis of a certain axis system. Does this vector necessarily have a component of zero along the x axis of another (rotated) axis system?

1.8 Addition of Vectors by Means of Components The components of a vector provide the most convenient and accurate way of adding (or sub- tracting) any number of vectors. For example, suppose that vector A→ is added to vector B→. The resultant vector is C→, where C→ = A→ + B→. Interactive Figure 1.21a illustrates this vector addition, along with the x and y vector components of A→ and B→. In part b of the drawing, the vectors A→ and B→ have been removed, because we can use the vector components of these vectors in place of them. The vector component B→x has been shifted downward and arranged tail to head with vector component A→ x. Similarly, the vector component A

→ y has been shifted to the right and arranged tail

to head with the vector component B→y. The x components are colinear and add together to give the x component of the resultant vector C→. In like fashion, the y components are colinear and add together to give the y component of C→. In terms of scalar components, we can write

Cx = Ax + Bx and Cy = Ay + By

The vector components C→x and C →

y of the resultant vector form the sides of the right triangle shown in Interactive Figure 1.21c. Thus, we can fi nd the magnitude of C→ by using the Pythagorean theorem:

C = √C 2x + C 2y The angle θ that C

→ makes with the x axis is given by θ = tan−1 (Cy/Cx). Example 9 illustrates how

to add several vectors using the component method.

C

A

B

Bx

By

Ax

Ay +x

+y

(a) (c)

Cy

Cx

C

C

Bx

By

Ax

Ay

(b)

A

θ

INTERACTIVE FIGURE 1.21 (a) The vectors A→ and B→ add together to give the resultant vector C→. The x and y components of A→ and B→ are also shown. (b) The drawing illustrates that C

→ x = A

→ x + B

→ x and C

→ y = A

→ y +

B→ y. (c) Vector C →

and its components form a right triangle.

+x

+y

A B

CYU FIGURE 1.1

16 CHAPTER 1 Introduction and Mathematical Concepts

Analyzing Multiple -Concept Problems

EXAMPLE 9 The Component Method of Vector Addition

A jogger runs 145 m in a direction 20.0° east of north (displacement vector A→) and then 105 m in a direction 35.0° south of east (displacement vector B→). Using components, determine the magnitude and direction of the resultant vector C→ for these two displacements.

Reasoning Figure 1.22 shows the vectors A→ and B→, assuming that the y axis corresponds to the dir- ection due north. The vectors are arranged in a tail-to-head fashion, with the resultant vector C→ drawn from the tail of A→ to the head of B→. The components of the vectors are also shown in the fi gure. Since C→ and its components form a right triangle (red in the drawing), we will use the Pythagorean theorem and trigonometry to express the magnitude and directional angle θ for C→ in terms of its components. The components of C→ will then be obtained from the components of A→ and B→ and the data given for these two vectors.

Knowns and Unknowns The data for this problem are listed in the table that follows:

Description Symbol Value Comment

Magnitude of vector A→ 145 m

Direction of vector A→ 20.0° east of north See Figure 1.22.

Magnitude of vector B→ 105 m

Direction of vector B→ 35.0° south of east See Figure 1.22.

Unknown Variable Magnitude of resultant vector C ?

Direction of resultant vector θ ?

FIGURE 1.22 The vectors A→ and B→ add together to give the resultant vector C→. The vector components of A→ and B→ are also shown. The resultant vector C→ can be obtained once its components have been found.

N

S

W E

20.0°

+x

35.0°

+y Bx

By

Ax

Ay

CyC

Cx

B A

θ

Modeling the Problem

STEP 1 Magnitude and Direction of C→ In Figure 1.22 the vector C→ and its components C→x and C→y form a right triangle, as the red arrows show. Applying the Pythagorean theorem to this right triangle shows that the magnitude of C→ is given by Equation 1a at the right. From the red triangle it also follows that the directional angle θ for the vector C→ is given by Equation 1b at the right.

STEP 2 Components of C→ Since vector C→ is the resultant of vectors A→ and B→, we have C→ = A→ + B→ and can write the scalar components of C→ as the sum of the scalar components of A→ and B→:

Cx = Ax + Bx and Cy = Ay + By

These expressions can be substituted into Equations 1a and 1b for the magnitude and direction of C→, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

C = √C 2x + C 2y = √(Ax + Bx)2 + (Ay + By )2

θ = tan−1 ( Cy Cx) = tan−1(

Ay + By Ax + Bx)

To use these results we need values for the individual components of A→ and B→.

C = √Cx 2 + Cy 2 (1a)

θ = tan−1 ( Cy Cx) (1b)

C = √Cx 2 + Cy 2 (1a)

Cx = Ax + Bx Cy = Ay + By (2)

θ = tan−1 ( Cy Cx) (1b)

Cx = Ax + Bx Cy = Ay + By (2)

STEP 1 STEP 2

STEP 1 STEP 2

1.8 Addition of Vectors by Means of Components 17

Referring to Figure 1.22, we fi nd these values to be

Ax = (145 m) sin 20.0° = 49.6 m and Ay = (145 m) cos 20.0° = 136 m

Bx = (105 m) cos 35.0° = 86.0 m and By = −(105 m) sin 35.0° = −60.2 m

Note that the component By is negative, because B →

y points downward, in the negative y direction in the drawing. Substituting these values into the results for C and θ gives

C = √(Ax + Bx)2 + (Ay + By)2

= √(49.6 m + 86.0 m)2 + (136 m − 60.2 m)2 = 155 m

θ = tan−1( Ay + By Ax + Bx)

= tan−1( 136 m − 60.2 m49.6 m + 86.0 m) = 29°

Math Skills According to the defi nitions given in Equations 1.1 and 1.2, the sine and

cosine functions are sin ϕ = ho h

and cos ϕ = ha h

, where ho is the length of the side of a

right triangle that is opposite the angle ϕ, ha is the length of the side adjacent to the angle ϕ, and h is the length of the hypotenuse (see Figure 1.23a). Applications of the sine and cosine functions to determine the scalar components of a vector occur frequently. In such applications we begin by identifying the angle ϕ. Figure 1.23b shows the relevant portion of Figure 1.22 and indicates that ϕ = 20.0° for the vector A→. In this case we have ho = Ax, ha = Ay, and h = A = 145 m; it follows that

sin 20.0° = ho h

= Ax A

or Ax = A sin 20.0° = (145 m) sin 20.0° = 49.6 m

cos 20.0° = ha h

= Ay A

or Ay = A cos 20.0° = (145 m) cos 20.0° = 136 m

(a) (b)

Ax

+y

+x

Ay ha

ho

h

ha

ho

h A

ϕ

20.0°

90°

FIGURE 1.23 Math Skills drawing.

Related Homework: Problems 45, 47, 50, 54

18 CHAPTER 1 Introduction and Mathematical Concepts

In later chapters we will often use the component method for vector addition. For future reference, the main features of the reasoning strategy used in this technique are summarized below.

REASONING STRATEGY The Component Method of Vector Addition 1. For each vector to be added, determine the x and y components relative to a conveniently

chosen x, y coordinate system. Be sure to take into account the directions of the compon- ents by using plus and minus signs to denote whether the components point along the positive or negative axes.

2. Find the algebraic sum of the x components, which is the x component of the resultant vector. Similarly, fi nd the algebraic sum of the y components, which is the y component of the resultant vector.

3. Use the x and y components of the resultant vector and the Pythagorean theorem to de- termine the magnitude of the resultant vector.

4. Use the inverse sine, inverse cosine, or inverse tangent function to fi nd the angle that specifi es the direction of the resultant vector.

Check Your Understanding

(1) (2) (3) (4)

Ax

Ay

R

R

R

RBy Bx

(The answer is given at the end of the book.) 18. Two vectors, A→ and B→, have vector components that are shown (to the same scale) in CYU Figure 1.2.

The resultant vector is labeled R→. Which drawing shows the correct vector sum of A→ + B→? (a) 1, (b) 2, (c) 3, (d) 4

joint, and vector B→ represents the position of the ball relative to the elbow joint. Use the component method of vector addition and the angles given in the fi gure to fi nd the magnitude and direction (θ) of vector C→, which represents the position of the ball relative to the shoulder joint. The mag- nitude of vector A→ is 35.6 cm, and the magnitude of vector B→ is 31.2 cm. The angle θ is measured relative to a vertical anatomical plane known as the frontal or coronal plane.

Reasoning Similar to Example 9, the vectors A→ and B→ in Figure 1.24 are drawn tail-to-head. Thus, the resultant vector C→ = A→ + B→. The com- ponents of vector C→ will be obtained from the components of vectors A→ and B→. Once Cx and Cy are known, we can calculate the magnitude of vector C→ using the Pythagorean theorem. The directional angle θ will be determined from the components of vector C→.

Solution Applying the component method of vector addition, we know that Cx = Ax + Bx, and Cy = Ay + By. From Figure 1.24, we see that Ax = (35.6 cm) cos 45° = 25.2 cm, Bx = – (31.2) cos 80° = – 5.42 cm, Ay = (35.6 cm) sin 45° = 25.2 cm, and By = (31.2 cm) sin 80° = 30.7 cm. Therefore, Cx = 25.2 cm + (– 5.42 cm) = 19.8 cm, and Cy = 25.2 cm + 30.7 cm = 55.9 cm. The magnitude of vector C→ is now found by the Pythagorean theorem: C = √C2x + C2y = √(19.8 cm)2 + (55.9 cm)2 = 59.3 cm The directional angle θ is found by using the tangent function:

θ = tan−1( Cy Cx) = tan−1(

55.9 cm 19.8 cm) = 70.5° .

EXAMPLE 10 BIO Multi-joint Movements

Figure 1.24 shows an example of a multi-joint movement involving the shoulder and elbow joints. The view from above shows a person holding a ball in a position that involves both shoulder fl exion and elbow extension. Vector A→ represents the position of the elbow joint relative to the shoulder

FIGURE 1.24 Top view of a multi-joint movement involving the shoulder and elbow joints. The vectors A→ and B→ add together to give the resultant vector C→, which represents the position vector of the ball with respect to the shoulder joint. The resultant vector C→ can be obtained, once its components are determined.

Shoulder joint

Elbow joint

Ball

Frontal plane

45.0°

80.0°

+y

B

A

C

+x θ

CYU FIGURE 1.2

Focus on Concepts 19

Concept Summary 1.2 Units The SI system of units includes the meter (m), the kilogram (kg), and the second (s) as the base units for length, mass, and time, respectively. One meter is the distance that light travels in a vacuum in a time of 1/299 792 458 second. One kilogram is the mass of a standard cylinder of platinum–iridium alloy kept at the International Bureau of Weights and Measures. One second is the time for a certain type of electromagnetic wave emitted by cesium-133 atoms to undergo 9 192 631 770 wave cycles.

1.3 The Role of Units in Problem Solving To convert a number from one unit to another, multiply the number by the ratio of the two units. For instance, to convert 979 meters to feet, multiply 979 meters by the factor (3.281 foot/1 meter).

The dimension of a quantity represents its physical nature and the type of unit used to specify it. Three such dimensions are length [L], mass [M], time [T]. Dimensional analysis is a method for checking mathematical rela- tions for the consistency of their dimensions.

1.4 Trigonometry The sine, cosine, and tangent functions of an angle θ are defi ned in terms of a right triangle that contains θ, as in Equations 1.1–1.3, where ho and ha are, respectively, the lengths of the sides opposite and adja- cent to the angle θ, and h is the length of the hypotenuse.

sin θ = ho h

(1.1)

cos θ = ha h

(1.2)

tan θ = ho ha

(1.3)

The inverse sine, inverse cosine, and inverse tangent functions are given in Equations 1.4–1.6.

θ = sin−1 ( ho h ) (1.4)

θ = cos−1 ( ha h ) (1.5)

θ = tan−1 ( ho ha) (1.6)

The Pythagorean theorem states that the square of the length of the hypot- enuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides, according to Equation 1.7.

h2 = ho2 + ha2 (1.7)

1.5 Scalars and Vectors A scalar quantity is described by its size, which is also called its magnitude. A vector quantity has both a magnitude and a direction. Vectors are often represented by arrows, the length of the arrow being proportional to the magnitude of the vector and the direction of the arrow indicating the direction of the vector.

1.6 Vector Addition and Subtraction One procedure for adding vectors utilizes a graphical technique, in which the vectors to be added are ar- ranged in a tail-to-head fashion. The resultant vector is drawn from the tail of the fi rst vector to the head of the last vector. The subtraction of a vector is treated as the addition of a vector that has been multiplied by a scalar factor of −1. Multiplying a vector by −1 reverses the direction of the vector.

1.7 The Components of a Vector In two dimensions, the vector compon- ents of a vector A→ are two perpendicular vectors A→ x and A

→ y that are parallel

to the x and y axes, respectively, and that add together vectorially so that A→ = A→ x + A

→ y. The scalar component Ax has a magnitude that is equal to that

of A→ x and is given a positive sign if A →

x points along the +x axis and a negative sign if A→ x points along the −x axis. The scalar component Ay is defi ned in a similar manner.

Two vectors are equal if, and only if, they have the same magnitude and direction. Alternatively, two vectors are equal in two dimensions if the x vector components of each are equal and the y vector components of each are equal. A vector is zero if, and only if, each of its vector components is zero.

1.8 Addition of Vectors by Means of Components If two vec- tors A→ and B→ are added to give a resultant C→ such that C→ = A→ + B→, then Cx = Ax + Bx and Cy = Ay + By, where Cx, Ax, and Bx are the scalar compon- ents of the vectors along the x direction, and Cy, Ay, and By are the scalar components of the vectors along the y direction.

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 1.6 Vector Addition and Subtraction 1. During a relay race, runner A runs a certain distance due north and then hands off the baton to runner B, who runs for the same distance in a direction south of east. The two displacement vectors A→ and B→ can be added together to give a resultant vector R→. Which drawing correctly shows the resultant vector? (a) 1 (b) 2 (c) 3 (d) 4

North

East (1)

North

East (2)

North

East (3)

North

East (4)

A A A

A

BBBB

RRRR

QUESTION 1

2. How is the magnitude R of the resultant vector R→ in the drawing related to the magnitudes A and B of the vectors A→ and B→? (a) The magnitude of the resultant vector R→ is equal to the sum of the magnitudes of A→ and B→, or R = A + B. (b) The magnitude of the resultant vector R→ is greater than the

Focus on Concepts

20 CHAPTER 1 Introduction and Mathematical Concepts

sum of the magnitudes of A→ and B→, or R > A + B. (c) The magnitude of the resultant vector R→ is less than the sum of the magnitudes of A→ and B→, or R < A + B. 5. The fi rst drawing shows three displacement vectors, A→, B→, and C→, which are added in a tail-to-head fashion. The resultant vector is labeled R→. Which of the following drawings shows the correct resultant vector for A→ + B→ − C→? (a) 1 (b) 2 (c) 3

(1) (2) (3)

A

A A

A B

C

R R R

R

QUESTION 5

6. The fi rst drawing shows the sum of three displacement vectors, A→, B→, and C→. The resultant vector is labeled R→. Which of the following drawings shows the correct resultant vector for A→ − B→ − C→? (a) 1 (b) 2 (c) 3

(1) (2) (3)

A

A A

A B

C

R R R

R

QUESTION 6

Section 1.7 The Components of a Vector 8. A person is jogging along a straight line, and her displacement is denoted by the vector A→ in the drawings. Which drawing represents the correct vector components, A→ x and A

→ y, for the vector A

→ ? (a) 1 (b) 2 (c) 3 (d) 4

+y

+x

+x

+x

+x

+y

+y +y

(1)

Ax

Ay Ay

Ay Ay

Ax

Ax Ax

A

(2)

A

(3)

A

(4)

A

QUESTION 8

11. A person drives a car for a distance of 450.0 m. The displacement A→ of the car is illustrated in the drawing. What are the scalar components of this displacement vector?

(a) Ax = 0 m and Ay = +450.0 m (b) Ax = 0 m and Ay = −450.0 m (c) Ax = +450.0 m and Ay = +450.0 m (d) Ax = −450.0 m and Ay = 0 m (e) Ax = −450.0 m and Ay = +450.0 m

12. Drawing a shows a displacement vector A→ (450.0 m along the −y axis). In this x, y coordinate system the scalar components are Ax = 0 m and Ay = −450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0°, but the magnitude (450.0 m) and direction of vector A→ remain unchanged, as in drawing b. What are the scalar components, Ax′ and Ay′, of the vector A

→ in the rotated x′, y′ coordinate system?

(a) Ax′ = −369 m and Ay′ = −258 m (b) Ax′ = +369 m and Ay′ = −258 m (c) Ax′ = +258 m and Ay′ = +369 m (d) Ax′ = +258 m and Ay′ = −369 m (e) Ax′ = −258 m and Ay′ = −369 m

QUESTION 12

+y

+x

A

(a)

A

+y´

+x´

35.0°

(b)

15. Suppose the vectors A→ and B→ in the drawing have magnitudes of 6.0 m and are directed as shown. What are Ax and Bx, the scalar components of A

→

and B→ along the x axis?

Ax Bx (a) +(6.0 m) cos 35° = +4.9 m −(6.0 m) cos 35° = −4.9 m

(b) +(6.0 m) sin 35° = +3.4 m −(6.0 m) cos 35° = −4.9 m

(c) −(6.0 m) cos 35° = −4.9 m +(6.0 m) sin 35° = +3.4 m

(d) −(6.0 m) cos 35° = −4.9 m +(6.0 m) cos 35° = +4.9 m

(e) −(6.0 m) sin 35° = −3.4 m +(6.0 m) sin 35° = +3.4 m

QUESTION 15

A B

35° 35°

6.0 m 6.0 m

+y

+x

Section 1.8 Addition of Vectors by Means of Components 17. Drawing a shows two vectors A→ and B→, and drawing b shows their com- ponents. The scalar components of these vectors are as follows:

Ax = −4.9 m Ay = +3.4 m Bx = +4.9 m By = +3.4 m

When the vectors A→ and B→ are added, the resultant vector is R→, so that R→ = A→ + B→. What are the values of Rx and Ry, the x and y components of R

→ ?

QUESTION 17

+y

+x

+x

+y

A B

35° 35°

6.0 m 6.0 m

A Ay By

BxAx

B

(a)

(b)

θ θ

A

B

R

QUESTION 2

QUESTION 11

+y

+x

A

Problems 21

18. The displacement vectors A→ and B→, when added together, give the res- ultant vector R→, so that R→ = A→ + B→. Use the data in the drawing to fi nd the magnitude R of the resultant vector and the angle θ that it makes with the +x axis.

QUESTION 18

A B

R

+y

6.0 m 8.0 m

23°

+xθ

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual MMH Problem-solving help GO Guided Online Tutorial V-HINT Video Hints CHALK Chalkboard Videos

BIO Biomedical application E Easy M Medium H Hard

Section 1.2 Units

Section 1.3 The Role of Units in Problem Solving 1. E GO A student sees a newspaper ad for an apartment that has 1330 square feet (ft2) of fl oor space. How many square meters of area are there?

2. E Bicyclists in the Tour de France reach speeds of 34.0 miles per hour (mi/h) on fl at sections of the road. What is this speed in (a) kilometers per hour (km/h) and (b) meters per second (m/s)? 3. E SSM Vesna Vulovic survived the longest fall on record without a para- chute when her plane exploded and she fell 6 miles, 551 yards. What is this distance in meters?

4. E Suppose a man’s scalp hair grows at a rate of 0.35 mm per day. What is this growth rate in feet per century?

5. E Given the quantities a = 9.7 m, b = 4.2 s, c = 69 m/s, what is the value of the quantity d = a3/(cb2)?

6. E Consider the equation 𝜐 = 13 zxt 2. The dimensions of the variables υ, x, and t are [L]/[T], [L], and [T], respectively. The numerical factor 3 is di- mensionless. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions? Show how you determined your answer.

7. E SSM A bottle of wine known as a magnum contains a volume of 1.5 liters. A bottle known as a jeroboam contains 0.792 U.S. gallons. How many magnums are there in one jeroboam?

8. E The CGS unit for measuring the viscosity of a liquid is the poise (P): 1 P = 1 g/(s · cm). The SI unit for viscosity is the kg/(s · m). The viscosity of water at 0 °C is 1.78 × 10−3 kg/(s · m). Express this viscosity in poise. 9. E BIO Azelastine hydrochloride is an antihistamine nasal spray. A standard size container holds one fl uid ounce (oz) of the liquid. You are searching for this medication in a European drugstore and are asked how many milliliters (mL) there are in one fl uid ounce. Using the following con- version factors, determine the number of milliliters in a volume of one fl uid ounce: 1 gallon (gal) = 128 oz, 3.785 × 10−3 cubic meters (m3) = 1 gal, and 1 mL = 10−6 m3.

10. M GO A partly full paint can has 0.67 U.S. gallons of paint left in it. (a) What is the volume of the paint in cubic meters? (b) If all the remaining paint is used to coat a wall evenly (wall area = 13 m2), how thick is the layer of wet paint? Give your answer in meters.

11. M SSM A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretch- ing the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T = 2π√m/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?

Section 1.4 Trigonometry 12. E You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway to the West arch. This monument rises to a height of 192 m. You estimate your line of sight with the top of the arch to be 2.0° above the horizontal. Approximately how far (in kilometers) are you from the base of the arch?

13. E SSM A highway is to be built between two towns, one of which lies 35.0 km south and 72.0 km west of the other. What is the shortest length of highway that can be built between the two towns, and at what angle would this highway be directed with respect to due west?

14. E GO A hill that has a 12.0% grade is one that rises 12.0 m vertically for every 100.0 m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?

15. E GO The corners of a square lie on a circle of diameter D = 0.35 m. Each side of the square has a length L. Find L.

16. E GO The drawing shows a person looking at a building on top of which an antenna is mounted. The horizontal distance between the per- son’s eyes and the building is 85.0 m. In part a the person is looking at the base of the antenna, and his line of sight makes an angle of 35.0° with the horizontal. In part b the person is looking at the top of the antenna, and his line of sight makes an angle of 38.0° with the horizontal. How tall is the antenna?

(a) (b)

35.0° 38.0°

85.0 m85.0 m

PROBLEM 16

Problems

22 CHAPTER 1 Introduction and Mathematical Concepts

Section 1.6 Vector Addition and Subtraction 23. E SSM (a) Two workers are trying to move a heavy crate. One pushes on the crate with a force A→, which has a magnitude of 445 newtons and is directed due west. The other pushes with a force B→, which has a magnitude of 325 newtons and is directed due north. What are the magnitude and direc- tion of the resultant force A→ + B→ applied to the crate? (b) Suppose that the second worker applies a force −B→ instead of B→. What then are the magnitude and direction of the resultant force A→ − B→ applied to the crate? In both cases express the direction relative to due west.

24. E A force vector F1 →

points due east and has a magnitude of 200 new- tons. A second force F2

→ is added to F1

→ . The resultant of the two vectors has

a magnitude of 400 newtons and points along the east/west line. Find the magnitude and direction of F2

→ . Note that there are two answers.

25. E SSM Consider the following four force vectors:

F1 →

= 50.0 newtons, due east

F2 →

= 10.0 newtons, due east

F3 →

= 40.0 newtons, due west

F4 →

= 30.0 newtons, due west

Which two vectors add together to give a resultant with the smallest mag- nitude, and which two vectors add to give a resultant with the largest mag- nitude? In each case specify the magnitude and direction of the resultant.

26. E GO Vector A→ has a magnitude of 63 units and points due west, while vector B→ has the same magnitude and points due south. Find the magnitude and direction of (a) A→ + B→ and (b) A→ − B→. Specify the directions relative to due west. 27. E Two bicyclists, starting at the same place, are riding toward the same campground by two diff erent routes. One cyclist rides 1080 m due east and then turns due north and travels another 1430 m before reaching the camp- ground. The second cyclist starts out by heading due north for 1950 m and then turns and heads directly toward the campground. (a) At the turning point, how far is the second cyclist from the campground? (b) In what direc- tion (measured relative to due east) must the second cyclist head during the last part of the trip?

28. E GO The drawing shows a triple jump on a checkerboard, starting at the center of square A and ending on the center of square B. Each side of a square measures 4.0 cm. What is the magnitude of the displacement of the colored checker during the triple jump?

B

A PROBLEM 28

29. E Given the vectors P→ and Q→ shown on the grid, sketch and calculate the magnitudes of the vectors (a) M

→ = P→ + Q→ and (b) K→ = 2P→ − Q→.

Use the tail-to-head method and express the mag- nitudes in centimeters with the aid of the grid scale shown in the drawing.

17. E CHALK The two hot-air balloons in the drawing are 48.2 and 61.0 m above the ground. A person in the left balloon observes that the right balloon is 13.3° above the horizontal. What is the horizontal distance x between the two balloons?

PROBLEM 17

48.2 m

13.3°

61.0 m

x

18. M Available on WileyPLUS. 19. M MMH The drawing shows sodium and chloride ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride (common table salt). The edges of the cube are each 0.281 nm (1 nm = 1 nanometer = 10−9 m) in length. What is the value of the angle θ in the drawing? 20. M GO A person is standing at the edge of the water and looking out at the ocean (see the drawing). The height of the person’s eyes above the water is h = 1.6 m, and the radius of the earth is R = 6.38 × 106 m. (a) How far is it to the horizon? In other words, what is the distance d from the person’s eyes to the horizon? (Note: At the horizon the angle between the line of sight and the radius of the earth is 90°.) (b) Express this distance in miles.

PROBLEM 20

Horizon

h

R

d

R

90°

21. M SSM Three deer, A, B, and C, are grazing in a fi eld. Deer B is located 62 m from deer A at an angle of 51° north of west. Deer C is located 77° north of east relative to deer A. The distance between deer B and C is 95 m. What is the distance between deer A and C? (Hint: Consider the law of cosines given in Appendix E.) 22. H An aerialist on a high platform holds on to a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

8.0 m

0.75 m

41°θ

PROBLEM 22

8.00 cm P

Q

PROBLEM 29

PROBLEM 19

Chloride ion Sodium ion

0.281 nanometers

θ

Problems 23

30. M MMH Vector A→ has a magnitude of 12.3 units and points due west. Vector B→ points due north. (a) What is the magnitude of B→ if A→ + B→ has a magnitude of 15.0 units? (b) What is the direction of A→ + B→ relative to due west? (c) What is the magnitude of B→ if A→ − B→ has a magnitude of 15.0 units? (d) What is the direction of A→ − B→ relative to due west? 31. M SSM A car is being pulled out of the mud by two forces that are ap- plied by the two ropes shown in the drawing. The dashed line in the drawing

bisects the 30.0° angle. The magnitude of the force applied by each rope is

2900 newtons. Arrange the force vectors tail to head and use the graphical

technique to answer the following questions. (a) How much force would a single rope need to apply to accomplish the same eff ect as the two forces

added together? (b) How would the single rope be directed relative to the dashed line?

30.0°

2900 newtons

2900 newtons

PROBLEM 31

32. M GO A jogger travels a route that has two parts. The fi rst is a displace- ment A→ of 2.50 km due south, and the second involves a displacement B→ that points due east. (a) The resultant displacement A→ + B→ has a magnitude of 3.75 km. What is the magnitude of B→, and what is the direction of A→ + B→ relative to due south? (b) Suppose that A→ − B→ had a magnitude of 3.75 km. What then would be the magnitude of B→, and what is the direction of A→ − B→ relative to due south?

33. M At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three directions. As a result, three forces act on

the ball, F1 →

, F2 →

, and F3 →

(see the drawing). The magnitudes of F1 →

and F2 → are

F1 = 50.0 newtons and F2 = 90.0 newtons. Using a scale drawing and the graphical technique, determine (a) the magnitude of F3

→ and (b) the angle θ

such that the resultant force acting on the ball is zero.

60.0°

F3

F1

F2

N

S

W E

θ

PROBLEM 33

Section 1.7 The Components of a Vector 34. E GO A force vector has a magnitude of 575 newtons and points at an angle of 36.0° below the positive x axis. What are (a) the x scalar component and (b) the y scalar component of the vector? 35. E SSM Vector A→ points along the +y axis and has a magnitude of 100.0 units. Vector B→ points at an angle of 60.0° above the +x axis and has a mag- nitude of 200.0 units. Vector C→ points along the +x axis and has a magnitude of 150.0 units. Which vector has (a) the largest x component and (b) the largest y component? 36. E Soccer player #1 is 8.6 m from the goal (see the drawing). If she kicks the ball directly into the net, the ball has a displacement labeled A→. If, on the other hand, she fi rst kicks it to player #2, who then kicks it into the net, the

ball undergoes two successive displacements, A→ y and A →

x. What are the mag-

nitudes and directions of A→ x and A →

y?

#1

Net

90°

30.0°

#2

N

S

W E

Ay A

Ax

PROBLEM 36

37. E GO The components of vector A→ are Ax and Ay (both positive), and the angle that it makes with respect to the positive x axis is θ. Find the angle θ if the components of the displacement vector A→ are (a) Ax = 12 m and Ay = 12 m, (b) Ax = 17 m and Ay = 12 m, and (c) Ax = 12 m and Ay = 17 m. 38. E During takeoff , an airplane climbs with a speed of 180 m/s at an angle of 34° above the horizontal. The speed and direction of the airplane constitute

a vector quantity known as the velocity. The sun is shining directly overhead.

How fast is the shadow of the plane moving along the ground? (That is, what

is the magnitude of the horizontal component of the plane’s velocity?)

39. E SSM The x vector component of a displacement vector r→ has a mag- nitude of 125 m and points along the negative x axis. The y vector component has a magnitude of 184 m and points along the negative y axis. Find the mag- nitude and direction of r→. Specify the direction with respect to the negative x axis. 40. E Your friend has slipped and fallen. To help her up, you pull with a force F→, as the drawing shows. The vertical component of this

force is 130 newtons, and the horizontal com-

ponent is 150 newtons. Find (a) the magnitude of F→ and (b) the angle θ. 41. E Available on WileyPLUS. 42. M GO Two racing boats set out from the same dock and speed away at the same con-

stant speed of 101 km/h for half an hour (0.500 h), the blue boat headed

25.0° south of west, and the green boat headed 37.0° south of west. During

this half hour (a) how much farther west does the blue boat travel, compared to the green boat, and (b) how much farther south does the green boat travel, compared to the blue boat? Express your answers in km.

43. M SSM MMH The magnitude of the force vector F→ is 82.3 newtons. The x component of this vector is directed along the +x axis and has a magnitude of 74.6 newtons. The y component points along the +y axis. (a) Find the direction of F→ relative to the +x axis. (b) Find the component of F→ along the +y axis. 44. H Available on WileyPLUS.

Section 1.8 Addition of Vectors by Means of Components 45. E SSM Consult Multiple-Concept Example 9 in preparation for this problem. A golfer, putting on a green, requires three strokes to “hole the ball.”

During the fi rst putt, the ball rolls 5.0 m due east. For the second putt, the

ball travels 2.1 m at an angle of 20.0° north of east. The third putt is 0.50 m

due north. What displacement (magnitude and direction relative to due east)

would have been needed to “hole the ball” on the very fi rst putt?

F

θ

PROBLEM 40

24 CHAPTER 1 Introduction and Mathematical Concepts

46. E CHALK The three displacement vectors in the drawing have mag- nitudes of A = 5.00 m, B = 5.00 m, and C = 4.00 m. Find the resultant (magnitude and directional angle) of the three vectors by means of the com- ponent method. Express the directional angle as an angle above the positive or negative x axis.

PROBLEM 46

+y

+x

A

B

60.0°20.0°

C

47. E MMH Multiple-Concept Example 9 reviews the concepts that play a role in this problem. Two forces are applied to a tree stump to pull it out of the ground. Force FA

→ has a magnitude of 2240 newtons and points 34.0°

south of east, while force FB →

has a magnitude of 3160 newtons and points due south. Using the component method, fi nd the magnitude and direction of the resultant force FA

→ + FB

→ that is applied to the stump. Specify the direction

with respect to due east.

48. E GO A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force FA

→ , as part a of the drawing shows. By

itself, however, force FA →

is insuffi cient. Therefore, two additional forces FB →

and FC

→ are applied, as in part b of the drawing. Each of these additional

forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b of the drawing is k times larger than that in part a. Find the ratio F/FA when k = 2.00.

20.0° 20.0°

FB

FA FA

FC

(a) (b)

PROBLEM 48

49. E Displacement vector A→ points due east and has a magnitude of 2.00 km. Displacement vector B→ points due north and has a magnitude of 3.75 km. Displacement vector C→ points due west and has a magnitude of 2.50 km. Displacement vector D→ points due south and has a magnitude of 3.00 km. Find the magnitude and direction (relative to due west) of the resultant vector A→ + B→ + C→ + D→. 50. E Multiple-Concept Example 9 provides background pertinent to this problem. The magnitudes of the four displacement vectors shown in the drawing are A = 16.0 m, B = 11.0 m, C = 12.0 m, and D = 26.0 m. Determine the magnitude and directional angle for the resultant that occurs when these vectors are added together.

B

A

D

50.0° +x

+y

35.0° 20.0°

C

PROBLEM 50

51. E MMH Available on WileyPLUS. 52. M GO Two geological fi eld teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the location of the fi rst team as 38 km away, 19° north of west, and the second team as 29 km away, 35° east of north. When the fi rst team uses its GPS to check the position of the second team, what does the GPS give for the second team’s (a) distance from them and (b) direction, measured from due east? 53. M SSM A sailboat race course con- sists of four legs, defi ned by the dis- placement vectors A→, B→, C→, and D→, as the drawing indicates. The mag- nitudes of the fi rst three vectors are A = 3.20 km, B = 5.10 km, and C = 4.80 km. The fi nish line of the course coincides with the starting line. Using the data in the drawing, fi nd the distance of the fourth leg and the angle θ. 54. M Multiple-Concept Example 9 deals with the concepts that are import- ant in this problem. A grasshopper makes four jumps. The displacement vec- tors are (1) 27.0 cm, due west; (2) 23.0 cm, 35.0° south of west; (3) 28.0 cm, 55.0° south of east; and (4) 35.0 cm, 63.0° north of east. Find the magnitude and direction of the resultant displacement. Express the direction with respect to due west.

55. M MMH Available on WileyPLUS. 56. M The route followed by a hiker consists of three displacement vec- tors A→, B→, and C→. Vector A→ is along a measured trail and is 1550 m in a direction 25.0° north of east. Vector B→ is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0° east of south. Similarly, the direction of vector C→ is 35.0° north of west. The hiker ends up back where she started. Therefore, it follows that the res- ultant displacement is zero, or A→ + B→ + C→ = 0. Find the magnitudes of (a) vector B→ and (b) vector C→.

57. E Available on WileyPLUS. 58. E A monkey is chained to a stake in the ground. The stake is 3.00 m from a vertical pole, and the chain is 3.40 m long. How high can the monkey climb up the pole?

59. E SSM Available on WileyPLUS. 60. E BIO The volume of liquid fl owing per second is called the volume fl ow rate Q and has the dimensions of [L]3/[T]. The fl ow rate of a liquid

through a hypodermic needle during an injection can be estimated with the following equation:

Q = πRn (P2 − P1)

8ηL

The length and radius of the needle are L and R, respectively, both of which have the dimension [L]. The pressures at opposite ends of the needle are P2

Additional Problems

23.0°

35.0°

40.0°

Finish Start

B

AD

C

θ

PROBLEM 53

Concepts and Calculations Problems 25

and P1, both of which have the dimensions of [M]/{[L][T]2}. The symbol 𝜂 represents the viscosity of the liquid and has the dimensions of [M]/{[L][T]}. The symbol 𝜋 stands for pi and, like the number 8 and the exponent n, has no dimensions. Using dimensional analysis, determine the value of n in the expression for Q. 61. E An ocean liner leaves New York City and travels 18.0° north of east for 155 km. How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ship’s displacement vector in the directions (a) due east and (b) due north? 62. E GO A pilot fl ies her route in two straight-line segments. The dis- placement vector A→ for the fi rst segment has a magnitude of 244 km and a direction 30.0° north of east. The displacement vector B→ for the second segment has a magnitude of 175 km and a direction due west. The resultant displacement vector is R→ = A→ + B→ and makes an angle θ with the direction due east. Using the component method, fi nd the magnitude of R→ and the directional angle θ. 63. E SSM Available on WileyPLUS. 64. M Available on WileyPLUS. 65. M SSM Vector A→ has a magnitude of 6.00 units and points due east. Vector B→ points due north. (a) What is the magnitude of B→, if the vector A→ + B→ points 60.0° north of east? (b) Find the magnitude of A→ + B→. 66. M GO Three forces act on an object, as indicated in the drawing. Force F1

→ has a mag-

nitude of 21.0 newtons (21.0 N) and is direc- ted 30.0° to the left of the +y axis. Force F2

→

has a magnitude of 15.0 N and points along the +x axis. What must be the magnitude and direction (specifi ed by the angle θ in the drawing) of the third force F3

→ such that the

vector sum of the three forces is 0 N?

67. M Available on WileyPLUS. 68. M CHALK You live in the building on the left in the drawing, and a friend lives in the other building. The two of you are having a discussion about the heights of the buildings, and your friend claims that the height of his building is more than 1.50 times the height of yours. To resolve the issue you climb to the roof of your building and estimate that your line of sight to the top edge of the other building makes an angle of 21° above the horizontal, whereas your line of sight to the base of the other building makes an angle of 52° below the horizontal. Determine the ratio of the height of the taller building to the height of the shorter building. State whether your friend is right or wrong.

21° 52°

PROBLEM 68

69. H Available on WileyPLUS.

+x

+y

30.0°

F1

F2

F3

θ

PROBLEM 66

This chapter has presented an introduction to the mathematics of trigono- metry and vectors, which will be used throughout the text. In this section we apply some of the important features of this mathematics, and review some concepts that can help in anticipating some of the characteristics of the numerical answers.

70. M CHALK The fi gure shows two dis- placement vectors A→ and B→. Vector A→ points at an angle of 22° above the x axis and has an unknown magnitude. Vector B→ has an x component Bx = 35.0 m and has an unknown y component By. These two vectors are equal. Concepts: (i) What does the condition that vector A→ equals B→ imply about the magnitudes and dir- ections of the vectors? (ii) What does the condition that vector A→ equals B→ imply about the x and y components of the vec- tors? Calculations: Find the magnitude of A→ and the value of By.

71. M CHALK SSM The fi gure shows three displacement vectors A→, B→, and C→. These vectors are arranged in tail-to-head fashion, and they add together to give a resultant displacement R→ , which lies along the x axis. Note that the vector B→ is parallel to the x axis. Concepts: (i) How is the magnitude of A→ related to its scalar components Ax and Ay? (ii) Do any of the vectors in the fi gure have a zero value for either their x or y components? (If so, which ones?) (iii) What does the fact that A→, B→, and C→ add together to give R→ tell you about the components of these vectors? Calculations: What is the mag- nitude of the vector A→ and its directional angle θ?

+y

B

A C

R +x

10.0 m

35.0 m

50.0°

23.0 m

θ

PROBLEM 71

+y

35.0 m

Bx

22.0°

By

B

+x

+y

A

+x

(b)

(a)

θ

PROBLEM 70

Concepts and Calculations Problems

26 CHAPTER 1 Introduction and Mathematical Concepts

72. M The Waterfall. You and your team are exploring a river in South America when you come to the bottom of a tall waterfall. You estimate the

cliff over which the water fl ows to be about 100 feet tall. You have to choose

between climbing the cliff or backtracking and taking another route, but

climbing the cliff would cut two hours off of your trip. There is only one

experienced climber in the group: she would climb the cliff alone and drop a

rope over the edge to lift supplies and allow the others to climb without packs.

The climber estimates it will take her 45 minutes to get to the top. However,

you are concerned that the rope might be too short to reach the bottom of the

cliff (it is exactly 30.0 m long). If it is too short, she’ll have to climb back

down (another 45 minutes) and you will be too far behind schedule to get to

your destination before dark. As you contemplate how to determine whether

the rope is long enough, you notice that the late afternoon shadow of the

cliff grows as the sun descends over its edge. You suddenly remember your

trigonometry. You measure the length of the shadow from the base of the cliff

to the shadow’s edge (144 ft), and the angle subtended between the base and

top of the cliff measured from the shadow’s edge. The angle is 38.1°. Do you

send the climber, or start backtracking to take another route?

73. M The Weather Monitor. Your South American expedition splits into two groups: one that stays at home base, and yours that goes off to set up a

sensor that will monitor precipitation, temperature, and sunlight through the

upcoming winter. The sensor must link up to a central communications sys-

tem at base camp that simultaneously uploads the data from numerous sensors

to a satellite. In order to set up and calibrate the sensor, you will have to com-

municate with base camp to give them specifi c location information. Unfor-

tunately, the group’s communication and navigation equipment has dwindled

to walkie-talkies and a compass due to a river-raft mishap, which means your

group must not exceed the range of the walkie-talkies (3.0 miles). However,

you do have a laser rangefi nder to help you measure distances as you navigate

with the compass. After a few hours of hiking, you fi nd the perfect plateau on

which to mount the sensor. You have carefully mapped your path from base

camp around lakes and other obstacles: 550 m West (W), 275 m S, 750 m W,

900 m NE, 800 m W, and 400 m 30.0° W of S. The fi nal leg is due south,

2.20 km up a constant slope and ending at a plateau that is 320 m above the

level of base camp. (a) How far are you from base camp? Will you be able to communicate with home base using the walkie-talkies? (b) What is the geographical direction from base camp to the sensor (expressed in the form

θ° south of west, etc.)? (c) What is the angle of inclination from base camp to the detector?

Team Problems

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

2.1 Define one-dimensional displacement.

2.2 Discriminate between speed and velocity.

2.3 Define one-dimensional acceleration.

2.4 Use one-dimensional kinematic equations to predict future or past values of variables.

2.5 Solve one-dimensional kinematic problems.

2.6 Solve one-dimensional free-fall problems.

2.7 Predict kinematic quantities using graphical analysis.

er n ie

d ec

k er

/i S

to ck

p h o to

CHAPTER 2

Kinematics in One Dimension

The pilots in the United States Navy’s Blue Angels can perform high-speed maneuvers in perfect unison.

They do so by controlling the displacement, velocity, and acceleration of their jet aircraft. These three

concepts and the relationships among them are the focus of this chapter.

2.1 Displacement There are two aspects to any motion. In a purely descriptive sense, there is the move-

ment itself. Is it rapid or slow, for instance? Then, there is the issue of what causes

the motion or what changes it, which requires that forces be considered. Kinematics deals with the concepts that are needed to describe motion, without any reference to

forces. The present chapter discusses these concepts as they apply to motion in one

dimension, and the next chapter treats two-dimensional motion. Dynamics deals with the eff ect that forces have on motion, a topic that is considered in Chapter 4. Together,

kinematics and dynamics form the branch of physics known as mechanics. We turn now to the fi rst of the kinematics concepts to be discussed, which is displacement.

To describe the motion of an object, we must be able to specify the location of

the object at all times, and Figure 2.1 shows how to do this for one-dimensional motion. In this drawing, the initial position of a car is indicated by the vector

labeled x0→ . The length of x0→ is the distance of the car from an arbitrarily chosen origin. At a later time the car has moved to a new position, which is indicated by the

vector x→. The displacement of the car ∆ x→ (read as “delta x” or “the change in x”) is a vector drawn from the initial position to the fi nal position. Displacement is a vec-

tor quantity in the sense discussed in Section 1.5, for it conveys both a magnitude 27

28 CHAPTER 2 Kinematics in One Dimension

(the distance between the initial and fi nal positions) and a direction. The displacement can be

related to x0→ and x→ by noting from the drawing that x0→ + ∆x→ = x→ or ∆x→ = x→ − x0→

Thus, the displacement ∆x→ is the diff erence between x→ and x0→ , and the Greek letter delta (∆) is used to signify this diff erence. It is important to note that the change in any variable is always the

fi nal value minus the initial value.

DEFINITION OF DISPLACEMENT The displacement is a vector that points from an object’s initial position to its fi nal position and has a magnitude that equals the shortest distance between the two positions. SI Unit of Displacement: meter (m)

The SI unit for displacement is the meter (m), but there are other units as well, such as the

centimeter and the inch. When converting between centimeters (cm) and inches (in.), remember

that 2.54 cm = 1 in.

Often, we will deal with motion along a straight line. In such a case, a displacement in one

direction along the line is assigned a positive value, and a displacement in the opposite direction is

assigned a negative value. For instance, assume that a car is moving along an east/west direction

and that a positive (+) sign is used to denote a direction due east. Then, ∆x→ = +500 m represents a displacement that points to the east and has a magnitude of 500 meters. Conversely, ∆x→ = −500 m is a displacement that has the same magnitude but points in the opposite direction, due west.

The magnitude of the displacement vector is the shortest distance between the initial and

fi nal positions of the object. However, this does not mean that displacement and distance are the

same physical quantities. In Figure 2.1, for example, the car could reach its fi nal position after going forward and backing up several times. In that case, the total distance traveled by the car

would be greater than the magnitude of the displacement vector.

Check Your Understanding

(The answer is given at the end of the book.) 1. A honeybee leaves the hive and travels a total distance of 2 km before returning to the hive. What is the

magnitude of the displacement vector of the bee?

2.2 Speed and Velocity

Average Speed One of the most obvious features of an object in motion is how fast it is moving. If a car travels

200 meters in 10 seconds, we say its average speed is 20 meters per second, the average speed being the distance traveled divided by the time required to cover the distance:

Average speed = Distance

Elapsed time (2.1)

Equation 2.1 indicates that the unit for average speed is the unit for distance divided by the unit for time,

or meters per second (m/s) in SI units. Example 1 illustrates how the idea of average speed is used.

Origin

Displacement = Δx

t0

x0

x

t

FIGURE 2.1 The displacement ∆x→ is a vector that points from the initial position x→ 0 to the fi nal position x→.

EXAMPLE 1 Distance Run by a Jogger

How far does a jogger run in 1.5 hours (5400 s) if his average speed is

2.22 m/s?

Reasoning The average speed of the jogger is the average distance per second that he travels. Thus, the distance covered by the jogger is equal

to the average distance per second (his average speed) multiplied by the

number of seconds (the elapsed time) that he runs.

Solution To fi nd the distance run, we rewrite Equation 2.1 as

Distance = (Average speed)(Elapsed time) = (2.22 m /s)(5400 s)

= 12 000 m

2.2 Speed and Velocity 29

Speed is a useful idea, because it indicates how fast an object is moving. However, speed does not

reveal anything about the direction of the motion. To describe both how fast an object moves and

the direction of its motion, we need the vector concept of velocity.

Average Velocity To defi ne the velocity of an object, we will use two concepts that we have already encountered,

displacement and time. The building of new concepts from more basic ones is a common theme

in physics. In fact the great strength of physics as a science is that it builds a coherent understand-

ing of nature through the development of interrelated concepts.

Suppose that the initial position of the car in Figure 2.1 is x→ 0 when the time is t0. A little later that car arrives at the fi nal position x→ at the time t. The diff erence between these times is the time required for the car to travel between the two positions. We denote this diff erence

by the shorthand notation ∆t (read as “delta t”), where ∆t represents the fi nal time t minus the initial time t0:

∆t = t − t0 Elapsed time

Note that ∆t is defi ned in a manner analogous to ∆x→, which is the fi nal position minus the initial position (∆x→ = x→ − x→ 0). Dividing the displacement ∆x→ of the car by the elapsed time ∆t gives the average velocity of the car. It is customary to denote the average value of a quantity by pla- cing a horizontal bar above the symbol representing the quantity. The average velocity, then, is

written as v→, as specifi ed in Equation 2.2:

DEFINITION OF AVERAGE VELOCITY

Average velocity = Displacement

Elapsed time

v→ = x→ − x→0 t − t0

= ∆ x→ ∆ t

(2.2)

SI Unit of Average Velocity: meter per second (m/s)

Equation 2.2 indicates that the unit for average velocity is the unit for length divided by the

unit for time, or meters per second (m/s) in SI units. Velocity can also be expressed in other units,

such as kilometers per hour (km/h) or miles per hour (mi/h).

Average velocity is a vector that points in the same direction as the displacement in Equa-

tion 2.2. Figure 2.2 illustrates that the velocity of an object confi ned to move along a line can point either in one direction or in the opposite direction. As with displacement, we will use

plus and minus signs to indicate the two possible directions. If the displacement points in the

positive direction, the average velocity is positive. Conversely, if the displacement points in

the negative direction, the average velocity is negative. Example 2 illustrates these features of

average velocity.

⏟

FIGURE 2.2 The boats in this photograph are traveling in opposite directions; in other

words, the velocity of one boat points in a

direction that is opposite to the velocity of

the other boat.

© Michael Dietrich/Age Fotostock

EXAMPLE 2 The World’s Fastest Jet-Engine Car

Andy Green in the car ThrustSSC set a world record of 341.1 m/s (763 mi/h) in 1997. The car was powered by two jet engines, and it was

the fi rst one offi cially to exceed the speed of sound. To establish such a

record, the driver makes two runs through the course, one in each direc-

tion, to nullify wind eff ects. Figure 2.3a shows that the car fi rst travels from left to right and covers a distance of 1609 m (1 mile) in a time of

4.740 s. Figure 2.3b shows that in the reverse direction, the car covers the same distance in 4.695 s. From these data, determine the average

velocity for each run.

Reasoning Average velocity is defi ned as the displacement divided by the elapsed time. In using this defi nition we recognize that the displace-

ment is not the same as the distance traveled. Displacement takes the dir-

ection of the motion into account, and distance does not. During both

runs, the car covers the same distance of 1609 m. However, for the fi rst

run the displacement is ∆ x→ = +1609 m, while for the second it is ∆ x→ = −1609 m. The plus and minus signs are essential, because the fi rst run is

to the right, which is the positive direction, and the second run is in the

opposite or negative direction.

30 CHAPTER 2 Kinematics in One Dimension

Instantaneous Velocity Suppose the magnitude of your average velocity for a long trip was 20 m/s. This value, being an

average, does not convey any information about how fast you were moving or the direction of

the motion at any instant during the trip. Both can change from one instant to another. Surely

there were times when your car traveled faster than 20 m/s and times when it traveled more

slowly. The instantaneous velocity v→ of the car indicates how fast the car moves and the direc- tion of the motion at each instant of time. The magnitude of the instantaneous velocity is called

the instantaneous speed, and it is the number (with units) indicated by the speedometer. The instantaneous velocity at any point during a trip can be obtained by measuring the time

interval ∆t for the car to travel a very small displacement ∆x→. We can then compute the average velocity over this interval. If the time ∆t is small enough, the instantaneous velocity does not change much during the measurement. Then, the instantaneous velocity v→ at the point of interest is approximately equal to (≈) the average velocity v→ computed over the interval, or v→ ≈ v→ = ∆x→/∆t (for suffi ciently small ∆t). In fact, in the limit that ∆t becomes infi nitesimally small, the instant- aneous velocity and the average velocity become equal, so that

v→ = lim ∆t → 0

∆x→

∆t (2.3)

The notation lim ∆ t → 0

∆x→

∆t means that the ratio ∆x→/∆t is defi ned by a limiting process in

which smaller and smaller values of ∆t are used, so small that they approach zero. As smaller values of ∆t are used, ∆x→ also becomes smaller. However, the ratio ∆x→/∆t does not become zero but, rather, approaches the value of the instantaneous velocity. For brevity, we will use the word

velocity to mean “instantaneous velocity” and speed to mean “instantaneous speed.”

Check Your Understanding

(The answers are given at the end of the book.) 2. Is the average speed of a vehicle a vector or a scalar quantity? 3. Two buses depart from Chicago, one going to New York and one to San Francisco. Each bus travels at

a speed of 30 m/s. Do they have equal velocities?

4. One of the following statements is incorrect. (a) The car traveled around the circular track at a con- stant velocity. (b) The car traveled around the circular track at a constant speed. Which statement is incorrect?

Start Finish

Δx = +1609 m

t0 = 0 s

(a)

t = 4.740 s

Finish Start

Δx = –1609 m

t0 = 0 s

(b)

t = 4.695 s

– +

FIGURE 2.3 The arrows in the box at the top of the drawing indicate the positive and negative directions for the displacements of the car, as

explained in Example 2.

Solution According to Equation 2.2, the average velocities are

Run 1 v→ = ∆x →

∆t =

+1609 m

4.740 s = +339.5 m/s

Run 2 v→ = ∆x →

∆t =

−1609 m

4.695 s = −342.7 m/s

In these answers the algebraic signs convey the directions of the velocity

vectors. In particular, for run 2 the minus sign indicates that the average

velocity, like the displacement, points to the left in Figure 2.3b. The magnitudes of the velocities are 339.5 and 342.7 m/s. The average of

these numbers is 341.1 m/s, and this is recorded in the record book.

2.3 Acceleration 31

5. A straight track is 1600 m in length. A runner begins at the starting line, runs due east for the full length of the track, turns around and runs halfway back. The time for this run is fi ve minutes. What is the

runner’s average velocity, and what is his average speed?

6. The average velocity for a trip has a positive value. Is it possible for the instantaneous velocity at a point during the trip to have a negative value?

2.3 Acceleration In a wide range of motions, the velocity changes from moment to moment, such as in the case of

the sprinter in Photo 2.1. To describe the manner in which it changes, the concept of acceleration is needed. The velocity of a moving object may change in a number of ways. For example, it may

increase, as it does when the driver of a car steps on the gas pedal to pass the car ahead. Or it may

decrease, as it does when the driver applies the brakes to stop at a red light. In either case, the

change in velocity may occur over a short or a long time interval. To describe how the velocity

of an object changes during a given time interval, we now introduce the new idea of acceleration.

This idea depends on two concepts that we have previously encountered, velocity and time. Spe-

cifi cally, the notion of acceleration emerges when the change in the velocity is combined with the time during which the change occurs.

The meaning of average acceleration can be illustrated by considering a plane during takeoff . Figure 2.4 focuses attention on how the plane’s velocity changes along the runway. Dur- ing an elapsed time interval ∆t = t − t0, the velocity changes from an initial value of v→ 0 to a fi nal velocity of v→. The change ∆v→ in the plane’s velocity is its fi nal velocity minus its initial velocity, so that ∆v→ = v→ − v→ 0. The average acceleration a→ is defi ned in the following manner, to provide a measure of how much the velocity changes per unit of elapsed time.

DEFINITION OF AVERAGE ACCELERATION

Average acceleration = Change in velocity

Elapsed time

a→ =

v→ − v0→

t − t0 =

∆ v→ ∆t (2.4)

SI Unit of Average Acceleration: meter per second squared (m/s2)

The average acceleration a→ is a vector that points in the same direction as ∆v→, the change in the velocity. Following the usual custom, plus and minus signs indicate the two possible direc-

tions for the acceleration vector when the motion is along a straight line.

We are often interested in an object’s acceleration at a particular instant of time. The

instantaneous acceleration a→ can be defi ned by analogy with the procedure used in Section 2.2 for instantaneous velocity:

a→ = lim ∆ t → 0

∆v→

∆t (2.5)

Equation 2.5 indicates that the instantaneous acceleration is a limiting case of the average

acceleration. When the time interval ∆t for measuring the acceleration becomes extremely small

t0

v0 v

t– +

a

FIGURE 2.4 During takeoff , the plane accelerates from an initial velocity v→0 to a fi nal velocity v→ during the time interval ∆t = t − t0.

PHOTO 2.1 As this sprinter explodes out of the starting block, his velocity is changing,

which means that he is accelerating.

Jim Cummins/Taxi/Getty Images

32 CHAPTER 2 Kinematics in One Dimension

(approaching zero in the limit), the average acceleration and the instantaneous acceleration be-

come equal. Moreover, in many situations the acceleration is constant, so the acceleration has

the same value at any instant of time. In the future, we will use the word acceleration to mean “instantaneous acceleration.” Example 3 deals with the acceleration of a plane during takeoff .

EXAMPLE 3 Acceleration and Increasing Velocity

Suppose the plane in Figure 2.4 starts from rest (v→0 = 0 m/s) when t0 = 0 s. The plane accelerates down the runway and at t = 29 s attains a velocity of v→ = +260 km/h, where the plus sign indicates that the velocity points to the right. Determine the average acceleration of the plane.

Reasoning The average acceleration of the plane is defi ned as the change in its velocity divided by the elapsed time. The change in the plane’s

velocity is its fi nal velocity v→ minus its initial velocity v→0, or v→ − v→0. The elapsed time is the fi nal time t minus the initial time t0, or t − t0.

Problem-Solving Insight The change in any variable is the fi nal value minus the initial value: for example, the change in velocity is ∆v→ = v→ − v0→ , and the change in time is ∆ t = t − t0.

Solution The average acceleration is expressed by Equation 2.4 as

a→ = v→ − v→0 t − t0

= 260 km/h − 0 km/h

29 s − 0 s = +9.0

km /h

s

The average acceleration calculated in Example 3 is read as “nine kilometers per hour

per second.” Assuming the acceleration of the plane is constant, a value of +9.0 km/h

s

means the velocity changes by +9.0 km/h during each second of the motion. During the fi rst

second, the velocity increases from 0 to 9.0 km/h; during the next second, the velocity increases

by another 9.0 km/h to 18 km/h, and so on. Figure 2.5 illustrates how the velocity changes dur- ing the fi rst two seconds. By the end of the 29th second, the velocity is 260 km/h.

It is customary to express the units for acceleration solely in terms of SI units. One way to

obtain SI units for the acceleration in Example 3 is to convert the velocity units from km/h to m/s:

(260 kmh )( 1000 m

1 km )( 1 h

3600 s) = 72 m

s

The average acceleration then becomes

a→ = 72 m/s − 0 m/s 29 s − 0 s

= +2.5 m/s2

where we have used 2.5 m/s

s = 2.5

m

s ·s = 2.5 m

s2 . An acceleration of 2.5

m

s2 is read as

“2.5 meters per second per second” (or “2.5 meters per second squared”) and means that the

velocity changes by 2.5 m/s during each second of the motion.

Example 4 deals with a case where the motion becomes slower as time passes.

Δt = 0 s

Δt = 1.0 s

Δt = 2.0 s

v0 = 0 m/s

v = +9.0 km/h

v = +18 km/h

a = +9.0 km/h

s

FIGURE 2.5 An acceleration of +9.0 km /h

s

means that the velocity of the plane changes

by +9.0 km/h during each second of the

motion. The “+” direction for a→ and v → is to the right.

2.3 Acceleration 33

Figure 2.7 shows how the velocity of the dragster in Example 4 changes during the braking, assuming that the acceleration is constant throughout the motion. The acceleration

calculated in Example 4 is negative, indicating that the acceleration points to the left in the

drawing. As a result, the acceleration and the velocity point in opposite directions.

Problem-Solving Insight Whenever the acceleration and velocity vectors have opposite directions, the object slows down and is said to be “decelerating.”

In contrast, the acceleration and velocity vectors in Figure 2.5 point in the same direction, and the object speeds up.

Check Your Understanding

(The answers are given at the end of the book.) 7. At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The

car has a greater velocity than the truck has. Does the car necessarily have the greater acceleration?

EXAMPLE 4 Acceleration and Decreasing Velocity

A drag racer crosses the fi nish line, and the driver deploys a parachute and

applies the brakes to slow down, as Figure 2.6 illustrates. The driver be- gins slowing down when t0 = 9.0 s and the car’s velocity is v→ 0 = +28 m/s. When t = 12.0 s, the velocity has been reduced to v→ = +13 m/s. What is the average acceleration of the dragster?

Reasoning The average acceleration of an object is always specifi ed as its change in velocity, v→ − v→ 0, divided by the elapsed time, t − t0. This

is true whether the fi nal velocity is less than the initial velocity or greater

than the initial velocity.

Solution The average acceleration is, according to Equation 2.4,

a→ = v→ − v→0 t − t0

= 13 m/s − 28 m/s

12.0 s − 9.0 s = −5.0 m /s2

FIGURE 2.6 (a) To slow down, a drag racer deploys a parachute and applies the brakes. (b) The velocity of the car is decreasing, giving rise to an average acceleration a→ that points opposite to the velocity.

v0 = +28 m/s v = +13 m/s

a = –5.0 m/s2

t0 = 9.0 s t = 12 s

– +

(b)(a)

C h ri

s S

za g o la

/C al

S p o rt

M ed

ia /

N ew

sc o m

FIGURE 2.7 Here, an acceleration of −5.0 m/s2 means the velocity decreases by 5.0 m/s during each

second of elapsed time.

Δt = 0 s

Δt = 1.0 s

Δt = 2.0 s

v0 = +28 m/s

v = +23 m/s

v = +18 m/s

a = –5.0 m/s 2

Continued

34 CHAPTER 2 Kinematics in One Dimension

8. Two cars are moving in the same direction (the positive direction) on a straight road. The acceleration of each car also points in the positive direction. Car 1 has a greater acceleration than car 2 has. Which

one of the following statements is true? (a) The velocity of car 1 is always greater than the velocity of car 2. (b) The velocity of car 2 is always greater than the velocity of car 1. (c) In the same time interval, the velocity of car 1 changes by a greater amount than the velocity of car 2 does. (d) In the same time interval, the velocity of car 2 changes by a greater amount than the velocity of car 1 does.

9. An object moving with a constant acceleration slows down if the acceleration points in the direction opposite to the direction of the velocity. But can an object ever come to a permanent halt if its acceler-

ation truly remains constant?

10. A runner runs half the remaining distance to the fi nish line every ten seconds. She runs in a straight line and does not ever reverse her direction. Does her acceleration have a constant magnitude?

2.4 Equations of Kinematics for Constant Acceleration It is now possible to describe the motion of an object traveling with a constant acceleration

along a straight line. To do so, we will use a set of equations known as the equations of

kinematics for constant acceleration. These equations entail no new concepts, because they

will be obtained by combining the familiar ideas of displacement, velocity, and acceleration.

However, they will provide a very convenient way to determine certain aspects of the motion,

such as the fi nal position and velocity of a moving object.

In discussing the equations of kinematics, it will be convenient to assume that the object is

located at the origin x→0 = 0 m when t0 = 0 s. With this assumption, the displacement ∆x→ = x→ − x→0 becomes ∆x→ = x→. Furthermore, in the equations that follow, as is customary, we dispense with the use of boldface symbols overdrawn with small arrows for the displacement, velocity, and

acceleration vectors. We will, however, continue to convey the directions of these vectors with

plus or minus signs.

Consider an object that has an initial velocity of υ0 at time t0 = 0 s and moves for a time t with a constant acceleration a. For a complete description of the motion, it is also necessary to know the fi nal velocity and displacement at time t. The fi nal velocity υ can be obtained directly from Equation 2.4:

a = a = υ − υ0

t  or υ = υ0 + at (constant acceleration) (2.4)

The displacement x at time t can be obtained from Equation 2.2, if a value for the average velocity υ can be obtained. Considering the assumption that x→0 = 0 m at t0 = 0 s, we have

υ = x − x0 t − t0

= x t  or x = υt (2.2)

Because the acceleration is constant, the velocity increases at a constant rate. Thus, the average

velocity υ is midway between the initial and fi nal velocities:

υ = 12(υ0 + υ) (constant acceleration) (2.6)

Equation 2.6, like Equation 2.4, applies only if the acceleration is constant and cannot be used

when the acceleration is changing. The displacement at time t can now be determined as

x = υt = 12(υ0 + υ)t (constant acceleration) (2.7)

Notice in Equations 2.4 (υ = υ0 + at) and 2.7 [x = 1

2(υ0 + υ)t] that there are fi ve kinematic variables:

1. x = displacement 2. a = a = acceleration (constant) 3. υ = fi nal velocity at time t 4. υ0 = initial velocity at time t0 = 0 s 5. t = time elapsed since t0 = 0 s

2.4 Equations of Kinematics for Constant Acceleration 35

Each of the two equations contains four of these variables, so if three of them are known, the

fourth variable can always be found. Example 5 illustrates how Equations 2.4 and 2.7 are used to

describe the motion of an object.

Analyzing Multiple-Concept Problems

EXAMPLE 5 The Displacement of a Speedboat

The speedboat in Figure 2.8 has a constant acceleration of +2.0 m/s2. If the initial velocity of the boat is +6.0 m/s, fi nd the boat’s displacement

after 8.0 seconds.

Reasoning As the speedboat accelerates, its velocity is changing. The displacement of the speedboat during a given time interval is equal to

the product of its average velocity during that interval and the time. The

average velocity, on the other hand, is just one-half the sum of the boat’s

initial and fi nal velocities. To obtain the fi nal velocity, we will use the

defi nition of constant acceleration, as given in Equation 2.4.

Knowns and Unknowns The numerical values for the three known variables are listed in the table:

Description Symbol Value Comment Acceleration a +2.0 m/s2 Positive, because the boat is accelerating to the right, which is the positive

direction. See Figure 2.8b.

Initial velocity υ0 +6.0 m/s Positive, because the boat is moving to the right, which is the positive direction. See Figure 2.8b.

Time interval t 8.0 s

Unknown Variable Displacement of boat x ?

FIGURE 2.8 (a) An accelerating speedboat. (b) The boat’s displacement x can be determined if the boat’s acceleration, initial velocity, and time of travel are known.

t0 = 0 s

x

a = +2.0 m/s2

0 = +6.0 m/sυ υ

t = 8.0 s

(b)

– +

(a)

A n th

o n y B

ra d sh

aw /G

et ty

Im ag

es

Modeling the Problem

STEP 1 Displacement Since the acceleration is constant, the displacement x of the boat is given by Equation 2.7, in which υ0 and υ are the initial and fi nal velocities, respectively. In this equation, two of the variables, υ0 and t, are known (see the Knowns and Unknowns table), but the fi nal velocity υ is not. However, the fi nal velocity can be determined by employing the defi nition of acceleration, as Step 2 shows.

STEP 2 Acceleration According to Equation 2.4, which is just the defi nition of constant acceleration, the fi nal velocity υ of the boat is

𝜐 = 𝜐0 + at (2.4)

All the variables on the right side of the equals sign are known, and we can substitute this relation

for υ into Equation 2.7, as shown at the right.

?

x = 12(𝜐0 + 𝜐)t (2.7)

x = 12(𝜐0 + 𝜐)t (2.7)

𝜐 = 𝜐0 + at (2.4)

36 CHAPTER 2 Kinematics in One Dimension

In Example 5, we combined two equations [x = 12(υ0 + υ)t and υ = υ0 + at] into a single equation by algebraically eliminating the fi nal velocity υ of the speedboat (which was not known). The result was the following expression for the displacement x of the speedboat:

x = υ0t + 1

2 at 2 (constant acceleration) (2.8)

The fi rst term (υ0t) on the right side of this equation represents the displacement that would res- ult if the acceleration were zero and the velocity remained constant at its initial value of υ0. The second term (

1

2 at 2) gives the additional displacement that arises because the velocity changes (a is not zero) to values that are diff erent from its initial value.

Equation 2.8 allows us to calculate the displacement of an object undergoing constant accel-

eration, given its acceleration (a), the initial velocity (υ0), and the time (t). However, even if the time interval is not given, we can still calculate the displacement, provided we also know the fi nal

velocity (υ) of the object. We can solve Equation 2.4 for the time t, and then make this substitu- tion into Equation 2.8. The result is x = (υ2 − υ02) /2a. Solving for υ2 gives

υ2 = υ 20 + 2ax (constant acceleration) (2.9)

Equation 2.9 is often used when the time involved in the motion is unknown.

Table 2.1 presents a summary of the equations that we have been considering. These equa- tions are called the equations of kinematics. Each equation contains four of the fi ve kinematic variables, as indicated by the check marks (✓) in the table. The next section shows how to apply

the equations of kinematics.

Check Your Understanding

(The answers are given at the end of the book.) 11. The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity

of one rifl e with a short barrel is greater than the muzzle velocity of another rifl e that has a longer bar-

rel. In which rifl e is the acceleration of the bullet larger?

12. A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a dis- placement x and attains a fi nal velocity υ. Then t is increased so that the displacement is 3x. In this same increased time interval, what fi nal velocity does the motorcycle attain?

Solution Algebraically combining the results of Steps 1 and 2, we fi nd that

x = 12(υ0 + υ)t = 1

2[𝜐0 + (υ0 + at) ]t = υ0t + 1

2at 2

The displacement of the boat after 8.0 s is

x = υ0t + 1

2at 2 = (+6.0 m/s)(8.0 s) + 1

2 (+2.0 m/s 2)(8.0 s)2 = +112 m

Related Homework: Problems 39, 55

STEP 1 STEP 2

TABLE 2.1 Equations of Kinematics for Constant Acceleration

Variables Equation Number Equation x a υ υ0 t (2.4) 𝜐 = 𝜐0 + at — ✓ ✓ ✓ ✓

(2.7) x = 12(𝜐0 + 𝜐)t ✓ — ✓ ✓ ✓

(2.8) x = 𝜐0t + 1

2 at 2 ✓ ✓ — ✓ ✓

(2.9) 𝜐 2 = 𝜐0  2 + 2ax ✓ ✓ ✓ ✓ —

2.5 Applications of the Equations of Kinematics 37

2.5 Applications of the Equations of Kinematics The equations of kinematics can be applied to any moving object, as long as the acceleration of

the object is constant. However, remember that each equation contains four variables. Therefore,

numerical values for three of the four must be available if an equation is to be used to calculate the

value of the remaining variable. To avoid errors when using these equations, it helps to follow a

few sensible guidelines and to be alert for a few situations that can arise during your calculations.

Problem-Solving Insight Decide at the start which directions are to be called positive (+) and negative (−) relative to a conveniently chosen coordinate origin.

This decision is arbitrary, but important because displacement, velocity, and acceleration are vectors, and their directions must always be taken into account. In the examples that follow, the positive and negative directions will be shown in the drawings that accompany the problems. It does not matter which direction is chosen to be positive. However, once the choice is made, it should not be changed during the course of the calculation.

Problem-Solving Insight As you reason through a problem before attempting to solve it, be sure to interpret the terms “decelerating” or “deceleration” correctly, should they occur in the

problem statement.

These terms are the source of frequent confusion, and Conceptual Example 6 off ers help in understanding them.

Problem-Solving Insight Sometimes there are two possible answers to a kinematics problem, each answer corresponding to a diff erent situation.

Example 7 discusses one such case.

CONCEPTUAL EXAMPLE 6 Deceleration Versus Negative Acceleration

A car is traveling along a straight road and is decelerating. Which one of

the following statements correctly describes the car’s acceleration? (a) It must be positive. (b) It must be negative. (c) It could be positive or negative.

Reasoning The term “decelerating” means that the acceleration vector points opposite to the velocity vector and indicates that the car is slowing

down. One possibility is that the velocity vector of the car points to the

right, in the positive direction, as Figure 2.9a shows. The term “deceler- ating” implies, then, that the acceleration vector of the car points to the

left, which is the negative direction. Another possibility is that the car

is traveling to the left, as in Figure 2.9b. Now, since the velocity vector points to the left, the acceleration vector would point opposite, or to the

right, which is the positive direction.

Answers (a) and (b) are incorrect. The term “decelerating” means only that the acceleration vector points opposite to the velocity vector. It

is not specifi ed whether the velocity vector of the car points in the positive

or negative direction. Therefore, it is not possible to know whether the

acceleration is positive or negative.

Answer (c) is correct. As shown in Figure 2.9, the acceleration vector of the car could point in the positive or the negative direction, so that the

acceleration could be either positive or negative, depending on the direc-

tion in which the car is moving.

Related Homework: Problems 14, 73

FIGURE 2.9 When a car decelerates along a straight road, the acceleration

vector points opposite to the velocity

vector, as Conceptual Example 6

discusses.

(a)

a

v

(b)

a

v

– +

– +

38 CHAPTER 2 Kinematics in One Dimension

Problem-Solving Insight The motion of two objects may be interrelated, so that they share a common variable. The fact that the motions are interrelated is an important piece of inform-

ation. In such cases, data for only two variables need be specifi ed for each object.

See Interactive LearningWare 2.2 at www.wiley.com/college/cutnell for an example that illustrates this.

Problem-Solving Insight Often the motion of an object is divided into segments, each with a diff erent acceleration. When solving such problems, it is important to realize that

the fi nal velocity for one segment is the initial velocity for the next segment, as Example 8

illustrates.

EXAMPLE 7 The Physics of Spacecraft Retrorockets

The spacecraft shown in Figure 2.10a is traveling with a velocity of +3250 m/s. Suddenly the retrorockets are

fi red, and the spacecraft begins to slow down with an

acceleration whose magnitude is 10.0 m/s2. What is the

velocity of the spacecraft when the displacement of the

craft is +215 km, relative to the point where the ret-

rorockets began fi ring?

Reasoning Since the spacecraft is slowing down, the acceleration must be opposite to the velocity. The

velocity points to the right in the drawing, so the ac-

celeration points to the left, in the negative direction;

thus, a = −10.0 m/s2. The three known variables are listed as follows:

The fi nal velocity υ of the spacecraft can be calcu- lated using Equation 2.9, since it contains the four per-

tinent variables.

Solution From Equation 2.9 (υ2 = υ0 2 + 2ax), we fi nd that

υ = ±√υ  02 + 2ax

= ±√(3250 m /s)2 + 2(−10.0 m /s2)(215 000 m)

= +2500 m/s and −2500 m/s

Both of these answers correspond to the same displace- ment (x = +215 km), but each arises in a diff erent part of the motion. The answer υ = +2500 m/s corresponds to the situation in Figure 2.10a, where the spacecraft has slowed to a speed of 2500 m/s, but is still traveling

to the right. The answer υ = −2500 m/s arises because the retrorockets eventually bring the spacecraft to a

momentary halt and cause it to reverse its direction.

Then it moves to the left, and its speed increases due

to the continually fi ring rockets. After a time, the

velocity of the craft becomes υ = −2500 m/s, giving rise to the situation in Figure 2.10b. In both parts of the drawing the spacecraft has the same displacement,

but a greater travel time is required in part b compared to part a.

Spacecraft Data x a υ υ0 t

+215 000 m −10.0 m/s2 ? +3250 m/s

FIGURE 2.10 (a) Because of an acceleration of −10.0 m/s2, the spacecraft changes its velocity from υ0 to υ. (b) Continued fi ring of the retrorockets changes the direction of the craft’s motion.

0 = +3250 m/sυ

0 = +3250 m/sυ

= +2500 m/sυ

= 0 m/s υ

= 0 m/s υ

= –2500 m/sυ

(b)

(a)

υ

x = +215 km

x = +215 km

–

+

2.5 Applications of the Equations of Kinematics 39

Analyzing Multiple-Concept Problems

EXAMPLE 8 A Motorcycle Ride

A motorcycle ride consists of two segments, as shown in Interactive Figure 2.11a. During segment 1, the motorcycle starts from rest, has an acceleration of +2.6 m/s2, and has a displacement of +120 m. Immedi-

ately after segment 1, the motorcycle enters segment 2 and begins slow-

ing down with an acceleration of −1.5 m/s2 until its velocity is +12 m/s.

What is the displacement of the motorcycle during segment 2?

Reasoning We can use an equation of kinematics from Table 2.1 to fi nd the displacement x2 for segment 2. To do this, it will be necessary to have values for three of the variables that appear in the equation.

Values for the acceleration (a2 = −1.5 m/s2) and fi nal velocity (υ2 = +12 m/s) are given. A value for a third variable, the initial velocity υ02, can be obtained by noting that it is also the fi nal velocity of segment 1. The

fi nal velocity of segment 1 can be found by using an appropriate equation

of kinematics, since three variables (x1, a1, and υ01) are known for this part of the motion, as the following table reveals.

Knowns and Unknowns The data for this problem are listed in the table:

Description Symbol Value Comment Explicit Data Displacement for segment 1 x1 +120 m

Acceleration for segment 1 a1 +2.6 m/s2 Positive, because the motorcycle moves in the +x direction and speeds up.

Acceleration for segment 2 a2 −1.5 m/s2 Negative, because the motorcycle moves in the +x direction and slows down.

Final velocity for segment 2 υ2 +12 m/s

Implicit Data Initial velocity for segment 1 υ01 0 m/s The motorcycle starts from rest.

Unknown Variable Displacement for segment 2 x2 ?

x2

v2 = +12 m/s

v02 v01 = 0 m/s

v1

a1 = +2.6 m/s 2a2 = –1.5 m/s

2

(b) Segment 2

x1 = +120 m

(c) Segment 1

Segment 1

x2

x1

Segment 2

(a)

–

+

INTERACTIVE FIGURE 2.11 (a) This motorcycle ride consists of two segments, each with a diff erent acceleration. (b) The variables for segment 2. (c) The variables for segment 1.

40 CHAPTER 2 Kinematics in One Dimension

Modeling the Problem

STEP 1 Displacement During Segment 2 Interactive Figure 2.11b shows the situation during segment 2. Two of the variables—the fi nal velocity υ2 and the acceleration a2—are known, and for convenience we choose Equation 2.9 to fi nd the displacement x2 of the motorcycle:

υ 22 = υ 202 + 2a2 x2 (2.9)

Solving this relation for x2 yields Equation 1 at the right. Although the initial velocity υ02 of segment 2 is not known, we will be able to determine its value from a knowledge of the motion

during segment 1, as outlined in Steps 2 and 3.

STEP 2 Initial Velocity of Segment 2 Since the motorcycle enters segment 2 immediately after leaving segment 1, the initial velocity υ02 of segment 2 is equal to the fi nal velocity υ1 of segment 1, or υ02 = υ1. Squaring both sides of this equation gives

𝜐 202 = 𝜐

2 1

This result can be substituted into Equation 1 as shown at the right. In the next step υ1 will be determined.

STEP 3 Final Velocity of Segment 1 Interactive Figure 2.11c shows the motorcycle during segment 1. Since we know the initial velocity υ01, the acceleration a1, and the displacement x1, we can employ Equation 2.9 to fi nd the fi nal velocity υ1 at the end of segment 1:

𝜐 21 = 𝜐

2 01 + 2a1x1

This relation for υ12 can be substituted into Equation 2, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

x2 = υ 22 − υ 202

2a2 =

υ 22 − υ 21 2a2

= υ 22 − (υ 201 + 2a1x1)

2a2 The displacement x2 of the motorcycle during segment 2 is

x2 = υ 22 − (υ 201 + 2a1x1)

2a2

= (+12 m/s)2 − [(0 m/s)2 + 2(+2.6 m/s2)(+120 m)]

2(−1.5 m/s2) = +160 m

Related Homework: Problems 59, 80

STEP 2STEP 1 STEP 3

x2 = 𝜐 22 − 𝜐

2 02

2a2 (1)

?

Math Skills Interactive Figure 2.11a shows which direction has been chosen as the positive direction, and the choice leads to a value of x2 = +160 m for the displacement of the motorcycle during segment 2. This answer means that displacement is 160 m in the

positive direction, which is upward and to the right in the drawing. The choice for the pos-

itive direction is completely arbitrary, however. The meaning of the answer to the problem

will be the same no matter what choice is made. Suppose that the choice for the positive

direction were opposite to that shown in Interactive Figure 2.11a. Then, all of the data listed in the Knowns and Unknowns table would appear with algebraic signs opposite to

those specifi ed, and the calculation of the displacement x2 would appear as follows:

x2 = (−12 m/s)2 − [(0 m/s)2 + 2(−2.6 m/s2)(−120 m)]

2(+1.5 m/s2) = −160 m

The value for x2 has the same magnitude of 160 m as before, but it is now negative. The negative sign means that the displacement of the motorcycle during segment 2 is 160 m

in the negative direction. But the negative direction is now upward and to the right in Interactive Figure 2.11a, so the meaning of the result is exactly the same as it was before.

?

x2 = 𝜐 22 − 𝜐

2 02

2a2 (1)

𝜐 202 = 𝜐 2

1 (2)

x2 = 𝜐 22 − 𝜐

2 02

2a2 (1)

𝜐 202 = 𝜐 2

1 (2)

𝜐 21 = 𝜐 2

01 + 2a1x1 (3)

2.6 Freely Falling Bodies 41

Now that we have seen how the equations of kinematics are applied to various situations,

it’s a good idea to summarize the reasoning strategy that has been used. This strategy, which

is outlined below, will also be used when we consider freely falling bodies in Section 2.6 and

two-dimensional motion in Chapter 3.

REASONING STRATEGY Applying the Equations of Kinematics 1. Make a drawing to represent the situation being studied. A drawing helps us to see what’s

happening. 2. Decide which directions are to be called positive (+) and negative (−) relative to a con-

veniently chosen coordinate origin. Do not change your decision during the course of a calculation.

3. In an organized way, write down the values (with appropriate plus and minus signs) that are given for any of the fi ve kinematic variables (x, a, υ, υ0, and t). Be on the alert for implicit data, such as the phrase “starts from rest,” which means that the value of the initial velocity is υ0 = 0 m/s. The data summary tables used in the examples in the text are a good way to keep track of this information. In addition, identify the variables that you are being asked to determine.

4. Before attempting to solve a problem, verify that the given information contains values for at least three of the fi ve kinematic variables. Once the three known variables are identifi ed along with the desired unknown variable, the appropriate relation from Table 2.1 can be selected. Remember that the motion of two objects may be interrelated, so they may share a common variable. The fact that the motions are interrelated is an important piece of information. In such cases, data for only two variables need be specifi ed for each object.

5. When the motion of an object is divided into segments, as in Example 8, remember that the fi nal velocity of one segment is the initial velocity for the next segment.

6. Keep in mind that there may be two possible answers to a kinematics problem as, for instance, in Example 7. Try to visualize the diff erent physical situations to which the answers correspond.

2.6 Freely Falling Bodies Everyone has observed the eff ect of gravity as it causes objects to fall downward. In the absence

of air resistance, it is found that all bodies at the same location above the earth fall vertically with

the same acceleration. Furthermore, if the distance of the fall is small compared to the radius of

the earth, the acceleration remains essentially constant throughout the descent. This idealized

motion, in which air resistance is neglected and the acceleration is nearly constant, is known as

free-fall. Since the acceleration is constant in free-fall, the equations of kinematics can be used. The acceleration of a freely falling body is called the acceleration due to gravity, and its

magnitude (without any algebraic sign) is denoted by the symbol g. The acceleration due to gravity is directed downward, toward the center of the earth. Near the earth’s surface, g is approximately

g = 9.80 m/s2   

or  32.2 ft/s2

Unless circumstances warrant otherwise, we will use either of these values for g in subsequent cal- culations. In reality, however, g decreases with increasing altitude and varies slightly with latitude.

Figure 2.12a shows the well-known phenomenon of a rock falling faster than a sheet of paper. The eff ect of air resistance is responsible for the slower fall of the paper, for when air is removed

from the tube, as in Figure 2.12b, the rock and the paper have exactly the same acceleration due to gravity. In the absence of air, the rock and the paper both exhibit free-fall motion. Free-fall is closely

approximated for objects falling near the surface of the moon, where there is no air to retard the mo-

tion. A nice demonstration of free-fall was performed on the moon by astronaut David Scott, who

dropped a hammer and a feather simultaneously from the same height. Both experienced the same

acceleration due to lunar gravity and consequently hit the ground at the same time. The acceleration

due to gravity near the surface of the moon is approximately one-sixth as large as that on the earth.

When the equations of kinematics are applied to free-fall motion, it is natural to use the

symbol y for the displacement, since the motion occurs in the vertical or y direction. Thus, when using the equations in Table 2.1 for free-fall motion, we will simply replace x with y. There is no

Air-filled tube (a)

Evacuated tube (b)

FIGURE 2.12 (a) In the presence of air res- istance, the acceleration of the rock is greater

than that of the paper. (b) In the absence of air resistance, both the rock and the paper

have the same acceleration.

42 CHAPTER 2 Kinematics in One Dimension

signifi cance to this change. The equations have the same algebraic form for either the horizontal

or vertical direction, provided that the acceleration remains constant during the motion. We now

turn our attention to several examples that illustrate how the equations of kinematics are applied

to freely falling bodies.

EXAMPLE 9 A Falling Stone

A stone is dropped from rest from the top of a tall building, as Animated Figure 2.13 indicates. After 3.00 s of free-fall, what is the displacement y of the stone?

Reasoning The upward direction is chosen as the positive direction. The three known variables are shown in the table below. The initial ve-

locity υ0 of the stone is zero, because the stone is dropped from rest. The acceleration due to gravity is negative, since it points downward in the

negative direction.

Stone Data y a υ υ0 t

? −9.80 m/s2 0 m/s 3.00 s

Equation 2.8 contains the appropriate variables and off ers a direct solu-

tion to the problem. Since the stone moves downward, and upward is the

positive direction, we expect the displacement y to have a negative value.

Problem-Solving Insight It is only when values are avail- able for at least three of the fi ve kinematic variables (y, a, υ, υ0, and t) that the equations in Table 2.1 can be used to determine the fourth and fi fth variables.

Solution Using Equation 2.8, we fi nd that

y = υ0t + 1

2 at 2 = (0 m/s)(3.00 s) + 1

2 (−9.80 m/s 2) (3.00 s)2 = −44.1 m

The answer for y is negative, as expected.

ANIMATED FIGURE 2.13 The stone, starting with zero velocity at the top of

the building, is accelerated downward

by gravity.

+

–

υ

y

υ0 = 0 m/s

t = 3.00 s

EXAMPLE 10 The Velocity of a Falling Stone

After 3.00 s of free-fall, what is the velocity υ of the stone in Animated Figure 2.13?

Reasoning Because of the acceleration due to gravity, the magnitude of the stone’s downward velocity increases by 9.80 m/s during each second

of free-fall. The data for the stone are the same as in Example 9, and

Equation 2.4 off ers a direct solution for the fi nal velocity. Since the stone

is moving downward in the negative direction, the value determined for υ should be negative.

Solution Using Equation 2.4, we obtain

υ = υ0 + at = 0 m/s + (−9.80 m/s2)(3.00 s) = −29.4 m/s

The velocity is negative, as expected.

FIGURE 2.14 At the start of a football game, a referee tosses a

coin upward with an initial

velocity of υ0 = +5.00 m/s. The velocity of the coin is

momentarily zero when the coin

reaches its maximum height.

υ = 0 m/s

υ 0 = +5.00 m/s

y

–

+

The acceleration due to gravity is always a downward-pointing vector. It describes how the

speed increases for an object that is falling freely downward. This same acceleration also de-

scribes how the speed decreases for an object moving upward under the infl uence of gravity

alone, in which case the object eventually comes to a momentary halt and then falls back to earth.

Examples 11 and 12 show how the equations of kinematics are applied to an object that is moving

upward under the infl uence of gravity.

2.6 Freely Falling Bodies 43

EXAMPLE 11 How High Does It Go?

A football game customarily begins with a coin toss to determine who

kicks off . The referee tosses the coin up with an initial speed of 5.00 m/s.

In the absence of air resistance, how high does the coin go above its point

of release?

Reasoning The coin is given an upward initial velocity, as in Figure 2.14. But the acceleration due to gravity points downward. Since the velocity

and acceleration point in opposite directions, the coin slows down as it

moves upward. Eventually, the velocity of the coin becomes υ = 0 m/s at the highest point. Assuming that the upward direction is positive, the data

can be summarized as shown below:

Coin Data y a υ υ0 t

? −9.80 m/s2 0 m/s +5.00 m/s

With these data, we can use Equation 2.9 (υ2 = υ02 + 2ay) to fi nd the maximum height y.

Problem-Solving Insight Implicit data are important. In Example 11, for instance, the phrase “how high does the coin go” refers to the maximum height, which occurs when the fi nal velocity υ in the vertical direction is υ = 0 m/s.

Solution Rearranging Equation 2.9, we fi nd that the maximum height of the coin above its release point is

y = υ2 − υ02

2a =

(0 m/s)2 − (5.00 m/s)2

2(−9.80 m/s2) = 1.28 m

Math Skills The rearrangement of algebraic equations is a problem-solving step that occurs often. The guiding rule in such

a procedure is that whatever you do to one side of an equation,

you must also do to the other side. Here in Example 11, for

instance, we need to rearrange υ2 = υ02 + 2ay (Equation 2.9) in order to determine y. The part of the equation that contains y is 2ay, and we begin by isolating that part on one side of the equals sign. To do this we subtract υ02 from each side of the equation:

𝜐2 − 𝜐 20 = 𝜐0 2 + 2ay − 𝜐02 = 2ay

Thus, we see that 2ay = υ2 − υ02. Then, we eliminate the term 2a from the left side of the equals sign by dividing both sides of the

equation by 2a.

2ay 2a

= υ2 − υ02

2a   or  y =

υ2 − υ02

2a

The term 2a occurs both in the numerator and the denominator on the left side of this result and can be eliminated algebraically,

leaving the desired expression for y.

EXAMPLE 12 How Long Is It in the Air?

In Figure 2.14, what is the total time the coin is in the air before returning to its release point?

Reasoning During the time the coin travels upward, gravity causes its speed to decrease to zero. On the way down, however, gravity causes the

coin to regain the lost speed. Thus, the time for the coin to go up is equal to

the time for it to come down. In other words, the total travel time is twice

the time for the upward motion. The data for the coin during the upward trip

are the same as in Example 11. With these data, we can use Equation 2.4

(υ = υ0 + at) to fi nd the upward travel time.

Solution Rearranging Equation 2.4, we fi nd that

t = υ − υ0

a =

0 m/s − 5.00 m/s

−9.80 m/s2 = 0.510 s

The total up-and-down time is twice this value, or 1.02 s .

It is possible to determine the total time by another method. When

the coin is tossed upward and returns to its release point, the displace-

ment for the entire trip is y = 0 m. With this value for the displacement, Equation 2.8 (y = υ0t +

1

2at 2) can be used to fi nd the time for the entire trip directly.

Examples 11 and 12 illustrate that the expression “freely falling” does not necessarily mean

an object is falling down. A freely falling object is any object moving either upward or downward

under the infl uence of gravity alone. In either case, the object always experiences the same down- ward acceleration due to gravity, a fact that is the focus of the next example.

CONCEPTUAL EXAMPLE 13 Acceleration Versus Velocity

There are three parts to the motion of the coin in Figure 2.14, in which air resistance is being ignored. On the way up, the coin has an upward-

pointing velocity vector with a decreasing magnitude. At the top of its

path, the velocity vector of the coin is momentarily zero. On the way

down, the coin has a downward-pointing velocity vector with an increasing

magnitude. Compare the acceleration vector of the coin with the velocity

44 CHAPTER 2 Kinematics in One Dimension

The motion of an object that is thrown upward and eventually returns to earth has a sym-

metry that is useful to keep in mind from the point of view of problem solving. The calculations

just completed indicate that a time symmetry exists in free-fall motion, in the sense that the time

required for the object to reach maximum height equals the time for it to return to its starting

point.

A type of symmetry involving the speed also exists. Figure 2.15 shows the coin considered in Examples 11 and 12. At any displacement y above the point of release, the coin’s speed during the upward trip equals the speed at the same point during the downward trip. For instance, when

y = +1.04 m, Equation 2.9 gives two possible values for the fi nal velocity υ, assuming that the initial velocity is υ0 = +5.00 m/s:

υ2 = υ02 + 2ay = (5.00 m/s)2 + 2(−9.80 m /s2)(1.04 m) = 4.62 m2/s2

υ = ±2.15 m/s

The value υ = +2.15 m/s is the velocity of the coin on the upward trip, and υ = −2.15 m/s is the velocity on the downward trip. The speed in both cases is identical and equals 2.15 m/s.

Likewise, the speed just as the coin returns to its point of release is 5.00 m/s, which equals the

initial speed. This symmetry involving the speed arises because the coin loses 9.80 m/s in speed

each second on the way up and gains back the same amount each second on the way down. In

Conceptual Example 14, we use just this kind of symmetry to guide our reasoning as we analyze

the motion of a pellet shot from a gun.

vector. (a) Do the direction and magnitude of the acceleration vector be- have in the same fashion as the direction and magnitude of the velocity

vector or (b) does the acceleration vector have a constant direction and a constant magnitude throughout the motion?

Reasoning Since air resistance is being ignored, the coin is in free-fall motion. This means that the acceleration vector of the coin is the acceler-

ation due to gravity. Acceleration is the rate at which velocity changes and is not the same concept as velocity itself.

Answer (a) is incorrect. During the upward and downward parts of the motion, and also at the top of the path, the acceleration due to

gravity has a constant downward direction and a constant magnitude

of 9.80 m/s2. In other words, the acceleration vector of the coin does

not behave as the velocity vector does. In particular, the acceleration

vector is not zero at the top of the motional path just because the velo-

city vector is zero there. Acceleration is the rate at which the velocity

is changing, and the velocity is changing at the top even though at one

instant it is zero.

Answer (b) is correct. The acceleration due to gravity has a constant downward direction and a constant magnitude of 9.80 m/s2 at all times

during the motion.

υ

y = +1.04 m

+

υ–

FIGURE 2.15 For a given displacement along the motional path, the upward speed

of the coin is equal to its downward speed,

but the two velocities point in opposite

directions.

(a) (b) (c)

30 m/s

30 m/s

30 m/s

INTERACTIVE FIGURE 2.16 (a) From the edge of a cliff , a pellet is fi red straight

upward from a gun. The pellet’s initial speed

is 30 m/s. (b) The pellet is fi red straight downward with an initial speed of 30 m/s.

(c) In Conceptual Example 14 this drawing plays the central role in reasoning that is

based on symmetry.

CONCEPTUAL EXAMPLE 14 Taking Advantage of Symmetry

Interactive Figure 2.16a shows a pellet that has been fi red straight upward from a gun at the edge of a cliff . The initial speed of the pellet is

30 m/s. It goes up and then falls back down, eventually hitting the ground

beneath the cliff . In Interactive Figure 2.16b the pellet has been fi red straight downward at the same initial speed. In the absence of air resist-

ance, would the pellet in Interactive Figure 2.16b strike the ground with

(a) a smaller speed than, (b) the same speed as, or (c) a greater speed than the pellet in Interactive Figure 2.16a?

Reasoning In the absence of air resistance, the motion is that of free- fall, and the symmetry inherent in free-fall motion off ers an immediate

answer.

2.7 Graphical Analysis of Velocity and Acceleration 45

Check Your Understanding

(The answers are given at the end of the book.) 13. An experimental vehicle slows down and comes to a halt with an acceleration whose magnitude is

9.80 m/s2. After reversing direction in a negligible amount of time, the vehicle speeds up with an

acceleration of 9.80 m/s2. Except for being horizontal, is this motion (a) the same as or (b) diff erent from the motion of a ball that is thrown straight upward, comes to a halt, and falls back to earth? Ignore

air resistance.

14. A ball is thrown straight upward with a velocity v→0 and in a time t reaches the top of its fl ight path, which is a displacement y→ above the launch point. With a launch velocity of 2v→0, what would be the time required to reach the top of its fl ight path and what would be the displacement of the top point

above the launch point? (a) 4t and 2y→ (b) 2t and 4y→ (c) 2t and 2y→ (d) 4t and 4y→ (e) t and 2y→ . 15. Two objects are thrown vertically upward, fi rst one, and then, a bit later, the other. Is it (a) possible or

(b) impossible that both objects reach the same maximum height at the same instant of time? 16. A ball is dropped from rest from the top of a building and strikes the ground with a speed υf. From

ground level, a second ball is thrown straight upward at the same instant that the fi rst ball is dropped.

The initial speed of the second ball is υ0 = υf, the same speed with which the fi rst ball eventually strikes the ground. Ignoring air resistance, decide whether the balls cross paths (a) at half the height of the building, (b) above the halfway point, or (c) below the halfway point.

2.7 Graphical Analysis of Velocity and Acceleration Graphical techniques are helpful in understanding the concepts of velocity and acceleration. Sup-

pose a bicyclist is riding with a constant velocity of υ = +4 m/s. The position x of the bicycle can be plotted along the vertical axis of a graph, while the time t is plotted along the horizontal axis. Since the position of the bike increases by 4 m every second, the graph of x versus t is a straight line. Furthermore, if the bike is assumed to be at x = 0 m when t = 0 s, the straight line passes through the origin, as Figure 2.17 shows. Each point on this line gives the position of the bike at a particular time. For instance, at t = 1 s the position is 4 m, while at t = 3 s the position is 12 m.

In constructing the graph in Figure 2.17, we used the fact that the velocity was +4 m/s. Sup- pose, however, that we were given this graph, but did not have prior knowledge of the velocity.

The velocity could be determined by considering what happens to the bike between the times of

1 and 3 s, for instance. The change in time is ∆t = 2 s. During this time interval, the position of the bike changes from +4 to +12 m, and the change in position is ∆x = +8 m. The ratio ∆x/∆t is called the slope of the straight line.

Slope = ∆x ∆t

= +8 m

2 s = +4 m /s

Notice that the slope is equal to the velocity of the bike. This result is no accident, because ∆x/∆t is the defi nition of average velocity (see Equation 2.2). Thus, for an object moving with a con- stant velocity, the slope of the straight line in a position–time graph gives the velocity. Since the position–time graph is a straight line, any time interval ∆t can be chosen to calculate the velocity. Choosing a diff erent ∆t will yield a diff erent ∆x, but the velocity ∆x/∆t will not change. In the real world, objects rarely move with a constant velocity at all times, as the next example illustrates.

Answers (a) and (c) are incorrect. These answers are incorrect, be- cause they are inconsistent with the symmetry that is discussed next in

connection with the correct answer.

Answer (b) is correct. Interactive Figure 2.16c shows the pellet after it has been fi red upward and has fallen back down to its starting point. Sym-

metry indicates that the speed in Interactive Figure 2.16c is the same as in Interactive Figure 2.16a—namely, 30 m/s, as is also the case when the

pellet has been actually fi red downward. Consequently, whether the pellet is

fi red as in Interactive Figure 2.16a or b, it begins to move downward from the cliff edge at a speed of 30 m/s. In either case, there is the same accel-

eration due to gravity and the same displacement from the cliff edge to the

ground below. Given these conditions, when the pellet reaches the ground,

it has the same speed in both Interactive Figures 2.16a and b.

Related Homework: Problems 47, 54

0 1 2 3 4

P os

it io

n

(m )

x

t = 2 sΔ

x = + 8 m Δ

Time (s)t

+16

+12

+8

+4

0

FIGURE 2.17 Position–time graph for an object moving with a constant velocity of

υ = Δx/Δt = +4 m/s.

46 CHAPTER 2 Kinematics in One Dimension

EXAMPLE 15 A Bicycle Trip

Segment 2 υ = ∆ x ∆ t

= 1200 m − 1200 m

1000 s − 600 s =

0 m

400 s = 0 m/s

Segment 3 υ = ∆ x ∆ t

= 400 m − 800 m

1800 s − 1400 s =

−400 m

400 s = −1 m/s

In the second segment of the journey the velocity is zero, refl ecting the

fact that the bike is stationary. Since the position of the bike does not

change, segment 2 is a horizontal line that has a zero slope. In the third

part of the motion the velocity is negative, because the position of the

bike decreases from x = +800 m to x = +400 m during the 400-s interval shown in the graph. As a result, segment 3 has a negative slope, and the

velocity is negative.

A bicyclist maintains a constant velocity on the outgoing leg of a trip,

zero velocity while stopped, and another constant velocity on the way

back. Interactive Figure 2.18 shows the corresponding position–time graph. Using the time and position intervals indicated in the drawing,

obtain the velocities for each segment of the trip.

Reasoning The average velocity υ is equal to the displacement ∆x di- vided by the elapsed time ∆t, υ = ∆x /∆t. The displacement is the fi nal position minus the initial position, which is a positive number for segment 1

and a negative number for segment 3. Note for segment 2 that ∆x = 0 m, since the bicycle is at rest. The drawing shows values for ∆t and ∆x for each of the three segments.

Solution The average velocities for the three segments are

Segment 1 υ = ∆ x ∆t

= 800 m − 400 m

400 s − 200 s =

+400 m

200 s = +2 m/s

+1200

+800

+400

0 0 200 400 600 800

Time t (s)

1000 1200 1400 1600 1800

Δ t = 400 s

Δ t = 400 s Δ x = 0 m

Δ t = 200 s

Δ x = –400 mΔ x = +400 m

2

3

1

P os

it io

n x

(m )

Positive velocity

Zero velocity Negative velocity

INTERACTIVE FIGURE 2.18 This position–time graph consists of three straight-line segments, each corresponding to a diff erent constant velocity.

If the object is accelerating, its velocity is changing. When the velocity is changing, the

position–time graph is not a straight line, but is a curve, perhaps like that in Figure 2.19. This curve was drawn using Equation 2.8 (x = υ0t +

1

2at 2), assuming an acceleration of a = 0.26 m/s2 and an initial velocity of υ0 = 0 m/s. The velocity at any instant of time can be determined by measuring the slope of the curve at that instant. The slope at any point along the curve is defi ned

to be the slope of the tangent line drawn to the curve at that point. For instance, in Figure 2.19 a tangent line is drawn at t = 20.0 s. To determine the slope of the tangent line, a triangle is con- structed using an arbitrarily chosen time interval of ∆t = 5.0 s. The change in x associated with this time interval can be read from the tangent line as ∆x = +26 m. Therefore,

Slope of tangent line = ∆x ∆t

= +26 m

5.0 s = +5.2 m /s

The slope of the tangent line is the instantaneous velocity, which in this case is υ = +5.2 m/s. This graphical result can be verifi ed by using Equation 2.4 with υ0 = 0 m/s: υ = at = (+0.26 m/s2)(20.0 s) = +5.2 m/s.

Time (s)t

5.0 10.0 15.0 20.0 25.0

Δ t = 5.0 s

Δ x = +26 m

Tangent line80.0

60.0

40.0

20.0

0 0

P os

it io

n x

(m )

FIGURE 2.19 When the velocity is changing, the position–time graph is a

curved line. The slope Δx/Δt of the tangent line drawn to the curve at a given time is

the instantaneous velocity at that time.

Concept Summary 47

Insight into the meaning of acceleration can also be gained with the aid of a graphical rep-

resentation. Consider an object moving with a constant acceleration of a = +6 m/s2. If the object has an initial velocity of υ0 = +5 m/s, its velocity at any time is represented by Equation 2.4 as

υ = υ0 + at = 5 m/s + (6 m/s2)t

This relation is plotted as the velocity–time graph in Figure 2.20. The graph of υ versus t is a straight line that intercepts the vertical axis at υ0 = 5 m/s. The slope of this straight line can be calculated from the data shown in the drawing:

Slope = ∆υ ∆t

= +12 m/s

2 s = +6 m/s2

The ratio ∆υ/∆t is, by defi nition, equal to the average acceleration (Equation 2.4), so the slope of the straight line in a velocity–time graph is the average acceleration.

+36

+24

+12

0 0 1 2 3 4

Ve lo

ci ty

(m /s

) υ

5

= +5 m/s0

= 2 sΔ t

Δ

Time (s)t

= +12 m/s

υ

υ

FIGURE 2.20 A velocity–time graph that applies to an object with an acceleration of

Δυ/Δt = +6 m/s2. The initial velocity is υ0 = +5 m/s when t = 0 s.

EXAMPLE 16 BIO The Average Acceleration of a Sprinter

Figure 2.21 shows the velocity-time graph for a sprinter competing in a 100-m race. At the beginning of the race, usually initiated by the report of

a starter’s pistol, the sprinter accelerates from rest for a period of 4.00 s.

After accelerating, the sprinter runs at constant speed, before slowing

(decelerating) near the end of the race. Use the velocity and time intervals

shown in the graph to determine the sprinter’s average acceleration during

the fi rst 4.00 s of the race.

Reasoning The slope of the line during the fi rst 4-second interval of

the race will be given by ∆υ ∆t

, which is equivalent to the sprinter’s average

acceleration, a = ∆υ ∆t

.

Solution Using the fact that the average acceleration is given by

a = ∆υ ∆t

, we just need to calculate the slope of the velocity-time graph

over the period from 0 to 4.00 s.

a = ∆υ ∆t

= +12.5 m/s − 0

4.00 s − 0 = +3.13 m/s2

FIGURE 2.21 The velocity-time graph for a track and fi eld athlete running a 100-m sprint. At the start of the race, the sprinter accelerates from rest

for 4.00 s and then maintains a constant speed of 12.5 m/s for approxim-

ately four more seconds, before decelerating near the end of the race.

Time t (s)

6

4

2

0

8

10

12

0 1 2 3 4

Ve lo

ci ty

(m /s

) υ

5 6 7 8 9 10

Δt = +4.00 s

= + 12.5 m/sΔυ

Concept Summary 2.1 Displacement Displacement is a vector that points from an object’s initial position to its fi nal position. The magnitude of the displacement is the

shortest distance between the two positions.

2.2 Speed and Velocity The average speed of an object is the distance traveled by the object divided by the time required to cover the distance, as

shown in Equation 2.1.

The average velocity v→ of an object is the object’s displacement ∆x→ divided by the elapsed time ∆t, as shown in Equation 2.2. Average veloc- ity is a vector that has the same direction as the displacement. When the

elapsed time becomes infi nitesimally small, the average velocity becomes

equal to the instantaneous velocity v→, the velocity at an instant of time, as indicated in Equation 2.3.

Average speed = Distance

Elapsed time (2.1)

v→ = ∆x →

∆t (2.2)

v→ = lim ∆t → 0

∆x→

∆t (2.3)

2.3 Acceleration The average acceleration a→ is a vector. It equals the change ∆v→ in the velocity divided by the elapsed time ∆t, the change in the velocity being the fi nal minus the initial velocity; see Equation 2.4. When ∆t becomes infi nitesimally small, the average acceleration becomes equal to the

48 CHAPTER 2 Kinematics in One Dimension

instantaneous acceleration a→, as indicated in Equation 2.5. Acceleration is the rate at which the velocity is changing.

a→ = ∆v →

∆t (2.4)

a→ = lim ∆ t → 0

∆v→ ∆t (2.5)

2.4 Equations of Kinematics for Constant Acceleration/2.5 Applications of the Equations of Kinematics The equations of kinematics apply when an object moves with a constant acceleration along a straight line. These equa-

tions relate the displacement x − x0, the acceleration a, the fi nal velocity υ, the initial velocity υ0, and the elapsed time t − t0. Assuming that x0 = 0 m at t0 = 0 s, the equations of kinematics are as shown in Equations 2.4 and 2.7–2.9.

υ = υ0 + at (2.4) x = 12 (υ0 + υ)t (2.7) x = υ0t +

1

2 at 2 (2.8) υ2 = υ 20 + 2ax (2.9)

2.6 Freely Falling Bodies In free-fall motion, an object experiences neg- ligible air resistance and a constant acceleration due to gravity. All objects at

the same location above the earth have the same acceleration due to gravity.

The acceleration due to gravity is directed toward the center of the earth and

has a magnitude of approximately 9.80 m/s2 near the earth’s surface.

2.7 Graphical Analysis of Velocity and Acceleration The slope of a plot of position versus time for a moving object gives the object’s velocity. The

slope of a plot of velocity versus time gives the object’s acceleration.

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 2.1 Displacement 1. What is the diff erence between distance and displacement? (a) Distance is a vector, while displacement is not a vector. (b) Displacement is a vector, while distance is not a vector. (c) There is no diff erence between the two concepts; they may be used interchangeably.

Section 2.2 Speed and Velocity 3. A jogger runs along a straight and level road for a distance of 8.0 km and then runs back to her starting point. The time for this round-trip is 2.0 h. Which one

of the following statements is true? (a) Her average speed is 8.0 km/h, but there is not enough information to determine her average velocity. (b) Her average speed is 8.0 km/h, and her average velocity is 8.0 km/h. (c) Her average speed is 8.0 km/h, and her average velocity is 0 km/h.

Section 2.3 Acceleration 6. The velocity of a train is 80.0 km/h, due west. One and a half hours later its velocity is 65.0 km/h, due west. What is the train’s average acceleration?

(a) 10.0 km/h2, due west (b) 43.3 km/h2, due west (c) 10.0 km/h2, due east (d) 43.3 km/h2, due east (e) 53.3 km/h2, due east.

Section 2.4 Equations of Kinematics for Constant Acceleration 10. In which one of the following situations can the equations of kinematics not be used? (a) When the velocity changes from moment to moment, (b) when the velocity remains constant, (c) when the acceleration changes from moment to moment, (d) when the acceleration remains constant. 13. In a race two horses, Silver Bullet and Shotgun, start from rest and each maintains a constant acceleration. In the same elapsed time Silver Bullet runs

1.20 times farther than Shotgun. According to the equations of kinematics,

which one of the following is true concerning the accelerations of the horses?

(a) aSilver Bullet = 1.44 aShotgun (b) aSilver Bullet = aShotgun (c) aSilver Bullet = 2.40 aShotgun (d) aSilver Bullet = 1.20 aShotgun (e) aSilver Bullet = 0.72 aShotgun

Section 2.6 Freely Falling Bodies 19. A rocket is sitting on the launch pad. The engines ignite, and the rocket begins to rise straight upward, picking up speed as it goes. At about 1000 m

above the ground the engines shut down, but the rocket continues straight

upward, losing speed as it goes. It reaches the top of its fl ight path and then

falls back to earth. Ignoring air resistance, decide which one of the follow-

ing statements is true. (a) All of the rocket’s motion, from the moment the engines ignite until just before the rocket lands, is free-fall. (b) Only part of the rocket’s motion, from just after the engines shut down until just before it

lands, is free-fall. (c) Only the rocket’s motion while the engines are fi ring is free-fall. (d) Only the rocket’s motion from the top of its fl ight path until just before landing is free-fall. (e) Only part of the rocket’s motion, from just after the engines shut down until it reaches the top of its fl ight path, is

free-fall.

22. The top of a cliff is located a distance H above the ground. At a distance H/2 there is a branch that juts out from the side of the cliff , and on this branch a bird’s nest is located. Two children throw stones at the nest with the same

initial speed, one stone straight downward from the top of the cliff and the

other stone straight upward from the ground. In the absence of air resistance,

which stone hits the nest in the least amount of time? (a) There is insuffi cient information for an answer. (b) Both stones hit the nest in the same amount of time. (c) The stone thrown from the ground. (d) The stone thrown from the top of the cliff .

Section 2.7 Graphical Analysis of Velocity and Acceleration 24. The graph accompanying this prob- lem shows a three-part motion. For each

of the three parts, A, B, and C, identify

the direction of the motion. A positive ve-

locity denotes motion to the right. (a) A right, B left, C right (b) A right, B right, C left (c) A right, B left, C left (d) A left, B right, C left (e) A left, B right, C right

Focus on Concepts

Time

Position

A B

C

QUESTION 24

Problems 49

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual

MMH Problem-solving help

GO Guided Online Tutorial

V-HINT Video Hints

CHALK Chalkboard Videos

BIO Biomedical application

E Easy

M Medium

H Hard

Section 2.1 Displacement

Section 2.2 Speed and Velocity 1. E SSM The space shuttle travels at a speed of about 7.6 × 103 m/s. The blink of an astronaut’s eye lasts about 110 ms. How many football fi elds

(length = 91.4 m) does the shuttle cover in the blink of an eye?

2. E GO For each of the three pairs of positions listed in the following table, determine the magnitude and direction (positive or negative) of the

displacement.

Initial position x0 Final position x (a) +2.0 m +6.0 m (b) +6.0 m +2.0 m (c) −3.0 m +7.0 m

3. E SSM Due to continental drift, the North American and European con- tinents are drifting apart at an average speed of about 3 cm per year. At this

speed, how long (in years) will it take for them to drift apart by another 1500 m

(a little less than a mile)?

4. E BIO You step onto a hot beach with your bare feet. A nerve impulse, generated in your foot, travels through your nervous system at an average

speed of 110 m/s. How much time does it take for the impulse, which travels

a distance of 1.8 m, to reach your brain?

5. E GO The data in the following table describe the initial and fi nal pos- itions of a moving car. The elapsed time for each of the three pairs of posi-

tions listed in the table is 0.50 s. Review the concept of average velocity in

Section 2.2 and then determine the average velocity (magnitude and direc-

tion) for each of the three pairs. Note that the algebraic sign of your answers

will convey the direction.

Initial position x0 Final position x (a) +2.0 m +6.0 m (b) +6.0 m +2.0 m (c) −3.0 m +7.0 m

6. E CHALK One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 1.50 km. They start at the west side

of the lake and head due south to begin with. (a) What is the distance they travel? (b) What are the magnitude and direction (relative to due east) of the couple’s displacement?

7. E The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of 0.037 m/s, considerably

slower than the giant tortoise, which walks at 0.076 m/s. After 12 minutes of

walking, how much further would the tortoise have gone relative to the sloth?

8. E GO An 18-year-old runner can complete a 10.0-km course with an av- erage speed of 4.39 m/s. A 50-year-old runner can cover the same distance

with an average speed of 4.27 m/s. How much later (in seconds) should the

younger runner start in order to fi nish the course at the same time as the older runner?

9. E A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distance d away. The bear is 26 m behind the tourist and running at 6.0 m/s. The tourist reaches the car

safely. What is the maximum possible value for d? 10. M V-HINT In reaching her destination, a backpacker walks with an av- erage velocity of 1.34 m/s, due west. This average velocity results because

she hikes for 6.44 km with an average velocity of 2.68 m/s, due west, turns

around, and hikes with an average velocity of 0.447 m/s, due east. How far

east did she walk?

11. M SSM A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the fi rst part, she

rides for 22 minutes at an average speed of 7.2 m/s. During the second part,

she rides for 36 minutes at an average speed of 5.1 m/s. Finally, during the

third part, she rides for 8.0 minutes at an average speed of 13 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

12. M V-HINT A car makes a trip due north for three-fourths of the time and due south one-fourth of the time. The average northward velocity has

a magnitude of 27 m/s, and the average southward velocity has a mag-

nitude of 17 m/s. What is the average velocity (magnitude and direction)

for the entire trip?

13. H You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a

12° angle with the horizontal. As you face the window (0.90 m high, 2.0 m

wide) in your compartment, the train is moving to the left, as the drawing

indicates. The top edge of the wall fi rst appears at window corner A and

eventually disappears at window corner B. How much time passes between

appearance and disappearance of the upper edge of the wall?

3.0 m/s

A

12°

BB

A

PROBLEM 13

Section 2.3 Acceleration 14. E Review Conceptual Example 6 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel

remains the same throughout this problem. The car’s initial speed is 27.0 m/s,

and during a 5.0-s interval, it changes to a fi nal speed of (a) 29.0 m/s and (b) 23.0 m/s. In each case, fi nd the acceleration (magnitude and algebraic sign) and state whether or not the car is decelerating.

15. E GO (a) Suppose that a NASCAR race car is moving to the right with a constant velocity of +82 m/s. What is the average acceleration of the car?

Problems

50 CHAPTER 2 Kinematics in One Dimension

25. E SSM A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car acceler- ates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

26. E GO A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s2. (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to 60.0 mi/h in 0.600 s. Find the acceleration (in m/s2) of the dragster.

27. E BIO The left ventricle of the heart accelerates blood from rest to a velocity of +26 cm/s. (a) If the displacement of the blood during the accel- eration is +2.0 cm, determine its acceleration (in cm/s2). (b) How much time does blood take to reach its fi nal velocity?

28. E (a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.0 m/s when going down a slope for

5.0 s? (b) How far does the skier travel in this time? 29. E SSM A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m of runway in which to reduce its

speed to 6.1 m/s. Compute the average acceleration (magnitude and direc-

tion) of the plane during landing.

30. E V-HINT MMH The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in

length. One of the most famous horses to win this event was Secretariat. In

1973 he set a Derby record that would be hard to beat. His average accelera-

tion during the last four quarter-miles of the race was +0.0105 m/s2. His ve-

locity at the start of the fi nal mile (x = +1609 m) was about +16.58 m/s. The acceleration, although small, was very important to his victory. To assess its

eff ect, determine the diff erence between the time he would have taken to run

the fi nal mile at a constant velocity of +16.58 m/s and the time he actually

took. Although the track is oval in shape, assume it is straight for the purpose

of this problem.

31. E SSM MMH A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x = 0 m, with an initial velocity of +5.0 m/s and a constant acceleration due to the fan.

The direction to the right is positive. The cart reaches a maximum position

of x = +12.5 m, where it begins to travel in the negative direction. Find the acceleration of the cart.

32. E MMH Two rockets are fl ying in the same direction and are side by side at the instant their retrorockets fi re. Rocket A has an initial velocity

of +5800 m/s, while rocket B has an initial velocity of +8600 m/s. After a

time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is −15 m/s2. What is the acceleration of

rocket B?

33. M CHALK MMH A car is traveling at 20.0 m/s, and the driver sees a traffi c light turn red. After 0.530 s (the reaction time), the driver applies the brakes,

and the car decelerates at 7.00 m/s2. What is the stopping distance of the car,

as measured from the point where the driver fi rst sees the red light?

34. M GO MMH A race driver has made a pit stop to refuel. After refuel- ing, he starts from rest and leaves the pit area with an acceleration whose

magnitude is 6.0 m/s2; after 4.0 s he enters the main speedway. At the same

instant, another car on the speedway and traveling at a constant velocity of

70.0 m/s overtakes and passes the entering car. The entering car maintains

its acceleration. How much time is required for the entering car to catch the

other car?

35. M V-HINT In a historical movie, two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George’s

acceleration has a magnitude of 0.300 m/s2, while Sir Alfred’s has a mag-

nitude of 0.200 m/s2. Relative to Sir George’s starting point, where do the

knights collide?

(b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration

of the car?

16. E Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.9 km/s to −18.5 km/s.

Find (a) the total change in the planet’s velocity (in m/s) and (b) its aver- age acceleration (in m/s2) during this interval. Include the correct algebraic

sign with your answers to convey the directions of the velocity and the

acceleration.

17. E SSM A motorcycle has a constant acceleration of 2.5 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How

much time is required for the motorcycle to change its speed from (a) 21 to 31 m/s, and (b) 51 to 61 m/s? 18. E A sprinter explodes out of the starting block with an acceleration of +2.3 m/s2, which she sustains for 1.2 s. Then, her acceleration drops to zero

for the rest of the race. What is her velocity (a) at t = 1.2 s and (b) at the end of the race?

19. E GO The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the fi nal speed of each of the objects, assuming that the time elapsed since t = 0 s is 2.0 s.

Initial velocity υ0 Acceleration a (a) +12 m/s +3.0 m/s2

(b) +12 m/s −3.0 m/s2

(c) −12 m/s +3.0 m/s2

(d) −12 m/s −3.0 m/s2

20. E GO An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 10.6 m/s in 4.0 s. (a) What is the direction of the bird’s acceleration? (b) Assuming that the accelera- tion remains the same, what is the bird’s velocity after an additional 2.0 s

has elapsed?

21. E SSM For a standard production car, the highest road-tested acceler- ation ever reported occurred in 1993, when a Ford RS200 Evolution went

from zero to 26.8 m/s (60 mi/h) in 3.275 s. Find the magnitude of the car’s

acceleration.

22. M MMH A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For the next twelve seconds the car slows down,

and its average acceleration is a1. For the next six seconds the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the eighteen-second period is +28.0 m/s. The ratio of the average accel-

eration values is a1 / a2 = 1.50. Find the velocity of the car at the end of the initial twelve-second interval.

23. H Two motorcycles are traveling due east with diff erent velocities. However, four seconds later, they have the same velocity. During this four-

second interval, cycle A has an average acceleration of 2.0 m/s2 due east,

while cycle B has an average acceleration of 4.0 m/s2 due east. By how much

did the speeds diff er at the beginning of the four-second interval, and which motorcycle was moving faster?

Section 2.4 Equations of Kinematics for Constant Acceleration/Section 2.5 Applications of the Equations of Kinematics 24. E In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of 6.0 m/s in 1.5 s. Assuming that the player accel-

erates uniformly, determine the distance he runs.

Problems 51

46. E GO MMH A ball is thrown vertically upward, which is the posit- ive direction. A little later it returns to its point of release. The ball is in

the air for a total time of 8.0 s. What is its initial velocity? Neglect air

resistance.

47. E Review Conceptual Example 14 before attempting this problem. Two identical pellet guns are fi red simultaneously from the edge of a cliff . These

guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fi red straight

upward, with the pellet going up and then falling back down, eventually hit-

ting the ground beneath the cliff . Gun B is fi red straight downward. In the

absence of air resistance, how long after pellet B hits the ground does pellet

A hit the ground?

48. E MMH An astronaut on a distant planet wants to determine its acceler- ation due to gravity. The astronaut throws a rock straight up with a velocity of

+15 m/s and measures a time of 20.0 s before the rock returns to his hand.

What is the acceleration (magnitude and direction) due to gravity on this

planet?

49. E SSM A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.00 m above the ground, the balloonist

accidentally drops a compass over the side of the balloon. How much time

elapses before the compass hits the ground?

50. E GO A ball is thrown straight upward and rises to a maximum height of 16 m above its launch point. At what height above its launch point has the

speed of the ball decreased to one-half of its initial value?

51. E A diver springs upward with an initial speed of 1.8 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = −3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?

52. E GO A ball is thrown straight upward. At 4.00 m above its launch point, the ball’s speed is one-half its launch speed. What maximum height

above its launch point does the ball attain?

53. E SSM From her bedroom window a girl drops a water-fi lled balloon to the ground, 6.0 m below. If the balloon is released from rest, how long is

it in the air?

54. E Before working this problem, review Conceptual Example 14. A pellet gun is fi red straight downward from the edge of a cliff that is 15 m above

the ground. The pellet strikes the ground with a speed of 27 m/s. How far

above the cliff edge would the pellet have gone had the gun been fi red straight

upward?

55. E V-HINT Consult Multiple-Concept Example 5 in preparation for this problem. The velocity of a diver just before hitting the water is −10.1 m/s,

where the minus sign indicates that her motion is directly downward. What is

her displacement during the last 1.20 s of the dive?

56. M V-HINT A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 5.70 m before falling back

down again. A boy then catches the ball on the way down when it is 1.20 m

above the pavement. Ignoring air resistance, calculate the total amount of

time that the ball is in the air, from drop to catch.

57. M MMH A woman on a bridge 75.0 m high sees a raft fl oating at a constant speed on the river below. Trying to hit the raft, she drops a stone

from rest when the raft has 7.00 m more to travel before passing under the

bridge. The stone hits the water 4.00 m in front of the raft. Find the speed

of the raft.

58. M GO Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff .

The height of the cliff is 6.00 m. The stones are thrown with the same speed

36. M GO Two soccer players start from rest, 48 m apart. They run directly toward each other, both players accelerating. The fi rst player’s

acceleration has a magnitude of 0.50 m/s2. The second player’s accelera-

tion has a magnitude of 0.30 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the fi rst player run?

37. M MMH A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway

from the ramp. The second car starts from rest and has a constant accelera-

tion. What acceleration must it maintain, so that the two cars meet for the fi rst

time at the next exit, which is 2.5 km away?

38. M Available on WileyPLUS. 39. M GO Refer to Multiple-Concept Example 5 to review a method by which this problem can be solved. You are driving your car, and the traffi c

light ahead turns red. You apply the brakes for 3.00 s, and the velocity of

the car decreases to +4.50 m/s. The car’s deceleration has a magnitude of

2.70 m/s2 during this time. What is the car’s displacement?

40. H A Boeing 747 “Jumbo Jet” has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of the intersec-

tion is 25.0 m. The plane decelerates through the intersection at a rate of

5.70 m/s2 and clears it with a fi nal speed of 45.0 m/s. How much time is

needed for the plane to clear the intersection?

41. H SSM A locomotive is accelerating at 1.6 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.4 s. After the locomotive leaves the

crossing, how much time is required until its speed reaches 32 m/s?

42. H Available on WileyPLUS.

Section 2.6 Freely Falling Bodies 43. E SSM The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top of the

Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s

(88 mi/h). To assess the eff ects of air resistance, determine how fast he would

have been traveling on impact had air resistance been absent.

44. E A dynamite blast at a quarry launches a chunk of rock straight up- ward, and 2.0 s later it is rising at a speed of 15 m/s. Assuming air resistance

has no eff ect on the rock, calculate its speed (a) at launch and (b) 5.0 s after launch.

45. E BIO The drawing shows a device that you can make with a piece of cardboard, which can be used to measure a person’s reaction time. Hold

the card at the top and suddenly drop it. Ask a friend to try to catch the card

between his or her thumb and index fi nger. Initially, your friend’s fi ngers

must be level with the asterisks at the bottom. By noting where your friend

catches the card, you can determine his or her reaction time in milliseconds

(ms). Calculate the distances d1, d2, and d3.

PROBLEM 45

HOLD HERE

180 ms

120 ms

60.0 ms

d3

d1 d2

∗ ∗ ∗ ∗ ∗ ∗ ∗

52 CHAPTER 2 Kinematics in One Dimension

+100

+80

+60

+40

+20

0 0

Ve lo

ci ty

(m

/s )

υ

A

B

C

Time (s)t

10 20 30 40 50 60

PROBLEM 67

68. E A bus makes a trip according to the position–time graph shown in the drawing. What is the average velocity (magnitude and direction) of

the bus during each of the segments A, B, and C? Express your answers in km/h.

+50.0

+40.0

+30.0

+20.0

+10.0

0 0 0.5 1.0 1.5

Time t (h)

2.0 2.5 3.0

P os

it io

n x

(k m

)

A

B

C

PROBLEM 68

69. M GO A bus makes a trip according to the position–time graph shown in the illustration. What is the average acceleration (in km/h2) of the bus for

the entire 3.5-h period shown in the graph?

+40.0

+30.0

+20.0

+10.0

0 0

P os

it io

n

(k m

) x

0.5 1.0 1.5 2.0 2.5 3.0 3.5

A

B C

Time (h)t

PROBLEM 69

70. M A runner is at the position x = 0 m when time t = 0 s. One hundred meters away is the fi nish line. Every ten seconds, this runner runs half the re-

maining distance to the fi nish line. During each ten-second segment, the run-

ner has a constant velocity. For the fi rst forty seconds of the motion, construct

(a) the position–time graph and (b) the velocity–time graph. 71. H SSM Available on WileyPLUS.

of 9.00 m/s. Find the location (above the base of the cliff ) of the point where

the stones cross paths.

59. M SSM Available on WileyPLUS. 60. M GO Two arrows are shot vertically upward. The second arrow is shot after the fi rst one, but while the fi rst is still on its way up. The initial

speeds are such that both arrows reach their maximum heights at the same

instant, although these heights are diff erent. Suppose that the initial speed

of the fi rst arrow is 25.0 m/s and that the second arrow is fi red 1.20 s after

the fi rst. Determine the initial speed of the second arrow.

61. M CHALK SSM A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the

ground, a man, 2.00 m tall, looks up and notices that the block is directly

above him. How much time, at most, does the man have to get out of the

way?

62. M V-HINT A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2 for

1.70 seconds, at which point its fuel abruptly runs out. Air resistance has

no eff ect on its fl ight. What maximum altitude (above the ground) will the

rocket reach?

63. H While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 m, you throw a second stone

straight down. What initial velocity must you give the second stone if they

are both to reach the ground at the same instant? Take the downward direc-

tion to be the negative direction.

64. H Available on WileyPLUS.

Section 2.7 Graphical Analysis of Velocity and Acceleration 65. E SSM A person who walks for exercise produces the position–time graph given with this problem. (a) Without doing any calculations, decide which segments of the graph (A, B, C, or D) indicate positive, negative, and zero average velocities. (b) Calculate the average velocity for each segment to verify your answers to part (a).

+1.25

+1.00

+0.75

+0.50

+0.25

0 0

P os

it io

n

( km

) x

A B

C

Time (h)t 0.20 0.40 0.60 0.80 1.00

D

PROBLEM 65

66. E Starting at x = −16 m at time t = 0 s, an object takes 18 s to travel 48 m in the +x direction at a constant velocity. Make a position–time graph of the object’s motion and calculate its velocity.

67. E MMH A snowmobile moves according to the velocity–time graph shown in the drawing. What is the snowmobile’s average acceleration during

each of the segments A, B, and C?

Additional Problems 53

72. E GO The data in the following table represent the initial and fi nal ve- locities for a boat traveling along the x axis. The elapsed time for each of the four pairs of velocities in the table is 2.0 s. Review the concept of aver-

age acceleration in Section 2.3 and then determine the average acceleration

(magnitude and direction) for each of the four pairs. Note that the algebraic

sign of your answers will convey the direction.

Initial velocity υ0 Final velocity υ (a) +2.0 m/s +5.0 m/s

(b) +5.0 m/s +2.0 m/s

(c) −6.0 m/s −3.0 m/s

(d) +4.0 m/s −4.0 m/s

73. E SSM Available on WileyPLUS.

74. E In 1954 the English runner Roger Bannister broke the four-minute barrier for the mile with a time of 3:59.4 s (3 min and 59.4 s). In 1999 the

Moroccan runner Hicham el-Guerrouj set a record of 3:43.13 s for the mile.

If these two runners had run in the same race, each running the entire race at

the average speed that earned him a place in the record books, el-Guerrouj

would have won. By how many meters?

75. E SSM Available on WileyPLUS.

76. E MMH At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.6 m/s. A player cannot touch the ball until after

it reaches its maximum height and begins to fall down. What is the minimum

time that a player must wait before touching the ball?

77. E Electrons move through a certain electric circuit at an average speed of 1.1 × 10−2 m/s. How long (in minutes) does it take an electron to traverse

1.5 m of wire in the fi lament of a light bulb?

78. E Available on WileyPLUS.

79. E V-HINT MMH A cheetah is hunting. Its prey runs for 3.0 s at a constant velocity of +9.0 m/s. Starting from rest, what constant acceleration must

the cheetah maintain in order to run the same distance as its prey runs in the

same time?

80. M Available on WileyPLUS. 81. M SSM Available on WileyPLUS. 82. M GO The leader of a bicycle race is traveling with a constant velocity of +11.10 m/s and is 10.0 m ahead of the second-place cyclist. The second-

place cyclist has a velocity of +9.50 m/s and an acceleration of +1.20 m/s2.

How much time elapses before he catches the leader?

83. M V-HINT A golfer rides in a golf cart at an average speed of 3.10 m/s for 28.0 s. She then gets out of the cart and starts walking at an average speed of

1.30 m/s. For how long (in seconds) must she walk if her average speed for

the entire trip, riding and walking, is 1.80 m/s?

84. M GO Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and main-

tains a constant acceleration. Both cars cover a distance of 460 m in 210 s.

Assume that they are moving in the +x direction. Determine (a) the constant velocity of car A, (b) the fi nal velocity of car B, and (c) the acceleration of car B.

85. M V-HINT MMH A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. After

a reaction time of 0.800 s the policeman begins to pursue the speeder with an

acceleration of 5.00 m/s2. Including the reaction time, how long does it take

for the police car to catch up with the speeder?

86. H MMH Available on WileyPLUS. 87. H SSM Available on WileyPLUS. 88. H Available on WileyPLUS. 89. M GO SSM A jet is taking off from the deck of an aircraft carrier, as shown in the image. Starting from rest, the jet is catapulted with a constant

acceleration of +31 m/s2 along a straight line and reaches a velocity of +62 m/s.

Find the displacement of the jet.

Additional Problems

0 = 0 m/sυ = + 62 m/sυa = + 31 m/s 2

– +

(b)

x

(a) A jet is being launched from an aircraft carrier. (b) During the launch, a catapult accelerates the jet down the fl ight deck.

(a)

M ar

k u s

S ch

re ib

er /A

P /W

id e

W o rl

d

P h o to

s

PROBLEM 89

54 CHAPTER 2 Kinematics in One Dimension

92. M A Deadly Virus. On your fl ight out of South America your cargo plane emergency-lands on a deserted island somewhere in the South Pacifi c.

Because it was a beach landing, the pilot skids the plane on its belly rather

than lowering the landing gear. The landing is rough: an initial vertical drop

followed by an abrupt horizontal deceleration. When the dust settles, the pi-

lot and your scientist colleagues are all okay, but you know there might be

a big problem. Your team is transporting samples of a deadly virus from a

recent breakout in a small village. The samples were packed in a cryogen-

ically cooled container that has a special seal that may leak if it exceeds

an acceleration greater than 10 g. The container has a mechanical acceler-

ometer gauge that measures the maximum vertical deceleration in case the

container is dropped. It reads a value of a = 7.30 g. However, you are still worried: there was another component of the acceleration. You get out of

the plane and measure the length of the skid the plane made in the sand as

it landed: 154 ft. The pilot claims the plane’s horizontal speed on impact

was 160 m.p.h. Assuming the worst case scenario (i.e., that the vertical and

horizontal accelerations occurred simultaneously), and that the horizontal

deceleration was uniform, is it likely that the seal on the virus container is

compromised? Explain.

93. M The Lost Drone. You and your team are exploring the edge of an Ant- arctic mountain range and you send a drone ahead to help navigate. After takeoff

you lose sight of the drone and, a few seconds later, the controls malfunction and

the drone stops sending visual images and navigational information except for

speed and directional data. Changing speeds erratically, the drone heads west

until it makes a drastic turn at the 5-minute mark to 35.0° east of south. After

nearly ten minutes, the speed

drops to zero and the drone

stops sending data. It has

crashed. Using the speed/

directional data, the team

draws up the graph shown in

the drawing.. How far is the

drone from you, and in what

direction must you go to re-

trieve it? Express your result

as a geographical direction

(i.e., in the form 30° north of

east, etc.).

In this chapter we have studied the displacement, velocity, and acceleration

vectors. The concept questions are designed to review some of the important

concepts that can help in anticipating some of the characteristics of the nu-

merical answers.

90. M CHALK A skydiver is falling straight down, along the negative y dir- ection. (a) During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s, as in part a of the fi gure. (b) Later, her parachute opens, and her speed decreases from 48 to 26 m/s in 11 s, as in part b of the draw- ing. Concepts: (i) Is her average acceleration positive or negative when her speed is increasing in part a of the fi gure? (ii) Is her average acceleration positive or negative when her speed is decreasing in part b of the fi gure? Calculations: In both instances (parts a and b of the fi gure) determine the magnitude and direction of her average acceleration.

91. M CHALK SSM A top-fuel dragster starts from rest and has a constant acceleration of 40.0 m/s2. Concepts: (i) At time t the dragster has a certain velocity. Keeping in mind that the dragster starts from rest, when the time

doubles to 2t, does the velocity also double? (ii) When the time doubles to 2t, does the displacement of the dragster also double? Calculations: What are the (a) fi nal velocities and (b) displacements of the dragster at the end of 2.0 s and at the end of twice this time, or 4.0 s?

Team Problems

Concepts and Calculations Problems

(b)

0υ

υ

–

+

(a)

0υ

υ

–

+

PROBLEM 90

200

WEST

Time t (s)

15

10

5

0

20

0 100 300 400

Ve lo

ci ty

(m /s

) υ

500 600

35º East of South

PROBLEM 93

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

3.1 Define two-dimensional kinematic variables.

3.2 Use two-dimensional kinematic equations to predict future or past values of variables.

3.3 Analyze projectile motion to predict future or past values of variables.

3.4 Apply relative velocity equations.

© t

h re

es p ee

d jo

n es

/i S

to ck

p h o to

CHAPTER 3

Kinematics in Two Dimensions

The beautiful water fountain at the World War II Memorial in Washington, D.C., is illuminated at night.

The arching streams of water follow parabolic paths whose sizes depend on the launch velocity of the

water and the acceleration due to gravity, assuming that the eff ects of air resistance are negligible.

3.1 Displacement, Velocity, and Acceleration In Chapter 2 the concepts of displacement, velocity, and acceleration are used to

describe an object moving in one dimension. There are also situations in which the

motion is along a curved path that lies in a plane. Such two-dimensional motion can

be described using the same concepts. In Grand Prix racing, for example, the course

follows a curved road, and Figure 3.1 shows a race car at two diff erent positions along it. These positions are identifi ed by the vectors r→ and r→0, which are drawn from an arbitrary coordinate origin. The displacement Δr→ of the car is the vector drawn from the initial position r→0 at time t0 to the fi nal position r→ at time t. The magnitude of Δr→ is the shortest distance between the two positions. In the drawing, the vectors r→0 and Δr→ are drawn tail to head, so it is evident that r→ is the vector sum of r→0 and Δr→. (See Sections 1.5 and 1.6 for a review of vectors and vector addition.) This means that

r→ = r→0 + Δr→, or Displacement = Δr→ = r→ − r→0 The displacement here is defi ned as it is in Chapter 2. Now, however, the displacement

vector may lie anywhere in a plane, rather than just along a straight line. 55

56 CHAPTER 3 Kinematics in Two Dimensions

The average velocity v→ of the car in Figure 3.1 between the two positions is defi ned in a manner similar to that in Equation 2.2, as the displacement Δr→ = r→ − r→0 divided by the elapsed time Δt = t − t0:

v→ = r→ − r→0 t − t0

= ∆r→

∆t (3.1)

Since both sides of Equation 3.1 must agree in direction, the average velocity vector has the same

direction as the displacement Δr→. The velocity of the car at an instant of time is its instanta- neous velocity v→. The average velocity becomes equal to the instantaneous velocity v→ in the limit that Δt becomes infi nitesimally small (Δt → 0 s):

v→ = lim ∆ t → 0

∆ r→

∆ t

Figure 3.2 illustrates that the instantaneous velocity v→ is tangent to the path of the car. The draw- ing also shows the vector components v →x and v →y of the velocity, which are parallel to the x and y axes, respectively.

The average acceleration a→ is defi ned just as it is for one-dimensional motion—namely, as the change in velocity, Δv→ = v→ − v →0, divided by the elapsed time Δt:

a→ = v→ − v→0 t − t0

= ∆v→

∆ t (3.2)

The average acceleration has the same direction as the change in velocity Δv→. In the limit that the elapsed time becomes infi nitesimally small, the average acceleration becomes equal to the

instantaneous acceleration a→:

→a = lim ∆t → 0

v→

∆t

The acceleration has a vector component a→x along the x direction and a vector component a→y along the y direction.

Check Your Understanding

(The answer is given at the end of the book.) 1. Suppose you are driving due east, traveling a distance of 1500 m in 2 minutes. You then turn due north

and travel the same distance in the same time. What can be said about the average speeds and the aver-

age velocities for the two segments of the trip? (a) The average speeds are the same, and the average velocities are the same. (b) The average speeds are the same, but the average velocities are diff erent. (c) The average speeds are diff erent, but the average velocities are the same.

3.2 Equations of Kinematics in Two Dimensions To understand how displacement, velocity, and acceleration are applied to two-dimensional

motion, consider a spacecraft equipped with two engines that are mounted perpendicular to each

other. These engines produce the only forces that the craft experiences, and the spacecraft is

assumed to be at the coordinate origin when t0 = 0 s, so that r→0 = 0 m. At a later time t, the space- craft’s displacement is Δr→ = r→ − r→0 = r→. Relative to the x and y axes, the displacement r→ has vector components of x→ and y→, respectively.

In Figure 3.3 only the engine oriented along the x direction is fi ring, and the vehicle accelerates along this direction. It is assumed that the velocity in the y direction is zero, and it remains zero, since the y engine is turned off . The motion of the spacecraft along the x direction is described by the fi ve kinematic variables x, ax, υx, υ0x, and t. Here the symbol “x” reminds us

+y

+x

r0

r

Δr

t0

t

FIGURE 3.1 The displacement Δr→ of the car is a vector that points from the initial

position of the car at time t0 to the fi nal position at time t. The magnitude of Δr→ is the shortest distance between the two

positions.

+y

+x

vy

vx

θ

v

FIGURE 3.2 The instantaneous velocity v→ and its two vector components v→x and v→y.

3.2 Equations of Kinematics in Two Dimensions 57

that we are dealing with the x components of the displacement, velocity, and acceleration vectors. (See Sections 1.7 and 1.8 for a review of vector components.) The variables x, ax, υx, and υ0x are scalar components (or “components,” for short). As Section 1.7 discusses, these components are

positive or negative numbers (with units), depending on whether the associated vector compo-

nents point in the +x or the −x direction. If the spacecraft has a constant acceleration along the x direction, the motion is exactly like that described in Chapter 2, and the equations of kinematics

can be used. For convenience, these equations are written in the left column of Table 3.1. Figure 3.4 is analogous to Figure 3.3, except that now only the y engine is fi ring, and

the spacecraft accelerates along the y direction. Such a motion can be described in terms of the kinematic variables y, ay, υy, υ0y, and t. And if the acceleration along the y direction is constant, these variables are related by the equations of kinematics, as written in the right

column of Table 3.1. Like their counterparts in the x direction, the scalar components, y, ay, υy, and υ0y, may be positive (+) or negative (−) numbers (with units).

If both engines of the spacecraft are fi ring at the same time, the resulting motion takes place in part along the x axis and in part along the y axis, as Interactive Figure 3.5 illus- trates. The thrust of each engine gives the vehicle a corresponding acceleration component.

The x engine accelerates the craft in the x direction and causes a change in the x component of the velocity. Likewise, the y engine causes a change in the y component of the velocity.

Problem-Solving Insight It is important to realize that the x part of the motion occurs exactly as it would if the y part did not occur at all. Similarly, the y part of the motion occurs exactly as it would if the x part of the motion did not exist.

In other words, the x and y motions are independent of each other. The independence of the x and y motions lies at the heart of two-dimensional kinematics. It

allows us to treat two-dimensional motion as two distinct one-dimensional motions, one for the x direction and one for the y direction. Everything that we have learned in Chapter 2 about kinemat- ics in one dimension will now be applied separately to each of the two directions. In so doing, we

will be able to describe the x and y variables separately and then bring these descriptions together to understand the two-dimensional picture. Examples 1 and 2 take this approach in dealing with

a moving spacecraft.

+y

ax

+x

x

υ 0x υ x x

y

x

y FIGURE 3.3 The spacecraft is moving with a constant

acceleration ax parallel to the x axis. There is no motion in the y direction, and the y engine is turned off .

TABLE 3.1 Equations of Kinematics for Constant Acceleration in Two-Dimensional Motion

x Component Variable y Component x Displacement y

ax Acceleration ay

υx Final velocity υy

υ0x Initial velocity υ0y

t Elapsed time t

υx = υ0x + axt (3.3a) υy = υ0y + ayt (3.3b)

x = 12 (υ0x + υx) t (3.4a) y = 1

2(υ0y + υy)t (3.4b)

x = υ0 xt + 1

2 ax t2 (3.5a) y = υ0y t + 1

2 ay t 2 (3.5b)

υx 2 = υ0x2 + 2ax x (3.6a) υy 2 = υ 20y + 2ay y (3.6b)

ay

+x

y

+y

υ 0y

υ y

x

y

y

x

FIGURE 3.4 The spacecraft is moving with a constant acceleration ay parallel to the y axis. There is no motion in the x direction, and the x engine is turned off .

58 CHAPTER 3 Kinematics in Two Dimensions

+y

υ 0x +x

υ 0y

υ y

υ x

y x

y x

INTERACTIVE FIGURE 3.5 The two-dimensional motion of the spacecraft can be viewed as the combination of the

separate x and y motions.

EXAMPLE 1 The Displacement of a Spacecraft

In Interactive Figure 3.5, the directions to the right and upward are the positive directions. In the x direction, the spacecraft has an initial veloc- ity component of υ0x = +22 m/s and an acceleration component of ax = +24 m/s2. In the y direction, the analogous quantities are υ0y = +14 m/s and ay = +12 m/s2. At a time of t = 7.0 s, fi nd the x and y components of the spacecraft’s displacement.

Reasoning The motion in the x direction and the motion in the y direc- tion can be treated separately, each as a one-dimensional motion subject

to the equations of kinematics for constant acceleration (see Table 3.1). By following this procedure we will be able to determine x and y, which specify the spacecraft’s location after an elapsed time of 7.0 s.

Problem-Solving Insight When the motion is two- dimensional, the time variable t has the same value for both the x and y directions.

Solution The data for the motion in the x direction are listed in the fol- lowing table:

x-Direction Data x ax υx υ0x t ? +24 m/s2 +22 m/s 7.0 s

The x component of the craft’s displacement can be found by using Equation 3.5a:

x = υ0x t + 1

2 ax t 2 = (22 m /s)(7.0 s) + 1

2 (24 m /s 2) (7.0 s)2 = +740 m

The data for the motion in the y direction are listed in the following table:

y-Direction Data y ay υy υ0y t ? +12 m/s2 +14 m/s 7.0 s

The y component of the craft’s displacement can be found by using Equation 3.5b:

y = υ0yt + 1

2ayt 2 = (14 m/s)(7.0 s) + 1

2 (12 m/s 2)(7.0 s)2 = +390 m

After 7.0 s, the spacecraft is 740 m to the right and 390 m above the origin.

Analyzing Multiple-Concept Problems

EXAMPLE 2 The Velocity of a Spacecraft

This example also deals with the spacecraft in Interactive Fig ure 3.5. As in Example 1, the x components of the craft’s initial velocity and acceleration are υ0x = +22 m/s and ax = +24 m/s2, respectively. The corresponding y components are υ0y = +14 m/s and ay = +12 m/s2. At a time of t = 7.0 s, fi nd the spacecraft’s fi nal velocity (magnitude and direction).

Reasoning Figure 3.6 shows the fi nal velocity vector, which has com- ponents υx and υy and a magnitude υ. The fi nal velocity is directed at an angle θ above the +x axis. The vector and its components form a right triangle, the hypotenuse being the magnitude of the velocity and the

components being the other two sides. Thus, we can use the Pythagorean

theorem to determine the magnitude υ from values for the components υx and υy. We can also use trigonometry to determine the directional angle θ.

+y

υ x

υ y

+x

y

x

υ

θ

FIGURE 3.6 The velocity vector has components υx and υy and a magnitude υ. The magnitude gives the speed of the spacecraft, and the

angle θ gives the direction of travel relative to the positive x direction.

3.2 Equations of Kinematics in Two Dimensions 59

Knowns and Unknowns The data for this problem are listed in the table that follows:

Description Symbol Value Comment x component of acceleration ax +24 m/s2

x component of initial velocity υ0x +22 m/s

y component of acceleration ay +12 m/s2

y component of initial velocity υ0y +14 m/s

Time t 7.0 s Same time for x and y directions

Unknown Variables Magnitude of final velocity υ ?

Direction of final velocity θ ?

Modeling the Problem

STEP 1 Final Velocity In Figure 3.6 the fi nal velocity vector and its components υx and υy form a right triangle. Applying the Pythagorean theorem to this right triangle shows that the

magnitude υ of the fi nal velocity is given in terms of the components by Equation 1a at the right. From the right triangle in Figure 3.6 it also follows that the directional angle θ is given by Equa- tion 1b at the right.

STEP 2 The Components of the Final Velocity Values are given for the kinematic variables ax, υ0x, and t in the x direction and for the corresponding variables in the y direction (see the table of knowns and unknowns). For each direction, then, these values allow us to calculate the fi nal

velocity components υx and υy by using Equations 3.3a and 3.3b from the equations of kinematics.

υx = υ0x + axt (3.3a)

υy = υ0y + ayt (3.3b)

These expressions can be substituted into Equations 1a and 1b for the magnitude and direction of

the fi nal velocity, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

υ = √υx2 + υy2 = √(υ0x + ax t)2 + (υ0y + ay t)2

θ = tan−1( υy υx) = tan−1(

υ0y + ay t υ0x + ax t)

With the data given for the kinematic variables in the x and y directions, we fi nd that the magnitude and direction of the fi nal velocity of the spacecraft are

υ = √(υ0x + ax t)2 + (υ0y + ay t)2

= √[(22 m /s) + (24 m /s2)(7.0 s)]2 + [(14 m/s) + (12 m/s2)(7.0 s)]2 = 210 m/s

θ = tan−1 ( υ0y + ay t υ0x + ax t) = tan−1[

(14 m/s) + (12 m/s2) (7.0 s)

(22 m/s) + (24 m/s2) (7.0 s)]= 27° After 7.0 s, the spacecraft, at the position determined in Example 1, has a velocity of 210 m/s in

a direction of 27° above the positive x axis.

Related Homework: Problem 66

STEP 1 STEP 2

STEP 1 STEP 2

υ = √υx2 + υy2 (1a)

θ = tan−1 ( υy υx) (1b)

υ = √υx2 + υy2 (1a)

υx = υ0x + axt

υy = υ0y + ayt

(3.3a,b)

θ = tan−1 ( υy υx) (1b)

υx = υ0x + axt

υy = υ0y + ayt

(3.3a,b)

60 CHAPTER 3 Kinematics in Two Dimensions

The following Reasoning Strategy gives an overview of how the equations of kinematics are

applied to describe motion in two dimensions, as in Examples 1 and 2.

REASONING STRATEGY Applying the Equations of Kinematics in Two Dimensions 1. Make a drawing to represent the situation being studied. 2. Decide which directions are to be called positive (+) and negative (−) relative to a con-

veniently chosen coordinate origin. Do not change your decision during the course of a calculation.

3. Remember that the time variable t has the same value for the part of the motion along the x axis and the part along the y axis.

4. In an organized way, write down the values (with appropriate + and − signs) that are given for any of the fi ve kinematic variables associated with the x direction and the y direction. Be on the alert for implied data, such as the phrase “starts from rest,” which means that the values of the initial velocity components are zero: υ0x = 0 m/s and υ0y = 0 m/s. The data summary tables that are used in the examples are a good way of keeping track of this information. In addition, identify the variables that you are being asked to determine.

5. Before attempting to solve a problem, verify that the given information contains values for at least three of the kinematic variables. Do this for the x and the y direction of the motion. Once the three known variables are identifi ed along with the desired unknown variable, the appropriate relations from Table 3.1 can be selected.

6. When the motion is divided into segments, remember that the fi nal velocity for one seg- ment is the initial velocity for the next segment.

7. Keep in mind that a kinematics problem may have two possible answers. Try to visualize the diff erent physical situations to which the answers correspond.

Check Your Understanding

(The answer is given at the end of the book.) 2. A power boat, starting from rest, maintains a constant acceleration. After a certain time t, its

displacement and velocity are r→ and v→. At time 2t, what would be its displacement and velocity, assuming the acceleration remains the same? (a) 2 r→ and 2 v→ (b) 2 r→ and 4 v→ (c) 4 r→ and 2 v→ (d) 4 r→ and 4 v→

3.3 Projectile Motion The biggest thrill in baseball is a home run. The motion of the ball on its curving path into

the stands is a common type of two-dimensional motion called “projectile motion.” A good

description of such motion can often be obtained with the assumption that air resistance is

absent.

Using the equations in Table 3.1, we consider the horizontal and vertical parts of the motion separately. In the horizontal or x direction, the moving object (the projectile) does not slow down in the absence of air resistance. Thus, the x component of the velocity remains con- stant at its initial value or υx = υ0x, and the x component of the acceleration is ax = 0 m/s2. In the vertical or y direction, however, the projectile experiences the eff ect of gravity. As a result, the y component of the velocity υy is not constant and changes. The y component of the accelera- tion ay is the downward acceleration due to gravity. If the path or trajectory of the projectile is near the earth’s surface, ay has a magnitude of 9.80 m/s2. In this text, then, the phrase “projec- tile motion” means that ax = 0 m/s2 and ay equals the acceleration due to gravity. Example 3 and other examples in this section illustrate how the equations of kinematics are applied to

projectile motion.

3.3 Projectile Motion 61

The freely falling package in Example 3 picks up vertical speed on the way downward. The

horizontal component of the velocity, however, retains its initial value of υ0x = +115 m/s throughout the entire descent. Since the plane also travels at a constant horizontal velocity of +115 m/s, it

remains directly above the falling package. The pilot always sees the package directly beneath the

plane, as the dashed vertical lines in Figure 3.7 show. This result is a direct consequence of the fact that the package has no acceleration in the horizontal direction. In reality, air resistance would

slow down the package, and it would not remain directly beneath the plane during the descent.

Figure 3.8 illustrates what happens to two packages that are released simultaneously from the same height, in order to emphasize that the vertical and horizontal parts of the motion in

Example 3 occur independently. Package A is dropped from a stationary balloon and falls straight

downward toward the ground, since it has no horizontal velocity component (υ0x = 0 m/s). Pack- age B, on the other hand, is given an initial velocity component of υ0x = +115 m/s in the hori- zontal direction, as in Example 3, and follows the path shown in the fi gure. Both packages hit the

ground at the same time. Not only do the packages in Figure 3.8 reach the ground at the same time, but the y components of their velocities are also equal at all points on the way down. How- ever, package B does hit the ground with a greater speed than does package A. Remember, speed

y = –1050 m

+x

+y

υ y

x = 115 m/sυ

x = 115 m/sυ

x = 115 m/sυ

υ

υ y υ

FIGURE 3.7 The package falling from the plane is an example of projectile motion,

as Examples 3 and 4 discuss.

EXAMPLE 3 A Falling Care Package

Figure 3.7 shows an airplane moving horizontally with a constant ve- locity of +115 m/s at an altitude of 1050 m. The directions to the right

and upward have been chosen as the positive directions. The plane

releases a “care package” that falls to the ground along a curved tra-

jectory. Ignoring air resistance, determine the time required for the

package to hit the ground.

Reasoning The time required for the package to hit the ground is the time it takes for the package to fall through a vertical distance of 1050 m.

In falling, it moves to the right, as well as downward, but these two

parts of the motion occur independently. Therefore, we can focus solely

on the vertical part. We note that the package is moving initially in

the horizontal or x direction, not in the y direction, so that υ0y = 0 m/s. Furthermore, when the package hits the ground, the y component of its displacement is y = −1050 m, as the drawing shows. The acceleration is that due to gravity, so ay = −9.80 m/s2. These data are summarized as follows:

y-Direction Data y ay υy υ0y t

−1050 m −9.80 m/s2 0 m/s ?

With these data, Equation 3.5b (y = υ0y t + 1

2 ay t 2) can be used to fi nd the fall time.

Problem-Solving Insight The variables y, ay, υy, and υ0y are scalar components. Therefore, an algebraic sign (+ or −) must be included with each one to denote direction.

Solution Since υ0y = 0 m/s, it follows from Equation 3.5b that y = 12ayt2 and

t = √2yay = √ 2(−1050 m)

−9.80 m/s2 = 14.6 s

62 CHAPTER 3 Kinematics in Two Dimensions

is the magnitude of the velocity vector, and the velocity of B has an x component, whereas the velocity of A does not. The magnitude and direction of the velocity vector for package B at the

instant just before the package hits the ground is computed in Example 4.

BA

x = 115 m/sυ

0x = 115 m/sυ

υ yυ y

υ yυ y

υ yυ y

υ

θ FIGURE 3.8 Package A and package B are released

simultaneously at the same height and strike the ground at

the same time because their y variables (y, ay, and 𝜐0y) are the same.

Analyzing Multiple-Concept Problems

EXAMPLE 4 The Velocity of the Care Package

Figure 3.7 shows a care package falling from a plane, and Figure 3.8 shows this package as package B. As in Example 3, the directions to the right

and upward are chosen as the positive directions, and the plane is moving

horizontally with a constant velocity of +115 m/s at an altitude of 1050 m.

Ignoring air resistance, fi nd the magnitude υ and the directional angle θ of the fi nal velocity vector that the package has just before it strikes the ground.

Reasoning Figures 3.7 and 3.8 show the fi nal velocity vector, which has components υx and υy and a magnitude υ. The vector is directed at

an angle θ below the horizontal or x direction. We note the right triangle formed by the vector and its components. The hypotenuse of the triangle

is the magnitude of the velocity, and the components are the other two

sides. As in Example 2, we can use the Pythagorean theorem to express

the magnitude or speed υ in terms of the components υx and υy, and we can use trigonometry to determine the directional angle θ.

Knowns and Unknowns The data for this problem are listed in the table that follows:

Description Symbol Value Comment Explicit Data, x Direction x component of initial velocity υ0x +115 m/s Package has plane’s horizontal velocity at instant of release

Implicit Data, x Direction x component of acceleration ax 0 m/s2 No horizontal acceleration, since air resistance is ignored

Explicit Data, y Direction y component of displacement y −1050 m Negative, since upward is positive and package falls downward

Implicit Data, y Direction y component of initial velocity υ0y 0 m/s Package traveling horizontally in x direction at instant of release,

not in y direction

y component of acceleration ay −9.80 m/s2 Acceleration vector for gravity points downward in the negative direction

Unknown Variables Magnitude of final velocity υ ?

Direction of final velocity θ ?

3.3 Projectile Motion 63

CONCEPTUAL EXAMPLE 5 Shot a Bullet Into the Air . . .

Suppose you are driving in a convertible with the top down. The car is

moving to the right at a constant velocity. As Figure 3.9 illustrates, you point a rifl e straight upward and fi re it. In the absence of air resistance,

would the bullet land (a) behind you, (b) ahead of you, or (c) in the barrel of the rifl e?

Reasoning Because there is no air resistance to slow it down, the bul- let experiences no horizontal acceleration. Thus, the bullet’s horizontal

velocity component does not change, and it stays the same as that of the

rifl e and the car.

Modeling the Problem

STEP 1 Final Velocity Using the Pythagorean theorem to express the speed υ in terms of the components υx and υy (see Figure 3.7 or 3.8), we obtain Equation 1a at the right. Furthermore, in a right triangle, the cosine of an angle is the side adjacent to the angle divided by the hypotenuse.

With this in mind, we see in Figure 3.7 or Figure 3.8 that the directional angle θ is given by Equation 1b at the right.

STEP 2 The Components of the Final Velocity Reference to the table of knowns and un- knowns shows that, in the x direction, values are available for the kinematic variables υ0x and ax. Since the acceleration ax is zero, the fi nal velocity component υx remains unchanged from its initial value of υ0x, so we have

𝜐x = 𝜐0x

In the y direction, values are available for y, υ0y, and ay, so that we can determine the fi nal velocity component υy using Equation 3.6b from the equations of kinematics:

𝜐 2y = 𝜐

2 0y + 2ayy (3.6b)

These results for υx and υy can be substituted into Equations 1a and lb, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

υ = √υx2 + υy2 = √υ 20x + υ 20y + 2ayy

θ = cos−1 ( υx

√υ 2x + υy2 ) = cos−1 (

υ0x √υ0x2 + υ0y2 + 2ayy

)

With the data given for the kinematic variables in the x and y directions, we fi nd that the magni- tude and direction of the fi nal velocity of the package are

STEP 1 STEP 2

STEP 1 STEP 2

υ = √υ 20x + υ 20y + 2ayy = √(115 m/s)2 + (0 m/s)2 + 2 (−9.80 m/s2)(−1050 m) = 184 m/s

θ = cos−1 ( υ0x

√υ 20x + υ 20y + 2ayy) = cos−1[ 115 m/s

√(115 m/s)2 + (0 m/s)2 + 2(−9.80 m/s2)(−1050 m) ] = 51.3°

Related Homework: Problems 33, 38, 71

υ = √υx2 + υy2 (1a)

θ = cos−1( 𝜐x

𝜐) = cos−1( 𝜐x

√𝜐x2 + 𝜐y2 )

(1b)

υ = √υx2 + υy2 (1a)

𝜐x = 𝜐0x 𝜐 2y = 𝜐 2

0y + 2ayy (3.6b)

θ = cos−1( 𝜐x

√𝜐x2 + 𝜐y2 ) (1b)

𝜐x = 𝜐0x 𝜐 2y = 𝜐 2

0y + 2ayy (3.6b)

An important feature of projectile motion is that there is no acceleration in the horizontal, or

x, direction. Conceptual Example 5 discusses an interesting implication of this feature.

Problem-Solving Insight The speed of a projectile at any location along its path is the magnitude 𝜐 of its velocity at that location: 𝜐 = √𝜐x2 + 𝜐y2. Both the hori- zontal and vertical velocity components contribute to the speed.

64 CHAPTER 3 Kinematics in Two Dimensions

Often projectiles, such as footballs and baseballs, are sent into the air at an angle with respect

to the ground. From a knowledge of the projectile’s initial velocity, a wealth of information can

be obtained about the motion. For instance, Example 6 demonstrates how to calculate the maxi-

mum height reached by the projectile.

Answers (a) and (b) are incorrect. If air resistance were present, it would slow down the bullet and cause it to land behind you, toward the

rear of the car. However, air resistance is absent. If the bullet were to land

ahead of you, its horizontal velocity component would have to be greater

than that of the rifl e and the car. This cannot be, since the bullet’s horizon-

tal velocity component never changes.

Answer (c) is correct. Since the bullet’s horizontal velocity component does not change, it retains its initial value, and remains matched to that of

the rifl e and the car. As a result, the bullet remains directly above the rifl e

at all times and would fall directly back into the barrel of the rifl e. This

situation is analogous to that in Figure 3.7, where the care package, as it falls, remains directly below the plane.

Related Homework: Problem 39

FIGURE 3.9 The car is moving with a constant velocity to the right, and the rifl e is

pointed straight up. In the absence of air

resistance, a bullet fi red from the rifl e has no

acceleration in the horizontal direction.

Example 5 discusses what happens to the

bullet.

EXAMPLE 6 The Height of a Kickoff

A placekicker kicks a football at an angle of θ = 40.0° above the horizon- tal axis, as Animated Figure 3.10 shows. The initial speed of the ball is υ0 = 22 m/s. Ignore air resistance, and fi nd the maximum height H that the ball attains.

Reasoning The maximum height is a characteristic of the vertical part of the motion, which can be treated separately from the horizontal part. In

preparation for making use of this fact, we calculate the vertical compo-

nent of the initial velocity:

υ0y = υ0 sin θ = +(22 m/s) sin 40.0° = +14 m/s

The vertical component of the velocity, υy, decreases as the ball moves up- ward. Eventually, υy = 0 m/s at the maximum height H. The data below can be used in Equation 3.6b (υy2 = υ0y2 + 2ay y) to fi nd the maximum height:

y-Direction Data y ay υy υ0y t

H = ? −9.80 m/s2 0 m/s +14 m/s

Problem-Solving Insight When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero (υy = 0 m/s). However, the horizontal component of its velocity is not zero.

Solution From Equation 3.6b, we fi nd that

y = H = υy 2 − υ0 y 2

2ay =

(0 m /s) 2 − (14 m /s) 2

2(−9.80 m /s2) = +10 m

The height H depends only on the y variables; the same height would have been reached had the ball been thrown straight up with an initial velocity of υ0y = +14 m/s.

ANIMATED FIGURE 3.10 A football is kicked with an initial speed of υ0 at an angle of θ above the ground. The ball attains a maximum height H and a range R.

H = Maximum height

+y

+x

R = Range

υ 0y

υ 0x

υ 0

+x

+y

θ

3.3 Projectile Motion 65

It is also possible to fi nd the total time or “hang time” during which the football in Animated Figure 3.10 is in the air. Example 7 shows how to determine this time.

EXAMPLE 7 The Physics of the “Hang Time” of a Football

For the motion illustrated in Animated Figure 3.10, ignore air resistance and use the data from Example 6 to determine the time of fl ight between

kickoff and landing.

Reasoning Given the initial velocity, it is the acceleration due to gravity that determines how long the ball stays in the air. Thus, to fi nd the time

of fl ight we deal with the vertical part of the motion. Since the ball starts

at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is the same as that in Example 6; that is, υ0y = +14 m/s. Therefore, we have

y-Direction Data y ay υy υ0y t

0 m −9.80 m/s2 +14 m/s ?

The time of fl ight can be determined from Equation 3.5b ( y = υ0y t + 1

2 ay t 2).

Solution Using Equation 3.5b and the fact that y = 0 m, we fi nd that

y = 0 = υ0y t + 1

2 ay t 2 = (υ0y + 1

2 ay t)t

There are two solutions to this equation. One is given by υ0y + 1

2 ay t = 0, with the result that

t = −2υ0y

ay =

−2(14 m/s)

−9.80 m/s2 = 2.9 s

The other is given by t = 0 s. The solution we seek is t = 2.9 s , because t = 0 s corresponds to the initial kickoff .

EXAMPLE 8 The Range of a Kickoff

For the motion shown in Animated Figure 3.10 and discussed in Examples 6 and 7, ignore air resistance and calculate the range R of the projectile.

Reasoning The range is a characteristic of the horizontal part of the motion. Thus, our starting point is to determine the horizontal component

of the initial velocity:

υ0x = υ0 cos θ = +(22 m/s)cos 40.0° = +17 m/s

Recall from Example 7 that the time of fl ight is t = 2.9 s. Since there is no acceleration in the x direction, υx remains constant, and the range is simply the product of υx = υ0x and the time.

Solution The range is

x = R = υ0x t = +(17 m/s)(2.9 s) = +49 m

Another important feature of projectile motion is called the “range.” The range, as Animated Figure 3.10 shows, is the horizontal distance traveled between launching and landing, assuming the projectile returns to the same vertical level at which it was fi red. Example 8 shows how to obtain the range.

Math Skills To show that a projectile launched from and returning to ground level has its maximum range when θ = 45°, we begin with the expression for the range R from Example 8:

R = 𝜐0x t = (𝜐0 cos θ)t

The range in the previous example depends on the angle θ at which the projectile is fi red above the horizontal. When air resistance is absent, the maximum range results when

θ = 45°.

(Continued)

66 CHAPTER 3 Kinematics in Two Dimensions

In projectile motion, the magnitude of the acceleration due to gravity aff ects the trajectory in

a signifi cant way. For example, a baseball or a golf ball would travel much farther and higher on

the moon than on the earth, when launched with the same initial velocity. The reason is that the

moon’s gravity is only about one-sixth as strong as the earth’s.

Section 2.6 points out that certain types of symmetry with respect to time and speed are

present for freely falling bodies. These symmetries are also found in projectile motion, since pro-

jectiles are falling freely in the vertical direction. In particular, the time required for a projectile

to reach its maximum height H is equal to the time spent returning to the ground. In addition, Figure 3.11 shows that the speed υ of the object at any height above the ground on the upward part of the trajectory is equal to the speed υ at the same height on the downward part. Although the two speeds are the same, the velocities are diff erent, because they point in diff erent directions.

Conceptual Example 9 shows how to use this type of symmetry in your reasoning.

Example 7 shows that the time of fl ight of the projectile is

t = −2𝜐0y

ay =

−2𝜐0y

−9.80 m/s2 =

2𝜐0y

9.80 m/s2

According to Example 6, the velocity component υ0y in this result is υ0y = υ0 sin θ, so that

t = 2𝜐0 sin θ 9.80 m/s2

Substituting this expression for t into the range expression gives

R = (𝜐0 cos θ)t = (𝜐0 cos θ) 2𝜐0 sin θ 9.80 m/s2

= 𝜐0

2 2 sin θ cos θ 9.80 m/s2

Equation 6 (Other Trigonometric Identities) in Appendix E.2 shows that 2 sin θ cos θ = sin 2θ, so the range expression becomes

R = 𝜐 20 2 sin θ cos θ

9.80 m/s2 =

𝜐 20 sin 2θ 9.80 m/s2

In this result R has its maximum value when sin 2θ has its maximum value of 1. This occurs when 2θ = 90°, or when θ = 45°.

FIGURE 3.11 The speed 𝜐 of a projectile at a given height above the ground is the

same on the upward and downward parts of

the trajectory. The velocities are diff erent,

however, since they point in diff erent

directions.

Height

υ

υ

θ θ

θ θ θ P

2

1

v0

v0 v0

FIGURE 3.12 Two stones are thrown off the cliff with identical initial speeds υ0, but at equal angles θ that are below and above the horizontal. Conceptual Example 9 compares the velocities with which the stones hit

the water below.

CONCEPTUAL EXAMPLE 9 Two Ways to Throw a Stone

From the top of a cliff overlooking a lake, a person throws two stones.

The stones have identical initial speeds υ0, but stone 1 is thrown down- ward at an angle θ below the horizontal, while stone 2 is thrown upward at the same angle above the horizontal, as Figure 3.12 shows. Neglect air resistance and decide which stone, if either, strikes the water with the

greater velocity: (a) both stones strike the water with the same velocity, (b) stone 1 strikes with the greater velocity, (c) stone 2 strikes with the greater velocity.

Reasoning Note point P in the drawing, where stone 2 returns to its initial height; here the speed of stone 2 is υ0 (the same as its initial speed), but its velocity is directed at an angle θ below the horizontal. This is exactly the type of projectile symmetry illustrated in Figure 3.11, and this symmetry will lead us to the correct answer.

3.3 Projectile Motion 67

In all the examples in this section, the projectiles follow a curved trajectory. In general, if

the only acceleration is that due to gravity, the shape of the path can be shown to be a parabola.

Check Your Understanding

(The answers are given at the end of the book.) 3. A projectile is fi red into the air, and it follows the parabolic path shown in CYU Figure 3.1, landing on

the right. There is no air resistance. At any instant, the projectile has a velocity v→ and an acceleration a→. Which one or more of the drawings could not represent the directions for v→ and a→ at any point on the trajectory?

(a) (b) (c) (d)

v

v

v

a

a a av

CYU FIGURE 3.1

4. An object is thrown upward at an angle θ above the ground, eventually returning to earth. (a) Is there any place along the trajectory where the velocity and acceleration are perpendicular? If so, where?

(b) Is there any place where the velocity and acceleration are parallel? If so, where? 5. Is the acceleration of a projectile equal to zero when the projectile reaches the top of its trajectory? 6. In baseball, the pitcher’s mound is raised to compensate for the fact that the ball falls downward as it

travels from the pitcher toward the batter. If baseball were played on the moon, would the pitcher’s

mound have to be (a) higher than, (b) lower than, or (c) the same height as it is on earth? 7. A tennis ball is hit upward into the air and moves along an arc. Neglecting air resistance, where along

the arc is the speed of the ball (a) a minimum and (b) a maximum? 8. A wrench is accidentally dropped from the top of the mast on a sailboat. Air resistance is negligible.

Will the wrench hit at the same place on the deck whether the sailboat is at rest or moving with a con-

stant velocity?

9. A rifl e, at a height H above the ground, fi res a bullet parallel to the ground. At the same instant and at the same height, a second bullet is dropped from rest. In the absence of air resistance, which bullet, if

either, strikes the ground fi rst?

10. A stone is thrown horizontally from the top of a cliff and eventually hits the ground below. A second stone is dropped from rest from the same cliff , falls through the same height, and also hits the ground

below. Ignore air resistance. Is each of the following quantities diff erent or the same in the two cases?

(a) Displacement (b) Speed just before impact with the ground (c) Time of fl ight 11. A leopard springs upward at a 45° angle and then falls back to the ground. Air resistance is negligible.

Does the leopard, at any point on its trajectory, ever have a speed that is one-half its initial value?

12. Two balls are launched upward from the same spot at diff erent angles with respect to the ground. Both balls rise to the same maximum height. Ball A, however, follows a trajectory that has a greater

range than that of ball B. Ignoring air resistance, decide which ball, if either, has the greater launch

speed.

Answers (b) and (c) are incorrect. You might guess that stone 1, being hurled downward, would strike the water with the greater velocity. Or,

you might think that stone 2, having reached a greater height than stone 1,

would hit the water with the greater velocity. To understand why neither

of these answers is correct, see the response for answer (a) below.

Answer (a) is correct. Let’s follow the path of stone 2 as it rises to its maximum height and falls back to earth. When it reaches point P in the

drawing, stone 2 has a velocity that is identical to the velocity with which

stone 1 is thrown downward from the top of the cliff (see the drawing).

From this point on, the velocity of stone 2 changes in exactly the same

way as that for stone 1, so both stones strike the water with the same

velocity.

Related Homework: Problems 23, 75

68 CHAPTER 3 Kinematics in Two Dimensions

3.4 Relative Velocity To someone hitchhiking along a highway, two cars speeding by in adjacent lanes seem like a blur.

But if the cars have the same velocity, each driver sees the other remaining in place, one lane

away. The hitchhiker observes a velocity of perhaps 30 m/s, but each driver observes the other’s

velocity to be zero. Clearly, the velocity of an object is relative to the observer who is making

the measurement.

Figure 3.13 illustrates the concept of relative velocity by showing a passenger walking toward the front of a moving train. The people sitting on the train see the passenger walking with

a velocity of +2.0 m/s, where the plus sign denotes a direction to the right. Suppose the train

is moving with a velocity of +9.0 m/s relative to an observer standing on the ground. Then the

ground-based observer would see the passenger moving with a velocity of +11 m/s, due in part

to the walking motion and in part to the train’s motion. As an aid in describing relative velocity,

let us defi ne the following symbols:

v→ PT = velocity of the Passenger relative to the Train = +2.0 m/s v→ TG = velocity of the Train relative to the Ground = +9.0 m/s v→ PG = velocity of the Passenger relative to the Ground = +11 m/s

In terms of these symbols, the situation in Figure 3.13 can be summarized as follows: v→PG = v→PT + v→TG (3.7) or

v→PG = (2.0 m/s) + (9.0 m/s) = +11 m/s According to Equation 3.7*, v→PG is the vector sum of v→PT and v→TG, and this sum is shown in the drawing. Had the passenger been walking toward the rear of the train, rather than toward the

front, the velocity relative to the ground-based observer would have been v→PG = (−2.0 m/s) + (9.0 m/s) = +7.0 m/s.

Each velocity symbol in Equation 3.7 contains a two-letter subscript. The fi rst letter in the

subscript refers to the body that is moving, while the second letter indicates the object relative

to which the velocity is measured. For example,v→TG and v→PG are the velocities of the Train and Passenger measured relative to the Ground. Similarly, v→PT is the velocity of the Passenger measured by an observer sitting on the Train.

The ordering of the subscript symbols in Equation 3.7 follows a defi nite pattern. The fi rst

subscript (P) on the left side of the equation is also the fi rst subscript on the right side of the equa-

tion. Likewise, the last subscript (G) on the left side is also the last subscript on the right side. The

third subscript (T) appears only on the right side of the equation as the two “inner” subscripts.

The colored boxes below emphasize the pattern of the symbols in the subscripts:

v→ PG = v→ P T + v →

T G

In other situations such as in the case of a helicopter landing on an aircraft carrier (Photo 3.1), the subscripts will not necessarily be P, G, and T, but will be compatible with the names of the

objects involved in the motion.

PHOTO 3.1 In landing on a moving aircraft carrier, the pilot of the helicopter must match

the helicopter’s horizontal velocity to the

carrier’s velocity, so that the relative velocity

of the helicopter and the carrier is zero.

TAL COHEN/AFP/Getty Images

FIGURE 3.13 The velocity of the passenger relative to the ground-based observer is v→PG. It is the vector sum of the velocity v→PT of the passenger relative to the train and the velocity v→TG of the train relative to the ground: v→PG = v→PT + v→TG.

vTG = +9.0 m/s

Ground-based observer

vPT = +2.0 m/s

vPG = +11.0 m/s

vTGvPT

*This equation assumes that the train and the ground move in a straight line relative to one another.

3.4 Relative Velocity 69

Equation 3.7 has been presented in connection with one-dimensional motion, but the result

is also valid for two-dimensional motion. Figure 3.14 depicts a common situation that deals with relative velocity in two dimensions. Part a of the drawing shows a boat being carried downstream by a river; the engine of the boat is turned off . In part b, the engine is turned on, and now the boat moves across the river in a diagonal fashion because of the combined motion produced by the

current and the engine. The list below gives the velocities for this type of motion and the objects

relative to which they are measured:

v→ BW = velocity of the Boat relative to the Water v→ WS = velocity of the Water relative to the Shore v→ BS = velocity of the Boat relative to the Shore The velocity v→BW of the boat relative to the water is the velocity measured by an observer who, for instance, is fl oating on an inner tube and drifting downstream with the current. When the

engine is turned off , the boat also drifts downstream with the current, and v→BW is zero. When the engine is turned on, however, the boat can move relative to the water, and v→BW is no longer zero. The velocity v→WS of the water relative to the shore is the velocity of the current measured by an observer on the shore. The velocity v→BS of the boat relative to the shore is due to the com- bined motion of the boat relative to the water and the motion of the water relative to the shore.

In symbols,

v→ BS = v→ B W + v →

W S

The ordering of the subscripts in this equation is identical to that in Equation 3.7, although the letters

have been changed to refl ect a diff erent physical situation. Example 10 illustrates the concept of

relative velocity in two dimensions.

FIGURE 3.14 (a) A boat with its engine turned off is carried along by the current.

(b) With the engine turned on, the boat moves across the river in a diagonal fashion.

Current

Current

(a)

(b)

EXAMPLE 10 Crossing a River

The engine of a boat drives it across a river that is 1800 m wide. The

velocity v→BW of the boat relative to the water is 4.0 m/s, directed perpen- dicular to the current, as in Interactive Figure 3.15. The velocity v→WS of the water relative to the shore is 2.0 m/s. (a) What is the velocity v→BS of the boat relative to the shore? (b) How long does it take for the boat to cross the river?

Reasoning (a) The velocity of the boat relative to the shore is v→BS. It is the vector sum of the velocity v→BW of the boat relative to the water and the velocity v→WS of the water relative to the shore: v→BS = v→BW + v→WS. Since v→BW and v→WS are both known, we can use this relation among the veloci- ties, with the aid of trigonometry, to fi nd the magnitude and directional

angle of v→BS. (b) The component of v→BS that is parallel to the width of the river (see Interactive Figure 3.15) determines how fast the boat crosses the river; this parallel component is υBS sin θ = υBW = 4.0 m/s. The time for the boat to cross the river is equal to the width of the river divided by

the magnitude of this velocity component.

Solution (a) Since the vectors v→BW and v→WS are perpendicular (see Interactive Figure 3.15), the magnitude of v→BS can be determined by using the Pythagorean theorem:

υBS = √(υBW)2 + (υWS)2 = √(4.0 m/s)2 + (2.0 m/s)2 = 4.5 m/s

Thus, the boat moves at a speed of 4.5 m/s with respect to an observer on

shore. The direction of the boat relative to the shore is given by the angle

θ in the drawing:

tan θ = υBW υWS

or θ = tan−1 ( υBW υWS) = tan−1 (

4.0 m/s

2.0 m/s) = 63° (b) The time t for the boat to cross the river is

t = Width

υBS sin θ =

1800 m

4.0 m/s = 450 s

INTERACTIVE FIGURE 3.15 The velocity of the boat relative to the shore

is v→BS. It is the vector sum of the velocity v→BW of the boat relative to the water and the velocity v→WS of the water relative to the shore: v→BS = v→BW + v→WS.

vBW = 4.0 m/s

vWS = 2.0 m/s

vBS +y

+x

θ

70 CHAPTER 3 Kinematics in Two Dimensions

Sometimes, situations arise when two vehicles are in relative motion, and it is useful to know

the relative velocity of one with respect to the other. Example 11 considers this type of relative

motion.

EXAMPLE 11 Approaching an Intersection

Figure 3.16a shows two cars approaching an intersection along perpen- dicular roads. The cars have the following velocities:

v→ AG = velocity of car A relative to the Ground = 25.0 m/s, eastward

v→ BG = velocity of car B relative to the Ground = 15.8 m/s, northward

Find the magnitude and direction of v→AB, where

v→ AB = velocity of car A as measured by a passenger in car B

Reasoning To fi nd v→AB, we use an equation whose subscripts follow the order outlined earlier. Thus,

v→ AB = v→ A G + v→G B

In this equation, the term v→GB is the velocity of the ground relative to a passenger in car B, rather than v→BG, which is given as 15.8 m/s, north- ward. In other words, the subscripts are reversed. However, v→GB is related to v→BG according to

v→GB = −v→BG This relationship refl ects the fact that a passenger in car B, moving north-

ward relative to the ground, looks out the car window and sees the ground

moving southward, in the opposite direction. Therefore, the equation

v→AB = v→AG + v→GB may be used to fi nd v→AB, provided we recognize v→GB as a vector that points opposite to the given velocity v→BG. With this in mind, Figure 3.16b illustrates how v→AG and v→GB are added vectorially to give v→AB.

Problem-Solving Insight In general, the velocity of object R relative to object S is always the negative of the velocity of object S relative to R: v→RS = −v→SR.

Solution From the vector triangle in Figure 3.16b, the magnitude and direction of v→AB can be calculated as

υAB = √(υAG)2 + (υGB)2 = √(25.0 m/s)2 + (−15.8 m/s)2 = 29.6 m/s

and

cos θ = υAG υAB

 or θ = cos−1( υAG υAB) = cos−1(

25.0 m/s

29.6 m/s) = 32.4°

FIGURE 3.16 Two cars are approaching an intersection along perpendicular roads.

(b)

(a)

Car B

Car A

vAG = 25.0 m/s vGB = –vBG

vAG

vAB = vAG + vGB

vBG = 15.8 m/s

N

S

W E

θ

THE PHYSICS OF . . . raindrops falling on car windows. While driving a car, have you ever noticed that the rear window sometimes remains dry, even though rain is falling? This

phenomenon is a consequence of relative velocity, as Figure 3.17 helps to explain. Part a shows a car traveling horizontally with a velocity of v→CG and a raindrop falling vertically with a velocity of v→RG. Both velocities are measured relative to the ground. To determine whether the raindrop

3.4 Relative Velocity 71

hits the window, however, we need to consider the velocity of the raindrop relative to the car, not

to the ground. This velocity is v→RC, and we know that v→RC = v→RG + v→GC = v→RG − v→CG Here, we have used the fact that v→GC = −v→CG. Part b of the drawing shows the tail-to-head arrangement corresponding to this vector subtraction and indicates that the direction of v→RC is given by the angle θR. In comparison, the rear window is inclined at an angle θW with respect to the vertical (see the blowup in part a). When θR is greater than θW, the raindrop will miss the window. However, θR is determined by the speed υRG of the raindrop and the speed υCG of the car, according to θR = tan−1(υCG/υRG). At higher car speeds, the angle θR becomes too large for the drop to hit the window. At a high enough speed, then, the car simply drives out from under each

falling drop!

Check Your Understanding

(The answers are given at the end of the book.) 13. Three cars, A, B, and C, are moving along a straight section of a highway. The velocity of A relative

to B is v→AB, the velocity of A relative to C is v→AC, and the velocity of C relative to B is v→CB. Fill in the missing velocities in the table.

v→AB v→AC v→CB (a) ? +40 m/s +30 m/s

(b) ? +50 m/s −20 m/s

(c) +60 m/s +20 m/s ?

(d) −50 m/s ? +10 m/s

14. On a riverboat cruise, a plastic bottle is accidentally dropped overboard. A passenger on the boat estimates that the boat pulls ahead of the bottle by 5 meters each second. Is it possible to conclude

that the magnitude of the velocity of the boat with respect to the shore is 5 m/s?

15. A plane takes off at St. Louis, fl ies straight to Denver, and then returns the same way. The plane fl ies at the same speed with respect to the ground during the entire fl ight, and there are no head winds or

tail winds. Since the earth revolves around its axis once a day, you might expect that the times for

the outbound trip and the return trip diff er, depending on whether the plane fl ies against the earth’s

rotation or with it. Is this true, or are the two times the same?

16. A child is playing on the fl oor of a recreational vehicle (RV) as it moves along the highway at a constant velocity. He has a toy cannon, which shoots a marble at a fi xed angle and speed with respect to the fl oor.

The cannon can be aimed toward the front or the rear of the RV. Is the range toward the front the same

as, less than, or greater than the range toward the rear? Answer this question (a) from the child’s point of view and (b) from the point of view of an observer standing still on the ground.

17. Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Swimmer A swims perpendicular to the current and lands on the far

shore downstream, because the current has swept him in that direction. Swimmer B swims upstream at

an angle to the current, choosing the angle so that he lands on the far shore directly opposite the starting

point. Swimmer C swims downstream at an angle to the current in an attempt to take advantage of the

current. Who crosses the river in the least time?

FIGURE 3.17 (a) With respect to the ground, a car is traveling at a velocity of v→CG and a raindrop is falling at a velocity of v→RG. The rear window of the car is inclined at an angle θW with respect to the vertical. (b) This tail-to-head arrangement of vectors corresponds to the equation v→RC = v→RG − v→CG.

w Raindrop

R

–vCG

vRC

(b)(a)

vRG vCG

vRG

θ

θ

72 CHAPTER 3 Kinematics in Two Dimensions

Finally, 𝜐CG = 𝜐BC

tan 15° =

4.6 m/s

tan 15° = 17 m/s

EXAMPLE 12 BIO The Physics of Blood Splatter

The shape of the splatter made by a falling drop of blood on a surface can

provide information on the velocity of the drop at impact. By measuring

the physical dimensions of the pattern, forensic scientists can determine

the drop’s angle of impact with the surface. If a drop of blood falls straight

downward and strikes a horizontal surface, the pattern it leaves is fairly

circular (Figure 3.18a). However, if the drop strikes the surface at some angle less than 90°, then the pattern will be elongated with a tail (Fig- ure 3.18a). Consider the following situation where a chef at a restaurant accidentally cuts her fi nger while slicing vegetables. She quickly gets in

her car to drive to the closest emergency room for stitches, but she keeps

her hand outside the window, so as to not drip blood inside the car. Based

on the elongation of the drops of blood on the roadway, the impact angle

was determined to be 15°. Assume the drops of blood behave like a pro-

jectile released from a height of 1.1 m and determine the speed of the car.

Reasoning This problem is similar to the physics of falling raindrops. We can analyze the blood splatter by using relative velocity. Here, we

assume the chef’s car is traveling horizontally with a velocity relative

to ground of v→CG, and the blood drop is falling vertically and impacts the ground with a velocity relative to the car of v→BC (Figure 3.18b). The velocity of the blood drop relative to the ground, which determines its

elongation on impact, will be given by:

v→BG = v→BC + v→CG Notice how the subscripts follow the order outlined earlier. This equation

can be solved for the velocity of the car relative to ground (v→CG). The velocity of the blood drop relative to the car (v→BC) at impact can be deter- mined using the kinematic equations for freely falling bodies.

Solution From the vector triangle in Figure 3.18c, tan 15° = 𝜐BC

𝜐CG ,

therefore, 𝜐CG = 𝜐BC

tan 15° . To calculate this we have to fi nd v→BC. This

represents the speed of the blood drop relative to the car at impact. The speed of the blood drop at impact will be equal to the speed of an object

that falls from rest through a distance of 1.1 m. Using Equation 2.9, we

have υBC2 = 𝜐 2BC0 + 2ay or υBC = √υBC02 + 2ay. Therefore,

υBC = √(0 m/s)2 + 2(−9.8 m/s2)(−1.1 m) = 4.6 m/s

Concept Summary 3.1 Displacement, Velocity, and Acceleration The position of an object is located with a vector r→ drawn from the coordinate origin to the object. The displacement Δ r→ of the object is defi ned as Δ r→ = r→ − r→0, where r→ and r→0 specify its fi nal and initial positions, respectively.

The average velocity v→ of an object moving between two positions is defi ned as its displacement Δr→ = r→ − r→0 divided by the elapsed time ∆t = t − t0, as in Equation 3.1.

v→ = r→ − r→0 t − t0

= ∆r→

∆t (3.1)

The instantaneous velocity v→ is the velocity at an instant of time. The average velocity becomes equal to the instantaneous velocity in

the limit that Δt becomes infi nitesimally small (∆t → 0 s), as shown in Equation 1.

v→ = lim ∆t → 0

∆r→

∆ t (1)

The average acceleration a→ of an object is the change in its velocity, Δv→ = v→ − v→0 divided by the elapsed time ∆t = t − t0, as in Equation 3.2.

a→ = v→ − v→0 t − t0

= ∆v→

∆ t (3.2)

The instantaneous acceleration a→ is the acceleration at an instant of time. The average acceleration becomes equal to the instantaneous acceleration in

the limit that the elapsed time Δt becomes infi nitesimally small, as shown in Equation 2.

→a = lim ∆ t → 0

∆v→

∆t (2)

vBC

(b)

1.1 m

vCG

(c)

15°

vCG

vBG vBC

FIGURE 3.18 (a) The shape of a drop of blood splatter on a horizontal surface depends on the angle of impact (θ ) of the drop. (b) The velocity of the car relative to the ground is v→CG, and the velocity of the blood drop relative to the car is v→BC. (c) This fi gure illustrates how v→BC and v→CG add vectorially to give v→BG.

Blood Drop Elongation

Elongation of droplet occurs upon impact

How Bloodstain Pattern Analysis Works

(a)

90° 80° 70° 60°

10°20° 40°

θ Blood droplet

Angle of impact

Focus on Concepts 73

3.2 Equations of Kinematics in Two Dimensions Motion in two dimen- sions can be described in terms of the time t and the x and y components of four vectors: the displacement, the acceleration, and the initial and fi nal velo-

cities. The x part of the motion occurs exactly as it would if the y part did not occur at all. Similarly, the y part of the motion occurs exactly as it would if the x part of the motion did not exist. The motion can be analyzed by treating the x and y components of the four vectors separately and realizing that the time t is the same for each component.

When the acceleration is constant, the x components of the displacement, the acceleration, and the initial and fi nal velocities are related by the equa-

tions of kinematics, and so are the y components:

x Component 𝜐x = 𝜐0x + axt (3.3a)

x = 12 (𝜐0x + 𝜐x)t (3.4a)

x = 𝜐0xt + 1

2 axt2 (3.5a)

𝜐x  2 = 𝜐0x 

2 + 2ax x (3.6a)

y Component 𝜐y = 𝜐0y + ayt (3.3b)

y = 12 (𝜐0y + 𝜐y)t (3.4b)

y = 𝜐0yt + 1

2 ay t 2 (3.5b)

𝜐y  2 = 𝜐0y 

2 + 2ayy (3.6b)

The directions of the components of the displacement, the acceleration,

and the initial and fi nal velocities are conveyed by assigning a plus (+) or

minus (−) sign to each one.

3.3 Projectile Motion Projectile motion is an idealized kind of motion that occurs when a moving object (the projectile) experiences only the ac-

celeration due to gravity, which acts vertically downward. If the trajectory

of the projectile is near the earth’s surface, the vertical component ay of the acceleration has a magnitude of 9.80 m/s2. The acceleration has no

horizontal component (ax = 0 m/s2), the eff ects of air resistance being negligible.

There are several symmetries in projectile motion: (1) The time to reach

maximum height from any vertical level is equal to the time spent returning

from the maximum height to that level. (2) The speed of a projectile depends

only on its height above its launch point, and not on whether it is moving

upward or downward.

3.4 Relative Velocity The velocity of object A relative to object B is v→AB, and the velocity of object B relative to object C is v→BC. The velocity of A rel- ative to C is shown in Equation 3 (note the ordering of the subscripts). While

the velocity of object A relative to object B is v→AB, the velocity of B relative to A is v→BA = −v→AB.

v→AC = v→AB + v→BC (3)

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 3.3 Projectile Motion 1. The drawing shows projectile motion at three points along the trajectory. The speeds at the

points are υ1, υ2, and υ3. Assume there is no air resistance and rank the speeds, largest to smallest.

(Note that the symbol > means “greater than.”)

(a) υ1 > υ3 > υ2 (b) υ1 > υ2 > υ3 (c) υ2 > υ3 > υ1 (d) υ2 > υ1 > υ3 (e) υ3 > υ2 > υ1 3. Two balls are thrown from the top of a building, as in the drawing. Ball 1 is thrown straight down, and ball 2 is thrown with the same speed, but upward

at an angle θ with respect to the horizontal. Consider the motion of the balls after they are released. Which

one of the following statements is true? (a) The accel- eration of ball 1 becomes larger and larger as it falls,

because the ball is going faster and faster. (b) The ac- celeration of ball 2 decreases as it rises, becomes zero

at the top of the trajectory, and then increases as the

ball begins to fall toward the ground. (c) Both balls have the same acceleration at all times. (d) Ball 2 has an acceleration in both the horizontal and vertical

directions, but ball 1 has an acceleration only in the

vertical direction.

4. Each drawing shows three points along the path of a projectile, one on its way up, one at the top, and one on its way down. The launch point

is on the left in each drawing. Which drawing correctly represents the

acceleration a→ of the projectile at these three points? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

1 2 3

a = 0 m/s2 a = 0 m/s2 a = 0 m/s2

4 5

QUESTION 4 6. Ball 1 is thrown into the air and it follows the trajectory for projectile motion shown in the drawing. At the instant that ball 1 is at the top of its

trajectory, ball 2 is dropped from rest from the same height. Which ball reaches

the ground fi rst? (a) Ball 1 reaches the ground fi rst, since it is moving at the top of the trajectory, while ball 2 is dropped from rest. (b) Ball 2 reaches the ground fi rst, because it has the shorter distance to travel. (c) Both balls reach the ground at the same time. (d) There is not enough information to tell which ball reaches the ground fi rst.

QUESTION 6

Ball 1

Ball 2 xυ

Focus on Concepts

QUESTION 1

1

2

3

QUESTION 3

Ball 2

Ball 1

θ

74 CHAPTER 3 Kinematics in Two Dimensions

9. Two objects are fi red into the air, and the drawing shows the projectile mo- tions. Projectile 1 reaches the greater height, but projectile 2 has the greater

range. Which one is in the air for the greater amount of time? (a) Projectile 1, because it travels higher than projectile 2. (b) Projectile 2, because it has the greater range. (c) Both projectiles spend the same amount of time in the air. (d) Projectile 2, because it has the smaller initial speed and, therefore, travels more slowly than projectile 1.

QUESTION 9

1

2

Section 3.4 Relative Velocity 14. A slower-moving car is traveling behind a faster-moving bus. The velo- cities of the two vehicles are as follows:

v→CG = velocity of the Car relative to the Ground = +12 m/s

v→BG = velocity of the Bus relative to the Ground = +16 m/s

A passenger on the bus gets up and walks toward the front of the bus with a

velocity of v→PB, where v→PB = velocity of the Passenger relative to the Bus = +2 m/s. What is v→PC, the velocity of the Passenger relative to the Car?

(a) +2 m/s + 16 m/s + 12 m/s = +30 m/s (b) −2 m/s + 16 m/s + 12 m/s = +26 m/s (c) +2 m/s + 16 m/s − 12 m/s = +6 m/s (d) −2 m/s + 16 m/s − 12 m/s = +2 m/s

15. Your car is traveling behind a jeep. Both are moving at the same speed, so the velocity of the jeep relative to you is zero. A spare tire is strapped to

the back of the jeep. Suddenly the strap breaks, and the tire falls off the jeep.

Will your car hit the spare tire before the tire hits the road? Assume that air

resistance is absent. (a) Yes. As long as the car doesn’t slow down, it will hit the tire. (b) No. The car will not hit the tire before the tire hits the ground, no matter how close you are to the jeep. (c) If the tire falls from a great enough height, the car will hit the tire. (d) If the car is far enough behind the jeep, the car will not hit the tire.

16. The drawing shows two cars traveling in diff erent directions with diff er- ent speeds. Their velocities are:

v→AG = velocity of car A relative to the Ground = 27.0 m/s, due east

v→BG = velocity of car B relative to the Ground = 21.0 m/s, due north

The passenger of car B looks out the window and sees car A. What is the

velocity (magnitude and direction) of car A as observed by the passenger of

car B? In other words, what is the velocity v→AB of car A relative to car B? Give the directional angle of v→AB with respect to due east.

QUESTION 16

Car A

Car B

East

North

vBG

vAG

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual

MMH Problem-solving help

GO Guided Online Tutorial

V-HINT Video Hints

CHALK Chalkboard Videos

BIO Biomedical application

E Easy

M Medium

H Hard

Section 3.1 Displacement, Velocity, and Acceleration 1. E SSM Two trees have perfectly straight trunks and are both growing perpendicular to the fl at horizontal ground beneath them. The sides of

the trunks that face each other are separated by 1.3 m. A frisky squirrel

makes three jumps in rapid succession. First, he leaps from the foot of

one tree to a spot that is 1.0 m above the ground on the other tree. Then,

he jumps back to the fi rst tree, landing on it at a spot that is 1.7 m above

the ground. Finally, he leaps back to the other tree, now landing at a spot

that is 2.5 m above the ground. What is the magnitude of the squirrel’s

displacement?

2. E A meteoroid is traveling east through the atmosphere at 18.3 km/s while descending at a rate of 11.5 km/s. What is its speed, in km/s?

3. E GO In a football game a kicker attempts a fi eld goal. The ball remains in contact with the kicker’s foot for 0.050 s, during which time it experiences an ac-

celeration of 340 m/s2. The ball is launched at an angle of 51° above the ground.

Determine the horizontal and vertical components of the launch velocity.

4. E A baseball player hits a triple and ends up on third base. A baseball “dia- mond” is a square, each side of length 27.4 m, with home plate and the three

bases on the four corners. What is the magnitude of the player’s displacement?

5. E SSM In diving to a depth of 750 m, an elephant seal also moves 460 m due east of his starting point. What is the magnitude of the seal’s displacement?

6. E A mountain-climbing expedition establishes two intermediate camps, labeled A and B in the drawing, above the base camp. What is the magnitude Δr of the displacement between camp A and camp B?

PROBLEM 6

3200 m

4900 m

B

A

Base Camp

11 200 m

19 600 m

rΔ

Problems

Problems 75

7. E SSM A radar antenna is tracking a satellite orbiting the earth. At a cer- tain time, the radar screen shows the satellite to be 162 km away. The radar

antenna is pointing upward at an angle of 62.3° from the ground. Find the x and y components (in km) of the position vector of the satellite, relative to the antenna.

8. E GO In a mall, a shopper rides up an escalator between fl oors. At the top of the escalator, the shopper turns right and walks 9.00 m to a store. The

magnitude of the shopper’s displacement from the bottom of the escalator

to the store is 16.0 m. The vertical distance between the fl oors is 6.00 m. At

what angle is the escalator inclined above the horizontal?

9. E SSM A skateboarder, starting from rest, rolls down a 12.0-m ramp. When she arrives at the bottom of the ramp her speed is 7.70 m/s. (a) De- termine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 25.0° with respect to the ground, what is the component

of her acceleration that is parallel to the ground?

10. M V-HINT MMH E A bird watcher meanders through the woods, walking 0.50 km due east, 0.75 km due south, and 2.15 km in a direction 35.0° north

of west. The time required for this trip is 2.50 h. Determine the magnitude

and direction (relative to due west) of the bird watcher’s (a) displacement and (b) average velocity. Use kilometers and hours for distance and time, respectively.

11. M MMH The earth moves around the sun in a nearly circular orbit of radius 1.50 × 1011 m. During the three summer months (an elapsed time of

7.89 × 106 s), the earth moves one-fourth of the distance around the sun.

(a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?

Section 3.2 Equations of Kinematics in Two Dimensions/ Section 3.3 Projectile Motion 12. E A spacecraft is traveling with a velocity of υ0x = 5480 m/s along the +x direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.20 m/s2, while the other gives it an acceleration in the +y direction of ay = 8.40 m/s2. At the end of the fi ring, fi nd (a) υx and (b) υy. 13. E SSM A volleyball is spiked so that it has an initial velocity of 15 m/s dir- ected downward at an angle of 55° below the horizontal. What is the horizontal

component of the ball’s velocity when the opposing player fi elds the ball?

14. E CHALK As a tennis ball is struck, it departs from the racket horizont- ally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance

of 19.6 m from the racket. How far above the court is the tennis ball when it

leaves the racket?

15. E A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58° above the horizontal. The end of the ramp is 1.2 m above

the ground. Let the x axis be parallel to the ground, the +y direction be vertic- ally upward, and take as the origin the point on the ground directly below the

top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

16. E GO A puck is moving on an air hockey table. Relative to an x, y coordin- ate system at time t = 0 s, the x components of the puck’s initial velocity and acceleration are υ0x = +1.0 m/s and ax = +2.0 m/s2. The y components of the puck’s initial velocity and acceleration are υ0y = +2.0 m/s and ay = −2.0 m/s2. Find the magnitude and direction of the puck’s velocity at a time of t = 0.50 s. Specify the direction relative to the +x axis. 17. E SSM A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/s at an angle of 35.0°

above the table, and it lands on the magazine 0.0770 s after leaving the table.

Ignore air resistance. How thick is the magazine? Express your answer in

millimeters.

18. E A horizontal rifl e is fi red at a bull’s-eye. The muzzle speed of the bullet is 670 m/s. The gun is pointed directly at the center of the bull’s-eye, but the

bullet strikes the target 0.025 m below the center. What is the horizontal dis-

tance between the end of the rifl e and the bull’s-eye?

19. E MMH A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the

green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?

20. E GO A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s

at an angle of 40.0° above the horizontal. It rises to its maximum height and

then falls down to the green. Ignoring air resistance, fi nd the speed of the ball

just before it lands.

21. E SSM In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward slope

launches them into the air, where they perform acrobatic maneuvers. The

end of a launch ramp is directed 63° above the horizontal. With this launch

angle, a skier attains a height of 13 m above the end of the ramp. What is the

skier’s launch speed?

22. E GO A space vehicle is coasting at a constant velocity of 21.0 m/s in the +y direction relative to a space station. The pilot of the vehicle fi res a RCS (reaction control system) thruster, which causes it to accelerate at 0.320 m/s2

in the +x direction. After 45.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off , fi nd (a) the magnitude and (b) the direction of the vehicle’s velocity relative to the space station. Express the direction as an

angle measured from the +y direction. 23. E SSM As preparation for this prob- lem, review Conceptual Example 9. The

drawing shows two planes each about to

drop an empty fuel tank. At the moment of

release each plane has the same speed of

135 m/s, and each tank is at the same height

of 2.00 km above the ground. Although the

speeds are the same, the velocities are dif-

ferent at the instant of release, because one

plane is fl ying at an angle of 15.0° above

the horizontal and the other is fl ying at an

angle of 15.0° below the horizontal. Find

the magnitude and direction of the velocity

with which the fuel tank hits the ground if it

is from (a) plane A and (b) plane B. In each part, give the directional angles with respect

to the horizontal.

24. E GO A criminal is escaping across a rooftop and runs off the roof ho- rizontally at a speed of 5.3 m/s, hoping to land on the roof of an adjacent

building. Air resistance is negligible. The horizontal distance between the

two buildings is D, and the roof of the adjacent building is 2.0 m below the jumping-off point. Find the maximum value for D. 25. E On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has x and y components of υ0x = 4370 m/s and υ0y = 6280 m/s. While the engines are fi ring, the craft undergoes a displace- ment that has components of x = 4.11 × 106 m and y = 6.07 × 106 m. Find the x and y components of the craft’s acceleration. 26. E GO In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in

Animated Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fi red at the same angle above the ground. What

is the new range?

27. E SSM A fi re hose ejects a stream of water at an angle of 35.0° above the horizontal. The water leaves the nozzle with a speed of 25.0 m/s. Assuming

PROBLEM 23

15°

15°

Fuel tank

Plane A

Plane B

76 CHAPTER 3 Kinematics in Two Dimensions

that the water behaves like a projectile, how far from a building should the

fi re hose be located to hit the highest possible fi re?

28. E Available on WileyPLUS. 29. E A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time

it reaches a catcher who is 17.0 m away from the point of release?

30. E A quarterback claims that he can throw the football a horizontal dis- tance of 183 m (200 yd). Furthermore, he claims that he can do this by launch-

ing the ball at the relatively low angle of 30.0° above the horizontal. To evaluate

this claim, determine the speed with which this quarterback must throw the ball.

Assume that the ball is launched and caught at the same vertical level and that

air resistance can be ignored. For comparison, a baseball pitcher who can ac-

curately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

31. E SSM An eagle is fl ying horizontally at 6.0 m/s with a fi sh in its claws. It accidentally drops the fi sh. (a) How much time passes before the fi sh’s speed doubles? (b) How much additional time would be required for the fi sh’s speed to double again?

32. E GO The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0° above the horizontal. What is the

projectile’s launch speed?

33. E Consult Multiple-Concept Example 4 for background before begin- ning this problem. Suppose the water at the top of Niagara Falls has a hori-

zontal speed of 2.7 m/s just before it cascades over the edge of the falls. At

what vertical distance below the edge does the velocity vector of the water

point downward at a 75° angle below the horizontal?

34. E MMH On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth,

given the same initial velocities on both planets. The ball is launched at a

speed of 45 m/s at an angle of 29° above the horizontal. When the ball lands,

it is at the same level as the tee. On the distant planet, what are (a) the max- imum height and (b) the range of the ball? 35. E CHALK A rocket is fi red at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fi red toward an

11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch

speed in a negligibly short period of time, after which its engines shut down

and the rocket coasts. By how much does the rocket clear the top of the wall?

36. E A rifl e is used to shoot twice at a target, using identical cartridges. The fi rst time, the rifl e is aimed parallel to the ground and directly at the

center of the bull’s-eye. The bullet strikes the target at a distance of HA below the center, however. The second time, the rifl e is similarly aimed, but from

twice the distance from the target. This time the bullet strikes the target at a

distance of HB below the center. Find the ratio HB/HA. 37. M SSM An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane’s altitude is

732 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where

the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

38. M V-HINT Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown horizontally with a speed of 15 m/s

from the top of a building. When it strikes the ground, the marble has a velo-

city that makes an angle of 65° with the horizontal. From what height above

the ground was the marble thrown?

39. M V-HINT MMH Review Conceptual Example 5 before beginning this problem. You are traveling in a convertible with the top down. The car is

moving at a constant velocity of 25 m/s, due east along fl at ground. You

throw a tomato straight upward at a speed of 11 m/s. How far has the car

moved when you get a chance to catch the tomato?

40. M V-HINT A diver springs upward from a diving board. At the instant she contacts the water, her speed is 8.90 m/s, and her body is extended at

an angle of 75.0° with respect to the horizontal surface of the water. At this

instant her vertical displacement is −3.00 m, where downward is the negative

direction. Determine her initial velocity, both magnitude and direction.

41. M GO MMH A soccer player kicks the ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a speed of 16.0 m/s and an angle

of 28.0° above the ground. Find the speed of the ball when the goalie catches

it in front of the net.

42. M V-HINT In the javelin throw at a track-and-fi eld event, the javelin is launched at a speed of 29 m/s at an angle of 36° above the horizontal. As the

javelin travels upward, its velocity points above the horizontal at an angle

that decreases as time passes. How much time is required for the angle to be

reduced from 36° at launch to 18°?

43. M SSM An airplane is fl ying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the

plane is 2.4 km, a fl are is released from the plane. The fl are hits the target on

the ground. What is the angle θ?

PROBLEM 43 Target

2.4 km

Line of sight

30.0°

Path of flare

θ

0 = 240 m/sυ

44. M V-HINT A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car’s displacement 1.1 s after leaving

the dock has a magnitude of 7.0 m. What is the car’s speed at the instant it

drives off the edge of the dock?

45. M MMH After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.0 m horizontally from the end of the ramp. His

velocity, just before landing, is 23.0 m/s and points in a direction 43.0° below

the horizontal. Neglecting air resistance and any lift he experiences while

airborne, fi nd his initial velocity (magnitude and direction) when he left the

end of the ramp. Express the direction as an angle relative to the horizontal.

46. M GO Stones are thrown horizontally with the same velocity from the tops of two diff erent buildings. One stone lands twice as far from the base of

the building from which it was thrown as does the other stone. Find the ratio

of the height of the taller building to the height of the shorter building.

47. H SSM The drawing shows an exaggerated view of a rifl e that has been “sighted in” for a 91.4-meter target. If the muzzle speed of the bullet is

υ0 = 427 m/s, what are the two possible angles θ1 and θ2 between the rifl e barrel and the horizontal such that the bullet will hit the target? One of

these angles is so large that it is never used in target shooting. (Hint: The following trigonometric identity may be useful: 2 sin θ cos θ = sin 2θ.)

91.4 m

0

θ

υ

PROBLEM 47

48. H MMH A projectile is launched from ground level at an angle of 12.0° above the horizontal. It returns to ground level. To what value should the

launch angle be adjusted, without changing the launch speed, so that the

range doubles?

Additional Problems 77

49. H SSM From the top of a tall building, a gun is fi red. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the drawing shows,

the bullet puts a hole in a window of another building and hits the wall that

faces the window. Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fi red. Assume that the bullet does not slow down as it passes through the window.

340 m/s

H

D

Hole in window

Bullet in wall

0.50 m

6.9 m

PROBLEM 49

50. H Available on WileyPLUS. 51. H Available on WileyPLUS.

Section 3.4 Relative Velocity 52. E In a marathon race Chad is out in front, running due north at a speed of 4.00 m/s. John is 95 m behind him, running due north at a speed of 4.50

m/s. How long does it take for John to pass Chad?

53. E SSM A swimmer, capable of swimming at a speed of 1.4 m/s in still water (i.e., the swimmer can swim with a speed of 1.4 m/s relative to the wa-

ter), starts to swim directly across a 2.8-km-wide river. However, the current

is 0.91 m/s, and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swim- mer be upon reaching the other side of the river?

54. E GO Two friends, Barbara and Neil, are out rollerblading. With re- spect to the ground, Barbara is skating due south at a speed of 4.0 m/s. Neil

is in front of her. With respect to the ground, Neil is skating due west at a

speed of 3.2 m/s. Find Neil’s velocity (magnitude and direction relative to

due west), as seen by Barbara.

55. E Available on WileyPLUS. 56. E At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you

can either stand or walk. Suppose a speed ramp has a length of 105 m and is

moving at a speed of 2.0 m/s relative to the ground. In addition, suppose you

can cover this distance in 75 s when walking on the ground. If you walk at the

same rate with respect to the speed ramp that you walk on the ground, how

long does it take for you to travel the 105 m using the speed ramp?

57. E You are in a hot-air balloon that, relative to the ground, has a velocity of 6.0 m/s in a direction due east. You see a hawk moving directly away from

the balloon in a direction due north. The speed of the hawk relative to you is

2.0 m/s. What are the magnitude and direction of the hawk’s velocity relative

to the ground? Express the directional angle relative to due east.

58. E GO On a pleasure cruise a boat is traveling relative to the water at a speed of 5.0 m/s due south. Relative to the boat, a passenger walks toward

the back of the boat at a speed of 1.5 m/s. (a) What are the magnitude and direction of the passenger’s velocity relative to the water? (b) How long does it take for the passenger to walk a distance of 27 m on the boat? (c) How long does it take for the passenger to cover a distance of 27 m on the water?

59. E SSM Available on WileyPLUS. 60. E GO The captain of a plane wishes to proceed due west. The cruising speed of the plane is 245 m/s relative to the air. A weather report indicates

that a 38.0-m/s wind is blowing from the south to the north. In what direc-

tion, measured with respect to due west, should the pilot head the plane?

61. M V-HINT A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 5.0 m/s relative to the

ground. When the train moves at a constant velocity, the raindrops make an

angle of 25° when they move past the window, as the drawing shows. How

fast is the train moving?

25°

PROBLEM 61

62. M MMH Available on WileyPLUS. 63. M SSM Mario, a hockey player, is skating due south at a speed of 7.0 m/s relative to the ice. A teammate passes the puck to him. The puck has a speed

of 11.0 m/s and is moving in a direction of 22° west of south, relative to

the ice. What are the magnitude and direction (relative to due south) of the

puck’s velocity, as observed by Mario?

64. H A jetliner can fl y 6.00 hours on a full load of fuel. Without any wind it fl ies at a speed of 2.40 × 102 m/s. The plane is to make a round-trip by

heading due west for a certain distance, turning around, and then heading due

east for the return trip. During the entire fl ight, however, the plane encounters

a 57.8-m/s wind from the jet stream, which blows from west to east. What is

the maximum distance that the plane can travel due west and just be able to

return home?

65. H SSM Available on WileyPLUS.

66. E Useful background for this problem can be found in Multiple-Concept Example 2. On a spacecraft two engines fi re for a time of 565 s. One gives

the craft an acceleration in the x direction of ax = 5.10 m/s2, while the other produces an acceleration in the y direction of ay = 7.30 m/s2. At the end of the fi ring period, the craft has velocity components of υx = 3775 m/s and υy = 4816 m/s. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

67. E A dolphin leaps out of the water at an angle of 35° above the hori- zontal. The horizontal component of the dolphin’s velocity is 7.7 m/s. Find

the magnitude of the vertical component of the velocity.

68. E A hot-air balloon is rising straight up with a speed of 3.0 m/s. A ballast bag is released from rest relative to the balloon at 9.5 m above the

ground. How much time elapses before the ballast bag hits the ground?

Additional Problems

78 CHAPTER 3 Kinematics in Two Dimensions

69. E CHALK A golf ball rolls off a horizontal cliff with an initial speed of 11.4 m/s. The ball falls a vertical distance of 15.5 m into a lake below.

(a) How much time does the ball spend in the air? (b) What is the speed υ of the ball just before it strikes the water?

70. E GO When chasing a hare along a fl at stretch of ground, a greyhound leaps into the air at a speed of 10.0 m/s, at an angle of 31.0° above the hori-

zontal. (a) What is the range of his leap and (b) for how much time is he in the air?

71. E SSM Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of 1.20 m/s off a plat-

form that is 10.0 m above the water. What is his speed just before striking

the water?

72. E GO A ball is thrown upward at a speed υ0 at an angle of 52° above the horizontal. It reaches a maximum height of 7.5 m. How high would this ball

go if it were thrown straight upward at speed υ0? 73. M MMH A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. If the ball leaves her

club with a velocity of 46.0 m/s at an angle of 35.0° above the ground, fi nd

the time that the ball is in the air before it hits the green.

74. M V-HINT Available on WileyPLUS. 75. M SSM As preparation for this problem, review Conceptual Example 9. The two stones described there have identical initial speeds of υ0 = 13.0 m/s and are thrown at an angle θ = 30.0°, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?

76. M GO Relative to the ground, a car has a velocity of 16.0 m/s, directed due north. Relative to this car, a truck has a velocity of 24.0 m/s, directed

52.0° north of east. What is the magnitude of the truck’s velocity relative to

the ground?

77. M V-HINT The lob in tennis is an eff ective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to

move quickly away from the net (see the drawing). Suppose that you lob

the ball with an initial speed of 15.0 m/s, at an angle of 50.0° above the

horizontal. At this instant your opponent is 10.0 m away from the ball. He

begins moving away from you 0.30 s later, hoping to reach the ball and

hit it back at the moment that it is 2.10 m above its launch point. With

what minimum average speed must he move? (Ignore the fact that he can

stretch, so that his racket can reach the ball before he does.)

10.0 m

15.0 m/s 2.10 m

50.0°

PROBLEM 77

78. H Available on WileyPLUS. 79. H Available on WileyPLUS. 80. M GO A baseball player hits a home run, and the ball lands in the left- fi eld seats, 7.5 m above the point at which it was hit. It lands with a velocity of

36 m/s at an angle of 28° below the horizontal (see the drawing). The positive

directions are upward and to the right in the drawing. Ignoring air resistance,

fi nd the magnitude and direction of the initial velocity with which the ball

leaves the bat.

+y

+x

= 36 m/s

x

y y = 7.5 m

28° 0y

0x

0

θ

υ υ

υ

υ

υ

υ

PROBLEM 80

A primary focus of this chapter has been projectile motion. These additional

problems serve as a review of the basic features of this type of motion.

81. M CHALK SSM Three balls are launched simultaneously, as in the fi gure. The red ball is launched at an initial velocity υ→0 directed at an angle θ above the horizontal. The blue ball is launched vertically with a speed of 10.0 m/s,

and the yellow ball is launched horizontally with a speed of 4.6 m/s. The blue

and red balls reach identical maximum heights and return to the ground at the

same time, and the red ball returns to the ground to meet the yellow ball at the

same place along the horizontal. Concepts: (i) Since the blue and red balls reach the same maximum height, what can be said about the y components of their velocities? (ii) Since the red and yellow balls travel equal distances

along the x direction in the same amount of time, what can be said about the x components of their initial velocities? Calculations: Find the launch speed υ0 and launch angle θ for the red ball.

10.0 m/s

4.6 m/s

y

hmax

x

v0

θ

PROBLEM 80

Concepts and Calculations Problems

Team Problems 79

82. M CHALK A projectile is launched from and returns to ground level, as the fi gure shows. There is no air resistance. The horizontal range of the projectile

is measured to be R = 175 m, and the horizontal component of the launch ve- locity is υ0x = +25 m/s. Concepts: (i) What is the fi nal value of the horizontal component υx of the projectile’s velocity? (ii) Can the time be determined for the horizontal part of the motion? (iii) Is the time for the horizontal part

of the motion the same as that for the vertical part? (iv) For the vertical part of

the motion, what is the displacement of the projectile? Calculations: Find the vertical component υ0y of the projectile.

PROBLEM 82

v0

0y

0x = +25 m/s

R = 175 m

+x

v0

+y

υ

υ

83. M A Water Balloon Battle. You are launching water balloons at a rival team using a large slingshot. The other team is set up on the opposite side of a

fl at-topped building that is 30.0 ft tall and 50.0 ft wide. Your reconnaissance

team has reported that the opposition is set up 10.0 m from the wall of the

building. Your balloon launcher is calibrated for launch speeds that can reach

as high as 100 mph at angles between 0 and 80.0° from the horizontal. Since

a direct shot is not possible (the opposing team is on the opposite side of the

building), you plan to splash the other team by making a balloon explode on

the ground near them. If your launcher is located 55.0 m from the building

(opposite side as the opposing team), what should your launch velocity be

(magnitude and direction) to land a balloon 5.0 meters beyond the opposing

team with maximum impact (i.e., maximum vertical speed)?

84. M A Motorcycle Jump. You are planning to make a jump with your motorcycle by driving over a ramp that will launch you at an angle of 30.0°

with respect to the horizontal. The front edge of the ramp on which you

are supposed to land, however, is 25.0 ft lower than the edge of the launch

ramp (i.e., your launch height). (a) Assuming a launch speed of 65 mph, at what horizontal distance from your launch point should the landing ramp be

placed? (b) In order to land smoothly, the angle of the landing ramp should match the direction of your velocity vector when you touch down. What

should be the angle of the landing ramp?

Team Problems

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

4.1 Discuss the concepts of force and mass.

4.2 Define Newton’s first law of motion.

4.3 Define Newton’s second law of motion.

4.4 Apply Newton’s second law of motion in two dimensions.

4.5 Apply Newton’s third law of motion.

4.6 Identify types of forces.

4.7 Define Newton’s law of universal gravitation.

4.8 Solve problems using the normal force.

4.9 Solve problems involving friction.

4.10 Solve problems involving tension.

4.11 Apply Newton’s first law to equilibrium problems.

4.12 Apply Newton’s second law to nonequilibrium problems.

© S

to ck

tr ek

I m

ag es

CHAPTER 4

Forces and Newton’s Laws of Motion

In order to successfully land the planetary rover, Curiosity, on the surface of Mars, NASA scientists and engineers had to take into account many forces, such as the thrust from the “sky crane” rocket engines, the

tension in its towing cables, air resistance, and the weight of the rover due to the gravity on Mars, just to

name a few. This chapter will discuss how forces infl uence the motion of objects.

4.1 The Concepts of Force and Mass In common usage, a force is a push or a pull, as the examples in Figure 4.1 illustrate. In football, an off ensive lineman pushes against his opponent. The tow bar attached to

a speeding boat pulls a water skier. Forces such as those that push against the football

player or pull the skier are called contact forces, because they arise from the physical contact between two objects. There are circumstances, however, in which two objects

exert forces on one another even though they are not touching. Such forces are referred to

as noncontact forces or action-at-a-distance forces. One example of such a noncontact force occurs when a diver is pulled toward the earth because of the force of gravity. The

earth exerts this force even when it is not in direct contact with the diver. In Figure 4.1, arrows are used to represent the forces. It is appropriate to use arrows, because a force is

a vector quantity and has both a magnitude and a direction. The direction of the arrow

gives the direction of the force, and the length is proportional to its strength or magnitude.

The word mass is just as familiar as the word force. A massive supertanker, for instance, is one that contains an enormous amount of mass. As we will see in the next

section, it is diffi cult to set such a massive object into motion and diffi cult to bring it

to a halt once it is moving. In comparison, a penny does not contain much mass. The

80

4.2 Newton’s First Law of Motion 81

emphasis here is on the amount of mass, and the idea of direction is of no concern. Therefore,

mass is a scalar quantity.

During the seventeenth century, Isaac Newton, building on the work of Galileo, developed

three important laws that deal with force and mass. Collectively they are called “Newton’s laws of

motion” and provide the basis for understanding the eff ect that forces have on an object. Because

of the importance of these laws, a separate section will be devoted to each one.

4.2 Newton’s First Law of Motion

The First Law To gain some insight into Newton’s fi rst law, think about the game of ice hockey (Figure 4.2). If a player does not hit a stationary puck, it will remain at rest on the ice. After the puck is struck,

however, it coasts on its own across the ice, slowing down only slightly because of friction. Since

ice is very slippery, there is only a relatively small amount of friction to slow down the puck. In

fact, if it were possible to remove all friction and wind resistance, and if the rink were infi nitely

large, the puck would coast forever in a straight line at a constant speed. Left on its own, the puck

would lose none of the velocity imparted to it at the time it was struck. This is the essence of

Newton’s fi rst law of motion:

NEWTON’S FIRST LAW OF MOTION An object continues in a state of rest or in a state of motion at a constant velocity (constant speed in a constant direction), unless compelled to change that state by a net force.

In the fi rst law the phrase “net force” is crucial. Often, several forces act simultaneously on

a body, and the net force is the vector sum of all of them. Individual forces matter only to the extent that they contribute to the total. For instance, if friction and other opposing forces were ab-

sent, a car could travel forever at 30 m/s in a straight line, without using any gas after it had come

up to speed. In reality gas is needed, but only so that the engine can produce the necessary force

to cancel opposing forces such as friction. This cancellation ensures that there is no net force to

change the state of motion of the car.

When an object moves at a constant speed in a constant direction, its velocity is constant.

Newton’s fi rst law indicates that a state of rest (zero velocity) and a state of constant velocity

are completely equivalent, in the sense that neither one requires the application of a net force to

sustain it. The purpose served when a net force acts on an object is not to sustain the object’s velocity, but, rather, to change it.

Inertia and Mass A greater net force is required to change the velocity of some objects than of others. For instance,

a net force that is just enough to cause a bicycle to pick up speed will cause only an imperceptible

change in the motion of a freight train. In comparison to the bicycle, the train has a much greater

tendency to remain at rest. Accordingly, we say that the train has more inertia than the bicycle.

FIGURE 4.1 The arrow labeled F→ represents the force that acts on (a) the football player, (b) the water skier, and (c) the cliff diver.

F

D av

id D

er m

er /D

ia m

o n d I

m ag

es /G

et ty

I m

ag es

(a)

F

ag e

fo to

st o ck

/S u p er

S to

ck

(b)

F

Ik o n I

m ag

es /S

u p er

S to

ck

(c)

S co

tt G

ar d n er

/A P

/W id

e W

o rl

d P

h o to

s

FIGURE 4.2 The game of ice hockey can give some insight into Newton’s laws of motion.

82 CHAPTER 4 Forces and Newton’s Laws of Motion

Quantitatively, the inertia of an object is measured by its mass. The following defi nition of inertia and mass indicates why Newton’s fi rst law is sometimes called the law of inertia:

DEFINITION OF INERTIA AND MASS Inertia is the natural tendency of an object to remain at rest or in motion at a constant velocity. The mass of an object is a quantitative measure of inertia. SI Unit of Inertia and Mass: kilogram (kg)

The SI unit for mass is the kilogram (kg), whereas the units in the CGS system and the BE

system are the gram (g) and the slug (sl), respectively. Conversion factors between these units are

given on the page facing the inside of the front cover. Interactive Figure 4.3 gives the typical masses of various objects, ranging from a penny to a supertanker. The larger the mass, the greater

is the inertia. Often the words “mass” and “weight” are used interchangeably, but this is incorrect.

Mass and weight are diff erent concepts, and Section 4.7 will discuss the distinction between them.

THE PHYSICS OF . . . seat belts. Figure 4.4 shows a useful application of inertia. Auto- mobile seat belts unwind freely when pulled gently, so they can be buckled. But in an accident, they

hold you safely in place. One seat-belt mechanism consists of a ratchet wheel, a locking bar, and a

pendulum. The belt is wound around a spool mounted on the ratchet wheel. While the car is at rest

or moving at a constant velocity, the pendulum hangs straight down, and the locking bar rests ho-

rizontally, as the gray part of the drawing shows. Consequently, nothing prevents the ratchet wheel

from turning, and the seat belt can be pulled out easily. When the car suddenly slows down in an ac-

cident, however, the relatively massive lower part of the pendulum keeps moving forward because

of its inertia. The pendulum swings on its pivot into the position shown in color and causes the

locking bar to block the rotation of the ratchet wheel, thus preventing the seat belt from unwinding.

An Inertial Reference Frame Newton’s fi rst law (and also the second law) can appear to be invalid to certain observers. Sup-

pose, for instance, that you are a passenger riding in a friend’s car. While the car moves at a con-

stant speed along a straight line, you do not feel the seat pushing against your back to any unusual

extent. This experience is consistent with the fi rst law, which indicates that in the absence of a net

force you should move with a constant velocity. Suddenly the driver fl oors the gas pedal. Imme-

diately you feel the seat pressing against your back as the car accelerates. Therefore, you sense

that a force is being applied to you. The fi rst law leads you to believe that your motion should

change, and, relative to the ground outside, your motion does change. But relative to the car, you can see that your motion does not change, because you remain stationary with respect to the car. Clearly, Newton’s fi rst law does not hold for observers who use the accelerating car as a frame of

reference. As a result, such a reference frame is said to be noninertial. All accelerating reference frames are noninertial. In contrast, observers for whom the law of inertia is valid are said to be

using inertial reference frames for their observations, as defi ned below:

DEFINITION OF AN INERTIAL REFERENCE FRAME An inertial reference frame is one in which Newton’s law of inertia is valid.

The acceleration of an inertial reference frame is zero, so it moves with a constant velocity. All

of Newton’s laws of motion are valid in inertial reference frames, and when we apply these laws,

we will be assuming such a reference frame. In particular, the earth itself is a good approximation

of an inertial reference frame.

Check Your Understanding

(The answer is given at the end of the book.) 1. Which of the following statements can be explained by Newton’s fi rst law? (A): When your car sud-

denly comes to a halt, you lunge forward. (B): When your car rapidly accelerates, you are pressed

backward against the seat. (a) Neither A nor B (b) Both A and B (c) A but not B (d) B but not A

INTERACTIVE FIGURE 4.3 Typical masses of various objects.

IN GOD

WE TRUST

LIBERTY 1996

Penny (0.003 kg)

Book (2 kg)

Bicycle (15 kg)

Car (2000 kg)

Jetliner (1.2 x 105 kg)

Supertanker (1.5 x 108 kg)

Seat belt

Motion of car

Ratchet wheel

Locking bar

Pivots

Pendulum

FIGURE 4.4 Inertia plays a central role in one seat-belt mechanism. The gray part of the

drawing applies when the car is at rest or

moving at a constant velocity. The colored

parts show what happens when the car

suddenly slows down, as in an accident.

4.3 Newton’s Second Law of Motion 83

4.3 Newton’s Second Law of Motion Newton’s fi rst law indicates that if no net force acts on an object, then the velocity of the object

remains unchanged. The second law deals with what happens when a net force does act. Consider

a hockey puck once again. When a player strikes a stationary puck, he causes the velocity of the

puck to change. In other words, he makes the puck accelerate. The cause of the acceleration is the

force that the hockey stick applies. As long as this force acts, the velocity increases, and the puck

accelerates. Now, suppose another player strikes the puck and applies twice as much force as the

fi rst player does. The greater force produces a greater acceleration. In fact, if the friction between

the puck and the ice is negligible, and if there is no wind resistance, the acceleration of the puck is

directly proportional to the force. Twice the force produces twice the acceleration. Moreover, the

acceleration is a vector quantity, just as the force is, and points in the same direction as the force.

Often, several forces act on an object simultaneously. Friction and wind resistance, for in-

stance, do have some eff ect on a hockey puck. In such cases, it is the net force, or the vector sum

of all the forces acting, that is important. Mathematically, the net force is written as ∑F →

, where

the Greek capital letter ∑ (sigma) denotes the vector sum. Newton’s second law states that the

acceleration is proportional to the net force acting on the object.

In Newton’s second law, the net force is only one of two factors that determine the accel-

eration. The other is the inertia or mass of the object. After all, the same net force that imparts

an appreciable acceleration to a hockey puck (small mass) will impart very little acceleration to

a semitrailer truck (large mass). Newton’s second law states that for a given net force, the mag-

nitude of the acceleration is inversely proportional to the mass. Twice the mass means one-half

the acceleration, if the same net force acts on both objects. Thus, the second law shows how the

acceleration depends on both the net force and the mass, as given in Equation 4.1.

NEWTON’S SECOND LAW OF MOTION When a net external force ∑F

→ acts on an object of mass m, the acceleration a→ that results

is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force.

a→ = ΣF →

m   or  ΣF→ = ma→ (4.1)

SI Unit of Force: kg · m/s2 = newton (N)

Note that the net force in Equation 4.1 includes only the forces that the environment exerts on

the object of interest. Such forces are called external forces. In contrast, internal forces are forces that one part of an object exerts on another part of the object and are not included in Equation 4.1.

According to Equation 4.1, the SI unit for force is the unit for mass (kg) times the unit for

acceleration (m/s2), or

SI unit for force = (kg)(m s2) = kg · m

s2

The combination of kg · m/s2 is called a newton (N) and is a derived SI unit, not a base unit; 1 newton = 1 N = 1 kg · m/s2.

In the CGS system, the procedure for establishing the unit of force is the same as with SI

units, except that mass is expressed in grams (g) and acceleration in cm/s2. The resulting unit for

force is the dyne; 1 dyne = 1 g · cm/s2. In the BE system, the unit for force is defi ned to be the pound (lb),* and the unit for acceleration

is ft/s2. With this procedure, Newton’s second law can then be used to obtain the BE unit for mass:

BE unit for force = lb = (unit for mass)(fts2) Unit for mass =

lb · s2 ft

*We refer here to the gravitational version of the BE system, in which a force of one pound is defi ned to be the pull of

the earth on a certain standard body at a location where the acceleration due to gravity is 32.174 ft/s2.

84 CHAPTER 4 Forces and Newton’s Laws of Motion

The combination of lb · s2/ft is the unit for mass in the BE system and is called the slug (sl); 1 slug = 1 sl = 1 lb · s2/ft.

Table 4.1 summarizes the various units for mass, acceleration, and force. Conversion factors between force units from diff erent systems are provided on the page facing the inside of the front

cover.

When using the second law to calculate the acceleration, it is necessary to determine the net

force that acts on the object. In this determination a free-body diagram helps enormously. A free- body diagram is a diagram that represents the object and the forces that act on it. Only the forces

that act on the object appear in a free-body diagram. Forces that the object exerts on its environment are not included. Example 1 illustrates the use of a free-body diagram.

TABLE 4.1 Units for Mass, Acceleration, and Force

System Mass Acceleration Force SI kilogram (kg) meter/second2 (m/s2) newton (N)

CGS gram (g) centimeter/second2 (cm/s2) dyne (dyn)

BE slug (sl) foot/second2 (ft/s2) pound (lb)

Check Your Understanding

(The answers are given at the end of the book.) 2. The net external force acting on an object is zero. Which one of the following statements is true?

(a) The object can only be stationary. (b) The object can only be traveling with a constant velocity. (c) The object can be either stationary or traveling with a constant velocity. (d) The object can only be traveling with a velocity that is changing.

EXAMPLE 1 Pushing a Stalled Car

Two people are pushing a stalled car, as Animated Figure 4.5a indicates. The mass of the car is 1850 kg. One person applies a force of 275 N to the

car, while the other applies a force of 395 N. Both forces act in the same

direction. A third force of 560 N also acts on the car, but in a direction

opposite to that in which the people are pushing. This force arises because

of friction and the extent to which the pavement opposes the motion of the

tires. Find the acceleration of the car.

Reasoning According to Newton’s second law, the acceleration is the net force divided by the mass of the car. To determine the net force, we

use the free-body diagram in Animated Figure 4.5b. In this diagram, the car is represented as a dot, and its motion is along the +x axis. The dia- gram makes it clear that the forces all act along one direction. Therefore,

they can be added as colinear vectors to obtain the net force.

Problem-Solving Insight The direction of the acceleration is always the same as the direction of the net force.

Solution From Equation 4.1, the acceleration is a = (∑F)/m. The net force is

ΣF = +275 N + 395 N − 560 N = +110 N

The acceleration can now be obtained:

a = ΣF m

= +110 N

1850 kg = +0.059 m/s2 (4.1)

The plus sign indicates that the acceleration points along the +x axis, in the same direction as the net force.

ANIMATED FIGURE 4.5 (a) Two people push a stalled car. A force created by friction and the pavement opposes their eff orts. (b) A free-body diagram that shows the horizontal forces acting on the car.

+x

+y

560 N 395 N

275 N

(b) Free-body diagram of the car(a)

Opposing force = 560 N

275 N

395 N

4.4 The Vector Nature of Newton’s Second Law of Motion 85

3. In Case A an object is moving straight downward with a constant speed of 9.80 m/s, while in Case B an object is moving straight downward with a constant acceleration of magnitude 9.80 m/s2. Which one of the following is true? (a) A nonzero net external force acts on the object in both cases. (b) A nonzero net external force acts on the object in neither case. (c) A nonzero net external force acts on the object in Case A only. (d) A nonzero net external force acts on the object in Case B only.

4.4 The Vector Nature of Newton’s Second Law of Motion When a football player throws a pass, the direction of the force he applies to the ball is important.

Both the force and the resulting acceleration of the ball are vector quantities, as are all forces and

accelerations. The directions of these vectors can be taken into account in two dimensions by

using x and y components. The net force ∑F →

in Newton’s second law has components ∑Fx and ∑Fy, while the acceleration a→ has components ax and ay. Consequently, Newton’s second law, as expressed in Equation 4.1, can be written in an equivalent form as two equations, one for the x components and one for the y components:

ΣFx = max (4.2a)

ΣFy = may (4.2b)

This procedure is similar to that employed in Chapter 3 for the equations of two-

dimensional kinematics (see Table 3.1). The components in Equations 4.2a and 4.2b are scalar components and will be either positive or negative numbers, depending on whether they point

along the positive or negative x or y axis. The remainder of this section deals with examples that show how these equations are used.

EXAMPLE 2 Applying Newton’s Second Law Using Components

(a)

FIGURE 4.6 (a) A man is paddling a raft, as in Examples 2 and 3. (b) The free-body diagram shows the forces P→ and A→ that act on the raft. Forces acting on the raft in a direction perpendicular to the surface of the water play no role in the examples and are omitted for clarity. (c) The raft’s acceleration components ax and ay. (d) In 65 s, the components of the raft’s displacement are x = 48 m and y = 23 m.

A man is stranded on a raft (mass of man and raft = 1300 kg), as shown

in Figure 4.6a. By paddling, he causes an average force P →

of 17 N to be

applied to the raft in a direction due east (the +x direction). The wind also exerts a force A→ on the raft. This force has a magnitude of 15 N and points 67° north of east. Ignoring any resistance from the water, fi nd the x and y components of the raft’s acceleration.

Reasoning Since the mass of the man and the raft is known, Newton’s second law can be used to determine the acceleration components from the

given forces. According to the form of the second law in Equations 4.2a

and 4.2b, the acceleration component in a given direction is the com-

ponent of the net force in that direction divided by the mass. As an aid

in determining the components ∑Fx and ∑Fy of the net force, we use the

+x

+y

+x

+y

A sin 67°

A = 15 N

P = 17 N

A cos 67°

67°

A

Free-body diagram of the raft

N

S

EW

(b)

+x

ax

ay

+y

(c)

x = 48x m

y = 23 m

+x

+y

(d )

86 CHAPTER 4 Forces and Newton’s Laws of Motion

Check Your Understanding

(The answers are given at the end of the book.) 4. Newton’s second law indicates that when a net force acts on an object, it must accelerate. Does this mean

that when two or more forces are applied to an object simultaneously, it must accelerate?

5. All of the following, except one, cause the acceleration of an object to double. Which one is the excep- tion? (a) All forces acting on the object double. (b) The net force acting on the object doubles. (c) Both the net force acting on the object and the mass of the object double. (d) The net force acting on the object remains the same, while the mass of the object is reduced by a factor of two.

4.5 Newton’s Third Law of Motion Imagine you are in a football game. You line up facing your opponent, the ball is snapped, and

the two of you crash together. No doubt, you feel a force. But think about your opponent. He too

feels something, for while he is applying a force to you, you are applying a force to him. In other

words, there isn’t just one force on the line of scrimmage; there is a pair of forces. Newton was

the fi rst to realize that all forces occur in pairs and there is no such thing as an isolated force,

existing all by itself. His third law of motion deals with this fundamental characteristic of forces.

NEWTON’S THIRD LAW OF MOTION Whenever one object exerts a force on a second object, the second object exerts an oppositely directed force of equal magnitude on the fi rst object.

free-body diagram in Figure 4.6b. In this diagram, the directions due east and due north are the +x and +y directions, respectively.

Solution Figure 4.6b shows the force components:

Force x Component y Component

P→ +17 N 0 N

A→ +(15 N)cos 67° = +6 N +(15 N)sin 67° = +14 N

ΣFx = +17 N + 6 N = +23 N ΣFy = +14 N

The plus signs indicate that ∑Fx points in the direction of the +x axis and ∑Fy points in the direction of the +y axis. The x and y components of the acceleration point in the directions of ∑Fx and ∑Fy, respectively, and can now be calculated:

ax = ΣFx m

= +23 N

1300 kg = +0.018 m/s2 (4.2a)

ay = ΣFy

m =

+14 N

1300 kg = +0.011 m/s2 (4.2b)

These acceleration components are shown in Figure 4.6c.

EXAMPLE 3 The Displacement of a Raft

At the moment that the forces P→ and A→ begin acting on the raft in Example 2, the velocity of the raft is 0.15 m/s, in a direction due east

(the +x direction). Assuming that the forces are maintained for 65 s, fi nd the x and y components of the raft’s displacement during this time interval.

Reasoning Once the net force acting on an object and the object’s mass have been used in Newton’s second law to determine the acceleration, it

becomes possible to use the equations of kinematics to describe the res-

ulting motion. We know from Example 2 that the acceleration compon-

ents are ax = +0.018 m/s2 and ay = +0.011 m/s2, and it is given here that the initial velocity components are υ0x = +0.15 m/s and υ0y = 0 m/s. Thus,

Equation 3.5a (x = 𝜐0x t + 1

2axt2) and Equation 3.5b (y = 𝜐0y t + 1

2ay t2) can be used with t = 65 s to determine the x and y components of the raft’s displacement.

Solution According to Equations 3.5a and 3.5b, the x and y components of the displacement are

x = 𝜐0xt + 1

2 axt 2 = (0.15 m /s)(65 s) + 1

2(0.018 m /s 2)(65 s)2 = 48 m

y = 𝜐0yt + 1

2 ayt 2 = (0 m /s)(65 s) + 1

2 (0.011 m /s 2)(65 s)2 = 23 m

Figure 4.6d shows the fi nal location of the raft.

4.5 Newton’s Third Law of Motion 87

The third law is often called the “action–reaction” law, because it is sometimes quoted as follows:

“For every action (force) there is an equal, but opposite, reaction.”

Figure 4.7 illustrates how the third law applies to an astronaut who is drifting just outside a spacecraft and who pushes on the spacecraft with a force P→. According to the third law, the spacecraft pushes back on the astronaut with a force −P→ that is equal in magnitude but opposite in direction. In Example 4, we examine the accelerations produced by each of these forces.

EXAMPLE 4 The Accelerations Produced by Action and Reaction Forces

Suppose that the mass of the spacecraft in Figure 4.7 is mS = 11 000 kg and that the mass of the astronaut is mA = 92 kg. In addition, assume that the astronaut pushes with a force of P→ = +36 N on the spacecraft. Find the accelerations of the spacecraft and the astronaut.

Reasoning Consistent with Newton’s third law, when the astronaut applies the force P→ = +36 N to the spacecraft, the spacecraft applies a reaction force −P→ = −36 N to the astronaut. As a result, the spacecraft and the astronaut accelerate in opposite directions. Although the action and

reaction forces have the same magnitude, they do not create accelerations

of the same magnitude, because the spacecraft and the astronaut have

diff erent masses. According to Newton’s second law, the astronaut,

having a much smaller mass, will experience a much larger accelera-

tion. In applying the second law, we note that the net force acting on

the spacecraft is ∑F→ = P→ , while the net force acting on the astronaut is ∑F→ = −P→.

Problem-Solving Insight Even though the magnitudes of the action and reaction forces are always equal, these forces do not necessarily produce accelerations that have equal magnitudes, since each force acts on a diff erent object that may have a diff erent mass.

Solution Using the second law, we fi nd that the acceleration of the spacecraft is

a →S = P →

mS =

+36 N

11 000 kg = +0.0033 m/s2

The acceleration of the astronaut is

a →A = −P → mA

= −36 N

92 kg = −0.39 m/s2

–P +P

FIGURE 4.7 The astronaut pushes on the spacecraft with a force +P→. According to Newton’s third law, the spacecraft simultaneously pushes back on the astronaut with

a force −P→.

Mechanism for actuating trailer brakes

FIGURE 4.8 Some rental trailers include an automatic brake-actuating mechanism.

THE PHYSICS OF . . . automatic trailer brakes. There is a clever application of Newton’s third law in some rental trailers. As Figure 4.8 illustrates, the tow bar connecting the trailer to the rear bumper of a car contains a mechanism that can automatically actuate brakes on

the trailer wheels. This mechanism works without the need for electrical connections between the

car and the trailer. When the driver applies the car brakes, the car slows down. Because of inertia,

however, the trailer continues to roll forward and begins pushing against the bumper. In reaction,

the bumper pushes back on the tow bar. The reaction force is used by the mechanism in the tow

bar to “push a brake pedal” for the trailer.

88 CHAPTER 4 Forces and Newton’s Laws of Motion

Check Your Understanding

(The answer is given at the end of the book.) 6. A father and his seven-year-old daughter are facing each other on ice skates. With their hands, they

push off against one another. Which one or more of the following statements is (are) true? (a) Each experiences an acceleration that has a diff erent magnitude. (b) Each experiences an acceleration of the same magnitude. (c) Each experiences a pushing force that has a diff erent magnitude. (d) Each experi- ences a pushing force that has the same magnitude.

4.6 Types of Forces: An Overview Newton’s three laws of motion make it clear that forces play a central role in determining the

motion of an object. In the next four sections some common forces will be discussed: the grav-

itational force (Section 4.7), the normal force (Section 4.8), frictional forces (Section 4.9), and

the tension force (Section 4.10). In later chapters, we will encounter still others, such as electric

and magnetic forces. It is important to realize that Newton’s second law is always valid, regard-

less of which of these forces may act on an object. One does not have a diff erent law for every

type of common force. Thus, we need only to determine what forces are acting on an object, add

them together to form the net force, and then use Newton’s second law to determine the object’s

acceleration.

In nature there are two general types of forces, fundamental and nonfundamental. Funda-

mental forces are the ones that are truly unique, in the sense that all other forces can be explained

in terms of them. Only three fundamental forces have been discovered:

1. Gravitational force 2. Strong nuclear force 3. Electroweak force The gravitational force is discussed in the next section. The strong nuclear force plays a primary

role in the stability of the nucleus of the atom (see Section 31.2). The electroweak force is a

single force that manifests itself in two ways (see Section 32.6). One manifestation is the elec-

tromagnetic force that electrically charged particles exert on one another (see Sections 18.5,

21.2, and 21.8). The other manifestation is the so-called weak nuclear force that plays a role in

the radioactive disintegration of certain nuclei (see Section 31.5).

Except for the gravitational force, all of the forces discussed in this chapter are nonfunda-

mental, because they are related to the electromagnetic force. They arise from the interactions

between the electrically charged particles that comprise atoms and molecules. Our understanding

of which forces are fundamental, however, is continually evolving. For instance, in the 1860s and

1870s James Clerk Maxwell showed that the electric force and the magnetic force could be ex-

plained as manifestations of a single electromagnetic force. Then, in the 1970s, Sheldon Glashow

(1932– ), Abdus Salam (1926–1996), and Steven Weinberg (1933– ) presented the theory that

explains how the electromagnetic force and the weak nuclear force are related to the electroweak

force. They received a Nobel Prize in 1979 for their achievement. Today, eff orts continue that

have the goal of further reducing the number of fundamental forces.

4.7 The Gravitational Force

Newton’s Law of Universal Gravitation Objects fall downward because of gravity, and Chapters 2 and 3 discuss how to describe the

eff ects of gravity by using a value of g = 9.80 m/s2 for the downward acceleration it causes. However, nothing has been said about why g is 9.80 m/s2. The reason is fascinating, as we will see later in this section.

4.7 The Gravitational Force 89

The acceleration due to gravity is like any other acceleration, and Newton’s second law

indicates that it must be caused by a net force. In addition to his famous three laws of motion,

Newton also provided a coherent understanding of the gravitational force. His “law of universal gravitation” is stated as follows:

NEWTON’S LAW OF UNIVERSAL GRAVITATION Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. For two particles that have masses m1 and m2 and are separated by a distance r, the force that each exerts on the other is directed along the line joining the particles (see Figure 4.9) and has a magnitude given by

F = G m1m2

r2 (4.3)

The symbol G denotes the universal gravitational constant, whose value is found exper- imentally to be G = 6.674 × 10−11 N · m2/kg2

The constant G that appears in Equation 4.3 is called the universal gravitational constant, because it has the same value for all pairs of particles anywhere in the universe, no matter what

their separation. The value for G was fi rst measured in an experiment by the English scientist Henry Cavendish (1731–1810), more than a century after Newton proposed his law of universal

gravitation.

To see the main features of Newton’s law of universal gravitation, look at the two particles

in Figure 4.9. They have masses m1 and m2 and are separated by a distance r. In the picture, it is assumed that a force pointing to the right is positive. The gravitational forces point along the line

joining the particles and are

+F →

, the gravitational force exerted on particle 1 by particle 2

−F →

, the gravitational force exerted on particle 2 by particle 1

These two forces have equal magnitudes and opposite directions. They act on diff erent

bodies, causing them to be mutually attracted. In fact, these forces are an action–reaction pair, as

required by Newton’s third law. Example 5 shows that the magnitude of the gravitational force is

extremely small for ordinary values of the masses and the distance between them.

r

m1 m2–F+F

FIGURE 4.9 The two particles, whose masses are m1 and m2, are attracted by gravitational forces +F→ and −F→.

Moon

r

Earth

ME

ME

MM

MM +F –F

+F –F

FIGURE 4.10 The gravitational force that each uniform sphere of matter exerts on the

other is the same as if each sphere were a

particle with its mass concentrated at its

center. The earth (mass ME) and the moon (mass MM) approximate such uniform spheres.

EXAMPLE 5 Gravitational Attraction

What is the magnitude of the gravitational force that acts on each particle

in Figure 4.9, assuming m1 = 12 kg (approximately the mass of a bicycle), m2 = 25 kg, and r = 1.2 m?

Reasoning and Solution The magnitude of the gravitational force can be found using Equation 4.3:

F = G m1m2

r2 = (6.67 × 10 −11 N · m2 / kg2)

(12 kg)(25 kg)

(1.2 m)2 = 1.4 × 10−8 N

For comparison, you exert a force of about 1 N when pushing a doorbell,

so that the gravitational force is exceedingly small in circumstances such

as those here. This result is due to the fact that G itself is very small. However, if one of the bodies has a large mass, like that of the earth

(5.98 × 1024 kg), the gravitational force can be large.

As expressed by Equation 4.3, Newton’s law of gravitation applies only to particles. How-

ever, most familiar objects are too large to be considered particles. Nevertheless, the law of

universal gravitation can be applied to such objects with the aid of calculus. Newton was able

to prove that an object of fi nite size can be considered to be a particle for purposes of using

the gravitation law, provided the mass of the object is distributed with spherical symmetry

about its center. Thus, Equation 4.3 can be applied when each object is a sphere whose mass

is spread uniformly over its entire volume. Figure 4.10 shows this kind of application, as- suming that the earth and the moon are such uniform spheres of matter. In this case, r is the

90 CHAPTER 4 Forces and Newton’s Laws of Motion

distance between the centers of the spheres and not the distance between the outer surfaces. The gravitational forces that the spheres exert on each other are the same as if the entire mass of

each were concentrated at its center. Even if the objects are not uniform spheres, Equation 4.3

can be used to a good degree of approximation if the sizes of the objects are small relative to

the distance of separation r.

Weight The weight of an object exists because of the gravitational pull of the earth, according to the

following defi nition:

DEFINITION OF WEIGHT The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downward, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N)

Using W for the magnitude of the weight,* m for the mass of the object, and ME for the mass of the earth, it follows from Equation 4.3 that

W = G M E m

r 2 (4.4)

Equation 4.4 and Figure 4.11 both emphasize that an object has weight whether or not it is resting on the earth’s surface, because the gravitational force is acting even when the distance r is not equal to the radius RE of the earth. However, the gravitational force becomes weaker as r increases, since r is in the denominator of Equation 4.4. Figure 4.12, for example, shows how the weight of the Hubble Space Telescope becomes smaller as the distance r from the center of the earth increases. In Example 6 the telescope’s weight is determined when it is on earth and

in orbit.

EXAMPLE 6 The Hubble Space Telescope

The mass of the Hubble Space Telescope is 11 600 kg. Determine the

weight of the telescope (a) when it was resting on the earth and (b) as it is in its orbit 598 km above the earth’s surface.

Reasoning The weight of the Hubble Space Telescope is the gravit- ational force exerted on it by the earth. According to Equation 4.4, the

weight varies inversely as the square of the radial distance r. Thus, we ex- pect the telescope’s weight on the earth’s surface (r smaller) to be greater than its weight in orbit (r larger).

Problem-Solving Insight When applying Newton’s gravit- ation law to uniform spheres of matter, remember that the distance r is between the centers of the spheres, not between the surfaces.

Solution (a) On the earth’s surface, the weight is given by Equation 4.4 with r = 6.38 × 106 m (the earth’s radius):

W = G ME m

r 2 =

(6.67 × 10−11 N · m2/kg2)(5.98 × 1024 kg)(11 600 kg) (6.38 × 106 m)2

W = 1.14 × 105 N

(b) When the telescope is 598 km above the surface, its distance from the center of the earth is

r = 6.38 × 106 m + 598 × 103 m = 6.98 × 106 m

The weight now can be calculated as in part (a), except that the new value

of r must be used: W = 0.950 × 105 N . As expected, the weight is less in orbit.

*Often, the word “weight” and the phrase “magnitude of the weight” are used interchangeably, even though weight is a

vector. Generally, the context makes it clear when the direction of the weight vector must be taken into account.

r

W

Object of mass m

RE

Mass of earth = ME

FIGURE 4.11 On or above the earth, the

weight W →

of an object is the gravitational

force exerted on the object by the earth.

5

0.2

0

0.4

0.6

0.8

1.0

10 15 20

r (× 106 m)

RE = 6.38 x 10 6 m

W (×

1 0

5 N

)

FIGURE 4.12 The weight of the Hubble Space Telescope decreases as the telescope

gets farther from the earth. The distance from

the center of the earth to the telescope is r.

4.7 The Gravitational Force 91

The space age has forced us to broaden our ideas about weight. For instance, an astronaut

weighs only about one-sixth as much on the moon as on the earth. To obtain his weight on the

moon from Equation 4.4, it is only necessary to replace ME by MM (the mass of the moon) and let r = RM (the radius of the moon).

Relation Between Mass and Weight Although massive objects weigh a lot on the earth, mass and weight are not the same quantity.

As Section 4.2 discusses, mass is a quantitative measure of inertia. As such, mass is an intrinsic

property of matter and does not change as an object is moved from one location to another. Weight,

in contrast, is the gravitational force acting on the object and can vary, depending on how far the

object is above the earth’s surface or whether it is located near another body such as the moon.

The relation between weight W and mass m can be written in two ways:

W = G ME m r2  

(4.4)

W = m g (4.5)

Equation 4.4 is Newton’s law of universal gravitation, and Equation 4.5 is Newton’s second law

(net force equals mass times acceleration) incorporating the acceleration g due to gravity. These expressions make the distinction between mass and weight stand out. The weight of an object

whose mass is m depends on the values for the universal gravitational constant G, the mass ME of the earth, and the distance r. These three parameters together determine the acceleration g due to gravity. The specifi c value of g = 9.80 m/s2 applies only when r equals the radius RE of the earth. For larger values of r, as would be the case on top of a mountain, the eff ective value of g is less than 9.80 m/s2. The fact that g decreases as the distance r increases means that the weight likewise decreases. The mass of the object, however, does not depend on these eff ects and does

not change. Conceptual Example 7 further explores the diff erence between mass and weight. →

CONCEPTUAL EXAMPLE 7 Mass Versus Weight

A vehicle designed for exploring the moon’s surface (see Photo 4.1) is be- ing tested on earth, where it weighs roughly six times more than it will on

the moon. The acceleration of the vehicle along the ground is measured.

To achieve the same acceleration on the moon, will the required net force

be (a) the same as, (b) greater than, or (c) less than that on earth?

Reasoning Do not be misled by the fact that the vehicle weighs more on earth. The greater weight occurs only because the mass and radius of the

earth are diff erent from the mass and radius of the moon. In any event, in

Newton’s second law the net force is proportional to the vehicle’s mass,

not its weight.

Answers (b) and (c) are incorrect. According to Newton’s second law, for a given acceleration, the net force depends only on the mass. If the

required net force were greater or smaller on the moon than it is on

the earth, the implication would be that the vehicle’s mass is diff erent on

the moon than it is on earth, which is contrary to fact.

Answer (a) is correct. The net force ∑F →

required to accelerate the

vehicle is specifi ed by Newton’s second law as ∑F →

= ma→, where m is the vehicle’s mass and a→ is the acceleration. For a given acceleration, the net force depends only on the mass, which is the same on the moon as it is on

the earth. Therefore, the required net force is the same on the moon as it

is on the earth.

Related Homework: Problems 26, 105

PHOTO 4.1 On the moon the Lunar Roving Vehicle that astronaut Eugene Cernan is driving and the Lunar Excursion

Module (behind the Roving Vehicle) have the same masses

that they have on the earth. However, their weights are

diff erent on the moon than on the earth, as Conceptual

Example 7 discusses.

N A

S A

/J o h n so

n S

p ac

e C

en te

r

92 CHAPTER 4 Forces and Newton’s Laws of Motion

Check Your Understanding

(The answers are given at the end of the book.) 7. When a body is moved from sea level to the top of a mountain, what changes: (a) the body’s mass,

(b) its weight, or (c) both its mass and its weight? 8. Object A weighs twice as much as object B at the same spot on the earth. Would the same be true at a

given spot on Mars?

9. Three particles have identical masses. Each particle experiences only the gravitational forces due to the other two particles. How should the particles be arranged so each one experiences a net gravitational

force that has the same magnitude? (a) On the corners of an equilateral triangle (b) On three of the four corners of a square (c) On the corners of a right triangle

10. Two objects with masses m1 and m2 are separated by a distance 2d (see CYU Figure 4.1). Mass m2 is greater than mass m1. A third object has a mass m3. All three objects are located on the same straight line. The net gravitational force acting on the third object is zero. Which one of the drawings correctly

represents the locations of the objects?

m1

m1 m1

m1 m2

m2 m2

m2m3

m3 m3

m3

dd dd

dd

(a) (b)

(c) (d)

dd

CYU FIGURE 4.1

4.8 The Normal Force The Definition and Interpretation of the Normal Force In many situations, an object is in contact with a surface, such as a tabletop. Because of the con-

tact, there is a force acting on the object. The present section discusses only one component of this

force, the component that acts perpendicular to the surface. The next section discusses the com-

ponent that acts parallel to the surface. The perpendicular component is called the normal force.

DEFINITION OF THE NORMAL FORCE The normal force FN

→ is one component of the force that a surface exerts on an object with which it is in contact—namely, the component that is perpendicular to the surface.

Figure 4.13 shows a block resting on a horizontal table and identifi es the two forces that act on the block, the weight W→ and the normal force FN

→ . To understand how an inanimate object, such as a

tabletop, can exert a normal force, think about what happens when you sit on a mattress. Your weight

causes the springs in the mattress to compress. As a result, the compressed springs exert an upward

force (the normal force) on you. In a similar manner, the weight of the block causes invisible “atomic

springs” in the surface of the table to compress, thus producing a normal force on the block.

Newton’s third law plays an important role in connection with the normal force. In

Figure 4.13, for instance, the block exerts a force on the table by pressing down on it. Consistent with the third law, the table exerts an oppositely directed force of equal magnitude on the block.

This reaction force is the normal force. The magnitude of the normal force indicates how hard the

two objects press against each other.

If an object is resting on a horizontal surface and there are no vertically acting forces except the

object’s weight and the normal force, the magnitudes of these two forces are equal; that is, FN = W. This is the situation in Figure 4.13. The weight must be balanced by the normal force for the object

FN

W

FIGURE 4.13 Two forces act on the block,

its weight W →

and the normal force FN →

exerted by the surface of the table.

4.8 The Normal Force 93

to remain at rest on the table. If the magnitudes of these forces

were not equal, there would be a net force acting on the block, and

the block would accelerate either upward or downward, in accord

with Newton’s second law.

If other forces in addition to W→ and FN →

act in the vertical

direction, the magnitudes of the normal force and the weight

are no longer equal. In Figure 4.14a, for instance, a box whose weight is 15 N is being pushed downward against a table. The

pushing force has a magnitude of 11 N. Thus, the total downward

force exerted on the box is 26 N, and this must be balanced by the

upward-acting normal force if the box is to remain at rest. In this

situation, then, the normal force is 26 N, which is considerably

larger than the weight of the box.

Figure 4.14b illustrates a diff erent situation. Here, the box is being pulled upward by a rope that applies a force of 11 N. The

net force acting on the box due to its weight and the rope is only

4 N, downward. To balance this force, the normal force needs to

be only 4 N. It is not hard to imagine what would happen if the

force applied by the rope were increased to 15 N—exactly equal

to the weight of the box. In this situation, the normal force would

become zero. In fact, the table could be removed, since the block

would be supported entirely by the rope. The situations in Figure 4.14 are consistent with the idea that the magnitude of the normal force indicates how hard two objects press against each other.

Clearly, the box and the table press against each other harder in part a of the picture than in part b. Like the box and the table in Figure 4.14, various parts of the human body press against

one another and exert normal forces. Example 8 illustrates the remarkable ability of the human

skeleton to withstand a wide range of normal forces.

FIGURE 4.14 (a) The normal force FN →

is greater than the weight of the box,

because the box is being pressed downward with an 11-N force. (b) The normal force is smaller than the weight, because the rope supplies an upward force of

11 N that partially supports the box.

11 N

W = 15 N (b)

FN = 4 N

11 N

FN = 26 N

W = 15 N (a)

EXAMPLE 8 BIO The Physics of the Human Skeleton

+x

+y

+x

+y

FN

FN

FN

FN

50 N

50 N

50 N

490 N

490 N

50 N

(b)(a) (c)

Seventh cervical vertebra

Free-body diagram Free-body diagram

FIGURE 4.15 (a) A young woman keeps her balance during a performance by China’s Sichuan Acrobatic group. A free-body diagram is shown for the standing performer’s body above the shoulders (b) before the act and (c) during the act. For convenience, the scales used for the vectors in parts b and c are diff erent.

In a circus balancing act, a woman performs a headstand on top of a

standing performer’s head, as Figure 4.15a illustrates. The woman weighs 490 N, and the standing performer’s head and neck weigh 50 N. It

is primarily the seventh cervical vertebra in the spine that supports all the

weight above the shoulders. What is the normal force that this vertebra

exerts on the neck and head of the standing performer (a) before the act and (b) during the act?

S u p ri

/R E

U T

E R

S /N

ew sc

o m

94 CHAPTER 4 Forces and Newton’s Laws of Motion

In summary, the normal force does not necessarily have the same magnitude as the weight of the

object. The value of the normal force depends on what other forces are present. It also depends on

whether the objects in contact are accelerating. In one situation that involves accelerating objects, the

magnitude of the normal force can be regarded as a kind of “apparent weight,” as we will now see.

Apparent Weight Usually, the weight of an object can be determined with the aid of a scale. However, even though

a scale is working properly, there are situations in which it does not give the correct weight. In

such situations, the reading on the scale gives only the “apparent” weight, rather than the gravit-

ational force or “true” weight. The apparent weight is the force that the object exerts on the scale

with which it is in contact.

To see the discrepancies that can arise between true weight and apparent weight, consider

the scale in the elevator in Interactive Figure 4.16. The reasons for the discrepancies will be explained shortly. A person whose true weight is 700 N steps on the scale. If the elevator is at

rest or moving with a constant velocity (either upward or downward), the scale registers the true

weight, as Interactive Figure 4.16a illustrates. If the elevator is accelerating, the apparent weight and the true weight are not equal. When

the elevator accelerates upward, the apparent weight is greater than the true weight, as Interactive Figure 4.16b shows. Conversely, if the elevator accelerates downward, as in part c, the apparent weight is less than the true weight. In fact, if the elevator falls freely, so its acceleration is equal to

the acceleration due to gravity, the apparent weight becomes zero, as part d indicates. In a situation such as this, where the apparent weight is zero, the person is said to be “weightless.” The apparent

weight, then, does not equal the true weight if the scale and the person on it are accelerating.

Reasoning To begin, we draw a free-body diagram for the neck and head of the standing performer. Before the act, there are only two forces,

the weight of the standing performer’s head and neck, and the normal

force. During the act, an additional force is present due to the woman’s

weight. In both cases, the upward and downward forces must balance for

the head and neck to remain at rest. This condition of balance will lead us

to values for the normal force.

Solution (a) Figure 4.15b shows the free-body diagram for the stand- ing performer’s head and neck before the act. The only forces acting are

the normal force FN →

and the 50-N weight. These two forces must balance

for the standing performer’s head and neck to remain at rest. Therefore,

the seventh cervical vertebra exerts a normal force of FN = 50 N .

(b) Figure 4.15c shows the free-body diagram that applies during the act. Now, the total downward force exerted on the standing performer’s

head and neck is 50 N + 490 N = 540 N, which must be balanced by the

upward normal force, so that FN = 540 N .

(a) No acceleration (v = constant) (b) Upward acceleration (c) Downward acceleration (d) Free-fall

0

700

0

W = 700 N

0

400

0

W = 700 N

a a

a = g

1000

W = 700 N W = 700 N

INTERACTIVE FIGURE 4.16 (a) When the elevator is not accelerating, the scale registers the true weight (W = 700 N) of the person. (b) When the elevator accelerates upward, the apparent weight (1000 N) exceeds the true weight. (c) When the elevator accelerates downward, the apparent weight (400 N) is less than the true weight. (d) The apparent weight is zero if the elevator falls freely—that is, if it falls with the acceleration due to gravity.

4.9 Static and Kinetic Frictional Forces 95

The discrepancies between true weight and apparent weight can be understood with the aid

of Newton’s second law. Figure 4.17 shows a free-body diagram of the person in the elevator. The two forces that act on him are the true weight W→ = mg→ and the normal force FN

→ exerted by

the platform of the scale. Applying Newton’s second law in the vertical direction gives

ΣFy = +FN − mg = ma

where a is the acceleration of the elevator and person. In this result, the symbol g stands for the magnitude of the acceleration due to gravity and can never be a negative quantity. However, the

acceleration a may be either positive or negative, depending on whether the elevator is accelerat- ing upward (+) or downward (−). Solving for the normal force FN shows that

FN  =  mg  + ma  (4.6)

In Equation 4.6, FN is the magnitude of the normal force exerted on the person by the scale. But in accord with Newton’s third law, FN is also the magnitude of the downward force that the person exerts on the scale—namely, the apparent weight.

Equation 4.6 contains all the features shown in Interactive Figure 4.16. If the elevator is not accelerating, a = 0 m/s2, and the apparent weight equals the true weight. If the elevator acceler- ates upward, a is positive, and the equation shows that the apparent weight is greater than the true weight. If the elevator accelerates downward, a is negative, and the apparent weight is less than the true weight. If the elevator falls freely, a = −g, and the apparent weight is zero. The apparent weight is zero because when both the person and the scale fall freely, they cannot push against one another.

In this text, when the weight is given, it is assumed to be the true weight, unless stated otherwise.

Check Your Understanding

(The answers are given at the end of the book.) 11. A stack of books whose true weight is 165 N is placed on a scale in an elevator. The scale reads 165 N.

From this information alone, can you tell whether the elevator is moving with a constant velocity of

2 m/s upward, is moving with a constant velocity of 2 m/s downward, or is at rest?

12. A 10-kg suitcase is placed on a scale that is in an elevator. In which direction is the elevator accelerating when the scale reads 75 N and when it reads 120 N? (a) Downward when it reads 75 N and upward when it reads 120 N (b) Upward when it reads 75 N and downward when it reads 120 N (c) Downward in both cases (d) Upward in both cases

13. You are standing on a scale in an elevator that is moving upward with a constant velocity. The scale reads 600 N. The following table shows fi ve options for what the scale reads when the elevator slows

down as it comes to a stop, when it is stopped, and when it picks up speed on its way back down. Which

one of the fi ve options correctly describes the scale’s readings? Note that the symbol < means “less

than” and > means “greater than.”

Option Elevator slows down as it comes to a halt

Elevator is stopped

Elevator picks up speed on its way back down

(a) > 600 N > 600 N > 600 N

(b) < 600 N 600 N < 600 N

(c) > 600 N 600 N < 600 N

(d) < 600 N < 600 N < 600 N

(e) < 600 N 600 N > 600 N

4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface, there is a force acting on the object. The previous

section discusses the component of this force that is perpendicular to the surface, which is called

the normal force. When the object moves or attempts to move along the surface, there is also a

Apparent

weight

⏟ True

weight

⏟

+x

W = mg

FN

+y

FIGURE 4.17 A free-body diagram showing the forces acting on the person riding in the

elevator of Interactive Figure 4.16. W →

is the

true weight, and FN →

is the normal force exerted

on the person by the platform of the scale.

96 CHAPTER 4 Forces and Newton’s Laws of Motion

component of the force that is parallel to the surface. This parallel force component is called the

frictional force, or simply friction. In many situations considerable engineering eff ort is expended trying to reduce friction.

For example, oil is used to reduce the friction that causes wear and tear in the pistons and

cylinder walls of an automobile engine. Sometimes, however, friction is absolutely necessary.

Without friction, car tires could not provide the traction needed to move the car. In fact, the

raised tread on a tire is designed to maintain friction. On a wet road, the spaces in the tread

pattern (see Figure 4.18) provide channels for the water to collect and be diverted away. Thus, these channels largely prevent the water from coming between the tire surface and the road

surface, where it would reduce friction and allow the tire to skid.

Surfaces that appear to be highly polished can actually look quite rough when examined

under a microscope. Such an examination reveals that two surfaces in contact touch only at re-

latively few spots, as Figure 4.19 illustrates. The microscopic area of contact for these spots is substantially less than the apparent macroscopic area of contact between the surfaces—perhaps

thousands of times less. At these contact points the molecules of the diff erent bodies are close

enough together to exert strong attractive intermolecular forces on one another, leading to what

are known as “cold welds.” Frictional forces are associated with these welded spots, but the exact

details of how frictional forces arise are not well understood. However, some empirical relations

have been developed that make it possible to account for the eff ects of friction.

Interactive Figure 4.20 helps to explain the main features of the type of friction known as static friction. The block in this drawing is initially at rest on a table, and as long as there is no attempt to move the block, there is no static frictional force. Then, a horizontal force F→ is applied to the block by means of a rope. If F→ is small, as in part a, experience tells us that the block still does not move. Why? It does not move because the static frictional force f s

→ exactly cancels the

eff ect of the applied force. The direction of f s →

is opposite to that of F,→ and the magnitude of f s →

equals the magnitude of the applied force, f s = F. Increasing the applied force in Interactive Figure 4.20 by a small amount still does not cause the block to move. There is no movement because the static frictional force also increases by an amount that cancels out the increase in the

applied force (see part b of the drawing). If the applied force continues to increase, however, there comes a point when the block fi nally “breaks away” and begins to slide. The force just before

breakaway represents the maximum static frictional force f MAXs →

that the table can exert on the

block (see part c of the drawing). Any applied force that is greater than f MAXs →

cannot be balanced

by static friction, and the resulting net force accelerates the block to the right.

Experimental evidence shows that, to a good degree of approximation, the maximum

static frictional force between a pair of dry, unlubricated surfaces has two main character-

istics. It is independent of the apparent macroscopic area of contact between the objects,

provided that the surfaces are hard or nondeformable. For instance, in Figure 4.21 the max- imum static frictional force that the surface of the table can exert on a block is the same,

whether the block is resting on its largest or its smallest side. The other main characteristic

of f MAXs →

is that its magnitude is proportional to the magnitude of the normal force FN →

. As

Section 4.8 points out, the magnitude of the normal force indicates how hard two surfaces are

being pressed together. The harder they are pressed, the larger is f s MAX, presumably because the number of “cold-welded,” microscopic contact points is increased. Equation 4.7 expresses

the proportionality between f sMAX and FN with the aid of a proportionality constant μs, which is called the coeffi cient of static friction.

STATIC FRICTIONAL FORCE The magnitude f s of the static frictional force can have any value from zero up to a max- imum value of f sMAX, depending on the applied force. In other words, f s ≤ f sMAX, where the symbol ≤ is read as “less than or equal to.” The equality holds only when f s attains its maximum value, which is f sMAX = μs FN (4.7)

In Equation 4.7, μs is the coeffi cient of static friction, and FN is the magnitude of the normal force.

It should be emphasized that Equation 4.7 relates only the magnitudes of f MAXs →

and FN →

, not the vectors themselves. This equation does not imply that the directions of the vectors are the same. In fact, f MAXs

→ is parallel to the surface, while FN

→ is perpendicular to it.

Microscopic contact points

FIGURE 4.19 Even when two highly polished surfaces are in contact, they touch

only at relatively few points.

F

No movement (a)

No movement (b)

When movement just begins (c)

F

fsMAX

F

fs

fs

INTERACTIVE FIGURE 4.20 (a) and (b) Applying a small force F→ to the block produces no movement, because the static

frictional force f s →

exactly balances the applied

force. (c) The block just begins to move when the applied force is slightly greater than the

maximum static frictional force f MAXs →

.

FIGURE 4.18 This photo, shot parallel to the road surface, shows a tire rolling under

wet conditions. The channels in the tire

collect and divert water away from the

regions where the tire contacts the surface,

thus providing better traction.

S to

ck sn

ap p

er /S

h u

tt er

st o

ck

4.9 Static and Kinetic Frictional Forces 97

The coeffi cient of static friction, being the ratio of the magnitudes of two forces

(μs = f sMAX/FN), has no units. Also, it depends on the type of material from which each surface is made (steel on wood, rubber on concrete, etc.), the condition of the surfaces (polished, rough,

etc.), and other variables such as temperature. Table 4.2 gives some typical values of μs for vari- ous surfaces. Example 9 illustrates the use of Equation 4.7 for determining the maximum static

frictional force. FIGURE 4.21 The maximum static frictional

force f MAXs →

would be the same, no matter

which side of the block is in contact with the

table.TABLE 4.2 Approximate Values of the Coefficients of Friction for Various Surfaces a

Materials Coefficient of

Static Friction, μs Coefficient of

Kinetic Friction, μk Glass on glass (dry) 0.94 0.4

Ice on ice (clean, 0 °C) 0.1 0.02

Rubber on dry concrete 1.0 0.8

Rubber on wet concrete 0.7 0.5

Steel on ice 0.1 0.05

Steel on steel (dry hard steel) 0.78 0.42

Teflon on Teflon 0.04 0.04

Wood on wood 0.35 0.3

a The last column gives the coefficients of kinetic friction, a concept that will be discussed shortly.

Analyzing Multiple -Concept Problems

EXAMPLE 9 The Force Needed to Start a Skier Moving

A skier is standing motionless on a horizontal patch of snow. She is hold-

ing onto a horizontal tow rope, which is about to pull her forward (see

Figure 4.22a). The skier’s mass is 59 kg, and the coeffi cient of static friction between the skis and snow is 0.14. What is the magnitude of the

maximum force that the tow rope can apply to the skier without causing

her to move?

Reasoning When the rope applies a relatively small force, the skier does not accelerate. The reason is that the static frictional force opposes

the applied force and the two forces have the same magnitude. We can

apply Newton’s second law in the horizontal direction to this situation.

In order for the rope to pull the skier forward, it must exert a force large

enough to overcome the maximum static frictional force acting on the skis. The magnitude of the maximum static frictional force depends on

the coeffi cient of static friction (which is known) and on the magnitude

of the normal force. We can determine the magnitude of the normal force

by using Newton’s second law, along with the fact that the skier does not

accelerate in the vertical direction.

Knowns and Unknowns The data for this problem are as follows:

Description Symbol Value Mass of skier m 59 kg

Coefficient of static friction μs 0.14

Unknown Variable Magnitude of maximum horizontal

force that tow rope can apply F ?

(a)

+y

+x

(b)

FN

mg

fs MAX

F

FIGURE 4.22 (a) Two forces act on the skier in the horizontal direction just

before she begins to move. (b) Two vertical forces act on the skier.

98 CHAPTER 4 Forces and Newton’s Laws of Motion

BIO THE PHYSICS OF . . . rock climbing. Static friction is often essential, as it is to the rock climber in Figure 4.23, for instance. She presses outward against the walls of the rock formation with her hands and feet to create suffi ciently large normal forces, so that the static

frictional forces help support her weight.

Once two surfaces begin sliding over one another, the static frictional force is no longer of any

concern. Instead, a type of friction known as kinetic* friction comes into play. The kinetic fric- tional force opposes the relative sliding motion. If you have ever pushed an object across a fl oor, you

may have noticed that it takes less force to keep the object sliding than it takes to get it going in the

fi rst place. In other words, the kinetic frictional force is usually less than the static frictional force.

*The word “kinetic” is derived from the Greek word kinetikos, meaning “pertaining to motion.”

Modeling the Problem

STEP 1 Newton’s Second Law (Horizontal Direction) Figure 4.22a shows the two hori- zontal forces that act on the skier just before she begins to move: the force F→ applied by the tow rope and the maximum static frictional force f MAXs

→ . Since the skier is standing motionless, she is

not accelerating in the horizontal or x direction, so ax = 0 m/s2. Applying Newton’s second law (Equation 4.2a) to this situation, we have

ΣFx = max = 0

Since the net force ∑Fx in the x direction is ∑Fx = +F − f sMAX, Newton’s second law can be writ- ten as +F − f sMAX = 0. Thus,

F = f sMAX

We do not know f sMAX, but its value will be determined in Steps 2 and 3.

STEP 2 The Maximum Static Frictional Force The magnitude f sMAX of the maximum static frictional force is related to the coeffi cient of static friction μs and the magnitude FN of the normal force by Equation 4.7:

f sMAX = μs FN (4.7)

We now substitute this result into Equation 1, as indicated in the right column. The coeffi cient

of static friction is known, but FN is not. An expression for FN will be obtained in the next step.

STEP 3 Newton’s Second Law (Vertical Direction) We can fi nd the magnitude FN of the normal force by noting that the skier does not accelerate in the vertical or y direction, so ay = 0 m/s2. Figure 4.22b shows the two vertical forces that act on the skier: the normal force FN

→ and

her weight mg→. Applying Newton’s second law (Equation 4.2b) in the vertical direction gives ΣFy = may = 0

The net force in the y direction is ∑Fy = +FN − mg, so Newton’s second law becomes +FN − mg = 0. Thus,

FN = mg

We now substitute this result into Equation 4.7, as shown at the right.

Solution Algebraically combining the results of the three steps, we have

F = f sMAX  =  μsFN  =  μsmg

The magnitude F of the maximum force is

F = μs mg = (0.14)(59 kg)(9.80 m /s2) = 81 N

If the force exerted by the tow rope exceeds this value, the skier will begin to accelerate forward.

Related Homework: Problems 44, 109, 118

STEP 1 STEP 2 STEP 3

?

F = f sMAX (1)

F = f sMAX (1)

f sMAX = μs FN (4.7)

FN = mg

?

F = f sMAX (1)

f sMAX = μs FN (4.7)

4.9 Static and Kinetic Frictional Forces 99

Experimental evidence indicates that the kinetic frictional force f k →

has three main charac-

teristics, to a good degree of approximation. It is independent of the apparent area of contact

between the surfaces (see Figure 4.21). It is independent of the speed of the sliding motion, if the speed is small. And last, the magnitude of the kinetic frictional force is proportional to the

magnitude of the normal force. Equation 4.8 expresses this proportionality with the aid of a pro-

portionality constant 𝜇k, which is called the coeffi cient of kinetic friction.

KINETIC FRICTIONAL FORCE The magnitude fk of the kinetic frictional force is given by

fk = μkFN (4.8)

In Equation 4.8, 𝜇k is the coeffi cient of kinetic friction, and FN is the magnitude of the normal force.

Equation 4.8, like Equation 4.7, is a relationship between only the magnitudes of the fric-

tional and normal forces. The directions of these forces are perpendicular to each other. Moreover,

like the coeffi cient of static friction, the coeffi cient of kinetic friction is a number without units

and depends on the type and condition of the two surfaces that are in contact. As indicated in

Table 4.2, values for 𝜇k are typically less than those for 𝜇s, refl ecting the fact that kinetic friction is generally less than static friction. The next example illustrates the eff ect of kinetic friction.

Analyzing Multiple-Concept Problems

EXAMPLE 10 Sled Riding

A sled and its rider are moving at a speed of 4.0 m/s along a

horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the

sled, so the sled slows down and eventually comes to a s

top. The coeffi cient of kinetic friction is 0.050. What is the

displacement x of the sled?

Reasoning As the sled slows down, its velocity is decreas- ing. As our discussions in Chapters 2 and 3 indicate, the

changing velocity is described by an acceleration (which in

this case is a deceleration since the sled is slowing down).

Assuming that the acceleration is constant, we can use one

of the equations of kinematics from Chapter 3 to relate the

displacement to the initial and fi nal velocities and to the ac-

celeration. The acceleration of the sled is not given directly.

However, we can determine it by using Newton’s second

law of motion, which relates the acceleration to the net force

(which is the kinetic frictional force in this case) acting on

the sled and to the mass of the sled (plus rider).

Knowns and Unknowns The data for this problem are listed in the table:

0x = 4.0 m/sυ x = 0 m/sυ

+x

+y

FN

fk

W = mg

(b) Free-body diagram for the sled and rider

(a)

x

FIGURE 4.24 (a) The moving sled decelerates because of the kinetic frictional force. (b) Three forces act on the moving sled: the weight W

→ of the sled and its

rider, the normal force FN →

, and the kinetic frictional force fk →

. The free-body diagram

for the sled and rider shows these forces.

Description Symbol Value Comment Explicit Data Initial velocity υ0x +4.0 m/s Positive, because the velocity points in the +x direction. See the drawing.

Coefficient of kinetic friction μk 0.050

Implicit Data Final velocity υx 0 m/s The sled comes to a stop.

Unknown Variable Displacement x ?

FIGURE 4.23 In maneuvering her way up Devil’s Tower at Devil’s Tower National

Monument in Wyoming, this rock climber

uses the static frictional forces between her

hands and feet and the vertical rock walls to

help support her weight.

© René Robert/Collection Catherine Destivelle/Sygma/Getty

100 CHAPTER 4 Forces and Newton’s Laws of Motion

Modeling the Problem

STEP 1 Displacement To obtain the displacement x of the sled we will use Equation 3.6a from the equations of kinematics:

𝜐x 2 = 𝜐0x

2 + 2ax x

Solving for the displacement x gives the result shown at the right. This equation is useful because two of the variables, υ0x and υx, are known and the acceleration ax can be found by applying Newton’s second law to the accelerating sled (see Step 2).

STEP 2 Newton’s Second Law Newton’s second law, as given in Equation 4.2a, states that the acceleration ax is equal to the net force ΣFx divided by the mass m:

ax = ΣFx m

The free-body diagram in Figure 4.24b shows that the only force acting on the sled in the hori- zontal or x direction is the kinetic frictional force f k

→ . We can write this force as −fk, where f k is

the magnitude of the force and the minus sign indicates that it points in the −x direction. Since the net force is ΣFx = −f k, Equation 4.2a becomes

ax = −f k m

This result can now be substituted into Equation 1, as shown at the right.

STEP 3 Kinetic Frictional Force We do not know the magnitude f k of the kinetic frictional force, but we do know the coeffi cient of kinetic friction 𝜇k. According to Equation 4.8, the two

are related by

f k = μkFN (4.8)

where FN is the magnitude of the normal force. This relation can be substituted into Equation 2, as shown at the right. An expression for FN will be obtained in the next step.

STEP 4 Normal Force The magnitude FN of the normal force can be found by noting that the sled does not accelerate in the vertical or y direction (ay = 0 m/s2). Thus, Newton’s second law, as given in Equation 4.2b, becomes

ΣFy = may = 0 There are two forces acting on the sled in the y direction: the normal force FN

→ and its weight W

→

(see Figure 4.24b). Therefore, the net force in the y direction is

ΣFy = +FN − W

where W = mg (Equation 4.5). Thus, Newton’s second law becomes

+FN − mg = 0 or FN = mg

This result for FN can be substituted into Equation 4.8, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

x = 𝜐x

2 − 𝜐0x 2

2ax =

𝜐x 2 − 𝜐0x

2

2( −fk m )

= 𝜐x

2 − 𝜐0x 2

2( −μkFN

m ) =

𝜐x 2 − 𝜐0x

2

2( −μk mg

m ) =

𝜐x 2 − 𝜐0x

2

2(−μkg)

Note that the mass m of the sled and rider is algebraically eliminated from the fi nal result. Thus, the displacement of the sled is

x = 𝜐x

2 − 𝜐0x 2

2(−μkg) =

(0 m/s)2 − (+4.0 m /s)2

2[−(0.050)(9.80 m/s2)] = +16 m

Related Homework: Problems 48, 112, 115

STEP 2STEP 1 STEP 3 STEP 4

x = 𝜐x

2 − 𝜐0x 2

2ax (1)

?

?

x = 𝜐x

2 − 𝜐0x 2

2ax (1)

ax = −f k m

(2)

x = 𝜐x

2 − 𝜐0x 2

2ax (1)

ax = −f k m

(2)

f k = μkFN (4.8)?

x = 𝜐x

2 − 𝜐0x 2

2ax (1)

ax = −f k m

(2)

f k = μkFN

(4.8)

FN = mg

4.10 The Tension Force 101

Static friction opposes the impending relative motion between two objects, while kinetic fric-

tion opposes the relative sliding motion that actually does occur. In either case, relative motion is opposed. However, this opposition to relative motion does not mean that friction prevents or

works against the motion of all objects. For instance, consider what happens when you walk. BIO THE PHYSICS OF . . . walking. Your foot exerts a force on the earth, and the

earth exerts a reaction force on your foot. This reaction force is a static frictional force, and it

opposes the impending backward motion of your foot, propelling you forward in the process.

Kinetic friction can also cause an object to move, all the while opposing relative motion, as it does

in Example 10. In this example the kinetic frictional force acts on the sled and opposes the relative

motion of the sled and the earth. Newton’s third law indicates, however, that since the earth exerts

the kinetic frictional force on the sled, the sled must exert a reaction force on the earth. In response,

the earth accelerates, but because of the earth’s huge mass, the motion is too slight to be noticed.

Check Your Understanding

(The answers are given at the end of the book.) 14. Suppose that the coeffi cients of static and kinetic friction have values such that μs = 1.4 μk for a crate in

contact with a cement fl oor. Which one of the following statements is true? (a) The magnitude of the static frictional force is always 1.4 times the magnitude of the kinetic frictional force. (b) The magnitude of the kinetic frictional force is always 1.4 times the magnitude of the static frictional force. (c) The mag- nitude of the maximum static frictional force is 1.4 times the magnitude of the kinetic frictional force.

15. A person has a choice of either pushing or pulling a sled at a constant velocity, as CYU Figure 4.2 illustrates. Friction is present. If the angle θ is the same in both cases, does it require less force to push or to pull the sled?

θ θ

CYU FIGURE 4.2

16. A box has a weight of 150 N and is being pulled across a horizontal fl oor by a force that has a mag- nitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an

angle θ. When the pulling force points horizontally, the kinetic frictional force acting on the box is twice as large as when the pulling force points at the angle θ. Find θ.

17. A box rests on the fl oor of an elevator. Because of static friction, a force is required to start the box sliding across the fl oor when the elevator is (a) stationary, (b) accelerating upward, and (c) accelerating downward. Rank the forces required in these three situations in ascending order—that is, smallest fi rst.

4.10 The Tension Force Forces are often applied by means of cables or ropes that are used to pull an object. For instance,

Figure 4.25a shows a force T→ being applied to the right end of a rope attached to a box. Each particle in the rope in turn applies a force to its neighbor. As a result, the force is applied to the

box, as part b of the drawing shows. In situations such as that in Figure 4.25, we say that the force T→ is applied to the box be-

cause of the tension in the rope, meaning that the tension and the force applied to the box have

the same magnitude. However, the word “tension” is commonly used to mean the tendency of

the rope to be pulled apart. To see the relationship between these two uses of the word “tension,”

consider the left end of the rope, which applies the force T→ to the box. In accordance with Newton’s third law, the box applies a reaction force to the rope. The reaction force has the same

magnitude as T→ but is oppositely directed. In other words, a force −T→ acts on the left end of the rope. Thus, forces of equal magnitude act on opposite ends of the rope, as in Figure 4.25c, and tend to pull it apart.

In the previous discussion, we have used the concept of a “massless” rope (m = 0 kg) without saying so. In reality, a massless rope does not exist, but it is useful as an idealization when applying

T

(a)

T

(b)

T

–T

(c)

FIGURE 4.25 (a) A force T →

is being

applied to the right end of a rope. (b) The force is transmitted to the box. (c) Forces are applied to both ends of the rope. These

forces have equal magnitudes and opposite

directions.

102 CHAPTER 4 Forces and Newton’s Laws of Motion

Newton’s second law. According to the second law, a net force is required to accelerate an object

that has mass. In contrast, no net force is needed to accelerate a massless rope, since ΣF→ = ma→ and m = 0 kg. Thus, when a force T→ is applied to one end of a massless rope, none of the force is needed to accelerate the rope. As a result, the force T→ is also applied undiminished to the object attached at the other end, as we assumed in Figure 4.25.* If the rope had mass, however, some of the force T→ would have to be used to accelerate the rope. The force applied to the box would then be less than T→, and the tension would be diff erent at diff erent locations along the rope. In this text we will assume that a rope connecting one object to another is massless, unless stated otherwise.

The ability of a massless rope to transmit tension undiminished from one end to the other is not

aff ected when the rope passes around objects such as the pulley in Figure 4.26 (provided the pulley itself is massless and frictionless).

Check Your Understanding

(The answer is given at the end of the book.) 18. A rope is used in a tug-of-war between two teams of fi ve people each. Both teams are equally strong,

so neither team wins. An identical rope is tied to a tree, and the same ten people pull just as hard on the

loose end as they did in the contest. In both cases, the people pull steadily with no jerking. Which rope

sustains the greater tension, (a) the rope tied to the tree or (b) the rope in the tug-of-war, or (c) do the ropes sustain the same tension?

4.11 Equilibrium Applications of Newton’s Laws of Motion Have you ever been so upset that it took days to recover your “equilibrium”? In this context, the

word “equilibrium” refers to a balanced state of mind, one that is not changing wildly. In physics,

the word “equilibrium” also refers to a lack of change, but in the sense that the velocity of an

object isn’t changing. If its velocity doesn’t change, an object is not accelerating. Our defi nition

of equilibrium, then, is as follows:

DEFINITION OF EQUILIBRIUM†

An object is in equilibrium when it has zero acceleration.

Newton’s laws of motion apply whether or not an object is in equilibrium. For an

object in equilibrium the acceleration is zero (a→ = 0 m /s2) in Newton’s second law, and the present section presents several examples of this type. In the nonequilibrium case the acceleration

of the object is not zero (a→ ≠ 0 m /s2) in the second law, and Section 4.12 deals with these kinds of situations.

Since the acceleration is zero for an object in equilibrium, all of the acceleration components

are also zero. In two dimensions, this means that ax = 0 m/s2 and ay = 0 m/s2. Substituting these values into the second law (∑Fx = max and ∑Fy = may) shows that the x component and the y component of the net force must each be zero. In other words, the forces acting on an object in

equilibrium must balance. Thus, in two dimensions, the equilibrium condition is expressed by

two equations:

ΣFx = 0 (4.9a)

ΣFy = 0 (4.9b)

*If a rope is not accelerating, a→ is zero in the second law, and ΣF→ = ma→ = 0, regardless of the mass of the rope. Then, the rope can be ignored, no matter what mass it has.

†In this discussion of equilibrium we ignore rotational motion, which is discussed in Chapters 8 and 9. In

Section 9.2 a more complete treatment of the equilibrium of a rigid object is presented and takes into account the

concept of torque and the fact that objects can rotate.

T

–T

FIGURE 4.26 The force T→ applied at one end of a massless rope is transmitted

undiminished to the other end, even when

the rope bends around a pulley, provided the

pulley is also massless and friction is absent.

4.11 Equilibrium Applications of Newton’s Laws of Motion 103

In using Equations 4.9a and 4.9b to solve equilibrium problems, we will use the following

fi ve-step reasoning strategy:

REASONING STRATEGY Analyzing Equilibrium Situations 1. Select the object (often called the “system”) to which Equations 4.9a and 4.9b are to be

applied. It may be that two or more objects are connected by means of a rope or a cable. If so, it may be necessary to treat each object separately according to the following steps.

2. Draw a free-body diagram for each object chosen above. Be sure to include only forces that act on the object. Do not include forces that the object exerts on its environment.

3. Choose a set of x, y axes for each object and resolve all forces in the free-body diagram into components that point along these axes. Select the axes so that as many forces as possible point along one or the other of the two axes. Such a choice minimizes the calcu- lations needed to determine the force components.

4. Apply Equations 4.9a and 4.9b by setting the sum of the x components and the sum of the y components of the forces each equal to zero.

5. Solve the two equations obtained in Step 4 for the desired unknown quantities, remem- bering that two equations can yield answers for only two unknowns at most.

Example 11 illustrates how these steps are followed. It deals with a traction device in which three

forces act together to bring about the equilibrium.

Problem-Solving Insight Choose the orientation of the x, y axes for convenience. In Example 11, the axes have been ro- tated so the force F→ points along the x axis. Since F→ does not have a component along the y axis, the analysis is simplifi ed.

Solution Since the sum of the y components of the forces is zero, it follows that

ΣFy = +T1 sin 35° − T2 sin 35° = 0 (4.9b)

or T1 = T2. In other words, the magnitudes of the tension forces are equal. In addition, the sum of the x components of the forces is zero, so we have that

ΣFx = +T1 cos 35° + T2 cos 35° − F = 0 (4.9a)

Solving for F and letting T1 = T2 = T, we fi nd that F = 2T cos 35°. How- ever, the tension T in the rope is determined by the weight of the 2.2-kg object: T = mg, where m is its mass and g is the acceleration due to gravity. Therefore, the magnitude of F→ is F = 2T cos 35° = 2mg cos 35° = 2(2.2 kg)(9.80 m/s2) cos 35° = 35 N

EXAMPLE 11 BIO The Physics of Traction for the Foot

F

T1 35°

35°

2.2 kg

T2

F

T1

T1 s in 35

°

T2 s in 35

°

T2 c os 35

°

T1 c os 35

°

T1

T2

35° +x

+y

35°

T2

(a) (b) Free-body diagram for the foot pulley

FIGURE 4.27 (a) A traction device for the foot. (b) The free-body diagram for the pulley on the foot.

Figure 4.27a shows a traction device used with a foot injury. The weight of the 2.2-kg object creates a tension in the rope that passes around the

pulleys. Therefore, tension forces T1 →

and T2 →

are applied to the pulley on

the foot. (It may seem surprising that the rope applies a force to either side

of the foot pulley. A similar eff ect occurs when you place a fi nger inside

a rubber band and push downward. You can feel each side of the rubber

band pulling upward on the fi nger.) The foot pulley is kept in equilibrium

because the foot also applies a force F→ to it. This force arises in reaction (Newton’s third law) to the pulling eff ect of the forces T1

→ and T2

→ . Ignoring

the weight of the foot, fi nd the magnitude of F→.

Reasoning The forces T1 →

, T2 →

, and F→ keep the pulley on the foot at rest. The pulley, therefore, has no acceleration and is in equilibrium. As a res-

ult, the sum of the x components and the sum of the y components of the three forces must each be zero. Figure 4.27b shows the free-body diagram of the pulley on the foot. The x axis is chosen to be along the direction of force F,→ and the components of the forces T1

→ and T2

→ are indic-

ated in the drawing. (See Section 1.7 for a review of vector components.)

104 CHAPTER 4 Forces and Newton’s Laws of Motion

Example 12 presents another situation in which three forces are responsible for the equilib-

rium of an object. However, in this example all the forces have diff erent magnitudes.

The plus signs in the table denote components that point along the pos-

itive axes, and the minus signs denote components that point along the

negative axes. Setting the sum of the x components and the sum of the y components equal to zero leads to the following equations:

ΣFx = −T1 sin 10.0° + T2 sin 80.0° = 0 (4.9a)

ΣFy = +T1 cos 10.0° − T2 cos 80.0° − W = 0 (4.9b)

Solving the fi rst of these equations for T1 shows that

T1 = (sin 80.0°sin 10.0°)T2 Substituting this expression for T1 into the second equation gives

(sin 80.0°sin 10.0°) T2 cos 10.0° − T2 cos 80.0° − W = 0

T2 = W

(sin 80.0°sin 10.0°) cos 10.0° − cos 80.0° Setting W = 3150 N in this result yields T2 = 582 N .

Since T1 = (sin 80.0°sin 10.0°) T2 and T2 = 582 N, it follows that T1 = 3.30 × 103 N .

EXAMPLE 12 Replacing an Engine

An automobile engine has a weight W →

, whose magnitude is W = 3150 N. This engine is being positioned above an engine compartment, as

Figure  4.28a illustrates. To position the engine, a worker is using a rope. Find the tension T1

→ in the supporting cable and the tension T2

→ in the

positioning rope.

Reasoning Under the infl uence of the forces W →

, T1 →

, and T2 →

the ring in

Figure 4.28a is at rest and, therefore, in equilibrium. Consequently, the sum of the x components and the sum of the y components of these forces must each be zero; ∑Fx = 0 and ∑Fy = 0. By using these relations, we can fi nd T1 and T2. Figure 4.28b shows the free-body diagram of the ring and the force components for a suitable x, y axis system.

Problem-Solving Insight When an object is in equilibrium, as here in Example 12, the net force is zero, ∑F→ = 0. This does not mean that each individual force is zero. It means that the vector sum of all the forces is zero.

Solution The free-body diagram shows the components for each of the three forces, and the components are listed in the following table:

Force x Component y Component

T1 → −T1 sin 10.0° +T1 cos 10.0°

T2 → +T2 sin 80.0° −T2 cos 80.0°

W → 0 −W

10.0° 10.0°

80.0°

80.0°

(a) (b) Free-body diagram for the ring

Ring

T1 T1

T2

T2

WW

+y

+x

T1

T1 cos 10.0°

T1 sin 10.0°

T2

T2

sin 80.0°

T2 cos 80.0°

FIGURE 4.28 (a) The ring is in equilibrium because of the three forces T1

→ (the tension force in the supporting

cable), T2 →

(the tension force in the positioning rope), and W

→ (the weight of the engine). (b) The free-body

diagram for the ring.

An object can be moving and still be in equilibrium, provided there is no acceleration.

Example 13 illustrates such a case, and the solution is again obtained using the fi ve-step reason-

ing strategy summarized at the beginning of the section.

4.11 Equilibrium Applications of Newton’s Laws of Motion 105

Check Your Understanding

(The answers are given at the end of the book.) 19. In which one of the following situations could an object possibly be in equilibrium? (a) Three forces

act on the object; the forces all point along the same line but may have diff erent directions. (b) Two perpendicular forces act on the object. (c) A single force acts on the object. (d) In none of the situations described in (a), (b), and (c) could the object possibly be in equilibrium.

20. A stone is thrown from the top of a cliff . Air resistance is negligible. As the stone falls, is it (a) in equi- librium or (b) not in equilibrium?

21. During the fi nal stages of descent, a sky diver with an open parachute approaches the ground with a constant velocity. There is no wind to blow him from side to side. Which one of the following state-

ments is true? (a) The sky diver is not in equilibrium. (b) The force of gravity is the only force acting on the sky diver, so that he is in equilibrium. (c) The sky diver is in equilibrium because no forces are

Solution When determining the components of the weight, it is neces- sary to realize that the angle 𝛽 in Figure 4.29a is 30.0°. Part c of the drawing focuses attention on the geometry that is responsible for this fact.

There it can be seen that α + 𝛽 = 90° and α + 30.0° = 90°, with the result that 𝛽 = 30.0°. The following table lists the components of the forces acting on the jet.

Force x Component y Component

W → −W sin 30.0° −W cos 30.0°

L→ 0 +L

T→ +T 0

R→ −R 0

Setting the sum of the x component of the forces to zero gives

ΣFx = −W sin 30.0° + T − R = 0 (4.9a)

R = T − W sin 30.0° = 103 000 N − (86 500 N) sin 30.0° = 59 800 N

Setting the sum of the y component of the forces to zero gives

ΣFy = −W cos 30.0° + L = 0 (4.9b)

L = W cos 30.0° = (86 500 N) cos 30.0° = 74 900 N

EXAMPLE 13 Equilibrium at Constant Velocity

A jet plane is fl ying with a constant speed along a straight line, at an

angle of 30.0° above the horizontal, as Figure 4.29a indicates. The plane has a weight W

→ whose magnitude is W = 86 500 N, and its engines

provide a forward thrust T→ of magnitude T = 103 000 N. In addition, the lift force L→ (directed perpendicular to the wings) and the force R→ of air resistance (directed opposite to the motion) act on the plane. Find L→ and R→.

Reasoning Figure 4.29b shows the free-body diagram of the plane, including the forcesW

→ , L→, T→, and R→. Since the plane is fl ying with a

constant speed along a straight line, it is not accelerating, it is in equi-

librium, and the sum of the x components and the sum of the y com- ponents of these forces must be zero. The weight W

→ and the thrust T→

are known, so the lift force L→ and the force R→ of air resistance can be obtained from these equilibrium conditions. To calculate the components,

we have chosen axes in the free-body diagram that are rotated by 30.0°

from their usual horizontal–vertical positions. This has been done purely

for convenience, since the weight W →

is then the only force that does not

lie along either axis.

Problem-Solving Insight A moving object is in equilibrium if it moves with a constant velocity; then its acceleration is zero. A zero acceleration is the fundamental characteristic of an object in equilibrium.

T L

R

+y

+x +y

+x

W

(a) (c)(b) Free-body diagram

30.0° 90°

90°

β α

α β

W sin 30.0° W cos 30.0°

30.0°

30.0°

TL

R

W

FIGURE 4.29 (a) A plane moves with a constant velocity at an angle of 30.0° above the horizontal due to the action of four forces, the weight W

→ , the lift

L→, the engine thrust T→, and the air re- sistance R→ . (b) The free-body diagram for the plane. (c) This geometry occurs often in physics.

(Continued)

106 CHAPTER 4 Forces and Newton’s Laws of Motion

acting on him. (d) The sky diver is in equilibrium because two forces act on him, the downward-acting force of gravity and the upward-acting force

of the parachute.

22. A crate hangs from a ring at the middle of a rope, as CYU Figure 4.3 illustrates. A person is pulling on the right end of the rope to keep the

crate in equilibrium. Can the rope ever be made to be perfectly horizontal?

4.12 Nonequilibrium Applications of Newton’s Laws of Motion When an object is accelerating, it is not in equilibrium. The forces acting on it are not balanced,

so the net force is not zero in Newton’s second law. However, with one exception, the reasoning

strategy followed in solving nonequilibrium problems is identical to that used in equilibrium

situations. The exception occurs in Step 4 of the fi ve steps outlined at the beginning of the pre-

vious section. Since the object is now accelerating, the representation of Newton’s second law

in Equations 4.2a and 4.2b applies instead of Equations 4.9a and 4.9b:

ΣFx = max (4.2a) and ΣFy = may (4.2b)

Example 14 uses these equations in a situation where the forces are applied in directions similar

to those in Example 11, except that now an acceleration is present.

EXAMPLE 14 Towing a Supertanker

A supertanker of mass m = 1.50 × 108 kg is being towed by two tugboats, as in Figure 4.30a. The tensions in the towing cables apply the forces T1

→

and T2 →

at equal angles of 30.0° with respect to the tanker’s axis. In addi-

tion, the tanker’s engines produce a forward drive force D→, whose mag- nitude is D = 75.0 × 103 N. Moreover, the water applies an opposing force R→, whose magnitude is R = 40.0 × 103 N. The tanker moves forward with an acceleration that points along the tanker’s axis and has a magnitude of

2.00 × 10−3 m/s2. Find the magnitudes of the tensions T1 →

and T2 →

.

Reasoning The unknown forces T1 →

and T2 →

contribute to the net force

that accelerates the tanker. To determine T1 →

and T2 →

, therefore, we analyze

the net force, which we will do using components. The various force com-

ponents can be found by referring to the free-body diagram for the tanker

in Figure 4.30b, where the ship’s axis is chosen as the x axis. We will then use Newton’s second law in its component form, ∑Fx = max and ∑Fy = may, to obtain the magnitudes of T1

→ and T2

→ .

30.0° 30.0°

30.0° 30.0°

(b) Free-body diagram for the tanker(a)

T1

T2T2

T1

+y

+x R D

T2 sin 30.0°

T2 cos 30.0°

T1 cos 30.0°

T1 sin 30.0° T1

T2

D

a

R

FIGURE 4.30 (a) Four forces act on a supertanker: T1 →

and T2 →

are the tension forces due to the towing

cables, D→ is the forward drive force produced by the tanker’s engines, and R→ is the force with which the water opposes the tanker’s motion. (b) The free-body diagram for the tanker.

CYU FIGURE 4.3

4.12 Nonequilibrium Applications of Newton’s Laws of Motion 107

It often happens that two objects are connected somehow, perhaps by a drawbar like that

used when a truck pulls a trailer. If the tension in the connecting device is of no interest, the ob-

jects can be treated as a single composite object when applying Newton’s second law. However,

if it is necessary to fi nd the tension, as in the next example, then the second law must be applied

separately to at least one of the objects.

Solution The individual force components are summarized as follows:

Force x Component y Component

T1 → +T1 cos 30.0° +T1 sin 30.0°

T2 → +T2 cos 30.0° −T2 sin 30.0°

D→ +D 0

R→ −R 0

Since the acceleration points along the x axis, there is no y component of the acceleration (ay = 0 m/s2). Consequently, the sum of the y components of the forces must be zero:

ΣFy = +T1 sin 30.0° − T2 sin 30.0° = 0

This result shows that the magnitudes of the tensions in the cables are

equal, T1 = T2. Since the ship accelerates along the x direction, the sum of the x components of the forces is not zero. The second law indicates that

ΣFx = T1 cos 30.0° + T2 cos 30.0° + D − R = max

Since T1 = T2, we can replace the two separate tension symbols by a single symbol T, the magnitude of the tension. Solving for T gives

T = max + R − D 2 cos 30.0°

= (1.50 × 108 kg)(2.00 × 10−3 m /s2) + 40.0 × 103 N − 75.0 × 103 N

2 cos 30.0°

= 1.53 × 105 N

EXAMPLE 15 Hauling a Trailer

A truck is hauling a trailer along a level road, as Figure 4.32a illustrates. The mass of the truck is m1 = 8500 kg and that of the trailer is m2 = 27 000 kg. The two move along the x axis with an acceleration of ax =

0.78 m/s2. Ignoring the retarding forces of friction and air resistance,

determine (a) the tension T→ in the horizontal drawbar between the trailer and the truck and (b) the force D→ that propels the truck forward.

Trailer Truck

(b) Free-body diagrams

+x +xT D

T

Drawbar

(a)

m1 = 8500 kg m2 = 27 000 kg

a = 0.78 m/s2

DT´ 

DT´  FIGURE 4.32 (a) The force D →

acts on the

truck and propels it forward. The drawbar

exerts the tension force T→′ on the truck and the tension force T→ on the trailer. (b) The free-body diagrams for the trailer and the

truck, ignoring the vertical forces.

Math Skills The sine and cosine functions are defi ned in

Equations 1.1 and 1.2 as sin θ = ho h

and cos θ = ha h

, where ho is

the length of the side of a right triangle that is opposite the angle θ, ha is the length of the side adjacent to the angle θ, and h is the length of the hypotenuse (see Figure 4.31a). When using the sine and cosine functions to determine the scalar components of a vector, we begin

by identifying the angle θ. Figure 4.31b indicates that θ = 30.0° for the vector T1

→ . The components of T1

→ are T1x and T1y. Comparing the

shaded triangles in Figure 4.31, we can see that ho = T1y, ha = T1x, and h = T1. Therefore, we have

cos 30.0° = ha h

= T1x T1

or T1x = T1 cos 30.0°

sin 30.0° = ho h

= T1y T1

or T1y = T1 sin 30.0°

FIGURE 4.31 Math Skills drawing.

(b)

T1 T1y

T1x

30.0°

+y

+x

(a)

ha

h

θ

ho

108 CHAPTER 4 Forces and Newton’s Laws of Motion

Reasoning Since the truck and the trailer accelerate along the hori- zontal direction and friction is being ignored, only forces that have com-

ponents in the horizontal direction are of interest. Therefore, Figure 4.32 omits the weight and the normal force, which act vertically. To determine

the tension force T→ in the drawbar, we draw the free-body diagram for the trailer and apply Newton’s second law, ∑Fx = max. Similarly, we can determine the propulsion force D→ by drawing the free-body diagram for the truck and applying Newton’s second law.

Problem-Solving Insight A free-body diagram is very helpful when applying Newton’s second law. Always start a problem by drawing the free-body diagram.

Solution (a) The free-body diagram for the trailer is shown in Fig- ure 4.32b. There is only one horizontal force acting on the trailer, the tension force T→ due to the drawbar. Therefore, it is straightforward to

obtain the tension from ∑Fx = m2ax, since the mass of the trailer and the acceleration are known:

ΣFx = T = m2 ax = (27 000 kg)(0.78 m /s2) = 21 000 N

(b) Two horizontal forces act on the truck, as the free-body diagram in Figure 4.32b shows. One is the desired force D→. The other is the force T→′. According to Newton’s third law, T→′ is the force with which the trailer pulls back on the truck, in reaction to the truck pulling forward. If the

drawbar has negligible mass, the magnitude of T→′ is equal to the mag- nitude of T→—namely, 21 000 N. Since the magnitude of T→′, the mass of the truck, and the acceleration are known, ∑Fx = m1ax can be used to determine the drive force:

ΣFx = +D − T′ = m1ax

D = m1ax + T′ = (8500 kg)(0.78 m /s2) + 21 000 N = 28 000 N

In Section 4.11 we examined situations where the net force acting on an object is zero, and in

the present section we have considered two examples where the net force is not zero. Conceptual

Example 16 illustrates a common situation where the net force is zero at certain times but is not

zero at other times.

a. The skier is fl oating motionless in the water, so her velocity and acceleration are both zero. Therefore, the net force acting on her is

zero, and she is in equilibrium.

b. As the skier is being pulled up and out of the water, her velocity is increasing. Thus, she is accelerating, and the net force acting on her

is not zero. The skier is not in equilibrium. The direction of the net

force is shown in Figure 4.33b. c. The skier is now moving at a constant speed along a straight line

(Figure 4.33c), so her velocity is constant. Since her velocity is con- stant, her acceleration is zero. Thus, the net force acting on her is

zero, and she is again in equilibrium, even though she is moving.

d. After the skier lets go of the tow rope, her speed decreases, so she is decelerating. Thus, the net force acting on her is not zero, and she is not

in equilibrium. The direction of the net force is shown in Figure 4.33d.

Related Homework: Problem 75

CONCEPTUAL EXAMPLE 16 The Motion of a Water Skier

Figure 4.33 shows a water skier at four diff erent moments:

a. The skier is fl oating motionless in the water. b. The skier is being pulled out of the water and up onto the skis. c. The skier is moving at a constant speed along a straight line. d. The skier has let go of the tow rope and is slowing down.

For each moment, explain whether the net force acting on the skier is

zero.

Reasoning and Solution According to Newton’s second law, if an ob- ject has zero acceleration, the net force acting on it is zero. In such a case,

the object is in equilibrium. In contrast, if the object has an acceleration,

the net force acting on it is not zero. Such an object is not in equilibrium.

We will consider the acceleration in each of the four phases of the motion

to decide whether the net force is zero.

Net force Net force

(a) (c)(b) (d )

FIGURE 4.33 A water skier (a) fl oating in the water, (b) being pulled up by the boat, (c) moving at a constant velocity, and (d) slowing down.

The force of gravity is often present among the forces that aff ect the acceleration of an

object. Examples 17 and 18 deal with typical situations.

4.12 Nonequilibrium Applications of Newton’s Laws of Motion 109

assumption that block 1 moves up the incline. Now there are two equa-

tions in two unknowns, and they may be solved simultaneously (see

Appendix C) to give T and a:

T = 86.3 N and a = 5.89 m/s2

EXAMPLE 17 Accelerating Blocks

Block 1 (mass m1 = 8.00 kg) is moving on a frictionless 30.0° incline. This block is connected to block 2 (mass m2 = 22.0 kg) by a massless cord that passes over a massless and frictionless pulley (see Figure 4.34a). Find the acceleration of each block and the tension in the cord.

Reasoning Since both blocks accelerate, there must be a net force acting on each one. The key to this problem is to realize that Newton’s

second law can be used separately for each block to relate the net force

and the acceleration. Note also that both blocks have accelerations of the

same magnitude a, since they move as a unit. We assume that block 1 accelerates up the incline and choose this direction to be the +x axis. If block 1 in reality accelerates down the incline, then the value obtained for

the acceleration will be a negative number.

Problem-Solving Insight Mass and weight are dif- ferent quantities. They cannot be interchanged when solving problems.

Solution Three forces act on block 1: (1) W1 →

is its weight [W1 = m1g = (8.00 kg) (9.80 m/s2) = 78.4 N], (2) T→ is the force applied because of the tension in the cord, and (3) FN

→ is the normal force the incline exerts.

Figure 4.34b shows the free-body diagram for block 1. The weight is the only force that does not point along the x, y axes, and its x and y components are given in the diagram. Applying Newton’s second law

(∑Fx = m1ax) to block 1 shows that

ΣFx = −W1 sin 30.0° + T = m1a

where we have set ax = a. This equation cannot be solved as it stands, since both T and a are unknown quantities. To complete the solution, we next consider block 2.

Two forces act on block 2, as the free-body diagram in Figure 4.34b indicates: (1) W2

→ is its weight [W2 = m2g = (22.0 kg)(9.80 m/s2) = 216 N]

and (2) T→′ is exerted as a result of block 1 pulling back on the connecting cord. Since the cord and the frictionless pulley are massless, the mag-

nitudes of T→′ and T→ are the same: T ′ = T. Applying Newton’s second law (∑Fy = m2ay) to block 2 reveals that

ΣFy = T − W2 = m2 (−a)

The acceleration ay has been set equal to −a since block 2 moves down- ward along the −y axis in the free-body diagram, consistent with the

(a) (b) Free-body diagrams

T´  T´ 

m1

a

a 30.0°

30.0°

T

Block 1 Block 2

FN

FN

W1 W2

W1

W2

T

W1 cos 30.0°

30.0°

W1 sin 30.0°

+x

+x

+y

+y

m2

FIGURE 4.34 (a) Three forces act on block 1: its weight W1

→ , the normal force FN

→ , and the force T→

due to the tension in the cord. Two forces act on block 2: its weight W2

→ and the force T→′ due to the tension. The

acceleration is labeled according to its magnitude a. (b) Free-body diagrams for the two blocks.

Math Skills The two equations containing the unknown quantit- ies T and a are

−W1 sin 30.0° + T = m1a (1) and T − W2 = m2 (−a) (2)

Neither equation by itself can yield numerical values for T and a. However, the two equations can be solved simultaneously in the following manner. To begin with, we rearrange Equation (1)

to give T = m1a + W1 sin 30.0°. Next, we substitute this result into Equation (2) and obtain a result containing only the unknown

acceleration a:

m1a + W1 sin 30.0° − W2 = m2 (−a) ⏟⎵⎵⎵⏟⎵⎵⎵⏟

T

Rearranging this equation so that the terms m1a and m2a stand alone on the left of the equals sign, we have

m1a + m2a = W2 − W1 sin 30.0° or a = W2 − W1 sin 30.0°

m1 + m2

This result yields the value of a = 5.89 m /s2 . We can also substitute the expression for a into either one of the two starting equations and obtain a result containing only the unknown tension T. We choose Equation (1) and fi nd that

−W1 sin 30.0° + T = m1( W2 − W1 sin 30.0°

m1 + m2 ) ⏟⎵⎵⎵⎵⏟⎵⎵⎵⎵⏟

a Solving for T gives

T = W1 sin 30.0° + m1( W2 − W1 sin 30.0°

m1 + m2 ) This expression yields the value of T = 86.3 N .

110 CHAPTER 4 Forces and Newton’s Laws of Motion

EXAMPLE 18 Hoisting a Scaff old

A window washer on a scaff old is hoisting the scaff old up the side of a

building by pulling downward on a rope, as in Figure 4.35a. The mag- nitude of the pulling force is 540 N, and the combined mass of the worker

and the scaff old is 155 kg. Find the upward acceleration of the unit.

Reasoning The worker and the scaff old form a single unit, on which the rope exerts a force in three places. The left end of the rope exerts an

upward force T→ on the worker’s hands. This force arises because he pulls downward with a 540-N force, and the rope exerts an oppositely directed

force of equal magnitude on him, in accord with Newton’s third law. Thus,

the magnitude T of the upward force is T = 540 N and is the magnitude of the tension in the rope. If the masses of the rope and each pulley are

negligible and if the pulleys are friction-free, the tension is transmitted

undiminished along the rope. Then, a 540-N tension force T→ acts upward on the left side of the scaff old pulley (see part a of the drawing). A tension force is also applied to the point P, where the rope attaches to the roof. The roof pulls back on the rope in accord with the third law, and this pull

leads to the 540-N tension force T→ that acts on the right side of the scaf- fold pulley. In addition to the three upward forces, the weight of the unit

must be taken into account [W = mg = (155 kg)(9.80 m/s2) = 1520 N]. Part b of the drawing shows the free-body diagram.

Solution Newton’s second law (ΣFy = may) can be applied to calculate the acceleration ay:

ΣFy = +T + T + T − W = may

ay = 3T − W

m =

3(540 N) − 1520 N

155 kg = 0.65 m /s2

ay

P

T

T

T T

+x

+y

(a)

(b) Free-body diagram of the unit

T

T

W = mg

W = mg

FIGURE 4.35 (a) A window washer pulls down on the rope to hoist the scaff old up the side of a building. The force T

→ results from the eff ort of the

window washer and acts on him and the scaff old in three places, as discussed in Example 18. (b) The free-body diagram of the unit comprising the man and the scaff old.

Sled

Passenger

Rails

6.5°

Stop

2.3 m

(a)

a

(b) Free-body diagram

6.5°

W cos 6.5° +x

+y

W sin 6.5°

FN

W

EXAMPLE 19 BIO The Importance of Seatbelts

Police departments around the country use a device, called The Convincer, to educate the public on the safety of seatbelts (see Figure 4.36a). The device consists of a sled that is free to move on rails that are inclined at

6.5° above the horizontal. A passenger is placed in the sled and secured

with a seatbelt. The sled is released from rest and travels a distance of

2.3 m along the rails before slamming into a stop at the bottom, thereby

simulating a car collision. Assume the rails are frictionless and calculate

the speed of the sled at impact.

Reasoning The free-body diagram for the sled is shown in Figure 4.36b. The x-component of the sled’s weight is the only force acting on the sled in the x-direction. Once released, the sled will accelerate down the rails.

Thus, this is a nonequilibrium problem, and the sum of the forces acting

on the sled in the x-direction will be equal to its mass times its accelera- tion in that direction. Since the sled’s acceleration down the rails will be

constant, we can use our relationships from kinematics to calculate its

speed when it strikes the stop.

Solution Applying Newton’s second law to the sled in the x-direction, we have ΣFx = W sin 6.5° = max. We can solve this expression for the

acceleration of the sled in the x-direction: ax = W sin 6.5°

m =

mg sin 6.5° m

=

g sin 6.5° = (9.8 m/s2)(sin 6.5°) = 1.1 m/s2. Now that we have the accelera- tion of the sled, we can use Equation 3.6a to calculate its speed when it

FIGURE 4.36 (a) The passenger in the sled accel- erates down the rails and collides with a stop in order to simulate a car collision. (b) Free-body diagram showing the forces acting on the sled. The only force providing its acceleration down the rails is the x-component of its weight.

Concept Summary 111

Check Your Understanding

(The answers are given at the end of the book.) 23. A circus performer hangs stationary from a rope. She then begins to climb upward by pulling herself

up, hand over hand. When she starts climbing, is the tension in the rope (a) less than, (b) equal to, or (c) greater than it is when she hangs stationary?

24. A freight train is accelerating on a level track. Other things being equal, would the tension in the coup- ling between the engine and the fi rst car change if some of the cargo in the last car were transferred to

any one of the other cars?

25. Two boxes have masses m1 and m2, and m2 is greater than m1. The boxes are being pushed across a frictionless horizontal surface (see CYU Figure 4.4). As the drawing shows, there are two possible arrangements, and the pushing force is the same in each. In which arrangement, (a) or (b), does the force that the left box applies to the right box have a greater magnitude, or (c) is the magnitude the same in both cases?

m1 m1 m2

(a) (b)

m2

Pushing force

Pushing force

CYU FIGURE 4.4

strikes the stop: 𝜐x 2 = 𝜐0x

2 + 2axx. Solving for υx, and using the fact that the sled begins from rest (υ0x = 0), we get the following result for its speed at impact: υx = √2axx = √2(1.1 m/s2)(2.3 m) = 2.2 m/s . Consider the magnitude of this speed. This is approximately equal to 5 mi/h, which is

a relatively low speed for a car. Even so, the passenger still receives quite

a jolt when the sled strikes the stop. Without the seatbelt to decelerate the

passenger upon impact, she would continue to move in a straight line and

strike the front of the sled (Newton’s fi rst law). Since many car collisions

occur at speeds much larger than 5 mi/h, The Convincer demonstrates that seatbelt use is essential for passenger safety.

Concept Summary 4.1 The Concepts of Force and Mass A force is a push or a pull and is a vector quantity. Contact forces arise from the physical contact between

two objects. Noncontact forces are also called action-at-a-distance forces,

because they arise without physical contact between two objects.

Mass is a property of matter that determines how diffi cult it is to acceler-

ate or decelerate an object. Mass is a scalar quantity.

4.2 Newton’s First Law of Motion Newton’s fi rst law of motion, some- times called the law of inertia, states that an object continues in a state of rest

or in a state of motion at a constant velocity unless compelled to change that

state by a net force.

Inertia is the natural tendency of an object to remain at rest or in motion

at a constant velocity. The mass of a body is a quantitative measure of inertia

and is measured in an SI unit called the kilogram (kg). An inertial reference

frame is one in which Newton’s law of inertia is valid.

4.3 Newton’s Second Law of Motion/4.4 The Vector Nature of Newton’s Second Law of Motion Newton’s second law of motion states that when a net force ΣF

→ acts on an object of mass m, the acceleration a→ of the object can

be obtained from Equation 4.1. This is a vector equation and, for motion in

two dimensions, is equivalent to Equations 4.2a and 4.2b. In these equations

the x and y subscripts refer to the scalar components of the force and acceler- ation vectors. The SI unit of force is the newton (N).

Σ F →

= ma→ (4.1) ΣFx = max (4.2a) ΣFy = may (4.2b)

When determining the net force, a free-body diagram is helpful. A free-

body diagram is a diagram that represents the object and the forces acting on it.

4.5 Newton’s Third Law of Motion Newton’s third law of motion, often called the action–reaction law, states that whenever one object exerts a force

on a second object, the second object exerts an oppositely directed force of

equal magnitude on the fi rst object.

4.6 Types of Forces: An Overview Only three fundamental forces have been discovered: the gravitational force, the strong nuclear force, and the

electroweak force. The electroweak force manifests itself as either the elec-

tromagnetic force or the weak nuclear force.

4.7 The Gravitational Force Newton’s law of universal gravitation states that every particle in the universe exerts an attractive force on every other

112 CHAPTER 4 Forces and Newton’s Laws of Motion

particle. For two particles that are separated by a distance r and have masses m1 and m2, the law states that the magnitude of this attractive force is as given in Equation 4.3. The direction of this force lies along the line between the

particles. The constant G has a value of G = 6.674 × 10−11 N · m2/kg2 and is called the universal gravitational constant.

F = G m1m2

r2 (4.3)

The weight W of an object on or above the earth is the gravitational force that the earth exerts on the object and can be calculated from the mass m of the object and the magnitude g of the acceleration due to the earth’s gravity, according to Equation 4.5.

W = mg (4.5)

4.8 The Normal Force The normal force FN →

is one component of the force

that a surface exerts on an object with which it is in contact—namely, the

component that is perpendicular to the surface.

Apparent weight = mg + ma (4.6)

The apparent weight is the force that an object exerts on the platform of a

scale and may be larger or smaller than the true weight mg if the object and the scale have an acceleration a (+ if upward, − if downward). The apparent weight is given by Equation 4.6.

4.9 Static and Kinetic Frictional Forces A surface exerts a force on an object with which it is in contact. The component of the force perpendicular

to the surface is called the normal force. The component parallel to the surface

is called friction.

The force of static friction between two surfaces opposes any impending

relative motion of the surfaces. The magnitude of the static frictional force

depends on the magnitude of the applied force and can assume any value up

to the maximum specifi ed in Equation 4.7, where μs is the coeffi cient of static friction and FN is the magnitude of the normal force.

f sMAX = μs FN (4.7)

The force of kinetic friction between two surfaces sliding against one an-

other opposes the relative motion of the surfaces. This force has a magnitude

given by Equation 4.8, where mk is the coeffi cient of kinetic friction.

f k = μkFN (4.8)

4.10 The Tension Force The word “tension” is commonly used to mean the tendency of a rope to be pulled apart due to forces that are applied at each

end. Because of tension, a rope transmits a force from one end to the other.

When a rope is accelerating, the force is transmitted undiminished only if the

rope is massless.

4.11 Equilibrium Applications of Newton’s Laws of Motion An object is in equilibrium when the object has zero acceleration, or, in other words,

when it moves at a constant velocity. The constant velocity may be zero, in

which case the object is stationary. The sum of the forces that act on an object

in equilibrium is zero. Under equilibrium conditions in two dimensions, the

separate sums of the force components in the x direction and in the y direction must each be zero, as in Equations 4.9a and 4.9b.

ΣFx = 0 (4.9a) ΣFy = 0 (4.9b)

4.12 Nonequilibrium Applications of Newton’s Laws of Motion If an object is not in equilibrium, then Newton’s second law, as expressed in Equa-

tions 4.2a and 4.2b, must be used to account for the acceleration.

ΣFx = max (4.2a) ΣFy = may (4.2b)

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 4.2 Newton’s First Law of Motion 1. An object is moving at a constant velocity. All but one of the following statements could be true. Which one cannot be true? (a) No forces act on the object. (b) A single force acts on the object. (c) Two forces act simultan- eously on the object. (d) Three forces act simultaneously on the object. 3. A cup of coff ee is sitting on a table in a recreational vehicle (RV). The cup slides toward the rear of the RV. According to Newton’s fi rst law, which one

or more of the following statements could describe the motion of the RV?

(A) The RV is at rest, and the driver suddenly accelerates. (B) The RV is

moving forward, and the driver suddenly accelerates. (C) The RV is moving

backward, and the driver suddenly hits the brakes. (a) A (b) B (c) C (d) A and B (e) A, B, and C

Section 4.4 The Vector Nature of Newton’s Second Law of Motion 5. Two forces act on a moving object that has a mass of 27 kg. One force has a magnitude of 12 N and points due south, while the other force has

a magnitude of 17 N and points due west. What is the acceleration of the

object? (a) 0.63 m/s2 directed 55° south of west (b) 0.44 m/s2 directed 24° south of west (c) 0.77 m/s2 directed 35° south of west (d) 0.77 m/s2 directed 55° south of west (e) 1.1 m/s2 directed 35° south of west

Section 4.5 Newton’s Third Law of Motion 7. Which one of the following is true, according to Newton’s laws of motion? Ignore friction. (a) A sports utility vehicle (SUV) hits a stationary motorcycle. Since it is stationary, the motorcycle sustains a greater force

than the SUV does. (b) A semitrailer truck crashes all the way through a wall. Since the wall collapses, the wall sustains a greater force than the

truck does. (c) Sam (18 years old) and his sister (9 years old) go ice skating. They push off against each other and fl y apart. Sam fl ies off with the greater

acceleration. (d) Two astronauts on a space walk are throwing a ball back and forth between each other. In this game of catch the distance between

them remains constant. (e) None of the above is true, according to the third law.

8. Two ice skaters, Paul and Tom, are each holding on to opposite ends of the same rope. Each pulls the other toward him. The magnitude of Paul’s

acceleration is 1.25 times greater than the magnitude of Tom’s acceleration.

What is the ratio of Paul’s mass to Tom’s mass? (a) 0.67 (b) 0.80 (c) 0.25 (d) 1.25 (e) 0.50

Focus on Concepts

Focus on Concepts 113

Section 4.7 The Gravitational Force 9. In another solar system a planet has twice the earth’s mass and three times the earth’s radius. Your weight on this planet is ________ times your earth-

weight. Assume that the masses of the earth and of the other planet are uni-

formly distributed. (a) 0.667 (b) 2.000 (c) 0.111 (d) 0.444 (e) 0.222 11. What is the mass on Mercury of an object that weighs 784 N on the earth’s surface? (a) 80.0 kg (b) 48.0 kg (c) 118 kg (d) 26.0 kg (e) There is not enough information to calculate the mass.

Section 4.8 The Normal Force 12. The apparent weight of a passenger in an elevator is greater than his true weight. Which one of the following is true? (a) The elevator is either mov- ing upward with an increasing speed or moving upward with a decreasing

speed. (b) The elevator is either moving upward with an increasing speed or moving downward with an increasing speed. (c) The elevator is either moving upward with a decreasing speed or moving downward with a de-

creasing speed. (d) The elevator is either moving upward with an increasing speed or moving downward with a decreasing speed. (e) The elevator is either moving upward with a decreasing speed or moving downward with

an increasing speed.

13. The drawings show three examples of the force with which

someone pushes against a vertical

wall. In each case the magnitude

of the pushing force is the same.

Rank the normal forces that the

wall applies to the pusher in as-

cending order (smallest fi rst). (a) C, B, A (b) B, A, C (c) A, C, B (d) B, C, A (e) C, A, B

Section 4.9 Static and Kinetic Frictional Forces 15. The drawing shows three blocks, each with the same mass, stacked one upon the other. The bottom block rests on a frictionless horizontal surface

and is being pulled by a force F→ that is parallel to this surface. The sur- faces where the blocks touch each other have identical coeffi cients of static

friction. Which one of the following correctly de-

scribes the magnitude of the net force of static friction fs that acts on each block?

(a) f s,A = f s,B = fs,C (b) f s,A = f s,B = 12 fs,C (c) fs,A = 0 and fs,B = 12 fs,C (d) fs,C = 0 and fs,A = 12 fs,B (e) fs,A = f s,C = 12 fs,B

16. Three identical blocks are being pulled or pushed across a horizontal surface by a force F→, as shown in the drawings. The force F→ in each case has the same magnitude. Rank the kinetic frictional forces that act on the blocks

in ascending order (smallest fi rst).

(a) B, C, A (b) C, A, B (c) B, A, C (d) C, B, A (e) A, C, B

A

F

FF

B C

QUESTION 16

Section 4.10 The Tension Force 18. A heavy block is suspended from a ceiling using pulleys in three diff er- ent ways, as shown in the drawings. Rank the tension in the rope that passes

over the pulleys in ascending order (smallest fi rst).

(a) B, A, C (b) C, B, A (c) A, B, C (d) C, A, B (e) B, C, A

CBA

QUESTION 18

Section 4.11 Equilibrium Applications of Newton’s Laws of Motion 20. A certain object is in equilibrium. Which one of the following state- ments is not true? (a) The object must be at rest. (b) The object has a con- stant velocity. (c) The object has no acceleration. (d) No net force acts on the object.

23. Two identical boxes are being pulled across a horizontal fl oor at a con- stant velocity by a horizontal pulling force of 176 N that is applied to one

of the boxes, as the drawing shows. There is kinetic friction between each

box and the fl oor. Find the tension in the rope between the boxes. (a) 176 N (b) 88.0 N (c) 132 N (d) 44.0 N (e) There is not enough information to calculate the tension.

Section 4.12 Nonequilibrium Applications of Newton’s Laws of Motion 25. A man is standing on a platform that is connected to a pulley arrange- ment, as the drawing shows. By pulling upward on the rope with a force P→ the man can raise the platform and himself. The total mass of the man plus

the platform is 94.0 kg. What pulling force should the man apply to create an

upward acceleration of 1.20 m/s2?

A B C

QUESTION 13

A

B

C F

QUESTION 15

176 N

QUESTION 23

P

Platform

Ceiling

QUESTION 25

114 CHAPTER 4 Forces and Newton’s Laws of Motion

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual

MMH Problem-solving help

GO Guided Online Tutorial

V-HINT Video Hints

CHALK Chalkboard Videos

BIO Biomedical application

E Easy

M Medium

H Hard

Section 4.3 Newton’s Second Law of Motion 1. E An airplane has a mass of 3.1 × 104 kg and takes off under the infl u- ence of a constant net force of 3.7 × 104 N. What is the net force that acts on

the plane’s 78-kg pilot?

2. E MMH A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind exerts a force of 800 N due east and the

water exerts a resistive force of 1200 N due east. What are the magnitude and

direction of the boat’s acceleration?

3. E GO Two horizontal forces, F1 →

and F2 →

, are acting on a box, but only F1 →

is shown in the drawing. F2 →

can point either to the right or to the left. The

box moves only along the x axis. There is no friction between the box and the surface. Suppose that F1

→ = +9.0 N and the mass of the box is 3.0 kg.

Find the magnitude and direction of F2 →

when the acceleration of the box is (a) +5.0 m/s2, (b) −5.0 m/s2, and (c) 0 m/s2.

PROBLEM 3

+x F1

4. E In the amusement park ride known as Magic Mountain Superman, power- ful magnets accelerate a car and its riders from rest to 45 m/s (about 100 mi/h)

in a time of 7.0 s. The combined mass of the car and riders is 5.5 × 103 kg.

Find the average net force exerted on the car and riders by the magnets.

5. E SSM A person in a kayak starts paddling, and it accelerates from 0 to 0.60 m/s in a distance of 0.41 m. If the combined mass of the person and the

kayak is 73 kg, what is the magnitude of the net force acting on the kayak?

6. E Scientists are experimenting with a kind of gun that may eventually be used to fi re payloads directly into orbit. In one test, this gun accelerates a 5.0-kg

projectile from rest to a speed of 4.0 × 103 m/s. The net force accelerating

the projectile is 4.9 × 105 N. How much time is required for the projectile to

come up to speed?

7. E SSM MMH A 1580-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to

a halt in a distance of 50.0 m?

8. E GO The space probe Deep Space 1 was launched on October 24, 1998. Its mass was 474 kg. The goal of the mission was to test a new kind of engine

called an ion propulsion drive. This engine generated only a weak thrust, but

it could do so over long periods of time with the consumption of only small

amounts of fuel. The mission was spectacularly successful. At a thrust of

56 mN how many days were required for the probe to attain a velocity of

805 m/s (1800 mi/h), assuming that the probe started from rest and that the

mass remained nearly constant?

9. M SSM Two forces FA →

and FB →

are applied to an object whose mass is

8.0 kg. The larger force is FA →

. When both forces point due east, the object’s

acceleration has a magnitude of 0.50 m/s2. However, when FA →

points due east

and FB →

points due west, the acceleration is 0.40 m/s2, due east. Find (a) the magnitude of FA

→ and (b) the magnitude of FB

→ .

10. M V-HINT An electron is a subatomic particle (m = 9.11 × 10−31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +5.40 × 105 m/s to a fi nal velocity of +2.10 × 106 m/s

while traveling a distance of 0.038 m. The electron’s acceleration is due

to two electric forces parallel to the x axis: F1 →

= +7.50 × 10−17 N, and F2 →

,

which points in the −x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F2

→ .

Section 4.4 The Vector Nature of Newton’s Second Law of Motion/Section 4.5 Newton’s Third Law of Motion 11. E Only two forces act on an object (mass = 3.00 kg), as in the drawing. Find the magnitude and

direction (relative to the x axis) of the acceleration of the object.

12. E At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or

x component of the ball’s acceleration is 810 m/s2 and the vertical or y component of its acceleration is 1100 m/s2. The ball’s mass is 0.43 kg. What is the

magnitude of the net force acting on the soccer ball

at this instant?

13. E SSM MMH A rocket of mass 4.50 × 105 kg is in fl ight. Its thrust is directed at an angle of 55.0° above the horizontal and has a magnitude of

7.50 × 106 N. Find the magnitude and direction of the rocket’s acceleration.

Give the direction as an angle above the horizontal.

14. E GO A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg. The ball approaches

the cushion with a velocity of +2.1 m/s and rebounds with a velocity of

−2.0 m/s. The ball remains in contact with the cushion for a time of

3.3 × 10−3 s. What is the average net force (magnitude and direction) exerted

on the ball by the cushion?

15. E When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus,

slows him down. Suppose the weight of the sky diver is 915 N and the drag

force has a magnitude of 1027 N. The mass of the sky diver is 93.4 kg. What

are the magnitude and direction of his acceleration?

16. E GO Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the man is 82

kg, and the mass of the woman is 48 kg. The woman pushes on the man with

a force of 45 N due east. Determine the acceleration (magnitude and direc-

tion) of (a) the man and (b) the woman. 17. M V-HINT A space probe has two engines. Each generates the same amount of force when fi red, and the directions of these forces can be in-

dependently adjusted. When the engines are fi red simultaneously and each

applies its force in the same direction, the probe, starting from rest, takes 28 s

to travel a certain distance. How long does it take to travel the same distance,

again starting from rest, if the engines are fi red simultaneously and the forces

that they apply to the probe are perpendicular?

18. H At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500-kg space tug and a 6200-kg asteroid. Us-

ing their tug’s engine, they pull on the asteroid with a force of 490 N. Initially

the tug and the asteroid are at rest, 450 m apart. How much time does it take

for the tug and the asteroid to meet?

Problems

60.0 N

40.0 N

45.0°

+x

+y

PROBLEM 11

Problems 115

19. H SSM A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s.

During this time, three forces act on the boat: a 31.0-N force directed 15.0°

north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° south

of west (resistance due to the water), and FW →

(due to the wind). Find the

magnitude and direction of the force FW →

. Express the direction as an angle

with respect to due east.

Section 4.7 The Gravitational Force 20. E GO A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the sur- face of the earth. (a) Determine the magnitude of the gravitational force ex- erted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.

21. E Mars has a mass of 6.46 × 1023 kg and a radius of 3.39 × 106 m. (a) What is the acceleration due to gravity on Mars? (b) How much would a 65- kg person weigh on this planet?

22. E On earth, two parts of a space probe weigh 11 000 N and 3400 N. These parts are separated by a center-to-center distance of 12 m and may be treated

as uniform spherical objects. Find the magnitude of the gravitational force that

each part exerts on the other out in space, far from any other objects.

23. E GO A raindrop has a mass of 5.2 × 10−7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop. 24. E The weight of an object is the same on two diff erent planets. The mass of planet A is only sixty percent that of planet B. Find the ratio rA/rB of the radii of the planets.

25. E SSM A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

spheres. What is the magnitude of the maximum gravitational force that each

can exert on the other?

26. E Review Conceptual Example 7 in preparation for this problem. In tests on earth a lunar surface exploration vehicle (mass = 5.90 × 103 kg) achieves

a forward acceleration of 0.220 m/s2. To achieve this same acceleration on the

moon, the vehicle’s engines must produce a drive force of 1.43 × 103 N. What

is the magnitude of the frictional force that acts on the vehicle on the moon?

27. E SSM Synchronous communications satellites are placed in a circular orbit that is 3.59 × 107 m above the surface of the earth. What is the mag-

nitude of the acceleration due to gravity at this distance?

28. E The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force FSM

→ that the sun exerts on the moon is perpen-

dicular to the force FEM →

that the earth exerts on the moon. The masses are:

mass of sun = 1.99 × 1030 kg, mass of earth = 5.98 × 1024 kg, mass of moon =

7.35 × 1022 kg. The distances shown in the drawing are rSM = 1.50 × 1011 m and rEM = 3.85 × 108 m. Determine the magnitude of the net gravitational force on the moon.

Sun

Earth

Moon

rEM

rSM

FSM

FEM

PROBLEM 28

29. E The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 363 kg,

mB = 517 kg, and mC = 154 kg. Find the magnitude and direction of the net gravitational force acting on (a) particle A, (b) particle B, and (c) particle C.

PROBLEM 29 0.500 m

A B C

0.250 m

30. E A space traveler weighs 540.0 N on earth. What will the traveler weigh on another planet whose radius is twice that of earth and whose mass

is three times that of earth?

31. E SSM Available in WileyPLUS. 32. M CHALK A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted

on the spacecraft by the earth balance that exerted by the moon? This point

lies on a line between the centers of the earth and the moon. The distance

between the earth and the moon is 3.85 × 108 m, and the mass of the earth is

81.4 times as great as that of the moon.

33. M V-HINT Available in WileyPLUS. 34. M GO A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain).

Suppose an object falls from rest near the surface of such a star. How fast

would this object be moving after it had fallen a distance of 0.010 m?

(Assume that the gravitational force is constant over the distance of the fall

and that the star is not rotating.)

35. M SSM The sun is more massive than the moon, but the sun is farther from the earth. Which one exerts a greater gravitational force on a person

standing on the earth? Give your answer by determining the ratio Fsun/Fmoon of the magnitudes of the gravitational forces. Use the data on the inside of

the front cover.

36. M Available in WileyPLUS. 37. H Two particles are located on the x axis. Particle 1 has a mass m and is at the origin. Particle 2 has a mass 2m and is at x = +L. A third particle is placed between particles 1 and 2. Where on the x axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? Express your answer in terms of L.

Section 4.8 The Normal Force/Section 4.9 Static and Kinetic Frictional Forces 38. E A 35-kg crate rests on a horizontal fl oor, and a 65-kg person is stand- ing on the crate. Determine the magnitude of the normal force that (a) the fl oor exerts on the crate and (b) the crate exerts on the person. 39. E SSM A 60.0-kg crate rests on a level fl oor at a shipping dock. The coeffi cients of static and kinetic friction are 0.760 and 0.410, respectively.

What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed? 40. E A rocket blasts off from rest and attains a speed of 45 m/s in 15 s. An astronaut has a mass of 57 kg. What is the astronaut’s apparent weight

during takeoff ?

41. E GO A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the normal force to the

weight of the car when (a) θ = 15° and (b) θ = 35°. 42. E A woman stands on a scale in a moving elevator. Her mass is 60.0 kg, and the combined mass of the elevator and scale is an additional 815 kg.

Starting from rest, the elevator accelerates upward. During the acceleration,

the hoisting cable applies a force of 9410 N. What does the scale read during

the acceleration?

43. E A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

116 CHAPTER 4 Forces and Newton’s Laws of Motion

directed due west. What must be the magnitude and direction of the third

force, such that the object continues to move with a constant velocity?

54. E Available in WileyPLUS. 55. E BIO Available in WileyPLUS. 56. E Available in WileyPLUS. 57. E SSM A worker stands still on a roof sloped at an angle of 36° above the horizontal. He is prevented from slipping by a static frictional force of 390 N.

Find the mass of the worker.

58. E A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The

mass of the stuntman is 109 kg, and the coeffi cient of kinetic friction between

the road and him is 0.870. Find the tension in the cable.

59. E GO Available in WileyPLUS. 60. E GO The drawing shows a circus clown who weighs 890 N. The coeffi cient of static friction between

the clown’s feet and the ground is 0.53. He pulls vertic-

ally downward on a rope that passes around three pulleys

and is tied around his feet. What is the minimum pulling

force that the clown must exert to yank his feet out from

under himself?

61. M V-HINT Available in WileyPLUS. 62. M V-HINT MMH During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a

point that is not in the middle between two fence posts. The limb exerts a

downward force of 151 N on the wire. The left section of the wire makes an

angle of 14.0° relative to the horizontal and sustains a tension of 447 N. Find

the magnitude and direction of the tension that the right section of the wire

sustains.

63. M SSM Available in WileyPLUS. 64. M GO Available in WileyPLUS. 65. M GO MMH A toboggan slides down a hill and has a constant velocity. The angle of the hill is 8.00° with respect to the horizontal. What is the coeffi cient of

kinetic friction between the surface of the hill and the toboggan?

66. M BIO V-HINT The person in the drawing is standing on crutches. Assume that the force

exerted on each crutch by the ground is dir-

ected along the crutch, as the force vectors in

the drawing indicate. If the coeffi cient of static

friction between a crutch and the ground is

0.90, determine the largest angle θ MAX that the crutch can have just before it begins to slip on

the fl oor.

67. H SSM A bicyclist is coasting straight down a hill at a constant speed. The combined

mass of the rider and bicycle is 80.0 kg, and the

hill is inclined at 15.0° with respect to the ho-

rizontal. Air resistance opposes the motion of

the cyclist. Later, the bicyclist climbs the same

hill at the same constant speed. How much force

(directed parallel to the hill) must be applied to

the bicycle in order for the bicyclist to climb the

hill?

68. H MMH A kite is hovering over the ground at the end of a straight 43-m line. The tension in the line has a magnitude of 16 N. Wind blowing on the

kite exerts a force of 19 N, directed 56° above the horizontal. Note that the line

attached to the kite is not oriented at an angle of 56° above the horizontal. Find the height of the kite, relative to the person holding the line.

69. H Available in WileyPLUS.

44. E GO Consult Multiple-Concept Example 9 to explore a model for solving this problem. A person pushes on a 57-kg refrigerator with a hori-

zontal force of −267 N; the minus sign indicates that the force points in the

−x direction. The coeffi cient of static friction is 0.65. (a) If the refrigerator does not move, what are the magnitude and direction of the static frictional

force that the fl oor exerts on the refrigerator? (b) What is the magnitude of the largest pushing force that can be applied to the refrigerator before it just

begins to move?

45. E SSM A 6.00-kg box is sliding across the horizontal fl oor of an el- evator. The coeffi cient of kinetic friction between the box and the fl oor is

0.360. Determine the kinetic frictional force that acts on the box when the

elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 1.20 m/s2, and (c) accelerating downward with an acceleration whose magnitude is 1.20 m/s2.

46. E GO A cup of coff ee is on a table in an airplane fl ying at a constant altitude and a constant velocity. The coeffi cient of static friction between the

cup and the table is 0.30. Suddenly, the plane accelerates forward, its altitude

remaining constant. What is the maximum acceleration that the plane can

have without the cup sliding backward on the table?

47. E MMH An 81-kg baseball player slides into second base. The coeffi - cient of kinetic friction between the player and the ground is 0.49. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.6 s, what was his initial velocity?

48. M V-HINT Consult Multiple-Concept Example 10 in preparation for this problem. Traveling at a speed of 16.1 m/s, the driver of an automobile sud-

denly locks the wheels by slamming on the brakes. The coeffi cient of kinetic

friction between the tires and the road is 0.720. What is the speed of the auto-

mobile after 1.30 s have elapsed? Ignore the eff ects of air resistance.

49. M SSM A person is trying to judge whether a picture (mass = 1.10 kg) is properly positioned by temporarily pressing it against a wall. The pressing

force is perpendicular to the wall. The coeffi cient of static friction between

the picture and the wall is 0.660. What is the minimum amount of pressing

force that must be used?

50. M V-HINT Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical down- force, which helps the cars hug the track more securely. The coeffi cient of static friction between the track and the tires of a 690-kg race car is 0.87.

What is the magnitude of the maximum acceleration at which the car can

speed up without its tires slipping when a 4060-N downforce and an 1190-N

horizontal-air-resistance force act on it?

51. H While moving in, a new homeowner is pushing a box across the fl oor at a constant velocity. The coeffi cient of kinetic friction between the box and

the fl oor is 0.41. The pushing force is directed downward at an angle θ below the horizontal. When θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of θ.

Section 4.10 The Tension Force/ Section 4.11 Equilibrium Applications of Newton’s Laws of Motion 52. E MMH The helicopter in the drawing is moving horizontally to the right at a constant velocity v→. The weight of the helicopter is W = 53 800 N. The lift force L→ generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the mag- nitude of the lift force? (b) Determine the magnitude of the air resistance R→ that opposes the motion. 53. E SSM Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed

due north. Another has a magnitude of 60.0 N and is

L

R

W

21.0°

v

PROBLEM 52

F

θ

PROBLEM 66

PROBLEM 60

Problems 117

Section 4.12 Nonequilibrium Applications of Newton’s Laws of Motion 70. E A 1450-kg submarine rises straight up toward the surface. Seawater exerts both an upward buoyant force of 16 140 N on the submarine and a down-

ward resistive force of 1030 N. What is the submarine’s acceleration?

71. E SSM A 15-g bullet is fi red from a rifl e. It takes 2.50 × 10−3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of

715 m/s. Assuming that the acceleration of the bullet is constant, fi nd the

average net force exerted on the bullet.

72. E A fi sherman is fi shing from a bridge and is using a “45-N test line.” In other words, the line will sustain a maximum force of 45 N without break-

ing. What is the weight of the heaviest fi sh that can be pulled up vertically

when the line is reeled in (a) at a constant speed and (b) with an acceleration whose magnitude is 2.0 m/s2?

73. E SSM Only two forces act on an object (mass = 4.00 kg), as in the drawing. Find the magnitude and

direction (relative to the x axis) of the acceleration of the object.

74. E A helicopter fl ies over the arctic ice pack at a constant altitude, towing an airborne 129-kg laser

sensor that measures the thickness of the ice (see the

drawing). The helicopter and the sensor both move

only in the horizontal direction and have a horizontal

acceleration of magnitude 2.84 m/s2. Ignoring air

resistance, fi nd the tension in the cable towing the

sensor.

PROBLEM 74

75. E Review Conceptual Example 16 as background for this problem. The water skier there has a mass of 73 kg. Find the magnitude of the net force

acting on the skier when (a) she is accelerated from rest to a speed of 11 m/s in 8.0 s and (b) she lets go of the tow rope and glides to a halt in 21 s. 76. E A rescue helicopter is lifting a man (weight = 822 N) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10 m/s2? (b) What is the tension during the remainder of the rescue when he is pulled upward

at a constant velocity?

77. E A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +11 m/s in a time of 28 s. The combined mass of

the boat and trailer is 410 kg. The frictional force acting on the trailer can be

ignored. What is the tension in the hitch that connects the trailer to the car?

78. E GO A 292-kg motorcycle is accelerating up along a ramp that is in- clined 30.0° above the horizontal. The propulsion force pushing the motor-

cycle up the ramp is 3150 N, and air resistance produces a force of 250 N that

opposes the motion. Find the magnitude of the motorcycle’s acceleration.

79. E SSM A student is skateboarding down a ramp that is 6.0 m long and inclined at 18° with respect to the horizontal. The initial speed of the skate-

boarder at the top of the ramp is 2.6 m/s. Neglect friction and fi nd the speed

at the bottom of the ramp.

80. E A man seeking to set a world record wants to tow a 109 000-kg air- plane along a runway by pulling horizontally on a cable attached to the air-

plane. The mass of the man is 85 kg, and the coeffi cient of static friction

between his shoes and the runway is 0.77. What is the greatest acceleration

the man can give the airplane? Assume that the airplane is on wheels that

turn without any frictional resistance.

81. M V-HINT A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0° with respect

to the horizontal. The coeffi cient of kinetic friction between the log and the

ramp is 0.900, and the log has an acceleration of magnitude 0.800 m/s2. Find

the tension in the rope.

82. M GO To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree.

He then pulls downward on the free end of the rope with a force of 358 N.

Neglect any friction between the rope and the branch, and determine the

man’s upward acceleration.

83. M CHALK SSM Two objects (45.0 and 21.0 kg) are connected by a mass- less string that passes over a massless, frictionless pulley. The pulley hangs

from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.

84. M GO A train consists of 50 cars, each of which has a mass of 6.8 × 103 kg. The train has an acceleration of +8.0 × 10−2 m/s2. Ignore friction and

determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars. 85. M In Problem 80, an 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead

attempts the feat by pulling the cable at an angle of 9.0° above the horizontal. The

coeffi cient of static friction between his shoes and the runway is 0.77. What is the

greatest acceleration the man can give the airplane? Assume that the airplane is

on wheels that turn without any frictional resistance.

86. M MMH The drawing shows a large cube (mass = 25 kg) being ac- celerated across a horizontal frictionless surface by a horizontal force P→ . A small cube (mass = 4.0 kg) is in contact with the front surface of the large

cube and will slide downward unless P→ is suffi ciently large. The coeffi cient of static friction between the cubes is 0.71. What is the smallest magnitude that P→ can have in order to keep the small cube from sliding downward?

PROBLEM 86

Frictionless

P

μ s = 0.71

87. M V-HINT The alarm at a fi re station rings and an 86-kg fi reman, starting from rest, slides down a pole to the fl oor below (a distance of 4.0 m). Just

before landing, his speed is 1.4 m/s. What is the magnitude of the kinetic

frictional force exerted on the fi reman as he slides down the pole?

88. M GO Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.0 kg. In Case B the

mass of block 1 (the block behind) is 6.0 kg, and the mass of block 2 is 3.0 kg.

No frictional force acts on block 1 in either Case A or Case B. However, a

kinetic frictional force of 5.8 N does act on block 2 in both cases and opposes

the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.

Block 1 Block 2Block 1

Case A Case B

m1 = 3.0 kg m2 = 3.0 kg m1 = 6.0 kg m2 = 3.0 kg

Block 2

PROBLEM 88

60.0 N

40.0 N

+y

+x

PROBLEM 73

118 CHAPTER 4 Forces and Newton’s Laws of Motion

89. M SSM A person whose weight is 5.20 × 102 N is being pulled up vertic- ally by a rope from the bottom of a cave that is 35.1 m deep. The maximum ten-

sion that the rope can withstand without breaking is 569 N. What is the shortest

time, starting from rest, in which the person can be brought out of the cave?

90. M CHALK A girl is sledding down a slope that is inclined at 30.0° with respect to the horizontal. The wind is aiding the motion by providing a steady

force of 105 N that is parallel to the motion of the sled. The combined mass of

the girl and the sled is 65.0 kg, and the coeffi cient of kinetic friction between

the snow and the runners of the sled is 0.150. How much time is required for the

sled to travel down a 175-m slope, starting from rest?

91. H Available in WileyPLUS. 92. H A small sphere is hung by a string from the ceiling of a van. When the van is stationary, the sphere hangs vertically. However, when the van

accelerates, the sphere swings backward so that the string makes an angle

of θ with respect to the vertical. (a) Derive an expression for the magnitude a of the acceleration of the van in terms of the angle θ and the magnitude g of the acceleration due to gravity. (b) Find the acceleration of the van when θ = 10.0°. (c) What is the angle θ when the van moves with a constant velocity?

93. H SSM A penguin slides at a constant velocity of 1.4 m/s down an icy incline. The incline slopes above the horizontal at an angle of 6.9°. At the

bottom of the incline, the penguin slides onto a horizontal patch of ice. The

coeffi cient of kinetic friction between the penguin and the ice is the same for

the incline as for the horizontal patch. How much time is required for the

penguin to slide to a halt after entering the horizontal patch of ice?

94. H Available in WileyPLUS.

95. E SSM Available in WileyPLUS. 96. E GO A person with a black belt in karate has a fi st that has a mass of 0.70 kg. Starting from rest, this fi st attains a velocity of 8.0 m/s in 0.15 s. What is the

magnitude of the average net force applied to the fi st to achieve this level of

performance?

97. E SSM A 95.0-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80 m/s2, (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.30 m/s2? 98. E MMH Two forces, F1

→ and F2

→ , act on the 7.00-kg block shown in the

drawing. The magnitudes of the forces are F1 = 59.0 N and F2 = 33.0 N. What is the horizontal acceleration (magnitude and direction) of the block?

PROBLEM 98

F1

F2

70.0°

7.00 kg

99. E SSM A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the

front and back covers of the book are perpendicular to the

book and are horizontal. The book weighs 31 N. The coef-

fi cient of static friction between his hands and the book

is 0.40. To keep the book from falling, what is the mag-

nitude of the minimum pressing force that each hand

must exert?

100. E MMH The speed of a bobsled is increasing be- cause it has an acceleration of 2.4 m/s2. At a given instant in time, the forces

resisting the motion, including kinetic friction and air resistance, total 450

N. The combined mass of the bobsled and its riders is 270 kg. (a) What is the magnitude of the force propelling the bobsled forward? (b) What is the magnitude of the net force that acts on the bobsled?

101. E Available in WileyPLUS. 102. E In a European country a bathroom scale displays its reading in kilo- grams. When a man stands on this scale, it reads 92.6 kg. When he pulls down

on a chin-up bar installed over the scale, the reading decreases to 75.1 kg.

What is the magnitude of the force he exerts on the chin-up bar?

103. E SSM A 1380-kg car is moving due east with an initial speed of 27.0 m/s. After 8.00 s the car has slowed down to 17.0 m/s. Find the mag-

nitude and direction of the net force that produces the deceleration.

104. E When a 58-g tennis ball is served, it accelerates from rest to a speed of 45 m/s. The impact with the racket gives the ball a constant acceleration

over a distance of 44 cm. What is the magnitude of the net force acting on the

ball?

105. E In preparation for this problem, review Conceptual Example 7. A space traveler whose mass is 115 kg leaves earth. What are his weight and mass (a) on earth and (b) in interplanetary space where there are no nearby plan- etary objects?

106. E (a) Calculate the magnitude of the gravitational force exerted on a 425-kg satellite that is a distance of two earth radii from the center of the

earth. (b) What is the magnitude of the gravitational force exerted on the earth by the satellite? (c) Determine the magnitude of the satellite’s acceleration. (d) What is the magnitude of the earth’s acceleration? 107. M SSM MMH The drawing shows Robin Hood (mass = 77.0 kg) about to es-

cape from a dangerous situation. With one

hand, he is gripping the rope that holds up a

chandelier (mass = 195 kg). When he cuts

the rope where it is tied to the fl oor, the

chandelier will fall, and he will be pulled

up toward a balcony above. Ignore the fric-

tion between the rope and the beams over

which it slides, and fi nd (a) the accelera- tion with which Robin is pulled upward

and (b) the tension in the rope while Robin escapes.

108. M GO A skater with an initial speed of 7.60 m/s stops propelling himself and

begins to coast across the ice, eventually

coming to rest. Air resistance is negligible. (a) The coeffi cient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

109. M V-HINT Available in WileyPLUS. 110. M GO A mountain climber, in the process of crossing between two cliff s by a rope, pauses to rest. She weighs 535 N. As the drawing shows, she

is closer to the left cliff than to the right cliff , with the result that the tensions

Additional Problems

PROBLEM 99

PROBLEM 107

Concepts and Calculations Problems 119

in the left and right sides of the rope are not the same. Find the tensions in the

rope to the left and to the right of the mountain climber.

65.0° 80.0°

PROBLEM 110

111. M SSM Available in WileyPLUS. 112. M V-HINT Consult Multiple-Concept Example 10 for insight into solv- ing this type of problem. A box is sliding up an incline that makes an angle

of 15.0° with respect to the horizontal. The coeffi cient of kinetic friction

between the box and the surface of the incline is 0.180. The initial speed of

the box at the bottom of the incline is 1.50 m/s. How far does the box travel

along the incline before coming to rest?

113. M V-HINT Available in WileyPLUS. 114. M V-HINT Available in WileyPLUS. 115. M SSM Available in WileyPLUS. 116. H As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412 N, and

the other has a weight of 908 N. The rope and the pulleys are massless and

there is no friction. (a) What is the acceleration of the lighter block? (b) Sup- pose that the heavier block is removed, and a downward force of 908 N is

provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block. (c) Explain why the answers in (a) and (b) are diff erent.

412 N 412 N 908 N

(a) (b)

908 N

PROBLEM 116

117. H SSM The three objects in the drawing are connected by strings that pass over massless and friction-free pulleys. The objects move, and the coef-

fi cient of kinetic friction between the middle object and the surface of the

table is 0.100. (a) What is the acceleration of the three objects? (b) Find the tension in each of the two strings.

PROBLEM 117

25.0 kg

80.0 kg

10.0 kg

118. H Available in WileyPLUS. 119. H Available in WileyPLUS. 120. M GO A fl atbed truck is carrying a crate up a hill of angle of inclin- ation θ = 10.0°, as the fi gure illustrates. The coeffi cient of static friction between the truck bed and the crate is μs = 0.350. Find the maximum acceleration that the truck can attain before the crate begins to slip backward

relative to the truck.

FN

W = mg

fs MAX

= 10.0° = 10.0°θ

θ

(a)

= 10.0°

mg cos

FN

fs MAX

θ

θ

W = mg

mg sin

(b) Free-body diagram of the crate

+x

+y

θ

PROBLEM 120

Newton’s three laws of motion provide the basis for understanding the eff ect

of forces on the motion of an object. The second law is especially important,

because it provides the quantitative relationship between net force and accel-

eration. The following problems serve as a review of the essential features of

this relationship.

121. M CHALK SSM The fi gure shows two forces, F→1 = +3000 N and F →

2 =

+5000 N, acting on a spacecraft; the plus signs indicate that the forces are

directed along the +x axis. A third force F3 →

also acts on the spacecraft but is not

shown in the drawing. Concepts: (i) Suppose the spacecraft were stationary.

What would be the direction of F3 →

? (ii) When the spacecraft is moving at a

constant velocity of +850 m/s, what is the direction of F3 →

? Calculations: Find the direction and magnitude of F3

→ .

F2

F1 +x

PROBLEM 121

Concepts and Calculations Problems

120 CHAPTER 4 Forces and Newton’s Laws of Motion

122. M CHALK On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on the moon, where the acceleration due to gravity is

1.60 m/s2. As the fi gure shows, the block is being pulled by a horizontal rope

in which the tension is T = 24 N. The coeffi cient of kinetic friction between the block and the surface is μk = 0.20. Concepts: (i) Which of Newton’s laws of motion provides the best way to determine the acceleration of the block?

(ii) Does the net force in the x direction equal the tension T? Calculations: Determine the acceleration of the block.

T

ax

(a) (b) Free-body diagram for the block

+x

+y

FN

fk T

mg moon

PROBLEM 122

123. E Towing a Detector. You are in a helicopter towing a 129-kg laser detector that is mapping out the thickness of the Brunt Ice Shelf along the

coast of Antarctica. The original cable used to suspend the detector was

damaged and replaced by a lighter one with a maximum tension rating of

310 pounds, not much more than the weight of the detector. The replacement

cable would work without question in the case that the detector and heli-

copter were not accelerating. However, some acceleration of the helicopter

is inevitable. In order to monitor the tension force on the cable to make sure

the maximum is not exceeded (and therefore to not lose the very expensive

detector) you calculate the maximum angle the cable can make with the ver-

tical without the cable exceeding the tension limit. (a) Assuming straight and level fl ight of the helicopter, what is that maximum angle? (b) What is the corresponding acceleration? (c) Your colleague wants to add a 10-kg infrared camera to the detector. What is the maximum allowable angle now?

124. E Stuck on a Ledge. You and your friends pull your car off a moun- tain road onto an overlook to take pictures of the frozen lake below. You lose

your footing on the packed snow, fall over the cliff , and land in a soft snow

bank on a ledge, 10 feet below the edge. The cliff ’s wall is too smooth to

climb upwards and too steep and dangerous to climb the remaining 300 feet

to the bottom. While trying to think your way out of your predicament, you

realize one reason you had slipped was that you had misjudged how the cliff

sloped towards the edge (a decline of about 5–10 degrees). There is nothing

near the edge to which your friends can tie a rope, and they fear the footing

is too slippery for them to pull you to the top. However, they can get the car

close enough to the edge to tie the rope to the bumper and hang it over the

edge. You are worried, however, that the car might slip over the edge as well.

On your phone (which has survived the fall) you look up the coeffi cient of

static friction of snow tires on packed snow and fi nd that it ranges from 0.22

to 0.26, and that the model of your friends’ car weighs 3856 lb. Your own

weight fl uctuates between 120–130 lb. (a) What is the worst-case scenario in terms of the parameters given? (b) Assuming the worst-case scenario, should your friends bring the car close to the edge? (c) Next, should you risk tying the rope to the trailer hitch and trying to climb out? You assume the

hitch is low enough to the ground so that the tension force is parallel to the

slope near the edge of the cliff and that you climb up the rope at a constant

velocity.

Team Problems

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

5.1 Define uniform circular motion.

5.2 Solve uniform circular motion kinematic problems.

5.3 Solve uniform circular motion dynamic problems.

5.4 Solve problems involving banked curves.

5.5 Analyze circular gravitational orbits.

5.6 Solve application problems involving gravity and uniform circular motion.

5.7 Analyze vertical circular motion.

© G

ar i W

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ill ia

m s/

A ge

F ot

os to

ck

CHAPTER 5

Dynamics of Uniform Circular Motion

While performing a circular loop-the-loop stunt, the pilot of this small airplane experiences a net force and acceleration that points toward the center of the circle. We will now see how and why this net force and acceleration arise.

5.1 Uniform Circular Motion There are many examples of motion on a circular path. Of the many possibilities, we single out those that satisfy the following defi nition:

DEFINITION OF UNIFORM CIRCULAR MOTION Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path.

As an example of uniform circular motion, Figure 5.1 shows a model airplane on a guideline. The speed of the plane is the magnitude of the velocity vector v→, and since the speed is constant, the vectors in the drawing have the same magnitude at all points on the circle.

Sometimes it is more convenient to describe uniform circular motion by specify- ing the period of the motion, rather than the speed. The period T is the time required to travel once around the circle—that is, to make one complete revolution. There is a relationship between period and speed, since speed υ is the distance traveled (here, the circumference of the circle = 2πr) divided by the time T:

υ = 2πr T

(5.1) 121

υυ

υ υ

FIGURE 5.1 The motion of a model airplane fl ying at a constant speed on a horizontal circular path is an example of uniform circular motion.

122 CHAPTER 5 Dynamics of Uniform Circular Motion

If the radius is known, as in Example 1, the speed can be calculated from the period, or vice versa.

EXAMPLE 1 A Tire-Balancing Machine

The wheel of a car has a radius of r = 0.29 m and is being rotated at 830 revolutions per minute (rpm) on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving.

Reasoning The speed υ can be obtained directly from υ = 2πr/T, but fi rst the period T is needed. The period is the time for one revolution, and it must be expressed in seconds, because the problem asks for the speed in meters per second.

Solution Since the tire makes 830 revolutions in one minute, the number of minutes required for a single revolution is

1 830 revolutions/min

= 1.2 × 10−3 min/revolution

Therefore, the period is T = 1.2 × 10−3 min, which corresponds to 0.072 s. Equation 5.1 can now be used to fi nd the speed:

υ = 2πr T

= 2π (0.29 m)

0.072 s = 25 m/s

The defi nition of uniform circular motion emphasizes that the speed, or the magnitude of the velocity vector, is constant. It is equally signifi cant that the direction of the vector is not constant. In Figure 5.1, for instance, the velocity vector changes direction as the plane moves around the circle. Any change in the velocity vector, even if it is only a change in direction, means that an ac- celeration is occurring. This particular acceleration is called “centripetal acceleration,” because it points toward the center of the circle, as the next section explains.

5.2 Centripetal Acceleration In this section we determine how the magnitude ac of the centripetal acceleration depends on the speed υ of the object and the radius r of the circular path. We will see that ac = υ2/r.

In Figure 5.2a an object (symbolized by a dot •) is in uniform circular motion. At a time t0 the velocity is tangent to the circle at point O, and at a later time t the velocity is tangent at point P. As the object moves from O to P, the radius traces out the angle θ, and the velocity vector changes direction. To emphasize the change, part b of the picture shows the velocity vector removed from point P, shifted parallel to itself, and redrawn with its tail at point O. The angle β between the two vectors indicates the change in direction. Since the radii CO and CP are perpendicular to the tangents at points O and P, respectively, it follows that α + β = 90° and α + θ = 90°. Therefore, angle β and angle θ are equal.

As always, acceleration is the change ∆v→ in velocity divided by the elapsed time ∆t, or a→ = ∆v→/∆t. Figure 5.3a shows the two velocity vectors oriented at the angle θ with respect to one another, together with the vector ∆v→ that represents the change in velocity. The change ∆v→ is the increment that must be added to the velocity at time t0, so that the resultant velocity has the new direction after an elapsed time ∆t = t ‒ t0. Figure 5.3b shows the sector of the circle COP.

C

at time

O

P

(a)

t0

tv

at timev

θ

C

O

P

(b)

θ

α β

FIGURE 5.2 (a) For an object (•) in uniform circular motion, the velocity v→ has diff erent directions at diff erent places on the circle. (b) The velocity vector has been removed from point P, shifted parallel to itself, and redrawn with its tail at point O.

(b)

(a) P

O

C

r

r

at time

t0

t

t

v

v

at timev

θθ

Δ

Δ

υ

FIGURE 5.3 (a) The directions of the velocity vector at times t and t0 diff er by the angle θ. (b) When the object moves along the circle from O to P, the radius r traces out the same angle θ. Here, the sector COP has been rotated clockwise by 90° relative to its orientation in Figure 5.2.

5.2 Centripetal Acceleration 123

In the limit that ∆t is very small, the arc length OP is approximately a straight line whose length is the distance υ∆t traveled by the object. In this limit, COP is an isosceles triangle, as is the triangle in part a of the drawing. Note that both triangles have equal angles θ. This means that they are similar triangles, so that

∆υ υ =

υ ∆t r

This equation can be solved for ∆υ/∆t, to show that the magnitude ac of the centripetal accelera- tion is given by ac = υ2/r.

Centripetal acceleration is a vector quantity and, therefore, has a direction as well as a mag- nitude. The direction is toward the center of the circle, and Conceptual Example 2 helps us to set the stage for explaining this important fact.

CONCEPTUAL EXAMPLE 2 Which Way Will the Object Go?

In Figure 5.4 an object, such as a model airplane on a guideline, is in uni- form circular motion. The object is symbolized by a dot (•), and at point O it is released suddenly from its circular path. For instance, suppose that the guideline for a model plane is cut suddenly. Does the object move (a) along the straight tangent line between points O and A or (b) along the circular arc between points O and P?

Reasoning Newton’s fi rst law of motion (see Section 4.2) guides our reasoning. This law states that an object continues in a state of rest or in a state of motion at a constant velocity (i.e., at a constant speed along a straight line) unless compelled to change that state by a net force. When an object is suddenly released from its circular path, there is no longer a net force being applied to the object. In the case of the model airplane, the guideline cannot apply a force, since it is cut. Gravity certainly acts on the plane, but the wings provide a lift force that balances the weight of the plane.

Answer (b) is incorrect. An object such as a model airplane will remain on a circular path only if a net force keeps it there. Since there is no net force, it cannot travel on the circular arc.

Answer (a) is correct. In the absence of a net force, the plane or any object would continue to move at a constant speed along a straight line

in the direction it had at the time of release, consistent with Newton’s fi rst law. This speed and direction are given in Figure 5.4 by the velocity vector v→.

Related Homework: Problem 3

C

O

P

A

v

θ

FIGURE 5.4 If an object (•) moving on a circular path were released from its path at point O, it would move along the straight tangent line OA in the absence of a net force.

As Example 2 discusses, the object in Figure 5.4 would travel on a tangent line if it were released from its circular path suddenly at point O. It would move in a straight line to point A in the time it would have taken to travel on the circle to point P. It is as if, in the process of remain- ing on the circle, the object drops through the distance AP, and AP is directed toward the center of the circle in the limit that the angle θ is small. Thus, the object in uniform circular motion accelerates toward the center of the circle at every moment. Since the word “centripetal” means “moving toward a center,” the acceleration is called centripetal acceleration.

CENTRIPETAL ACCELERATION Magnitude: The centripetal acceleration of an object moving with a speed υ on a circular path of radius r has a magnitude ac given by

ac = υ 2

r (5.2)

Direction: The centripetal acceleration vector always points toward the center of the circle and continually changes direction as the object moves.

The following example illustrates the eff ect of the radius r on the centripetal acceleration.

124 CHAPTER 5 Dynamics of Uniform Circular Motion

In Section 4.11 we learned that an object is in equilibrium when it has zero acceleration. Conceptual Example 4 discusses whether an object undergoing uniform circular motion can ever be at equilibrium.

EXAMPLE 3 The Physics of a Bobsled Track

The bobsled track at the 1994 Olympics in Lillehammer, Norway, con- tained turns with radii of 33 m and 24 m, as Interactive Figure 5.5 illus- trates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event. Express the answers as multiples of g = 9.8 m/s2.

Reasoning In each case, the magnitude of the centripetal accelera- tion can be obtained from ac = υ2/r. Since the radius r is in the denomi- nator on the right side of this expression, we expect the acceleration to be smaller when r is larger.

Solution From ac = υ2/r it follows that

Radius = 33 m ac = (34 m/s)2

33 m = 35 m/s2 = 3.6 g

Radius = 24 m ac = (34 m/s)2

24 m = 48 m/s2 = 4.9 g

The centripetal acceleration is indeed smaller when the radius is larger. In fact, with r in the denominator on the right of ac = υ2/r, the accel- eration approaches zero when the radius becomes very large. Uniform circular motion along the arc of an infi nitely large circle entails no acceleration, because it is just like motion at a constant speed along a straight line.

r = 33 m

= 34 m/s

r = 24 m

υ

INTERACTIVE FIGURE 5.5 This bobsled travels at the same speed around two curves with diff erent radii. For the turn with the larger radius, the sled has a smaller centripetal acceleration.

CONCEPTUAL EXAMPLE 4 Uniform Circular Motion and Equilibrium

A car moves at a constant speed along a straight line as it approaches a circular turn. In which of the following parts of the motion is the car in equilibrium? (a) As it moves along the straight line toward the circular turn, (b) as it is go- ing around the turn, (c) as it moves away from the turn along a straight line.

Reasoning An object is in equilibrium when it has no acceleration, according to the defi nition given in Section 4.11. If the object’s velocity remains constant, both in magnitude and direction, its acceleration is zero.

Answer (b) is incorrect. As the car goes around the turn, the direction of travel changes, so the car has a centripetal acceleration that is characteristic

of uniform circular motion. Because of this acceleration, the car is not in equilibrium during the turn.

Answers (a) and (c) are correct. As the car either approaches the turn or moves away from the turn it is traveling along a straight line, and both the speed and direction of the motion are constant. Thus, the velocity vec- tor does not change, and there is no acceleration. Consequently, for these parts of the motion, the car is in equilibrium.

Related Homework: Problem 13

We have seen that going around tight turns (smaller r) and gentle turns (larger r) at the same speed entails diff erent centripetal accelerations. And most drivers know that such turns “feel” diff erent. This feeling is associated with the force that is present in uniform circular motion, and we turn to this topic in the next section.

5.3 Centripetal Force 125

Check Your Understanding

(The answers are given at the end of the book.) 1. The car in CYU Figure 5.1 is moving clockwise around a

circular section of road at a constant speed. What are the di- rections of its velocity and acceleration at (a) position 1 and (b) position 2? Specify your responses as north, east, south, or west.

2. The speedometer of your car shows that you are traveling at a constant speed of 35 m/s. Is it possible that your car is accelerating?

3. Consider two people, one on the earth’s surface at the equator and the other at the north pole. Which has the larger centripetal acceleration?

4. Which of the following statements about centripetal accelera- tion is true? (a) An object moving at a constant velocity cannot have a centripetal acceleration. (b) An object moving at a con- stant speed may have a centripetal acceleration.

5. A car is traveling at a constant speed along the road ABCDE shown in CYU Figure 5.2. Sections AB and DE are straight. Rank the accelerations in each of the four sections according to magnitude, listing the smallest fi rst.

5.3 Centripetal Force Newton’s second law indicates that whenever an object accelerates, there must be a net force to create the acceleration. Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. The second law gives this net force as the product of the object’s mass m and its acceleration υ2/r. The net force causing the centripetal acceleration is called the centripetal force Fc

→ and points in the same direction as the acceleration—that is, toward the

center of the circle.

CENTRIPETAL FORCE Magnitude: The centripetal force is the name given to the net force required to keep an object of mass m, moving at a speed υ, on a circular path of radius r, and it has a mag- nitude of

Fc = mυ 2

r (5.3)

Direction: The centripetal force always points toward the center of the circle and con- tinually changes direction as the object moves.

The phrase “centripetal force” does not denote a new and separate force created by nature. The phrase merely labels the net force pointing toward the center of the circular path, and this net force is the vector sum of all the force components that point along the radial direction. Some- times the centripetal force consists of a single force such as tension (see Example 5), friction (see Example 7), the normal force or a component thereof (see Examples 8 and 13), or the gravita- tional force (see Examples 9–11). However, there are circumstances when a number of diff erent forces contribute simultaneously to the centripetal force (see Section 5.7).

In some cases, it is easy to identify the source of the centripetal force, as when a model airplane on a guideline fl ies in a horizontal circle. The only force pulling the plane inward is the tension in the line, so this force alone (or a component of it) is the centripetal force. Example 5 illustrates the fact that higher speeds require greater tensions.

N

S

W E

1

2

A B

C

D E

CYU FIGURE 5.1

CYU FIGURE 5.2

126 CHAPTER 5 Dynamics of Uniform Circular Motion

Conceptual Example 6 deals with another case where it is easy to identify the source of the centripetal force.

When a car moves at a steady speed around an unbanked curve, the centripetal force keeping the car on the curve comes from the static friction between the road and the tires, as Figure 5.8 in-

dicates. It is static, rather than kinetic friction, because the tires are not slipping with respect to the radial direction. If the static frictional force is insuffi cient, given the speed and the radius of the turn, the car will skid off the road. Example 7 shows how an icy road can limit safe driving.

Static frictional

force FIGURE 5.8 When the car moves without skidding around a curve, static friction between the road and the tires provides the centripetal force to keep the car on the road.

EXAMPLE 5 The Eff ect of Speed on Centripetal Force

The model airplane in Figure 5.6 has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension in the guideline (length = 17 m) for speeds of 19 and 38 m/s.

Reasoning Since the plane fl ies on a circular path, it experiences a centripetal acceleration that is directed toward the center of the circle. According to Newton’s second law of motion, this acceleration is pro- duced by a net force that acts on the plane, and this net force is called the centripetal force. The centripetal force is also directed toward the center of the circle. Since the tension in the guideline is the only force pulling the plane inward, it must be the centripetal force.

Solution Equation 5.3 gives the tension directly: Fc = T = mυ2/r. Scale records tension

0 N

FIGURE 5.6 The scale records the tension in the guideline. See Example 5.

Speed = 19 m/s T = (0.90 kg)(19 m /s) 2

17 m = 19 N

Speed = 38 m/s T = (0.90 kg)(38 m /s) 2

17 m = 76 N

CONCEPTUAL EXAMPLE 6 The Physics of a Trapeze Act

In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward, holding his partner (see Figure 5.7). Is it harder for the man to hold his partner (a) when the partner hangs straight down and is stationary or (b) when the partner is swinging through the straight-down position?

Reasoning Whenever an object moves on a circular path, it experiences a centripetal acceleration that is directed toward the center of the path. A net force, known as the centripetal force, is required to produce this acceleration.

Answer (a) is incorrect. When the man and his partner are stationary, the man’s arms must support only his partner’s weight. When the two are swinging, however, the man’s arms must provide the additional force required to produce the partner’s centripetal acceleration. Thus, it is easier, not harder, for the man to hold his partner when the partner hangs straight down and is stationary.

Answer (b) is correct. When the two are swinging, the partner is mov- ing on a circular arc and, therefore, has a centripetal acceleration. The man’s arms must support the partner’s weight and must simultaneously exert an additional pull to provide the centripetal force that produces this

FIGURE 5.7 See Example 6 for a discussion of the role of centripetal force in this trapeze act. To

m R

os en

th al

/S up

er St

oc k

acceleration. Because of the additional pull, it is harder for the man to hold his partner while swinging.

Related Homework: Problems 17, 19

5.3 Centripetal Force 127

Modeling the Problem

STEP 1 Speed and Centripetal Force According to Equation 5.3, we have

Fc = mυ 2

r

Solving for the speed υ gives Equation 1 at the right. To use this result, it is necessary to consider the centripetal force Fc, which we do in Step 2.

STEP 2 Static Friction The force of static friction is the centripetal force, and the greater its value, the greater is the speed at which the car can negotiate the turn. The maximum force f sMAX of static friction is given by f sMAX = μsFN (Equation 4.7). Thus, the maximum available centripetal force is

Fc = f sMAX = μs FN

which we can substitute into Equation 1, as shown at the right. The next step considers the normal force FN.

STEP 3 The Normal Force The fact that the car does not accelerate in the vertical direction tells us that the normal force balances the car’s weight mg, or

FN = mg

This result for the normal force can now be substituted into Equation 2, as indicated at the right.

Solution Algebraically combining the results of each step, we fi nd that

υ = √rFcm = √ rμs FN

m = √rμs mgm

The mass m of the car is eliminated algebraically from this result, and we fi nd that the maximum speeds are

Dry road ( μs = 0.95) υ = √rμs g = √(51 m)(0.95)(9.8 m/s2) = 22 m/s

Icy road (μs = 0.10) υ = √rμs g = √(51 m)(0.10)(9.8 m/s2) = 7.1 m/s

As expected, the dry road allows a greater maximum speed.

Related Homework: Problems 15, 20, 25

Analyzing Multiple-Concept Problems

EXAMPLE 7 Centripetal Force and Safe Driving

At what maximum speed can a car safely negotiate a horizontal unbanked turn (radius = 51 m) in dry weather (coeffi cient of static friction = 0.95) and icy weather (coeffi cient of static friction = 0.10)?

Reasoning The speed υ at which the car of mass m can negotiate a turn of radius r is related to the centripetal force Fc that is available, according to Fc = mυ2/r (Equation 5.3). Static friction provides the centripetal force and can pro- vide a maximum force fsMAX given by fsMAX = μsFN (Equation 4.7), in which μs is the coeffi cient of static friction and FN is the normal force. Thus, we will use the fact that Fc = fsMAX to obtain the maximum speed from Equation 5.3. It will also be necessary to evaluate the normal force, which we will do by con- sidering that it must balance the weight of the car. Experience indicates that the maximum speed should be greater for the dry road than for the icy road.

Description Symbol Value Comment Radius of turn r 51 m

Coefficient of static friction μs 0.95 Dry conditions

Coefficient of static friction μs 0.10 Icy conditions

Unknown Variable Speed of car υ ?

Knowns and Unknowns The data for this problem are as follows:

?

υ = √rFcm (1)

STEP 1 STEP 2 STEP 3 Problem-Solving Insight When using an equation to obtain a numerical answer, algebraically solve for the unknown vari- able in terms of the known variables. Then substitute in the numbers for the known variables, as this example shows.

?

υ = √rFcm (1)

Fc = μs FN (2)

υ = √rFcm (1)

Fc = μs FN (2)

FN = mg

128 CHAPTER 5 Dynamics of Uniform Circular Motion

A passenger in Figure 5.8 must also experience a centripetal force to remain on the circular path. However, if the upholstery is very slippery, there may not be enough static friction to keep him in place as the car makes a tight turn at high speed. Then, when viewed from inside the car, he appears to be thrown toward the outside of the curve. What really happens is that the passenger slides off on a tangent to the circle, until he encounters a source of centripetal force to keep him in place while the car turns. This occurs when the passenger bumps into the side of the car, which pushes on him with the necessary force.

THE PHYSICS OF . . . flying an airplane in a banked turn. Sometimes the source of the centripetal force is not obvious. A pilot making a turn, for instance, banks or tilts the plane at an angle to create the centripetal force. As a plane fl ies, the air pushes upward on the wing surfaces with a net lifting force L

→ that is perpendicular to the wing surfaces, as Animated Figure 5.9a shows. Part b of the drawing illustrates that when the plane is banked at an angle θ, a component L sin θ of the lifting force is directed toward the center of the turn. It is this component that pro- vides the centripetal force. Greater speeds and/or tighter turns require greater centripetal forces. In such situations, the pilot must bank the plane at a larger angle, so that a larger component of the lift points toward the center of the turn. The technique of banking into a turn also has an application in the construction of high-speed roadways, where the road itself is banked to achieve a similar eff ect, as the next section discusses.

Check Your Understanding

(The answers are given at the end of the book.) 6. A car is traveling in uniform circular motion on a section of road whose radius is r (see CYU Figure 5.3).

The road is slippery, and the car is just on the verge of sliding. (a) If the car’s speed were doubled, what would be the smallest radius at which the car does not slide? Express your answer in terms of r. (b) What would be your answer to part (a) if the car were replaced by one that weighed twice as much, the car’s speed still being doubled?

7. Other things being equal, would it be easier to drive at high speed around an unbanked horizontal curve on the moon than to drive around the same curve on the earth?

8. What is the chance of a light car safely rounding an unbanked curve on an icy road as compared to that of a heavy car: worse, the same, or better? Assume that both cars have the same speed and are equipped with identical tires.

9. A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centrip- etal force to remain in place: at the center of the turntable or at the edge of the turntable?

r

CYU FIGURE 5.3

L

Path of plane

L sin

(a) (b)

1 2 –L

1 2 –L

θ

θ

ANIMATED FIGURE 5.9 (a) The air exerts an upward lifting force 12 L

→ on each

wing. (b) When a plane executes a circular turn, the plane banks at an angle θ. The lift component L sin θ is directed toward the center of the circle and provides the centripetal force.

5.4 Banked Curves 129

5.4 Banked Curves When a car travels without skidding around an unbanked curve, the static frictional force between the tires and the road provides the centripetal force. The reliance on friction can be eliminated completely for a given speed, however, if the curve is banked at an angle relative to the horizontal, much in the same way that a plane is banked while making a turn.

Ineractive Figure 5.10a shows a car going around a friction-free banked curve. The radius of the curve is r, where r is measured parallel to the horizontal and not to the slanted surface. Part b shows the normal force FN

→ that the road applies to the car, the normal force being perpendicular

to the road. Because the roadbed makes an angle θ with respect to the horizontal, the normal force has a component FN sin θ that points toward the center C of the circle and provides the centripetal force:

Fc = FN sin θ = mυ 2

r

The vertical component of the normal force is FN cos θ and, since the car does not accelerate in the vertical direction, this component must balance the weight mg of the car. Therefore, FN cos θ = mg. Dividing this equation into the previous one shows that

FN sin θ FN cos θ

= mυ2/r

mg

tan θ = υ2

rg (5.4)

Equation 5.4 indicates that, for a given speed υ, the centripetal force needed for a turn of radius r can be obtained from the normal force by banking the turn at an angle θ, independent of the mass of the vehicle. Greater speeds and smaller radii require more steeply banked curves— that is, larger values of θ. At a speed that is too small for a given θ, a car would slide down a frictionless banked curve; at a speed that is too large, a car would slide off the top. The next example deals with a famous banked curve.

EXAMPLE 8 The Physics of the Daytona International Speedway

The Daytona 500 is the major event of the NASCAR (National Association for Stock Car Auto Racing) season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316 m and are banked steeply, with θ = 31° (see Interactive Figure 5.10). Suppose these maximum-radius turns were fric- tionless. At what speed would the cars have to travel around them?

Reasoning In the absence of friction, the horizontal component of the normal force that the track exerts on the car must provide the centripetal force. Therefore, the speed of the car is given by Equation 5.4.

Solution From Equation 5.4, it follows that

υ = √rg tan θ = √(316 m)(9.80 m /s2) tan 31° = 43 m /s (96 mph)

Drivers actually negotiate the turns at speeds up to 195 mph, however, which requires a greater centripetal force than that implied by Equa- tion 5.4 for frictionless turns. Static friction provides the additional force.

(a)

C r

θ

C r

mg

FN sin

FN

FN cos

(b)

θ

θ θ

θ

INTERACTIVE FIGURE 5.10 (a) A car travels on a circle of radius r on a frictionless banked road. The banking angle is θ, and the center of the circle is at C. (b) The forces acting on the car are its weight mg→ and the normal force FN

→ .

A component FN sin θ of the normal force provides the centripetal force.

Check Your Understanding

(The answer is given at the end of the book.) 10. Go to Concept Simulation 5.2 at www.wiley.com/college/cutnell to review the concepts involved

in this question. Two cars are identical, except for the type of tread design on their tires. The cars are driven at the same speed and enter the same unbanked horizontal turn. Car A cannot negotiate the turn, but car B can. Which tread design, the one on car A or the one on car B, yields a larger coeffi cient of static friction between the tires and the road?

130 CHAPTER 5 Dynamics of Uniform Circular Motion

5.5 Satellites in Circular Orbits Today there are many satellites in orbit about the earth. The ones in circular orbits are examples of uniform circular motion. Like a model airplane on a guideline, each satellite is kept on its cir- cular path by a centripetal force. The gravitational pull of the earth provides the centripetal force and acts like an invisible guideline for the satellite.

There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fi xed radius. To see how this fundamental characteristic arises, consider the gravitational force acting on the satellite of mass m in Figure 5.11. Since the gravitational force is the only force acting on the satellite in the radial direction, it alone provides the centripetal force. Therefore, using Newton’s law of gravitation (Equation 4.3), we have

Fc = G mME

r 2 =

mυ2

r

where G is the universal gravitational constant, ME is the mass of the earth, and r is the distance from the center of the earth to the satellite. Solving for the speed υ of the satellite gives

υ = √GMEr (5.5) If the satellite is to remain in an orbit of radius r, the speed must have precisely this value. Note that the radius r of the orbit is in the denominator in Equation 5.5. This means that the closer the satellite is to the earth, the smaller is the value for r and the greater the orbital speed must be.

The mass m of the satellite does not appear in Equation 5.5, having been eliminated algebra- ically. Consequently, for a given orbit, a satellite with a large mass has exactly the same orbital speed as a satellite with a small mass. However, more eff ort is certainly required to lift the larger- mass satellite into orbit. The orbital speed of one famous artifi cial satellite is determined in the following example.

r

Gravitational force

FIGURE 5.11 For a satellite in circular orbit around the earth, the gravitational force provides the centripetal force.

EXAMPLE 9 The Physics of the Hubble Space Telescope

Determine the speed of the Hubble Space Telescope (see Figure 5.12) orbiting at a height of 598 km above the earth’s surface.

Reasoning Before Equation 5.5 can be applied, the orbital radius r must be determined relative to the center of the earth. Since the radius of the earth is approximately 6.38 × 106 m, and the height of the telescope above the earth’s surface is 0.598 × 106 m, the orbital radius is r = 6.98 × 106 m.

Problem-Solving Insight The orbital radius r that appears in the relation υ = √GME /r is the distance from the satellite to the center of the earth (not to the surface of the earth).

Solution The orbital speed is

υ = √GMEr = √ (6.67 × 10−11 N ·m2/kg2)(5.98 × 1024 kg)

6.98 × 106 m

υ = 7.56 × 103 m /s (16 900 mi/h) FIGURE 5.12 The Hubble Space Telescope orbiting the earth.

Fr an

k W

hi tn

ey /G

et ty

Im ag

es , I

nc .

5.5 Satellites in Circular Orbits 131

THE PHYSICS OF . . . the Global Positioning System. Many applications of satellite technology aff ect our lives. An increasingly important application is the network of 24 satel- lites called the Global Positioning System (GPS), which can be used to determine the position of an object to within 15 m or less. Figure 5.13 illustrates how the system works. Each GPS satellite carries a highly accurate atomic clock, whose time is transmitted to the ground con- tinually by means of radio waves. In the drawing, a car carries a computerized GPS receiver that can detect the waves and is synchronized to the satellite clock. The receiver can therefore determine the distance between the car and a satellite from a knowledge of the travel time of the waves and the speed at which they move. This speed, as we will see in Chapter 24, is the speed of light and is known with great precision. A measurement using a single satellite locates the car somewhere on a circle, as Figure 5.13a shows, while a measurement using a second satellite locates the car on another circle. The intersection of the circles reveals two possible positions for the car, as in Figure 5.13b. With the aid of a third satellite, a third circle can be established, which intersects the other two and identifi es the car’s exact position, as in Figure 5.13c. The use of ground-based radio beacons to provide additional reference points leads to a system called Diff erential GPS, which can locate objects even more accurately than the satellite-based system alone. Navigational systems for automobiles and portable systems that tell hikers and people with visual impairments where they are located are two of the many uses for the GPS technique. GPS applications are so numerous that they have developed into a multibillion dollar industry.

Equation 5.5 applies to human-made satellites or to natural satellites like the moon. It also applies to circular orbits about any astronomical object, provided ME is replaced by the mass of the object on which the orbit is centered. Example 10, for instance, shows how scientists have applied this equation to conclude that a supermassive black hole is probably located at the center of the galaxy known as M87. This galaxy is located at a distance of about 50 million light-years away from the earth. (One light-year is the distance that light travels in a year, or 9.5 × 1015 m.)

(a) (b) (c)

FIGURE 5.13 The Navstar Global Positioning System (GPS) of satellites can be used with a GPS receiver to locate an object, such as a car, on the earth. (a) One satellite identifi es the car as being somewhere on a circle. (b) A second places it on another circle, which identifi es two possibilities for the exact spot. (c) A third provides the means for deciding where the car is.

EXAMPLE 10 The Physics of Locating a Black Hole

The Hubble Telescope has detected the light being emitted from diff erent regions of galaxy M87, which is shown in Figure 5.14. The black circle identifi es the center of the galaxy. From the characteristics of this light,

astronomers have determined that the orbiting speed is 7.5 × 105 m/s for matter located at a distance of 5.7 × 1017 m from the center. Find the mass M of the object located at the galactic center.

132 CHAPTER 5 Dynamics of Uniform Circular Motion

The period T of a satellite is the time required for one orbital revolution. As in any uniform circular motion, the period is related to the speed of the motion by υ = 2πr/T. Substituting υ from Equation 5.5 shows that

√GMEr = 2πr T

Solving this expression for the period T gives

T = 2πr3/2

√GME (5.6)

Although derived for earth orbits, Equation 5.6 can also be used for calculating the periods of those planets in nearly circular orbits about the sun, if ME is replaced by the mass MS of the sun and r is interpreted as the distance between the center of the planet and the center of the sun. The fact that the period is proportional to the three-halves power of the orbital radius is known as Kepler’s third law, and it is one of the laws discovered by Johannes Kepler (1571–1630) during his studies of planetary motion. Kepler’s third law also holds for elliptical orbits, which will be discussed in Chapter 9.

An important application of Equation 5.6 occurs in the fi eld of communications, where “syn- chronous satellites” are put into a circular orbit that is in the plane of the equator, as Figure 5.15 shows. The orbital period is chosen to be one day, which is also the time it takes for the earth to turn once about its axis. Therefore, these satellites move around their orbits in a way that is synchronized with the rotation of the earth. For earth-based observers, synchronous satellites have the useful char- acteristic of appearing in fi xed positions in the sky and can serve as “stationary” relay stations for communication signals sent up from the earth’s surface.

THE PHYSICS OF . . . digital satellite system TV. This is exactly what is done in the digital satellite systems that are a popular alternative to cable TV. As the blowup in Figure 5.15 indicates, a small “dish” antenna on your house picks up the digital TV signals relayed back to earth by the satellite. After being decoded, these signals are delivered to your TV set. All synchronous satellites are in orbit at the same height above the earth’s surface, as Example 11 shows.

Reasoning and Solution Replacing ME in Equation 5.5 with M gives υ = √GM /r, which can be solved to show that

M = υ2r G

= (7.5 × 105 m /s) 2 (5.7 × 1017 m)

6.67 ×10−11 N ·m2/kg2

= 4.8 × 1039 kg

Since the mass of our sun is 2.0 × 1030 kg, matter equivalent to 2.4 bil- lion suns is located at the center of galaxy M87. The volume of space in which this matter is located contains relatively few visible stars, so researchers believe that the data provide strong evidence for the existence of a supermassive black hole. The term “black hole” is used because the tremendous mass prevents even light from escaping. The light that forms the image in Figure 5.14 comes not from the black hole itself, but from matter that surrounds it.*

*More complete analysis shows that the mass of the black hole is equivalent to 6.6 billion suns (K. Gebhardt et al., The Astrophysical Journal, Vol. 729, 2011, p. 119).

Equator

Small dish antenna

TV signals from synchronous

satellite

FIGURE 5.15 A synchronous satellite orbits the earth once per day on a circular path in the plane of the equator. Digital satellite system television uses such satellites as relay stations. TV signals are sent up from the earth’s surface and then rebroadcast down to your own small dish antenna.

FIGURE 5.14 This image of the ionized gas (yellow) at the heart of galaxy M87 was obtained by the Hubble Space Telescope. The circle identifi es the center of the galaxy, at which a black hole is thought to exist.Co

ur te

sy N

A SA

a nd

S pa

ce T

el es

co pe

Sc

ie nc

e In

st itu

te

5.6 Apparent Weightlessness and Artificial Gravity 133

Check Your Understanding

(The answer is given at the end of the book.) 11. Two satellites are placed in orbit, one about Mars and the other about Jupiter, such

that the orbital speeds are the same. Mars has the smaller mass. Is the radius of the satellite in orbit about Mars less than, greater than, or equal to the radius of the satel- lite orbiting Jupiter?

5.6 Apparent Weightlessness and Artificial Gravity

The idea of life on board an orbiting satellite conjures up visions of astronauts fl oating around in a state of “weightlessness,” as in Figure 5.16. Actually, this state should be called “apparent weightlessness,” because it is similar to the condition of zero apparent weight that occurs in an elevator during free-fall. Conceptual Example 12 explores this similarity.

FIGURE 5.16 In a state of apparent weightlessness in orbit, as- tronaut and pilot Guy S. Gardner and mission specialist William M. Shepherd (in chair) fl oat, with Gardner appearing to “bal- ance” the chair on his nose.

C ou

rte sy

N A

SA

*Successive appearances of the sun defi ne the solar day of 24 h or 8.64 × 104 s. The sun moves against the background of the stars, however, and the time required for the earth to turn once on its axis relative to the fi xed stars is 23 h 56 min, which is called the sidereal day. The sidereal day should be used in Example 11, but the neglect of this eff ect introduces an error of less than 0.4% in the answer.

EXAMPLE 11 The Orbital Radius for Synchronous Satellites

What is the height H above the earth’s surface at which all synchronous satellites (regardless of mass) must be placed in orbit?

Reasoning The period T of a synchronous satellite is one day, so we can use Equation 5.6 to fi nd the distance r from the satellite to the center of the earth. To fi nd the height H of the satellite above the earth’s surface we will have to take into account the fact that the earth itself has a radius of 6.38 × 106 m.

Solution A period of one day* corresponds to T = 8.64 × 104 s. In using this value it is convenient to rearrange the equation T = 2πr3/2/√GME as follows:

r 3/2 = T√GME

2π =

(8.64 × 104 s)√(6.67 ×10−11 N ·m2/kg2)(5.98 ×1024 kg) 2π

By squaring and then taking the cube root, we fi nd that r = 4.23 × 107 m. Since the radius of the earth is approximately 6.38 × 106 m, the height of the satellite above the earth’s surface is

H = 4.23 × 107 m ‒ 0.64 × 107 m = 3.59 × 107 m (22 300 mi)

Math Skills In solving T = 2πr 3/2

√GME (Equation 5.6) for r, we

remember that whatever is done to one side of the equals sign must also be done to the other side. To isolate the term r3/2 on one side we multiply both sides by √GME and divide both sides by 2π:

T √GME

2π =

2πr 3/2

√GME √GME

2π  or r 3/2 =

T √GME 2π

Next, we square both sides of the result for r3/2:

(r 3/2)2 = ( T√GME

2π ) 2

 or r 3 = T 2GME

4π 2

Finally, taking the cube root of the result for r3 gives

√3 r 3 = √3 T 2GME 4π2

or (r 3)1/3 = ( T 2GME

4π2 ) 1/3

or r = T 2/3G 1/3M 1/3E

41/3π 2/3

Using values for T, G, and ME in the result for r shows that r = 4.23 × 107 m.

134 CHAPTER 5 Dynamics of Uniform Circular Motion

THE PHYSICS OF . . . artifi cial gravity. The physiological eff ects of prolonged appar- ent weightlessness are only partially known. To minimize such eff ects, it is likely that artifi cial gravity will be provided in large space stations of the future. To help explain artifi cial gravity, Figure 5.18 shows a space station rotating about an axis. Because of the rotational motion, any

object located at a point P on the interior surface of the station experiences a centripetal force directed toward the axis. The surface of the station provides this force by pushing on the feet of an astronaut, for instance. The centripetal force can be adjusted to match the astronaut’s earth-weight by properly selecting the rotational speed of the space station, as Example 13 illustrates.

CONCEPTUAL EXAMPLE 12 The Physics of Apparent Weightlessness

Figure 5.17 shows a person on a scale in a freely falling elevator and in a satellite in a circular orbit. Assume that when the person is standing stationary on the earth, his weight is 800 N (180 lb). In each case, what apparent weight is recorded by the scale? (a) The scale in the elevator records 800 N while that in the satellite records 0 N. (b) The scale in the elevator records 0 N while that in the satellite records 800 N. (c) Both scales record 0 N.

Reasoning As Section 4.8 discusses, apparent weight is the force that an object exerts on the platform of a scale. This force depends on whether or not the object and the platform are accelerating together.

Answer (a) is incorrect. The scale and the person in the free-falling elevator are accelerating toward the earth at the same rate. Therefore, they cannot push against one another, and so the scale does not record an apparent weight of 800 N.

Answer (b) is incorrect. The scale and the person in the satellite are accelerating toward the center of the earth at the same rate (they have the same centripetal acceleration). Therefore, they cannot push against one another, and so the scale does not record an apparent weight of 800 N.

Answer (c) is correct. The scale and the person in the elevator fall together and, therefore, they cannot push against one another. There- fore, the scale records an apparent weight of 0 N. In the orbiting satellite in Figure 5.17b, both the person and the scale are in uniform circular motion. Objects in uniform circular motion continually accelerate or “fall” toward the center of the circle in order to remain on the circular path. Consequently, both the person and the scale “fall” with the same acceleration toward the center of the earth and cannot push against one another. Thus, the apparent weight in the satellite is zero, just as it is in the freely falling elevator.

Earth

Orbit

Free-fall (a)

Free-fall (b)

FIGURE 5.17 (a) During free-fall, the elevator accelerates downward with the acceleration due to gravity, and the apparent weight of the person is zero. (b) The orbiting space station is also in free-fall toward the center of the earth.

P Fc

FIGURE 5.18 The surface of the rotating space station pushes on an object with which it is in contact and thereby provides the centripetal force that keeps the object moving on a circular path.

5.6 Apparent Weightlessness and Artificial Gravity 135

Analyzing Multiple-Concept Problems

EXAMPLE 13 Artificial Gravity

At what speed must the interior surface of the space station (r = 1700 m) move in Figure 5.18, so that the astronaut located at point P experiences a push on his feet that equals his weight on earth?

Reasoning The fl oor of the rotating space station exerts a normal force on the feet of the astronaut. This is the centripetal force (Fc = mυ2/r) that keeps the astronaut moving on a circular path. Since the magnitude of the normal force equals the astronaut’s weight on earth, we can determine the speed υ of the space station’s fl oor.

Knowns and Unknowns The data for this problem are given in the following table:

Description Symbol Value Radius of space station r 1700 m

Unknown Variable Speed of space station’s floor υ ?

Modeling the Problem

STEP 1 Speed and Centripetal Force The centripetal force acting on the astronaut is given by Equation 5.3 as

Fc = mυ 2

r

Solving for the speed υ gives Equation 1 at the right. Step 2 considers the centripetal force Fc that appears in this result.

STEP 2 Magnitude of the Centripetal Force The normal force applied to the astronaut’s feet by the fl oor is the centripetal force and has a magnitude equal to the astronaut’s earth-weight. This earth-weight is given by Equation 4.5 as the astronaut’s mass m times the magnitude g of the acceleration due to the earth’s gravity. Thus, we have for the centripetal force that

Fc = mg

which can be substituted into Equation 1, as indicated at the right.

Solution Algebraically combining the results of each step, we fi nd that

υ = √rFcm = √ rmg m

The astronaut’s mass m is eliminated algebraically from this result, so the speed of the space station fl oor is

υ = √rg = √(1700 m)(9.80 m/s2) = 130 m/s

Related Homework: Problem 34

STEP 1 STEP 2

Check Your Understanding

(The answer is given at the end of the book.) 12. The acceleration due to gravity on the moon is one-sixth that on earth. (a) Is the true weight of a person

on the moon less than, greater than, or equal to the true weight of the same person on earth? (b) Is the apparent weight of a person in orbit about the moon less than, greater than, or equal to the apparent weight of the same person in orbit about the earth?

υ = √rFcm (1) ?

υ = √rFcm (1)

Fc = mg

136 CHAPTER 5 Dynamics of Uniform Circular Motion

5.7 *Vertical Circular Motion Motorcycle stunt drivers perform a feat in which they drive their cycles around a vertical cir- cular track, as in Interactive Figure 5.19a. Usually, the speed varies in this stunt. When the speed of travel on a circular path changes from moment to moment, the motion is said to be nonuniform. Nonetheless, we can use the concepts that apply to uniform circular motion to gain considerable insight into the motion that occurs on a vertical circle.

There are four points on a vertical circle where the centripetal force can be identifi ed easily, as Interactive Figure 5.19b indicates. As you look at Interactive Figure 5.19b, keep in mind that the centripetal force is not a new and separate force of nature. Instead, at each point the centripetal force is the net sum of all the force components oriented along the radial direction, and it points toward the center of the circle. The drawing shows only the weight of the cycle plus rider (magnitude = mg) and the normal force pushing on the cycle (magnitude = FN). The propulsion and braking forces are omitted for simplicity, because they do not act in the radial direction. The magnitude of the centrip- etal force at each of the four points is given as follows in terms of mg and FN:

(1) FN1 − mg = mυ1 2

r (3) FN3 + mg = mυ3 2

r ⏟⎵⏟⎵⏟ ⏟⎵⏟⎵⏟

(2) FN2 = mυ2 2

r (4) FN4 = mυ4 2

r

Problem-Solving Insight Centripetal force is the name given to the net force that points toward the center of a circular path. As shown here, there may be more than one force that contributes to this net force.

THE PHYSICS OF . . . the loop-the-loop motorcycle stunt. As the cycle goes around, the magnitude of the normal force changes. It changes because the speed changes and because the weight does not have the same eff ect at every point. At the bottom, the normal force and the weight oppose one another, giving a centripetal force of magnitude FN1 ‒ mg. At the top, the normal force and the weight reinforce each other to provide a centripetal force whose magnitude is FN3 + mg. At points 2 and 4 on either side, only FN2 and FN4 provide the centripetal force. The weight is tangent to the circle at points 2 and 4 and has no component pointing toward the center. If the speed at each of the four places is known, along with the mass and the radius, the normal forces can be determined.

Riders who perform the loop-the-loop trick must have at least a minimum speed at the top of the circle to remain on the track. This speed can be determined by considering the centripetal force at point 3. The speed υ3 in the equation FN3 + mg = mυ 23 /r is a minimum when FN3 is zero. Then, the speed is given by υ3 = √rg. At this speed, the track does not exert a normal force to keep the cycle on the circle, because the weight mg provides all the centripetal force. Under these conditions, the rider experiences an apparent weightlessness like that discussed in Section 5.6, because for an instant the rider and the cycle are falling freely toward the center of the circle.

Check Your Understanding

(The answers are given at the end of the book.) 13. Would a change in the earth’s mass aff ect (a) the banking of airplanes as they turn, (b) the banking of

roadbeds, (c) the speeds with which satellites are put into circular orbits, and (d) the performance of the loop-the-loop motorcycle stunt?

14. A stone is tied to a string and whirled around in a circle at a constant speed. Is the string more likely to break when the circle is horizontal or when it is vertical? Assume that the constant speed is the same in each case.

= Fc1 = Fc3

⏟= Fc2 ⏟= Fc4

r

(a)

(b)

1

3

mg

mg

mg

mg

FN3

FN1

FN4 FN24 2

INTERACTIVE FIGURE 5.19 (a) A vertical loop-the-loop motorcycle stunt. (b) The normal force F→N and the weight mg→ of the cycle and the rider are shown here at four locations.

Concept Summary 137

EXAMPLE 14 BIO The Physics of g-Suits

Solution Since both forces on the pilot in Figure 5.20 point in toward the center of the circular motion, the centripetal force will be equal to the sum of the normal force and the weight: FC = FN + mg. Equa- tion 5.3 tells us that the centripetal force is equal to mυ2/r. Therefore, we

fi nd that mυ2

r = FN + mg. The problem tells us we want the normal

force on the pilot to be equal to his weight. Thus, we make that substi-

tution and fi nd the following: mυ 2

r = mg + mg = 2mg. Notice that the

mass of the pilot drops out, and we can solve for the speed of the plane: υ = √2rg = √2(1200 m)(9.8 m/s2) = 150 m/s

In addition to the importance of g-suits in this type of maneuver, this particular combination of forces can be very alarming for young pilots. Since the normal force on the pilot is equal to his weight, he feels as if he is sitting right side up, when in fact, he is upside down. This can lead to spatial disorientation.

Jet fi ghter pilots and astronauts often experience large accelerations from centripetal forces during fl ight. This is commonly referred to as “pulling g’s”. For example, pilots fl ying along the path of a vertical circle are sus- ceptible to the loss of consciousness (blackout) due to blood fl owing from the top of their head and pooling in their lower body. To help prevent this potentially dangerous situation, the pilot wears a device called a g-suit. The suit contains air-fi lled bladders that expand and squeeze on a pilot’s legs and abdomen. This helps prevent the blood from draining from the upper part of the body, thereby allowing the pilot to remain conscious during higher levels of acceleration. Consider the following situation shown in Figure 5.20, where a fi ghter pilot is fl ying upside down along a vertical circle whose radius is 1.2 km. What must be the speed of the plane, so that the normal force acting on the pilot is equal to his weight?

Reasoning The forces acting on the pilot are shown in Figure 5.20. Since we have circular motion, we should be able to identify the cen- tripetal force. We can then apply Equation 5.3 to calculate the speed of the plane.

Concept Summary 5.1 Uniform Circular Motion Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path. The period T is the time required for the object to travel once around the circle. The speed υ of the object is related to the period and the radius r of the circle by Equation 5.1.

υ = 2πr T

(5.1)

5.2 Centripetal Acceleration An object in uniform circular motion exper- iences an acceleration, known as centripetal acceleration. The magnitude ac of the centripetal acceleration is given by Equation 5.2, where υ is the speed of the object and r is the radius of the circle. The direction of the centripetal acceleration vector always points toward the center of the circle and continu- ally changes as the object moves.

ac = υ 2

r (5.2)

5.3 Centripetal Force To produce a centripetal acceleration, a net force pointing toward the center of the circle is required. This net force is called the

centripetal force, and its magnitude Fc is given by Equation 5.3, where m and υ are the mass and speed of the object, and r is the radius of the circle. The direction of the centripetal force vector, like that of the centripetal accelera- tion vector, always points toward the center of the circle.

Fc = mυ 2

r (5.3)

5.4 Banked Curves A vehicle can negotiate a circular turn without relying on static friction to provide the centripetal force, provided the turn is banked at an angle relative to the horizontal. The angle θ at which a friction- free curve must be banked is related to the speed υ of the vehicle, the radius r of the curve, and the magnitude g of the acceleration due to gravity by Equation 5.4.

tan θ = υ 2

rg (5.4)

5.5 Satellites in Circular Orbits When a satellite orbits the earth, the gravitational force provides the centripetal force that keeps the satellite mov- ing in a circular orbit. The speed υ and period T of a satellite depend on the

υ

FN mg

FIGURE 5.20 The normal force and weight of a pilot both point down toward the center of the circular motion while fl ying upside down at the top of a vertical circle. The speed of the plane will change the magnitude of the normal force (i.e., how much the seat pushes against the pilot).

138 CHAPTER 5 Dynamics of Uniform Circular Motion

mass ME of the earth and the radius r of the orbit according to Equations 5.5 and 5.6, where G is the universal gravitational constant.

υ = √GMEr (5.5)

T = 2πr 3/2

√GME (5.6)

5.6 Apparent Weightlessness and Artifi cial Gravity The apparent weight of an object is the force that it exerts on a scale with which it is in

contact. All objects, including people, on board an orbiting satellite are in free-fall, since they experience negligible air resistance and they have an ac- celeration that is equal to the acceleration due to gravity. When a person is in free-fall, his or her apparent weight is zero, because both the person and the scale fall freely and cannot push against one another.

5.7 Vertical Circular Motion Vertical circular motion occurs when an object, such as a motorcycle, moves on a vertical circular path. The speed of the object often varies from moment to moment, and so do the magnitudes of the centripetal acceleration and centripetal force.

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 5.2 Centripetal Acceleration 1. Two cars are traveling at the same constant speed υ. As the drawing in- dicates, car A is moving along a straight section of the road, while car B is rounding a circular turn. Which statement is true about the accelerations of the cars? (a) The acceleration of both cars is zero, since they are traveling at a constant speed. (b) Car A is accelerating, but car B is not accelerating. (c) Car A is not accelerating, but car B is accelerating. (d) Both cars are accelerating.

QUESTION 1

B

A

υ

υ

3. Two cars are driving at the same constant speed υ around a racetrack. However, they are traveling through turns that have diff erent radii, as shown in the drawing. Which statement is true about the magnitude of the centripetal acceleration of each car? (a) The magnitude of the centripetal acceleration of each car is the same, since the cars are moving at the same speed. (b) The magnitude of the centripetal acceleration of the car at A is greater than that of the car at B, since the radius of the circular track is smaller at A. (c) The magnitude of the centripetal acceleration of the car at A is greater than that of the car at B, since the radius of the circular track is greater at A. (d) The magnitude of the centripetal acceleration of the car at A is less than that of the car at B, since the radius of the circular track is smaller at A.

QUESTION 3

BA

υ

υ

Section 5.3 Centripetal Force 7. The drawing shows two identical stones attached to cords that are be- ing whirled on a tabletop at the same speed. The radius of the larger circle is twice that of the smaller circle. How is the tension T1 in the longer cord related to the tension T2 in the shorter cord? (a) T1 = T2 (b) T1 = 2T2 (c) T1 = 4T2 (d) T1 = 12T2 (e) T1 =

1 4T2

QUESTION 7 Path 1

Identical stones

Path 2

υ

υ

8. Three particles have the following masses (in multiples of m0) and move on three diff erent circles with the following speeds (in multiples of υ0) and radii (in multiples of r0):

Particle Mass Speed Radius 1 m0 2υ0 r0 2 m0 3υ0 3r0 3 2m0 2υ0 4r0

Rank the particles according to the magnitude of the centripetal force that acts on them, largest fi rst. (a) 1, 2, 3 (b) 1, 3, 2 (c) 2, 1, 3 (d) 2, 3, 1 (e) 3, 2, 1

Section 5.4 Banked Curves 10. Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same radius r. There are two forces acting on each car, its weight mg and the normal force FN exerted by the road. Recall that the weight of an object on the moon is about one- sixth of its weight on earth. How does the centripetal force on the moon com- pare with that on the earth? (a) The centripetal forces are the same. (b) The centripetal force on the moon is less than that on the earth. (c) The centripetal force on the moon is greater than that on the earth.

Section 5.5 Satellites in Circular Orbits 11. Two identical satellites are in orbit about the earth. As the drawing shows, one orbit has a radius r and the other 2r. The centripetal force on the

Focus on Concepts

Problems 139

satellite in the larger orbit is ________ as that on the satellite in the smaller orbit. (a) the same (b) twice as great (c) four times as great (d) half as great (e) one-fourth as great

QUESTION 11

r

2r

Earth

Section 5.7 Vertical Circular Motion 15. The drawing shows an extreme skier at the bottom of a ski jump. At this point the track is circular with a radius r. Two forces act on the skier, her weight mg and the normal force FN. Which relation describes how the net force acting on her is related to her mass m and speed υ and to the radius r? Assume that “up” is the positive direction.

(a) FN + mg = mυ 2

r (b) FN − mg =

mυ 2

r

(c) FN = mυ 2

r (d) −mg = mυ

2

r

QUESTION 15

FN

r

mg

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual MMH Problem-solving help GO Guided Online Tutorial V-HINT Video Hints CHALK Chalkboard Videos

BIO Biomedical application E Easy M Medium H Hard

Section 5.1 Uniform Circular Motion,

Section 5.2 Centripetal Acceleration 1. E SSM A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?

2. E GO The following table lists data for the speed and radius of three examples of uniform circular motion. Find the magnitude of the centripetal acceleration for each example.

Radius Speed Example 1 0.50 m 12 m/s

Example 2 Infinitely large 35 m/s

Example 3 1.8 m 2.3 m/s

3. E Available in WileyPLUS. 4. E GO Speedboat A negotiates a curve whose radius is 120 m. Speedboat B negotiates a curve whose radius is 240 m. Each boat experiences the same centripetal acceleration. What is the ratio υA/υB of the speeds of the boats? 5. E SSM How long does it take a plane, traveling at a constant speed of 110 m/s, to fl y once around a circle whose radius is 2850 m?

6. E BIO The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of

the body. The curved artery can be approximated as a semicircular arch whose diameter is 5.0 cm. If blood fl ows through the aortic arch at a speed of 0.32 m/s, what is the magnitude (in m/s2) of the blood’s centripetal acceleration?

7. E MMH The blade of a windshield wiper moves through an angle of 90.0° in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade?

8. E V-HINT There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cyl- inder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 12 cm. When the cylinder is rotating at 2.0 revolutions per second, what is the mag- nitude of the centripetal acceleration at the outer wall?

9. E SSM Computer-controlled display screens provide drivers in the Indi- anapolis 500 with a variety of information about how their cars are performing. For instance, as a car is going through a turn, a speed of 221 mi/h (98.8 m/s) and centripetal acceleration of 3.00 g (three times the acceleration due to gravity) are displayed. Determine the radius of the turn (in meters).

10. M V-HINT A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 120 m/s2. What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

11. M BIO A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is 6.25 × 103 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 5.00 cm from the axis of rotation?

12. M CHALK The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 × 106 m, de- termine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0° north of the equator.

Problems

140 CHAPTER 5 Dynamics of Uniform Circular Motion

Section 5.3 Centripetal Force 13. E SSM Review Example 3, which deals with the bobsled in Interactive Figure 5.5. Also review Conceptual Example 4. The mass of the sled and its two riders in Interactive Figure 5.5 is 350 kg. Find the magnitude of the centripetal force that acts on the sled during the turn with a radius of (a) 33 m and (b) 24 m. 14. E At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s, and an 83-kg person feels a 560-N force pressing against his back. What is the radius of the chamber?

15. E MMH Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for which the coeffi cient of static fric- tion is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coeffi - cient of static friction is 0.85 on the same curve. What is the maximum speed at which car B can negotiate the curve?

16. E A speed skater goes around a turn that has a radius of 31 m. The skater has a speed of 14 m/s and experiences a centripetal force of 460 N. What is the mass of the skater?

17. E For background pertinent to this problem, review Conceptual Example 6. In Figure 5.7 the man hanging upside down is holding a partner who weighs 475 N. Assume that the partner moves on a circle that has a radius of 6.50 m. At a swinging speed of 4.00 m/s, what force must the man apply to his partner in the straight-down position?

18. E MMH A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

19. E See Conceptual Example 6 to review the concepts involved in this problem. A 9.5-kg monkey is hanging by one arm from a branch and is swinging on a vertical circle. As an approximation, assume a radial distance of 85 cm between the branch and the point where the monkey’s mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 2.8 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey’s arm. 20. E GO Multiple-Concept Example 7 deals with the concepts that are important in this problem. A penny is placed at the outer edge of a disk (radius = 0.150 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.80 s. Find the minimum coef- fi cient of friction necessary to allow the penny to rotate along with the disk.

21. M CHALK SSM The hammer throw is a track-and-fi eld event in which a 7.3-kg ball (the “hammer”) is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion and eventually returns to earth some distance away. The world record for this distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball is released above the ground rather than at ground level. Furthermore, assume that the ball is whirled on a circle that has a radius of 1.8 m and that its velocity at the instant of release is directed 41° above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.

22. M V-HINT An 830-kg race car can drive around an unbanked turn at a maximum speed of 58 m/s without slipping. The turn has a radius of curvature of 160 m. Air fl owing over the car’s wing exerts a downward-pointing force (called the downforce) of 11 000 N on the car. (a) What is the coeffi cient of static friction between the track and the car’s tires? (b) What would be the maximum speed if no downforce acted on the car?

23. M CHALK MMH A “swing” ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a vertical rotating pole, as

the drawing shows. Suppose the total mass of a chair and its occupant is 179 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.

PROBLEM 23

60.0° 15.0 m

Section 5.4 Banked Curves 24. E On a banked race track, the smallest circular path on which cars can move has a radius of 112 m, while the largest has a radius of 165 m, as the drawing illustrates. The height of the outer wall is 18 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

PROBLEM 24

18 m

165 m

112 m

25. E V-HINT Before attempting this problem, review Examples 7 and 8. Two curves on a highway have the same radii. However, one is unbanked and the other is banked at an angle θ. A car can safely travel along the unbanked curve at a maximum speed υ0 under conditions when the coeffi cient of static friction between the tires and the road is μs = 0.81. The banked curve is fric- tionless, and the car can negotiate it at the same maximum speed υ0. Find the angle θ of the banked curve. 26. E GO A woman is riding a Jet Ski at a speed of 26 m/s and notices a seawall straight ahead. The farthest she can lean the craft in order to make a turn is 22°. This situation is like that of a car on a curve that is banked at an angle of 22°. If she tries to make the turn without slowing down, what is the minimum distance from the seawall that she can begin making her turn and still avoid a crash?

27. E Two banked curves have the same radius. Curve A is banked at an angle of 13°, and curve B is banked at an angle of 19°. A car can travel around curve A without relying on friction at a speed of 18 m/s. At what speed can this car travel around curve B without relying on friction?

28. M MMH A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in circles that are parallel to the ground. For a speed of 34.0 m/s, at what value of the distance d should a driver locate his car if he wishes to stay on a circular path without depending on friction?

PROBLEM 28 40.0° d

40.0°

29. M GO A jet fl ying at 123 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is 2.00 × 105 kg. Calculate the magnitude of the necessary lifting force.

30. H The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 11.0 m) as the carousel turns. The coeffi cient of static friction between the suitcase and the carousel is 0.760, and the angle θ in the drawing is 36.0°. How much time is required for your suitcase to go around once?

Additional Problems 141

PROBLEM 30

r θ

Section 5.5 Satellites in Circular Orbits,

Section 5.6 Apparent Weightlessness and Artificial Gravity

31. E GO Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 360 km above the earth’s surface, while that for satellite B is at a height of 720 km. Find the orbital speed for each satellite.

32. E A rocket is used to place a synchronous satellite in orbit about the earth. What is the speed of the satellite in orbit?

33. E SSM A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.70 × 104 m/s, and the radius of the orbit is 5.25 × 106 m. A second satellite also has a circular orbit around this same planet. The orbit of this second satellite has a radius of 8.60 × 106 m. What is the orbital speed of the second satellite?

34. E Available in WileyPLUS. 35. E GO A satellite is in a circular orbit about the earth (ME = 5.98 × 1024 kg). The period of the satellite is 1.20 × 104 s. What is the speed at which the satellite travels?

36. E A satellite circles the earth in an orbit whose radius is twice the earth’s radius. The earth’s mass is 5.98 × 1024 kg, and its radius is 6.38 × 106 m. What is the period of the satellite?

37. M SSM Available in WileyPLUS. 38. M GO A satellite has a mass of 5850 kg and is in a circular orbit 4.1 × 105 m above the surface of a planet. The period of the orbit is 2.00 hours. The radius of the planet is 4.15 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?

39. M V-HINT Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 43.3 km/s and 58.6 km/s. The slower planet’s orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

40. H Available in WileyPLUS.

Section 5.7 Vertical Circular Motion 41. E SSM A motorcycle has a constant speed of 25.0 m/s as it passes over the top of a hill whose radius of curvature is 126 m. The mass of the motor- cycle and driver is 342 kg. Find the magnitudes of (a) the centripetal force and (b) the normal force that acts on the cycle. 42. E BIO V-HINT Pilots of high-performance fi ghter planes can be sub- jected to large centripetal accelerations during high-speed turns. Because of these accelerations, the pilots are subjected to forces that can be much greater than their body weight, leading to an accumulation of blood in the abdomen and legs. As a result, the brain becomes starved for blood, and the pilot can lose consciousness (“black out”). The pilots wear “anti-G suits” to help keep the blood from draining out of the brain. To appreciate the forces that a fi ghter pilot must endure, consider the magnitude FN of the normal force that the pilot’s seat exerts on him at the bottom of a dive. The magnitude of the pilot’s weight is W. The plane is traveling at 230 m/s on a vertical circle of radius 690 m. Determine the ratio FN/W. For comparison, note that blackout can occur for values of FN/W as small as 2 if the pilot is not wearing an anti-G suit. 43. E SSM For the normal force in Interactive Figure 5.19 to have the same magnitude at all points on the vertical track, the stunt driver must ad- just the speed to be diff erent at diff erent points. Suppose, for example, that the track has a radius of 3.0 m and that the driver goes past point 1 at the bottom with a speed of 15 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?

44. E GO A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 21 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 770 N. At the top of the loop, the rider is upside down and moving, and the sensor reads 350 N. What is the speed of the rider at the top of the loop?

45. E GO A 0.20-kg ball on a stick is whirled on a vertical circle at a con- stant speed. When the ball is at the three o’clock position, the stick tension is 16 N. Find the tensions in the stick when the ball is at the twelve o’clock and at the six o’clock positions.

46. M GO MMH A stone is tied to a string (length = 1.10 m) and whirled in a circle at the same constant speed in two diff erent ways. First, the circle is hori- zontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 15.0% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

47. M GO A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of 45.0 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

48. H Available in WileyPLUS.

49. E SSM In a skating stunt known as crack-the-whip, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of 80.0 kg and is 6.10 m from the pivot. He is skat- ing at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him.

50. E BIO Available in WileyPLUS. 51. E GO MMH The second hand and the minute hand on one type of clock are the same length. Find the ratio (ac, second/ac, minute) of the centripetal acceler- ations of the tips of the second hand and the minute hand.

52. E MMH A child is twirling a 0.0120-kg plastic ball on a string in a hori- zontal circle whose radius is 0.100 m. The ball travels once around the circle in 0.500 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

53. E MMH Available in WileyPLUS. 54. E GO MMH Two cars are traveling at the same speed of 27 m/s on a curve that has a radius of 120 m. Car A has a mass of 1100 kg, and car B has a mass of 1600 kg. Find the magnitude of the centripetal acceleration and the magnitude of the centripetal force for each car.

Additional Problems

142 CHAPTER 5 Dynamics of Uniform Circular Motion

In uniform circular motion the concepts of acceleration and force play central roles. Problem 63 deals with acceleration, particularly in connection with the equations of kinematics. Problem 64 deals with force, and it stresses that centripetal force is the net force along the radial direction. This means that all forces along the radial direction must be taken into account when identifying the centripetal force.

63. M CHALK SSM At time t = 0 s, automobile A is traveling at a speed of 18 m/s along a straight road and is picking up speed with an acceleration that has a magnitude of 3.5 m/s2 as in part a of the fi gure. At t = 0 s, automobile B is traveling at a speed of 18 m/s in uniform circular motion as it negotiates a turn (part b of the fi gure). It has a centripetal acceleration whose magnitude

is also 3.5 m/s2. Concepts: (i) Which automobile has a constant acceleration? (ii) For which automobile do the equations of kinematics apply? Calcula- tions: Determine the speed of each automobile when t = 2 s. 64. M CHALK Ball A is attached to one end of a rigid massless rod, while an identical ball B is attached to the center of the rod, as shown in the fi gure. Each ball has a mass of m = 0.50 kg, and the length of each half of the rod is L = 0.40 m. This arrangement is held by the empty end and is whirled around in a horizontal circle at a constant rate, so that each ball is in uniform circular motion. Ball A travels at a constant speed of υA = 5.0 m/s. Concepts: (i) How many tension forces contribute to the centripetal force that acts on ball A? (ii) How many tension forces contribute to the centripetal force that acts on ball B? (iii) Is the speed of ball A the same as that of ball B? Calculations: Find the tension in each half of the rod.

Concepts and Calculations Problems

55. E SSM A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing up- ward on her with a force equal to twice her weight as she goes through the dip. If r = 20.0 m, how fast is the roller coaster traveling at the bottom of the dip? 56. E BIO The National Aeronautics and Space Administration (NASA) studies the physiological eff ects of large accelerations on astronauts. Some of these studies use a machine known as a centrifuge. This machine con- sists of a long arm, to one end of which is attached a chamber in which the astronaut sits. The other end of the arm is connected to an axis about which the arm and chamber can be rotated. The astronaut moves on a cir- cular path, much like a model airplane fl ying in a circle on a guideline. The chamber is located 15 m from the center of the circle. At what speed must the chamber move so that an astronaut is subjected to 7.5 times the acceleration due to gravity?

57. M SSM Available in WileyPLUS. 58. M V-HINT The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6.7 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 3.0 m from the center of the circle.

59. M A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 28 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 150 m), the block swings toward the outside of the curve. Then the string makes an angle θ with the vertical. Find θ.

60. M GO A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass m1 attached to the center of the rod and a particle of mass m2 attached to the outer end of the rod. The inner section of the rod sustains a tension that is three times as great as the tension that the outer section sustains. Find the ratio m1/m2. 61. H SSM Available in WileyPLUS. 62. M GO A space laboratory is rotating to create artifi cial gravity, as the fi gure indicates. Its period of rotation is chosen so that the outer ring (rO = 2150 m) simulates the acceleration due to gravity on earth (9.80 m/s2). What should be the radius r1 of the inner ring, so it simulates the acceleration due to gravity on the surface of Mars (3.72 m/s2)?

PROBLEM 62

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PROBLEM 64

Team Problems 143

65. M Multilevel Artifi cial Gravity. You and your team have the task of designing a space station that will mimic the force due to gravity on the earth,

the moon, and Mars. The station’s purpose is to physically prepare astronauts

for the transition to the diff erent gravitational forces they will experience on

Mars and the moon, but also to acclimate them to Earth’s gravity when they

return. The space station should be constructed of concentric cylinders, with

a maximum allowed diameter of 300 m. (a) Draw a sketch of the space station and label the levels (i.e., the cylinders) according to the celestial body whose

gravitational force each cylinder simulates. (b) Determine the required radii of each cylinder. Assume the outermost cylinder has the maximum allowable

diameter. (c) What is the period of rotation of the station? (gEarth = 9.80 m/s2, gMars = 3.76 m/s

2, and gmoon = 1.62 m/s 2)

66. M A Dangerous Ride. You and your exploration team are stuck on a steep slope in the Andes Mountains in Argentina. A deadly winter storm is

approaching and you must get down the mountain before the storm hits. Your

path leads you around an extremely slippery, horizontal curve with a diameter

of 90 m and banked at an angle of 40.0° relative to the horizontal. You get

the idea to unpack the toboggan that you have been using to haul supplies,

load your team upon it, and ride it down the mountain to get enough speed to

get around the banked curve. You must be extremely careful, however, not to

slide down the bank: At the bottom of the curve is a steep cliff . (a) Neglecting friction and air resistance, what must be the speed of your toboggan in order

to get around the curve without sliding up or down its bank? Express your

answer in m/s and m.p.h. (b) You will need to climb up the mountain and ride the toboggan down in order to attain the speed you need to safely navigate

the curve (from part (a)). The mountain slope leading into the curve is at an

angle of 30.0° relative to the horizontal, and the coeffi cient of kinetic friction

between the toboggan and the surface of the slope is (μmountain = 0.15). How far up the mountain (distance along the slope, not elevation) from the curve

should you start your ride? Note: the path down the mountain levels off at the

bottom so that the toboggan enters the curve moving in the horizontal plane

(i.e., in the same plane as the curve).

Team Problems

144

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

6.1 Determine the work done by a constant force.

6.2 Relate work and kinetic energy.

6.3 Use gravitational potential energy.

6.4 Identify conservative and nonconservative forces.

6.5 Use the law of conservation of mechanical energy.

6.6 Solve problems involving nonconservative forces.

6.7 Solve problems involving power.

6.8 Employ the law of conservation of energy.

6.9 Determine the work done by a variable force.

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CHAPTER 6

Work and Energy

On the popular Behemoth roller coaster at Canada’s Wonderland amusement park, riders cresting the top of the hill will pick up speed on the way down, as they convert gravitational potential energy into kinetic energy. This conversion is governed by the principle of conservation of energy, a central topic in this chapter.

6.1 Work Done by a Constant Force Work is a familiar concept. For example, it takes work to push a stalled car. In fact, more work is done when the pushing force is greater or when the displacement of the car is greater. Force and displacement are, in fact, the two essential elements of work, as Figure 6.1 illustrates. The drawing shows a constant pushing force F

→ that points

in the same direction as the resulting displacement s→.* In such a case, the work W is defi ned as the magnitude F of the force times the magnitude s of the displacement: W = Fs. The work done to push a car is the same whether the car is moved north to south or east to west, provided that the amount of force used and the distance moved are the same. Since work does not convey directional information, it is a scalar quantity.

The equation W = Fs indicates that the unit of work is the unit of force times the unit of distance, or the newton · meter in SI units. One newton · meter is referred to

*When discussing work, it is customary to use the symbol s→ for the displacement, rather than x→ or y→ .

FF

s

FIGURE 6.1 Work is done when a force F→ pushes a car through a displacement s→.

6.1 Work Done by a Constant Force 145

as a joule (J) (rhymes with “cool”), in honor of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table 6.1 summarizes the units for work in several systems of measurement.

The defi nition of work as W = Fs does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work.

Often, the force and displacement do not point in the same direction. For instance, Figure 6.2a shows a suitcase-on-wheels being pulled to the right by a force that is applied along the handle. The force is directed at an angle 𝜃 relative to the displacement. In such a case, only the component of the force along the displacement is used in defi ning work. As Figure 6.2b shows, this compo- nent is F cos 𝜃, and it appears in the general defi nition below:

DEFINITION OF WORK DONE BY A CONSTANT* FORCE The work done on an object by a constant force F→ is

W = (F cos θ) s (6.1)

where F is the magnitude of the force, s is the magnitude of the displacement, and 𝜽 is the angle between the force and the displacement. SI Unit of Work: newton · meter = joule (J)

When the force points in the same direction as the displacement, then 𝜃 = 0°, and Equation 6.1 reduces to W = Fs. Example 1 shows how Equation 6.1 is used to calculate work.

TABLE 6.1 Units of Measurement for Work

System Force × Distance = Work

SI newton (N) meter (m) joule (J)

CGS dyne (dyn) centimeter (cm) erg

BE pound (lb) foot (ft) foot · pound (ft · 1b)

*Section 6.9 considers the work done by a variable force.

EXAMPLE 1 Pulling a Suitcase-on-Wheels

Find the work done by a 45.0-N force in pulling the suitcase in Figure 6.2a at an angle 𝜃 = 50.0° for a distance s = 75.0 m.

Reasoning The pulling force causes the suitcase to move a distance of 75.0 m and does work. However, the force makes an angle of 50.0° with the displacement, and we must take this angle into account by using the defi nition of work given by Equation 6.1.

Solution The work done by the 45.0-N force is

W = (F cos θ)s = [(45.0 N)cos 50.0°](75.0 m) = 2170 J

The answer is expressed in newton · meters or joules (J).

FF

(a) (b)

F cos