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Study Unit Ill Engineerin M

Part4 an1cs

Andrew Pytel, Ph.D. Associate Professor, Engineering Mechanics

The Pennsylvania State University

When you complete this study unit, you'll be able to

• Calculate the mass moment of inertia

• Calculate the kinetic energy of a body

• Determine the linear impulse and momentum of a body

• Analyze the equations and conditions used to determine the forces involving rectilinear translation

• Describe centripetal and centrifugal force

• Describe the forces that impact the rotation of a rigid body without translation

• Explain the motion of a wheel, and calculate the magnitude of the linear acceleration and friction forces

• Analyze the work-energy method as it applies to the motion and action of a body

iii

PRELIMINARY EXPLANATIONS PERTAINING TO KINETICS .

FORCE-MASS-ACCELERATION METHOD .....

Translation of Rigid Body Rotation of Rigid Body without Translation General Plane Motion of Rigid Body

WORK-ENERGY METHOD . . . . . . . . . . . . . . . . . . . . . . . . . 53

Application of Method for Translation Other Applications of Work-Energy Method

IMPULSE-MOMENTUM METHOD . . . . . . .

Rectilinear Translation of Single Body Collision of Two Bodies

PRACTICE PROBLEMS ANSWERS

EXAMINATION . . . . . . . . . .

. ........... 77

Engineering Nlechanics, Part 4

PRELIMINARY EXPLANATIONS PERTAINING TO KINETICS

Scope of This Text

1 • In the preceding texts on engineering mechanics, we have discussed separately the relations of forces in a system and the conditions of mo- tion of bodies. In this text, we shall consider the relation between the motion of a body and the force or forces acting on the body to produce the motion. The basis for the relationship between motion and force is Newton's second law of motion. However, there are three different methods of applying this law. These are commonly called the force- mass acceleration method, the work-energy method, and the impulse- momentum method. Each method is most useful for solving certain types of problems.

Statement of Newton's Second Law of Motion

2 • In Engineering Mechanics, Part 1, Newton's second law of motion was stated as follows:

If a resultant force acts upon a particle, the particle will be accelerated in the direction of the force. Furthermore, the magnitude of the accel- eration will be directly proportional to the magnitude of the resultant force and inversely proportional to the mass of the particle.

Newton's second law can be expressed mathematically by the following equation:

F a=k-

in which a = magnitude of the acceleration of a particle k = a numerical factor F = magnitude of the force acting upon the particle m = mass of the particle

(1)

The mass of a particle is a measure of the exact amount of matter in the particle. Any body is composed of a number of particles, and the mass of a body is the sum of the masses of all the particles in the body.

2 Engineering Mechanics, Part 4

The acceleration and the force are actually vector quantities, because their directions must be considered. However, since the acceleration and the force must have the same direction, it is permissible to consider only the magnitudes of these quantities in a particular problem. In Engineering Mechanics, Part 3, it was explained that these magnitudes

can be indicated by the notations fdl and 111. The simpler notations a and F will be used in this text, but you must remember that accelera- tions and forces are really vector quantities. The mass of a particle is a scalar quantity.

The value of the factor k depends on the units in which the other quantities in the equation are expressed. However, it is the usual prac- tice to choose the units in such a manner that the value of k will be 1. When suitable units are used, the preceding equation becomes

F a=-

This relation is generally given in the following form:

F=ma

Newton's Law of Universal Gravitation

(2)

(3)

3 • Another law that was formulated by Newton is known as the law of universal gravitation. According to this law, any two particles of matter are attracted to each other with a force which acts along the straight line between the two particles and which has a magnitude expressed by the following relationship:

in which F = magnitude of the force of attraction K = a numerical factor

m1 and m2 = masses of the two particles r = length of the straight line between the particles

(1)

The force F is actually a vector quantity. However, since it acts in one direction on one particle and it acts in the opposite direction on the other particle, and since the position of its line of action is fixed by the locations of the particles, it is permissible to consider only the magni- tude of the force in a particular problem. The factor K is commonly called the universal gravitation constant. The value used for this constant depends on the units in which the force, the masses, and the distance are expressed. These units must be consistent.

Engineering Mechanics, Part 4

The magnitude of the force of attraction between two bodies is the sum of the magnitudes of the forces of attraction between the particles of one body and the particles of the other body. In ordinary practical work, the only force of attraction that need be considered is the force between the earth and some other body. This force is commonly called the weight of the body. If W denotes the weight of a body, m the mass of the body, me the mass of the earth, and re the distance from the center of gravity of the earth to the center of gravity of the other body, the relation given by equation 1 becomes

(2)

Since the weight W is a force, it follows from equation 3, Article 2, that Km

the factor __ e in equation 2 must represent an acceleration. This fac- re

tor is really the acceleration due to gravity mentioned in Engineering Mechanics, Part 3, and is usually denoted by g. Therefore, equation 2 is usually written in the following form:

W=gm (3)

The distance re is not exactly the same for every point on the earth's surface. Therefore, the value of g varies slightly in different localities on the earth. However, for ordinary practical purposes, it may be taken as 32.2 fps per sec (feet per second per second), or 9.81 meters per sec per sec.

Relation Between Mass and Weight of a Body

4 • Although the amount of matter contained in a body is indicated exactly by the mass of the body, it is not practicable to determine the mass of a body directly. Since the weight of a body can usually be measured quite easily, the amount of matter in a body is generally indicated by its weight. When it is necessary to consider the mass of a body, the following relationship (obtained from equation 3, Article 3) may be used:

w m=-g

in which m =mass of a body W =weight of the body g = acceleration due to gravity

The units for mass will be discussed next.

4 Engineering Mechanics, Part 4

Systems of Units

5 • Two general categories of systems of units are used in problems in which Newton's second law of motion is applied. Systems in one cate- gory are called absolute systems, and those in the other category are called gravitational systems. In a system in either category, there are three fundamental units. Also, in any system, two of these fundamen- tal units are a unit of length, or distance, and a unit of time. However, in an absolute system, the third fundamental unit is a unit of mass; whereas, in a gravitational system, the third unit is a unit of force.

Acceleration is defined as a change in velocity in a unit of time; velocity is defined as a displacement in a unit of time. When Newton's second law is applied, it is necessary to use the same unit of time for the change in velocity and for the displacement. Therefore, acceleration involves a unit of distance and a unit of time. In an absolute system of units, there is a fundamental unit of mass, and the unit of force needed to make the factor kin equation 1, Article 2, equal to 1 depends on the units used for length, time, and mass. In a gravitational system of units, there is a fundamental unit of force, and the unit of mass needed to make the factor k equal to 1 depends on the units used for length, time, and force.

For English units, it is the common practice to use a gravitational system, known as the British Engineering System, in which the unit of length is the foot, the unit of time is the second, and the unit of force is the pound. Then the unit of mass is called the slug. To deter- mine the mass of a body in slugs from the weight of the body in pounds by applying the relationship in Article 4, it is necessary to di- vide the weight by 32.2. For instance, if the weight of a body is 10 lb (pounds), the mass of the body is 1%2.2 = 0.311 slug.

For metric units, there are three common systems. One such system is an absolute system, called the CGS (Centimeter-Gram-Second) System, in which the unit of length is the centimeter, the unit of time is the sec- ond, and the unit of mass is the gram. Then the unit of force is called the dyne. When equation 3, Article 2, is applied, the value of a force in dynes is obtained by multiplying the mass in grams by the acceleration in centimeters per second per second.

In a gravitational system for metric units, called the MKS (Meter- Kilogram-Second) System, the unit of length is the meter, the unit of time is the second, and the unit of force is the kilogram. Then the unit of mass is called the metric slug. To determine the mass of a body in metric slugs from the weight of the body in kilograms by applying the relationship in Article 4, it is necessary to divide the weight by 9.81.

The third system for metric units is an absolute system called the International System of Units, abbreviated "SI." In this system, the unit of length is the meter, the unit of time is the second, and the unit

Engineering Mechanics, Part 4

of mass is the kilogram. Then the unit of force is called the newton. When equation 3, Article 2, is applied, the value of a force in newtons is obtained by multiplying the mass in kilograms by the acceleration in meters per second per second. You must be careful to distinguish between a kilogram of mass used in the International System and a kilogram of force used in the MKS System.

Mass Moment of Inertia of Particle

6 • The mass moment of inertia of a particle with respect to a certain straight reference line is defined by the following relationship:

in which Ix = mass moment of inertia of a particle m = mass of the particle r x = perpendicular distance from the reference line to the

particle

In Figure lA are shown a horizontal reference line, a vertical reference line, and a particle. This particle is so located that the vertical distance from the horizontal reference line to the particle is r1 and the horizontal distance from the vertical reference line to the particle is r2. If the mass of the particle is m1r its mass moment of inertia with respect to the horizontal reference line is m1r/ and its mass moment of inertia with respect to the vertical reference line is m1rt In Figure lB are shown a selected reference line and a particle which has a mass m2 and is so located that the distance from the reference line to the particle in a direction at right angles to the line is r3• The mass moment of inertia of this particle with respect to the reference line is m1rt

VERTICAL REFERENCE LINE

r2 - f - - - -~ PARTICLE 90°

I r I 1

90~1 I

HORIZONTAL REFERENCE LINE

FIGURE 1-Mass Moment of Inertia of Particle

' ' 90o,.)W"' ' r3 ' ' :-

PARTICLE

6 Engineering Mechanics, Part 4

The term moment of inertia has been adopted for convenience to refer to the quantity that would be obtained by multiplying the mass of a particle by the square of the distance from a line to the particle. Since the "moment" of the mass of a particle could be the value of the quantity mrx, the quantity mrx2 is often called a "second moment."

In the English system of units, the unit commonly used for mass moment of inertia is slug-feet2 if the distance rx is in feet or slug-inches2 if the distance r x is in inches.

In the metric system, the unit used for mass moment of inertia depends on the units used for mass and length. Common units are kilogram- meters2 and gram-centimeters2.

Formulas for Mass Moment of Inertia of Basic Bodies

7 • It is usually necessary to consider the mass moment of inertia of a body, rather than the mass moment of inertia of a single particle. Since a body is composed of a number of particles, the mass moment of inertia of a body with respect to any selected reference line is equal to the sum of the moments of inertia of all the particles in the body with respect to the same reference line. When a body has a simple regular shape and the reference line passes through the center of gravity of the body, the mass moment of inertia of the body can be computed by applying a simple formula. In this text, we shall consider only a few special cases.

When the body is a sphere composed of uniform material and the reference line is any diameter of the sphere (a diameter must pass through the center of gravity of the sphere), the formula for the mass momentofinertiais

2 Ic=-mr

5 (1)

in which Ic = mass moment of inertia of a sphere with respect to any diameter

m = mass of the entire sphere r =radius of the sphere

When the body is a right circular cylinder and the reference line is the line called the axis of the cylinder, as in Figure 2, the formula for the mass moment of inertia of the body is

(2)

in which Ic = mass moment of inertia of a right circular cylinder with respect to the axis of the cylinder

m = mass of the entire cylinder r =radius of a circular cross section of the cylinder

Engineering Mechanics, Part 4

1., 2. Diameters through center of gravity

FIGURE 2-Dimensions of Right Circular Cylinder

When the body is a right circular cylinder and the reference line is any diameter of the circular cross section through the center of gravity of the cylinder, as either line 1 or line 2 in Figure 2, the formula for the mass moment of inertia of the body is

1 2 1 2 Ic= 4mr + 12mz

in which Ic = mass moment of inertia of a right circular cylinder

(3)

with respect to any diameter through its center of gravity m = mass of the entire cylinder r =radius of a cross section of the cylinder l = length of the cylinder

When the body is a rectangular solid, like that represented in Figure 3, and the reference line is the longitudinal line 1 through the center of gravity, the formula for the mass moment of inertia is

1. longitudinal line through center of gravity 2. 3. Traverse lines through center of gravity

FIGURE 3-Dimensions of Rectangular Solid

8 Engineering Mechanics, Part 4

1 2 2 Ic = 12 m(a + b ) (4)

in which a and b are the dimensions of a cross section of the body.

When the body is a rectangular solid and the reference line is the transverse line 2 in Figure 3 through the center of gravity, the mass moment of inertia can be computed by the formula

1 2 2 Ic = 12 m(l + a ) (5)

If the reference line is the transverse line 3 in Figure 3, the formula for a rectangular solid is

1 2 2 I =-- m(l +b) G 12 (6)

Mass Moment of Inertia of Body with Respect to Any Line

8 • It is sometimes necessary to know the mass moment of inertia of a body with respect to a line that does not pass through the center of gravity of the body. Examples are the bodies represented in Figures 4 and 5. In Figure 4, the body is a right circular cylinder, but the reference line for the moment of inertia is located at one end of the cylinder. In Figure 5 is shown a side view of a body that consists of a sphere and a right circular cylinder which are welded together. The reference line for the moment of inertia of the entire body is located at the free end of the cylinder. There is a relatively simple relationship between the mass moment of inertia of a body with respect to any reference line and the mass moment of inertia of the same body with respect to a par- allel line through the center of gravity of the body. This relationship, which is given here_ without proof, is as follows:

10 =lc+md 2

in which 10 =mass moment of inertia of a body with respect to a reference line through any point 0

Ic = mass moment of inertia of the body with respect to a parallel line through the center of gravity of the body

m =mass of the body d = perpendicular distance from the specified reference line

to the center of gravity of the body

It is important to remember that the line through the center of gravity of the body must be parallel to the specified reference line. The method of applying the relationship just given is illustrated in the following example problems.

Engineering Mechanics, Part 4

Example Problems

FIGURE 4-cy/inder with Reference Line at One End

CYLINDER

FIGURE 5-connected Cylinder and Sphere

REFERENCE LINE

At intervals throughout this text you will find one or more example problems solved to illustrate clearly the application of a principle, a rule, or a formula. Read each problem carefully and study the solution until you understand it thoroughly.

Problem 1: The diameter of the cylinder represented in Figure 4 is 1 ft (foot), and its length is 6ft. If the cylinder weighs 64.4lb, what is its mass moment of inertia with respect to the specified reference line which is perpendicular to the axis of the cylinder and passes through the centroid of the cross section at one end of the cylinder?

Solution: In this problem, the value to be used for Ic in the formula for computing 10 is the mass moment of inertia of the cylinder with respect to a horizontal line that passes through the center of gravity and is perpendicular to the axis. This line would correspond to the line 2 in Figure 3, and Ic can be computed by applying formula 3 in Article 7. The mass m of the cylinder must be obtained by applying the formula in Article 4, in which W = 64.4lb and g = 32.2 fps per sec. Hence,

64.4 2 1 m=--= sugs 32.2

10 Engineering Mechanics, Part 4

Since r = ~ x 1 = 0.5 ft and 1 = 6 ft,

1 2 1 2 1 2 1 2 .2 Ic = 4mr + 12mz = 4 x 2 x 0.5 + 12 x 2 x 6 = 6.13 slug-ft

The distance d from the specified reference line to the center of gravity of the cylinder is one-half the length of the cylinder, as indicated in Figure 2. This distance is 1;2 x 6 = 3 ft. The required mass moment of inertia is

10 = Ic + md 2 = 6.13 + 2 x 32 = 24.1 slug-ft2

Problem 2: A body like that in Figure 5 consists of a sphere whose diameter is 2 ft, and a cylinder whose diameter is 6 in. and whose length is 5 ft. The weight of the sphere is 2200 lb and the weight of the cylinder is 500 lb. What is the mass moment of inertia of the entire body with respect to a vertical reference line through the centroid of the cross section of the cylinder at its right hand end?

Solution: The moment of inertia of the entire body is equal to the sum of the moment of inertia of the cylinder and the moment of inertia of the sphere. For the cylinder, the value of Ic is the moment of inertia with respect to a vertical line through the center of gravity of the cylinder, or a line corresponding to the line 2 in Figure 2. Hence, Ic can be com- puted by formula 3, Article 7. The mass of the cylinder is 50%2.2 = 15.53 slugs, and the distance from the specified reference line to the center of gravity of the cylinder is h x 5 = 2.5 ft. For the sphere, the value of 67 and can be computed by formula 1, Article 7. The mass of the sphere is 2200/32.2 = 68. 3 slugs, and the distance from the reference line to its cen- ter of gravity is equal to the sum of the length of the cylinder and the ra- dius of the sphere, or 5 + h x 2 = 6 ft.

For computing the moment of inertia, all distances will be expressed in feet. The calculations are as follows:

For the cylinder,

1 2 1 2 .2 lc = 4 X 15.53 X 0.25 + 12 X 15.53 X 5 = 0.24 + 32.35 = 32.6 slug-ft

10 = 32.6 + 15.53 x 2.52 = 130 slug-ft2

For the sphere,

2 2 2 2 2 lc = Smr = S X 68.3 X 1 = 27.3 slug-ft 10 = 27.3 + 68.3 x 62 = 2490 slug-ft2

For the entire body, the mass moment of inertia is

130 + 2490 = 2620 slug--fr

Engineering Mechanics, Part 4

Practice Problems 1 Pradice problems are included in this text to test your ability to apply a rule or a formula. Work each problem carefully, and check your answer against the answer given.

1. If a body weighs 200 lb, what is its mass in slugs?

2. The diameter of a sphere is 9 in. and its weight is 250 lb. What is the mass moment of inertia of the sphere with respect to a diameter of the sphere?

3. A right circular cylinder has a diameter equal to 6 in. and is 5 ft long. If the cylinder weighs 500 lb, what is its mass moment of inertia with respect to a diameter of a cross section through its center of gravity?

4. A body like that in Figure 5 consists of a sphere attached to the end of a right circular cylinder. The diameter of the sphere is 6 in. and its weight is 80 lb. The diameter of the cylinder is 3 in., its length is 4ft, and its weight is 100 lb. What is the mass moment of inertia of the entire body with respect to a horizontal reference line through the centroid of a cross section of the cylinder at its right hand end?

Check your answers with those on page 93.

12 Engineering Mechanics, Part 4

Work Done by a force

9 ° In engineering mechanics, the term work has a specific meaning. When a force acts on a body, the force tends to cause a change in the magni- tude or the direction of the velocity of the body. However, when a body is under the action of a system of forces, the body may be in equilib- rium. A body that is in equilibrium either remains at rest or moves with uniform velocity along a straight line. When there is a change in either the magnitude or the direction of the velocity of a body (or in both the magnitude and the direction of the velocity), the body is not in equilibrium. In such a case, the body moves so that there is a com- ponent of the displacement in the direction in which at least one of the forces acts. When there is a component of the displacement in the direction in which a force acts, it is said that the force does work on the body.

The amount of work done during an interval of time by a force that moves a body is determined by multiplying the magnitude of the force by the distance through which the body moves in the direction of the force during that interval of time. It is assumed that the magnitude of the force remains constant during the interval of time considered. If the magnitude of a force varies during a certain interval of time, it is necessary to divide that time interval into a suitable number of shorter intervals during each of which the force remains constant; and to de- termine the amount of work done during the entire interval by adding together the amounts of work done during the several shorter intervals. Also, the distance by which the magnitude of a force is multiplied must be measured in the direction of the line of action of the force. Although a force is a vector quantity, it is permissible to assume that the work done by a force is a scalar quantity because the direction of the distance used in computing the work must be the same as the direction of the force.

When a body is acted upon by a system of forces and all the forces do not have a common line of action, the path along which the body moves may not coincide with the line of action of a certain force in the system. To determine the amount of work done by that force, it is necessary to multiply the magnitude of the force by the component of the displace- ment of the body which has the same direction as the line of action of the force. Typical conditions are represented in Figure 6. It is as- sumed that a body is acted upon by a system of forces which includes

the constant force represented by the vector F. During a certain interval of time, the body moves from position 1 to position 2 along some path. For this motion, the displacement vector is represented by a

straight line. The magnitude of this vector is denoted by i"& I and the ~

angle between the line of action of the force F and the displacement vector is denoted by A. Then the component of the displacement

Engineering Mechanics, Part 4 13

~ vector in the direction of the line of action of the force is I & I x cos A,

~ ? ~ and the amount of work done by the force F is I F I x I & I x cos A.

FIGURE 6-Disp/acement for Work Done by Force

Since work is computed by multiplying a force by a distance, the unit for work combines a unit for force and a unit for distance. In the English system of units, work is usually expressed either in foot-pounds or in inch-pounds. In the metric system, a unit for work is the kilogram-meter. It is permissible to give the units of force and distance in either order.

The symbol used for work in this text will be U, because the symbol W is used for weight.

Facts about Work

• The following facts about the work done by a force should be kept in mind:

1. Work is a scalar quantity, and only its magnitude is important. The direction of work should never be mentioned.

2. The distance by which the magnitude of the force is multiplied must be measured along the line of action of the force. This distance is often called the work-absorbing component of the displacement.

3. When all forces acting on a body remain constant, both in magni- tude and in direction, the body will move along a straight path. The velocity of the body may not remain constant, but velocity is not considered when work is computed.

4. Work is always considered to be a positive quantity. If a body moves at right angles to the line of action of a force, or the angle A in Figure 6 is exactly 90° (degrees), the work done by that force

Example Problem

1 1

Engineering Mechanics, Part 4

would be zero. In a practical problem in which work is computed, it is never necessary to consider the work done by a force which is so located that the angle between the line of action of the force and the displacement vector will be greater than 90° and cos A would be negative.

5. Work involves a displacement during a certain interval of time. It is not practicable to consider the work at a particular instant.

Problem: A body moves along a straight path under the action of a system of forces. The constant magnitude of one force of the system is 100 lb, and the angle between the line of action of this force and the path along which the body moves remains 30° .If the body moved 5 ft along its path during a certain interval of time, how much work was done on the body by the specified force during that time interval?

Solution: The conditions may be represented in Figure 6, in which the --7

magnitude of the force F is 100 lb, the angle A is 30°, and the magnitude --7

of the displacement & is 5 ft. The required amount of work is

u = IFI X 11s I X cos A = 100 X 5 X cos 30° = 433 ft-lb

Kinetic Energy of Translation

• Every moving body possesses what is called kinetic energy. Energy is often defined in a general way as the "ability to do work," and en- ergy can be converted into work. A body may possess kinetic energy either because of translation or because of rotation. The magnitude of the kinetic energy possessed by a moving body at a particular instant because of translation can be computed by the following formula:

1 w 2 Et=--V

2 g

in which £1 =kinetic energy of a body at a particular instant because of translation

W =weight of the body g =acceleration due to gravity v = magnitude of the velocity of the body at the instant

In many cases, a moving body possesses kinetic energy because work has been done by forces to produce the motion and to give the body its energy. In such a case, work has been converted into energy.

Engineering Mechanics, Part 4

Example Problems

Although the kinetic energy of a moving body because of translation depends on the magnitude of the velocity of the body, the direction of the velocity need not be considered. Therefore, kinetic energy is considered to be a scalar quantity. It is necessary to consider the ki- netic energy at a certain instant of time. Unless the velocity remains con- stant, the kinetic energy varies from instant to instant.

The unit for kinetic energy is determined from the units for the weight W, the acceleration g, and the velocity v. However, acceleration is determined by dividing velocity by time; and velocity is determined

by dividing length by time. Therefore, ifF stands for force, L for length or distance, and T for time, the quantities involved in kinetic energy can be indicated as follows:

Weight=F

Velocity=~

Acceleration = ~ + T = ~

Therefore, each unit for kinetic energy combines a unit for force and a unit for distance, and the units used for kinetic energy are the same as the units used for work. For the English system, it is usually best to express Win pounds, gin feet per second per second, and v in feet per second. Then, the kinetic energy will be in foot-pounds.

Even though work and kinetic energy are expressed in the same units, these two quantities are basically different. A force does work, whereas a body or a mass possesses kinetic energy. Furthermore, work is actu- ally done during a certain interval of time, whereas kinetic energy is considered at a particular instant.

Problem 1: A body that weighs 200 lb is falling vertically through the air. At a certain instant, the magnitude of the velocity is 50 fps (feet per second). What is the kinetic energy of the body at that instant because of translation?

Solution: In this problem, it would be convenient to express the kinetic energy in foot-pounds. For the given values,

1 200 Et = 2 x 32_2 x 502 = 7764 ft-lb (foot-pounds)

Engineering Mechanics, Part 4

Problem 2: An iron ball that weighs 56 lb is attached to one end of a steel chain that is 6 ft long. A man holding a handle at the other end of the chain swings the ball through the air with increasing speed un- til it travels along a circular path with a velocity of 40 fps. What is the kinetic energy of the ball at that instant because of translation?

Solution: Even though the direction of the velocity is changing continu- ally, the only characteristic of the velocity that has to be considered is its magnitude. The kinetic energy of the ball at the specified instant is

1 56 Et = 2 x 32.2 x 402 = 1392 ft-lb (foot-pounds)

Kinetic Energy of Rotation

• The kinetic energy of a body that is rotating about a stationary axis through its center of gravity can be computed by the formula

1 2 Er=2. Ic ro (1)

in which Er =kinetic energy of a body at a particular instant because of rotation about a line through its center of gravity

Ic =mass moment of inertia of the body with respect to the line in the body that remains stationary

ro =angular velocity of any rotating line through the center of gravity of the body (ro is the Greek letter omega)

For instance, if a cylinder is rotating about a stationary longitudinal axis through its center of gravity, as indicated in Figure 7, the kinetic energy of the cylinder would be computed by applying formula 1. For this case, the value of Ic would be determined by formula 2, Article 7.

Engineering Mechanics, Part 4 17

FIGURE 7-Cylinder Rotating about Axis through Center

If a rigid body is being rotated about a stationary axis that does not pass through the center of gravity of the body, the formula for computing the kinetic energy produced by rotation is

1 2 E, =2. I0 ro

in which I0 =mass moment of inertia of the body with respect to the axis of rotation

ro = angular velocity of a rotating line through the axis of rotation

For instance, if a bar is rotating about a stationary axis near one end as indicated in Figure 8, the kinetic energy of the bar would be com- puted by applying formula 2. For this case, the value of Ic would be determined by formula 5 or 6, Article 7, and the value of I0 would be computed by applying the relationship in Article 8.

AXIS

FIGURE 8-Bar Rotating about Axis near One End

Example Problems

Engineering Mechanics, Part 4

linear Impulse of a Force

• When a force having a constant magnitude acts on a body that is in translation for a certain period of time, it is said that the force exerts a linear impulse on the body during that time interval. The magnitude of the impulse exerted by the force is found by multiplying the magnitude of the force by the time interval. Thus,

~ in which Ill = magnitude of the linear impulse exerted by a force

~ I F I = magnitude of the force ,1.t =interval of time during which the force acts (,1. is the

Greek letter delta)

Although the letter I is used to represent both moment of inertia and impulse, there will be no confusion in a practical problem.

When the linear impulse of a force is considered, it is assumed that neither the magnitude nor the direction of the force changes during the time interval M. Since a force is a vector quantity, an impulse also must be a vector quantity. However, the direction of an impulse exerted by a force must be the same as the direction of the force. The unit for impulse combines a unit for force and a unit for time. In the English system, the unit commonly used is the pound-second.

A force may exert a linear impulse on a body without doing work For instance, a body may remain at rest under the action of a system of forces. As a result, none of the forces does any useful work. However, each force exerts a linear impulse during the entire interval of time during which it acts. Also, it is possible to consider components of a linear impulse by applying the same principles that were explained for components of a force.

A discussion of impulse due to rotation, or angular impulse, is beyond the scope of this text.

Problem 1: As indicated in Figure 9, a body having the shape of a rectangular solid is being pulled along a very smooth horizontal sur- face by a horizontal force P. If the magnitude of this force is 5 lb and the force is applied for 3 sec, what is the linear impulse of this force during that interval?

Engineering Mechanics, Part 4 19

p = 51b

FIGURE 9-Force in Example Problem 1, Article 13

Solution: Regardless of the effects of any other forces that act on the body, such as its own weight and the force exerted by the supporting surface, the magnitude of the linear impulse of the force P during the specified time interval is

l11 = 5 x 3 = 15lb-sec (pound-seconds) Since the body in Figure 9 moves along the supporting surface in the direction in which the force Pacts, the force P does work on the body.

The weight of the body in Figure 9, which acts vertically downward, exerts an impulse on the body during the specified time interval and also at any other time. However, since the body in Figure 9 would not be displaced in a vertical direction, the weight of this body would not do any work on the body.

Problem 2: One force of a system of forces that acts on a body is a force F, whose characteristics are indicated in Figure 10. The magnitude of the force F is 40 lb, and the angle between the line of action of the force and a horizontal reference line is 53°10' (minutes). If this force acts on the body for 5 sec, what are the magnitudes of the horizontal and ver- tical components of the linear impulse of the force during the specified time interval?

FIGURE lo-Force in Example Problem 2, Article 13

53°1 0'

---~--------·

Solution: The magnitude of the linear impulse exerted by the given force is

l11 = 40 x 5 = 200 lb-sec

20 Engineering Mechanics, Part 4

The magnitude of the horizontal component of this impulse is

~ llx I = 200 cos 53°10' = 120 lb-sec

The magnitude of the vertical component is

,.-,t lly I = 200 sin 53°10' = 160 lb-sec

Linear Momentum of a Body

14 • Another property of a moving body that is useful in solving some problems is known as the linear momentum of the body. The magni- tude of the linear momentum of a moving body in translation at a particular instant can be computed by the formula

~ w~ IMI=-Ivl

~ in which I M I =magnitude of the linear momentum of a body at a

particular instant W =weight of the body g = acceleration due to gravity

rifl = magnitude of the velocity of the body at the instant Although the letter Misused to represent both moment and momen- tum, there will be no confusion in a practical problem.

Since velocity is a vector quantity, linear momentum is also a vector quantity. However, the direction of the linear momentum of a body must be the same as the direction of the velocity of the body. The unit for linear momentum is determined from the units for the weight W, the acceleration g, and the velocity v by using the notation F, L, and Tin the manner described in Article 11. Since weight= F, velocity= Lfr, and acceleration= L;f,

Momentum~r+ :,)x ~ =FxT2 xL-FxT

L T-

Engineering Mechanics, Part 4

Example Problem

In words, a unit for linear momentum combines a unit for force and a unit for time, and a unit for momentum is the same as a unit for im- pulse. When the weight is in pounds, the acceleration due to gravity is in feet per second per second, and the velocity is in feet per second, then the linear momentum will be in pound-seconds. Even though impulse and momentum are expressed in the same units, these two quantities differ in the same respects as do work and kinetic energy. A force exerts an impulse, whereas a body has momentum. Also, an impulse is exerted during a certain interval of time, whereas momen- tum is considered at a particular instant. It is sometimes convenient to consider the horizontal and vertical components of the linear mo- mentum of a body.

A discussion of momentum due to rotation, or angular momentum, is beyond the scope of this text.

Problem: An automobile that weighs 3200 lb is traveling along a hori- zontal highway with a velocity whose magnitude is 55 mph (miles per hour). What is the magnitude of the automobile's linear momentum?

Solution: The velocity must be expressed in feet per second. Thus, the magnitude of the velocity is 55 x 1.467 = 80.7 fps, and the magnitude of the linear momentum is

~ 3200 I M I = 32.2 x 80.7 = 8020 lb-sec

22 Engineering Mechanics, Part 4

Practice Problems 2 1. A body that is under the action of a system of forces moves along a straight path. One

force of the system has a constant magnitude equal to 200 lb, and the angle between the line of action of this force and the path of the moving body is 20°. If the body moved 3 ft along its path during 5 sec, how much work was done on the body by the specified force during that time interval?

2. What is the kinetic energy of translation of a body at a certain instant if the body weighs 250 lb and its linear velocity at the specified instant is 120 fpm (feet per minute)?

3. What was the linear impulse of the force described in problem 1 during the specified interval of time?

4. What was the linear momentum of the body in problem 2 at the specified instant?

Check your answers with those on page 93.

Engineering Mechanics, Part 4 23

FORCE--MASS--ACCELERATION METHOD

Translation of Rigid Body

Basic Equation or Equations for Translation

• When the motion of a rigid body involves translation alone, every particle in the body moves in exactly the same way. The paths of the various particles are parallel. Also, at any instant, each particle has the same linear acceleration. Therefore, when Newton's second law is applied to a rigid body that is being translated, it is the usual practice to consider the motion of the center of gravity of the body.

In most problems on rectilinear translation, or translation along a straight path, the basic equation is

F=ma

in which F =resultant of all the forces acting on a body m =mass of the entire body a = acceleration of the center of gravity of the body

(1)

In some problems on rectilinear translation and in problems on curvi- linear translation, or translation along a curved path, it is convenient to consider rectangular components Fx and Fy of the resultant force, and also to consider corresponding components ax and ay of the accelera- tion of the center of gravity of the body. The single equation F = ma is then replaced by the following two equations:

(2)

(3)

In the solution of a problem, the mass m of a body is replaced by the quantity wig, in which W is the weight of the body and g is the accel- eration due to gravity.

When a body is subjected to a constant resultant force, the acceleration of the body remains constant. As a result, the velocity of the body changes continually, but the change in velocity in each unit of time is constant. The kinematics of motion involving translation with constant acceleration was discussed in Engineering Mechanics, Part 3.

24 Engineering Mechanics, Part 4

Types of Problems

16 • As the name force-mass-acceleration method implies, any problem in which this method is used is solved simply by applying either equa- tion 1 or equations 2 and 3, Article 15. In some problems, the known quantities are either the weight or the mass of the body and a desired acceleration, and it is required to compute the force needed to pro- duce that acceleration. In other problems, the known quantities are the weight or the mass of the body and an available force, and it is required to compute the acceleration that will be produced. In still other problems, the known quantities are an available force and a de- sired acceleration, and it is required to compute either the weight or the mass of the heaviest body that can be moved in the specified manner by the available force.

Steps in Solution of Problem

• The usual procedure in solving any problem in kinetics by applying the force-mass-acceleration method is to carry out the following three steps:

1. Draw a free-body diagram, abbreviated FBD, indicating the characteristics of all the forces acting on the body.

2. Analyze the conditions in the problem, and substitute suitable values either in equation 1, Article 15, or in equations 2 and 3.

3. Solve the equation or equations obtained in step 2.

Typical Problems Involving Rectilinear Translation

• The use of equation 1 or equations 2 and 3 in Article 15 for solving problems involving rectilinear translation is illustrated in the follow- ing example problems. In each problem, it will be assumed that hori- zontal components of forces acting toward the right are positive, and horizontal components acting toward the left are negative. Also, verti- cal components acting upward will be considered positive, and verti- cal components acting downward will be considered negative.

Example Problems

Problem 1: An elevator car in a building is expected to carry a total weight (including the weight of the car itself) of 3500 lb. If it is de- sired to move the car upward with an acceleration equal to 8 fps per sec and the effect of friction is neglected, what force would be required to lift the car?

Engineering Mechanics, Part 4 25

Solution: The FBD for the car is shown in Figure 11. The car is repre- sented by a small square. The only two forces acting on the car that are considered in this problem are 1) the weight Wof the car and its con- tents, which acts vertically downward, and 2) the force T repre- senting the pull exerted by a cable to which the car is attached and which controls the movement of the car. It is evident that the force T must act vertically upward.

FIGURE 11-FBD for Elevator Car in Example Problem 1, Article 18

w = 3500 lb

The conditions may be analyzed as follows: Since the forces Wand T are collinear and act in opposite directions along their common line of action, the line of action of the resultant force acting on the elevator must coincide with the line of action of Wand T, and the magnitude of this resultant force must be T -W. Then, by equation 1, Article 15,

w T- W=-xa

T- 3500 = 3500 X 8 = 870 lb 32.2

The magnitude of the required force T can be found by solving the equation just obtained. Thus, T = 3500 + 870 = 4370 lb.

Problem 2: A man who weighs 200 lb is riding in the elevator car in problem 1 when the upward acceleration of the car is 8 fps per sec. What downward force does this man exert on the floor of the elevator?

26 Engineering Mechanics, Part 4

Solution: The FBD for the part of the floor of the car on which the man stands is shown in Figure 12. The weight of the man is represented by a force W which acts vertically downward. When the man exerts a force on the floor of the car, the floor exerts an equal opposite force on the man. This reaction of the floor is represented by the force N. The force N must act vertically upward.

w = 200 lb

FLOOR '

OFCARifl '••••••···

FIGURE 12-FBD for Example Problem 2, Article 18

The analysis of the conditions follows: The magnitude of the force N must be greater than the weight of the man, because the man is being moved upward with the specified acceleration. Since the forces of W and N are collinear and act in opposite directions along their com- mon line of action, the resultant of the forces N and W must also act along the same line, and its magnitude is N-W. By equation 1, Article 15,

200 N -200 = 32.2 X 8 = 49.7lb

The magnitude of the required force N, which can be computed by solving the foregoing equation, is 250 lb.

Problem 3: It is desired to move an elevator car in a building upward with an acceleration equal to 12 fps per sec. If the magnitude of the available upward force is 6200 lb and the effect of friction is neglected, what would be the greatest weight (including the weight of the car itself) that could be lifted?

Solution: The FBD for the car is shown in Figure 13. By equation 1, Article 15,

w 6200- W = 32.2 X 12 = 0.372W

Hence, 1.372 w = 6200 and W=4520 lb

Engineering Mechanics, Part 4

FIGUilE 13-FBD for Example Problem 3, Article 18

T = 6200 lb

CAR

Problem 4: As indicated in Figure 14A, a body having the shape of a rectangular solid is sliding along a horizontal surface under the action of a horizontal force P. The weight of the body is 50 lb, and the magnitude of the force P is 15lb. If the coefficient of friction for the materials used for the surface and the body is 0.2, what is the linear acceleration of the body?

w = 501b

BODY.....r----, p p = 151 b

A. Position of pulling force B. FBD

FIGUilE 14-conditions in Example Problem 4, Article 18

28 Engineering Mechanics, Part 4

Solution: The FBD for the body is shown in Figure 14B. The forces that must be considered are the weight W of the body, the horizontal force P, the vertical reaction N exerted by the supporting surface on the body, and the friction force F m· In this problem, it may be assumed that the line of action of the force N coincides with the line of action of the weight W. Also, since Lfy = 0, the magnitude of the force N must be the same as the magnitude of the weight W, or 50 lb.

The maximum friction force that can be developed is equal to the product of the coefficient of friction and the weight of the body. Thus, F m = 0.2 X 50 = 10 lb. Since the force pacts toward the right, the friction force must act toward the left. Also, since the magnitude of the force Pis greater than the magnitude ofF m' the body will slide along the supporting surface. Although the lines of action of the forces P and F m do not coincide, these lines are parallel. It may be assumed that the force causing motion of the body is a horizontal force that is the resul- tant of the parallel forces P and F m· The magnitude of this resultant is P - F m = 15 - 10 = 5 lb, and its direction along the line of action is the same as the direction of the force P, or toward the right. When equation 1, Article 15, is applied, the result is

50 5 = 32.2 x a = 1.553 a

Hence, the required acceleration is 3.22 fps per sec.

Problem 5: As indicated in Figure 15A, a body that weighs 40 lb and is supported on a horizontal surface is sliding along that surface under the action of a force P whose location and characteristics are as shown. The coefficient of friction for the materials used for the body and the surface is 0.25. If it may be assumed that the line of action of the up- ward force N exerted on the body by the supporting surface coincides with the line of action of the weight W, and that there is no tendency for the body to rotate, what would be the magnitude of the linear acceleration of the body along the supporting surface?

Solution: The FBD for the body is shown in Figure 15B. The applied inclined force P is replaced by its horizontal and vertical components, the magnitudes of which are 50 cos 40° = 38.3lb and 50 sin 40° = 32.11b. If the equation Lfy = 0 is applied and it is assumed that N acts upward, the result is

Hence,

N -40-32.1 =0

N = + 72.1, or 72.1lb i

Engineering Mechanics, Part 4

SUPPORTING SURFACE A. Applied forces

1 ~ a:- ...---.lir---i

B. FBD

Px = 38.31b

FIGURE 15-conditions in Example Problem 5, Article 18

The direction of the friction force must be opposite to the direction of the horizontal component of the force P. Also, the magnitude of the maximum friction force F m that can be developed is the product of the coefficient of friction and the magnitude of N, or F m = 025 x 72.1 = 18.0 lb. Since this result is less than the horizontal component of P, which tends to cause the body to slide, the body will move.

To determine the magnitude of the linear acceleration of the body, it is only necessary to apply equation 1 or equation 2 in Article 15. The horizontal component of the resultant force acting on the body is - P x + F m = -38.3 + 18.0 = - 20.3, or 20.3 lb.

40 Hence, 20.3 = 32.2 x a

and a = 16.3 fps per sec

30 Engineering Mechanics, Part 4

Since the horizontal component of the resultant force acts toward the left, the linear acceleration is 16.3 fps per sec toward the left.

Curvilinear Translation

19 • In order that a body may be moved along a curved path, forces must act on the body in two directions that are at right angles to one another. For curvilinear translation, it is usually necessary to consider the con- ditions at some particular instant. The most convenient directions for the reference axes at the selected instant are along the tangent to the path and perpendicular to the tangent. These directions are commonly called the tangential direction and the normal direction. In almost all practi- cal problems involving curvilinear translation, the body is moving along a circular path.

Translation of a rigid body along a circular path is similar to the motion of a particle on a rotating line. As explained in Engineering Mechanics, Part 3, the magnitude of the tangential component of the linear accel- eration of such a particle is

in which 1"'l41 =magnitude of the tangential component of the linear acceleration

R = radius of the curved path at the selected instant

(1)

a = angular acceleration of the body at the selected instant (a is the Greek letter alpha)

Also, the magnitude of the normal component of the linear acceleration of a particle on a rotating line is

(2)

in which 1"Z I =magnitude of the normal component of the linear acceleration

ro = angular velocity of the body at the selected instant

The direction of the normal component of the linear acceleration is always toward the center of the circle.

It is often more convenient to use the magnitude of the linear velocity

d instead of the angular velocity. Since I~ I = R I oil and Rro2 = v1R, ~ v2 Ia I=-

n R (3)

Engineering Mechanics, Part 4

Example Problem

The use of equations 2 and 3 in Article 15 for solving problems involving curvilinear translation is illustrated in the following example problem.

Problem: An automobile that weighs 3000 lb is moving along a circular path whose radius is 500 ft, with a uniform linear velocity of 40 mph. a) What is the magnitude of the normal component of the linear accel- eration? b) What is the magnitude of the force that acts along the radius of the curve and causes the automobile to travel along the circular path?

Solution: a) The FBD for the automobile is shown in Figure 16. Here, the automobile is represented by a small square, and the circular path is represented by the broken line. Since the linear velocity of the auto- mobile along the curve is uniform, the tangential component of the linear acceleration must be zero. Theoretically, once the automobile ac- quires the velocity of 40 mph, no force need be applied in the tangential direction to keep the automobile moving with that velocity. Actually, however, a relatively small tangential force must be applied to overcome the effects of friction and other forces that tend to resist the movement of the automobile. The magnitude of this tangential force need not be considered in this problem. Since the direction of the normal compo- nent of the linear acceleration must be toward the center of the circle along which the automobile travels, the direction of the normal com- ponent Fn of the force producing that acceleration must be toward the center of the curve, as indicated in Figure 16.

FIGURE 16-Forces Acting on Automobile in Example Problem, Article 19

AUTOMOBILE

It is convenient to determine the magnitude of the normal component of the linear acceleration by applying formula 3, Article 19. Since 40 mph = 40 x 1.467 = 58.7 fps,

--7 v2 58.72 I an I= R = 500 = 6.89 fpsper sec

b) The magnitude of the force required to produce the normal compo- nent of the acceleration can be computed by applying equation 3, Article 15. Thus,

~ 3000 I F n I = 32.2 x 6.89 = 642 lb

Engineering Mechanics, Part 4

Centripetal Force and Centrifugal Force

• According to Newton's first law of motion, a body that is moving along a straight path will continue to move in the same direction unless it is acted upon by some force which compels a change in direction. Therefore, in order to cause a body to move along a circular path, it is necessary to apply a force continuously to a body that acts toward the cen- ter of the circle along a radial line. Such a force is known as a centripe- tal force. Even when a body moves along a circular path under the action of a suitable centripetal force, there is at every instant a ten- dency for the body to follow a straight path which is tangent to the cir- cular path at that instant. This tendency is the result of the reaction to the centripetal force. Such a reaction, which is known as a centrifugal force, acts along a radial line in a direction away from the center of the circle. In other words, when a body moves along a circular path un- der the action of a centripetal force, the body develops a centrifugal force which is equal and opposite to the centripetal force. At any instant, the centripetal force and the centrifugal force must act along a radial line.

There are many familiar examples of centrifugal force. If you fasten a comparatively small object to one end of a piece of string a few feet long, and you put the object in motion along a circular path by holding the other end of the string in your hand and whirling the object, a cen- tripetal force is applied to one end of the string by your hand and a centrifugal force is exerted on the other end of the string by the moving object. If the string should break or if you should let go of the string, the object will continue its motion along a straight path that is tan- gent to the circular path at the instant when the centripetal force ceases to act on the string.

When a train runs along a curved track, the outer rail of the track applies a centripetal force to the wheels of the train that are in contact with that rail and compels the train to proceed around the curve. At the same time, the wheels exert a centrifugal force on the outer rail. If the train were to go too fast, the train would leave the track and would move along a path that is tangent to the curve at the point of derailment.

The magnitude of the centrifugal force that is developed at any instant by a body which is moving along a circular path is the magnitude of a force that would produce the normal component of the linear accel- eration of the body at the selected instant. By formula 3, Article 19, the magnitude of this component of the acceleration is vYR. By equation 3, Article 15, the magnitude of the centrifugal force is

in which I~ I =magnitude of the centrifugal force exerted by a body at a particular instant

Engineering Mechanics, Part 4

Example Problem

W =weight of the body v =magnitude of the linear velocity of the body at the

chosen instant g = acceleration due to gravity R =radius of the circular path along which the body

is moving

Problem: What is the magnitude of the centrifugal force that would be developed by the automobile in the example problem in Article 19?

Solution: In this problem, W = 3000 lb, v = 58.7 fps, and R =500ft. Hence, the magnitude of the centrifugal force would be

IFI = 3000 X 58.72 = 642lb c 32.2 X 500

This result is the same as that obtained in the example problem in Article 19 for the magnitude of the force that caused the automobile to travel along the curved path. The reason is that the force in Article 19 is really a centripetal force, and the magnitude of the centrifugal force in Article 20 is equal to that of the centripetal force. However, the centrifugal force acts away from the center of the circular path.

34 Engineering Mechanics, Part 4

Practice Problems 3 1. The magnitude of the upward force that is available to lift an elevator car in a building

is 5000 lb. The desired upward acceleration of the elevator is 10 fps per sec. If the effect of friction is neglected, what is the greatest weight (including the weight of the elevator car itself) that can be lifted by the upward force?

2. A body that has the shape of a rectangular solid and weighs 100 lb is sliding toward the left along a horizontal surface under the action of a horizontal force whose magnitude is 50 lb.lf the coefficient of friction for the materials used for the surface and the body is 0.3, what is the linear acceleration of the body?

3. A railway car that weighs 40,000 lb is traveling along a curved track with a velocity whose magnitude is 30 mph. If the radius of the curve is 800 ft, what is the magnitude of the centrifugal force exerted by the car on the outer rail of the track?

Check your answers with those on page 93.

Engineering Mechanics, Part 4 35

Rotation of Rigid Body without Translation

Rotation about Axis through Center of Gravity

• In order to produce rotation of a rigid body without translation, the resultant of the system of forces acting on the body must be a couple. As explained in Engineering Mechanics, Part 1, a couple consists of two forces which have the same magnitude, have parallel lines of action (but are not collinear), and act in opposite directions along their lines of action. The only effect of a couple on a rigid body is to produce a mo- ment which has a specific magnitude and a specific direction (clock- wise or counterclockwise) with respect to any point in the body. This moment causes the body to rotate about a straight line which remains stationary and is called the axis of rotation. The axis of rotation may or may not pass through the center of gravity of the body. The moment of the couple causing rotation is equivalent to the resultant moment of all the forces that act on the body.

When a body is rotating under the action of a system of forces and the resultant moment of all the forces does not change, the body is subjected to a uniform angular acceleration. If the body is rotating about an axis through the center of gravity, the relation between the resultant moment and the angular acceleration is as follows:

(1)

in which Me= resultant moment, with respect to an axis of rotation through the center of gravity of the body, of all the forces that act on the body

Ie = mass moment of inertia of the body with respect to the axis through the center of gravity of the body

a= angular acceleration of any rotating straight line through the center of gravity of the body

The same unit of length must be used in Me and in Ie. In problems in this text, it will be assumed that clockwise moments are positive and counterclockwise moments are negative.

A typical body that rotates without translation about an axis through its center of gravity is a flywheel on a gas engine. In Figure 17 A the flywheel is represented as a solid cylinder, although an actual flywheel may consist of a rim and a number of spokes connecting the hub to the rim. The wheel rotates on a shaft that passes through the center of gravity of the wheel, and this shaft is supported in bearings in such a manner that it can rotate. A free-body diagram for the flywheel is shown in Figure 17B. Of course, the flywheel is acted upon by its own weight, which is represented by the vertical force W through its cen- ter of gravity G. It is assumed that the force which causes rotation in

Example Problems

Engineering Mechanics, Part 4

this case is a force P applied at the rim of the wheel. This force has a con- stant magnitude, but the direction of its line of action changes from in- stant to instant. One position is shown in Figure 17B. The horizontal and vertical components of the force P at any instant will be denoted by P x and P y· In order that the shaft of the flywheel will not move in translation, the bearings must exert a reaction on the shaft. This reaction must have a horizontal component and a vertical component, which are designated in Figure 17B as Rx and Ry.

FLYWHEEL

A. Support of flywheel B. FBD

FIGURE 17-conditions for Flywheel

Since the center of gravity of the wheel remains at rest under the action of the forces W, P x1 P Y' Rx, and Ry, these forces must be related in the following manner:

"Lfx=O=Px 4Rx

"Lfx=O= W 4 Py 4Ry

in which rF x= vector sum of the horizontal components of forces rF y= vector sum of the vertical components of forces

(1)

(2)

The symbol~ indicates vector addition. In equations 2 and 3, each force must be given its proper sign.

Problem 1: The radius of a cylindrical flywheel is 1.5 ft, and its weight is 1000 lb. If the constant magnitude of a force applied to the rim of the wheel along a tangent to the rim is 80 lb, what would be the magni- tude of the angular acceleration of the wheel?

Engineering Mechanics, Part 4 37

Solution: It may be assumed that the mass moment of inertia of the wheel is, by formula 2, Article 7,

1 1000 2 .2 Ic = 2 x 32.2 x 1.5 = 34.9 slug-fr

The magnitude of the moment of the force is 80 x 1.5 = 120ft-lb. Using formula 1, Article 21, we obtain

120 =34.9a

a= 3.44 rad (radians) per sec per sec

Problem 2: If the force applied to the rim of the flywheel in problem 1 is located as indicated in Figure 17B, and the angle between the line of action of this force and a horizontal reference line is 60°, what are the horizontal and vertical components of the reaction exerted by the bearings on the flywheel?

Solution: The horizontal and vertical components of the applied force are

and

Px = + 80 COS 60° =+40.0 lb

P y = - 80 sin 60° = - 69.3 lb

If it is assumed that Rx acts toward the left and Ry acts upward, the results obtained by applying equations 2 and 3 are as follows:

0=+40.0 -Rx

0 =-1000- 69.3 +Ry

Hence, Rx = +40.0 lb and Ry = + 1069lb. Since both of these results are positive, the assumed directions for Rx and Ry are correct.

Rotation about Any Axis

• If the axis about which a body rotates does not pass through the center of gravity of the body, the basic relationship for the angular acceleration is

(1)

in which M0 = resultant moment, with respect to the axis of rotation, of all the external forces that act on the body

10 = mass moment of inertia of the body with respect to the axis of rotation

a= angular acceleration of any rotating straight line through the axis of rotation

38 Engineering Mechanics, Part 4

It was explained in Engineering Mechanics, Part 3, that when a line is rotating with an angular acceleration, each particle on the line has a linear acceleration. The magnitude of the tangential component of the linear acceleration of the center of gravity of the body is

in which I (a;)t I =magnitude of the tangential component of the linear acceleration of the center of gravity

(2)

Rc = radius of the circular path of the center of gravity a= angular acceleration of any rotating line through

the axis

~ Also, the magnitude of the normal component (ac)n of the linear acceleration of the center of gravity is

(3)

in which ro = angular velocity of any rotating line through the axis. The direction of the tangential component corresponds to the direction of the angular acceleration. The direction of the normal component is always toward the axis of rotation.

A typical body that rotates without translation about a line near one end is a part of a mechanism called a crank. In Figure 18 is shown a free-body diagram of a crank supported near one end by a stationary pinned connection at 0. At the instant under consideration, the crank is in the position indicated. But it may have any position with respect to the point 0. It will be assumed that rotation of the crank is pro- duced by a force P which is applied near the moving end of the crank and the line of action of which is tangent to the circular path of the moving end.

The crank is acted upon by the following external forces: the weight of the crank, which is represented by the vertical force W through the center of gravity G of the crank; the force P, which causes rotation of the crank and acts in the direction indicated at the instant under consideration; the horizontal component Rx of the reaction exerted by the supporting connection at 0; and the vertical component Ry of that reaction.

Engineering Mechanics, Part 4

.,..,.-------. ........... ,.. ....

,.. ' ,.. ' / '

/ \

I R \ y I

....

FIGURE 18-FBD for Rotating Crank

...... ____ .,. ....

/ /

It is assumed that the crank is rotating with a certain angular velocity (either clockwise or counterclockwise) at the instant under considera- tion. The direction of the angular acceleration of the crank is determined by the direction of the resultant moment of the external forces. The direction of the tangential component of the linear acceleration of the center of gravity of the crank corresponds to the direction of the angular acceleration, and the direction of the normal component of the linear acceleration of the center of gravity is toward the point 0. The magni- tude of the force required to produce either component of the linear acceleration of the center of gravity is equal to the mass of the crank times the magnitude of the corresponding component of the linear acceleration.

To determine the components of the reaction exerted by the pin on the crank, it is necessary to apply the following equations:

(4)

(5)

in which LF x = vector sum of the horizontal components of the forces acting on the crank

W =weight of the crank g =acceleration due to gravity

(ac)x =horizontal component of the linear acceleration of the center of gravity of the crank

LF y = vector sum of the vertical components of the forces acting on the crank

(ac)y =vertical component of the linear acceleration of the center of gravity of the crank

Example Problem

Engineering Mechanics, Part 4

Problem: The position of a rotating crank at a certain instant is indi- cated in Figure 19A. The weight of the crank is 80 lb. The distance from the pin connection at 0 in Figure 19B, about which the crank rotates, to the applied force P at the moving end of the crank is 1.5 ft; and the constant magnitude of this force is 32lb. At the instant under consideration, the crank is rotating counterclockwise with an angular velocity equal to 2.5 rad per sec. It may be assumed that the length of the crank is 1.5 ft, and the distance along the crank from the pin con- nection to the center of gravity of the crank is Yz x 1.5 = 0.75 ft. Deter- mine a) the magnitude and direction of the angular acceleration of the crank; b) the magnitude and direction of the horizontal component of the reaction of the pin at 0; and c) the magnitude and direction of the vertical component. of the pin reaction.

A. Conditions B. FBD

C. Directions of components of linear acceleration

FIGURE 19-crank in Example Problem, Article 22

Engineering Mechanics, Part 4 41

Solution: a) The FBD of the crank for the external forces is shown in Figure 19B. The external forces acting on the crank are its weight, which is represented by the vertical force W through its center of grav- ity; the applied force P; and the components Rx and Ry of the reaction exerted by the pin. The resultant moment of these forces with respect to the pin is

M0 =-P x 1.5 + Wx 0.75x cos 45° =- 32 X 1.5 + 80 X 0.75 X 0.707 =- 5.58 ft-lb

To apply equation 1, Article 22, it is necessary to determine the mass moment of inertia. The mass of the crank is

80 m = 32.2 = 2.48 slugs

Since the dimensions of the cross section of the crank are relatively small in comparison to the crank's length, the term i or if in formula 5 or 6, Article 7, can be neglected and it may be assumed that

1 2 1 2 2 Ic = 12 ml = 12 x 2.48 x 1.5 = 0.465 slug-ft

By the formula in Article 8

10 = 0.465 + 2.48 x 0.752 = 1.860 slug-~

By equation 1, Article 22,

- 5.58 = 1.86a a= -3.00

Hence, a= 3.00 rad per sec per sec, counterclockwise.

b) and c) The magnitudes of the tangential and normal components of the linear acceleration of the center of gravity of the crank are, by for- mulas 2 and 3, Article 22,

~ I (ac)1 I = 0.75 x 3.00 = 2.25 fps per sec

~ I (ac)n I = 0.75 x 2.52 = 4.69 fps per sec

The directions of these components are indicated in Figure 19C. The horizontal component of (ac)x of the linear acceleration of the center of gravity of the crank is determined by combining the horizontal com-

~ ~ ponents of (ac)1 and (ac)n. Also, the vertical component (ac)y of that acceleration is determined by combining the vertical components of ~ ~ (ac)t and (ac)n·

42 Engineering Mechanics, Part 4

If it is assumed that the horizontal component Rx of the pin reaction acts toward the right and the vertical componentRy acts upward, the results obtained by applying equations 4 and 5, Article 22, are as follows:

For horizontal forces,

+ 32 COS 45° + Rx = + 2.48 X 2.25 COS 45° - 2.48 X 4.69 COS 45°

For vertical forces,

- 80 + 32 sin 45° + Ry = + 2.48 x 2.25 sin 45° + 2.48 x 4.69 sin 45°

Hence,

and

Rx=-26.9lb

Ry=+ 69.6lb

Since the value of Rx is negative and the value of Ry is positive, Rx is 26.9lb toward the left and Ry is 69.6lb upward.

Engineering Mechanics, Part 4

Practice Problems 4 1. A cylindrical flywheel has a radius of 2ft and weighs 1610 lb. If a constant force whose

magnitude is 100 lb is applied to the rim of the wheel along a tangent to the rim and causes clockwise rotation of the wheel, what would be the magnitude of the angular acceleration of the wheel?

2. When the applied force in problem 1 acts toward the right and upward and the angle

between a horizontal reference line and the line of action of the force is 50°, what are a) the magnitude and direction of the horizontal component of the reaction exerted on the wheel by the bearings and b) the magnitude and direction of the vertical component of that reaction?

3. At a certain instant, a rotating crank is in the position indicated in Figure 20. The crank weighs 60 lb, and the distance from the pin connection, about which the crank rotates, to the constant force P applied at the moving end of the crank is 2.5 ft. The magnitude of P is 30 lb, and the angular velocity of the crank at the instant under consideration is 2.0 rad per sec, clockwise. It may be assumed that the distance along the crank from the pin connection to the center of gravity of the crank is 1.25 ft and that the dimensions of the cross section of the crank can be neglected when its mass moment of inertia is computed. Determine a) the magnitude and direction of the angular acceleration of the crank at the instant under consideration; b) the magnitude and direction of the tangential component of the linear acceleration of the center of gravity of the crank; and c) the magnitude and direction of the normal component of that linear acceleration.

FIGURE 20-Position of Crank in Practice Problems 3 and 4

4. For the crank in Problem 3, determine a) the magnitude and direction of the horizontal component of the reaction exerted on the crank by the pin and b) the magnitude and direction of the vertical component of that reaction.

Check your answers with those on page 93.

44 Engineering Mechanics, Part 4

General Plane Motion of Rigid Body

Simple Conditions for General Plane Motion

• As explained in Engineering Mechanics, Part 3, it is said that a body moves with general plane motion when it moves in translation while it rotates about an axis. A common example is a wheel that rolls along a horizontal supporting surface. In the simplest case, the axis of rotation passes through the center of gravity of the body, and the transla- tion of the center of gravity occurs along a horizontal path. Also, it is as- sumed that the body rolls along the supporting surface without slipping.

Typical conditions for a case like that just described are represented in Figure 21A. It is assumed that a horizontal force Pis applied to the body in such a way that the line of action of the force passes through the center of gravity G of the body and the body can rotate about an axis through the center of gravity. The method of applying the force is indicated schematically in Figure 21A, where the force is applied by a cable which forms a loop around the axle of the wheel. The dis- placement of the center of gravity of the body in translation must be in the direction in which the applied force Pacts. The direction of the angular acceleration of the wheel must correspond to the direction of the linear acceleration of its center of gravity in translation. In Figure 21 the direction of the angular acceleration must be clockwise.

As a wheel rotates on its axis and rolls along the supporting surface, a different point on the rim of the wheel is in contact with the surface at each instant. In a sense, the wheel rotates about the point of contact at an instant under consideration. Such a point is called an instantaneous center. As was shown in Engineering Mechanics, Part 3, the magnitude of the actual linear velocity of the instantaneous center while it is in contact with the supporting surface is zero.

In order that a rotating wheel may roll along a supporting surface, rather than slide over the surface without translation, an adequate friction force must be developed between the wheel and the surface. When a wheel of an automobile rests on a slippery pavement and the engine is started, the wheel often turns without gripping the pave- ment and there is no translation of the automobile. (This is one of the conditions under which friction is helpful.)

Engineering Mechanics, Part 4

SUPPORTING SURFACE

A. Conditions

B. FBD

FIGURE 21-Wheel Rolling along Horizontal Surface

46 Engineering Mechanics, Part 4

Wheel Rolling along Horizontal Surface

24 • When a wheel rolls along a supporting horizontal surface without slipping, it is usually assumed that the forces acting on the wheel are those represented in the free-body diagram in Figure 21B. They are the force W, which represents the weight of the wheel and which acts vertically downward through the center of gravity of the wheel; the applied force P, which is considered to be horizontal and which causes translation of the center of gravity of the wheel; the vertical reaction N, which is exerted on the wheel by the supporting surface and which acts vertically upward; and the friction force F, which acts horizontally at the area of contact between the wheel and the supporting surface and which acts in the direction opposite to that of the applied force P.

When the supporting surface is horizontal and the applied force P also is horizontal, the magnitude of N will be equal to the magnitude of W. The magnitude of the force F cannot be greater than the value obtained by multiplying the magnitude of W or N by the coefficient of sliding friction. The magnitude ofF will be less than that of P.

When a body moves with general plane motion along a horizontal surface, it is usually most convenient to consider the motion of the center of gravity of the body. For rotation about an axis through the center of gravity, equation 1, Article 21, can be applied. The moment that causes rotation is the moment of the force F with respect to the center of gravity of the wheel.

Since the wheel does not slip on the surface, the distance traveled by the center of gravity of the wheel must be equal to the distance moved by a particle on the rim of the wheel. Therefore, the magnitude of the linear acceleration ac of the center of gravity must be equal to the radius of the wheel times the angular acceleration. Thus,

ac=Ra (1)

Also, by Newton's second law of motion, the magnitude of the vector sum ITx of the horizontal forces acting on the wheel is

(2)

In Figure 21, the magnitude of the vector sum of the horizontal forces

is P- F. Since the direction of ad is horizontal, the vertical component of ad is zero and the vector sum of the vertical forces acting on the wheel must be zero. The magnitude of the angular velocity of the wheel at any instant has no effect on the linear acceleration of the center of gravity and the forces acting on the wheel.

Engineering Mechanics, Part 4

Example Problem

Problem: A solid wheel weighing 40 lb has a diameter equal to 3 ft and is pulled along a horizontal surface by a horizontal force whose magnitude is 10 lb. The friction force is sufficient to prevent the slipping of the wheel on the surface. Determine a) the magnitude of the vertical force N exerted on the wheel by the supporting surface; b) the linear acceleration of the center of gravity of the wheel; and c) the magnitude of the friction force F.

Solution: The FBD for this problem is shown in Figure 21B, in which it is specified that W = 40 lb, P = 10 lb, and the radius of the wheel is R = % x 3 = 1.5 ft. It is necessary first to determine the mass of the wheel and its mass moment of inertia with respect to the axis of rotation. Thus,

40 m = 32.i = 1.242 slugs

By formula 2, Article 7,

1 2 2 Ic = 2 x 1.242 x 1.5 = 1.397 slug-ft

a) Since the vector sum of the vertical forces must be zero, the forces Wand N must be equal and opposite. Hence, N = 40 lb upward.

b) From formula 1, Article 24, the magnitude of the linear acceleration of the center of gravity of the wheel is

~ lac I = 1.5 a

From equation 1, Article 21,

Mc=Ica

But, Me = 1.5 F and Ic = 1.397

Substituting, we obtain

1.5 F = 1.397 a

and F=0.931a (A)

From equation 2, Article 24,

10 - F = 1.242 x 1.5 a= 1.863 a

and F = 10 - 1.863a (B)

When the two expressions for F are equated, the result is

0.931a = 10 -1.863 a

Hence, 10

a= 2.794 = 3.58 rad per sec per sec

Engineering Mechanics, Part 4

The linear acceleration of the center of gravity is

1.5 a = 1.5 x 3.58 = 5.37 fps per sec

c) Also, by equation A, the magnitude of the friction force is

F = 0.931 X 3.58 = 3.33lb

Wheel Rolling along an Inclined Plane Surface

• In Figure 22A a wheel is represented on an inclined plane surface. It is assumed that the wheel is rolling upward along the surface under the action of a force P, which overcomes the resistance of the weight W of the wheel and the friction force between the wheel and the in- clined surface. The direction of the linear acceleration of the center of gravity G of the wheel is indicated by the arrow to the right of the wheel. In Figure 23A, it is assumed that a wheel is rolling downward along an inclined surface under the action of its weight W, a clock- wise moment with magnitude MR, and the friction force. In either Fig- ure 22 or Figure 23, there is also a force representing the reaction which is exerted on the wheel by the inclined surface and which acts at right angles to that surface.

A. Conditions B. FBD

FIGURE 22-Wheel Rolling upward along Inclined Surface

Engineering Mechanics, Part 4 49

v w w

A. Conditions B. FBD

FIGURE 23-Wheel Rolling downward along Inclined Surface

Example Problems

In Figure 22A, the angular acceleration of the wheel must be clockwise, and the friction force applied at the rim of the wheel must act down- ward along the inclined surface. The free-body diagram for the wheel is shown in Figure 22B. In Figure 23A, the angular acceleration must be counterclockwise and the friction force must act upward along the plane. The free-:body diagram for this case is shown in Figure 23B.

The conditions in Figure 22 or Figure 23 are basically similar to those in Figure 21, but the directions of the reference axes should be paral- lel and perpendicular to the inclined surface (instead of horizontal and vertical). Inclined axes are represented near the FBD by x' andy'.

Problem 1: A solid wheel is being rolled upward along a plane surface that slopes upward toward the right and makes an angle with a hori- zontal reference plane equal to 20°. The wheel weighs 50 lb and its diameter is 2.5 ft. The force that causes movement is parallel to the inclined surface and its magnitude is 30 lb. Determine the linear accel- eration of the center of gravity of the wheel in a direction parallel to the inclined surface and the magnitude of the friction force.

50 Engineering Mechanics, Part 4

Solution: The FBD for the wheel is similar to that in Figure 22B. In this problem, W =50 lb, P = 30 lb, the angle between a horizontal reference plane and the inclined surface is 20°, and the radius of the wheel is % x 2.5 = 1.25 ft. The mass of the wheel is

50 m = 32_2 = 1.553 slugs

and its mass moment of inertia is

1 2 2 Ic = 2 x 1.553 x 1.25 = 1.213 slug-ft

By formula 1, Article 24,

ac = 1.25 a

By equation 1, Article 21,

and

1.25F= 1.213 a

F=0.970 a

To apply the equation rFx' = m(ac)x', it is necessary to determine the component of the weight Win the desired direction. This component acts toward the left and downward, and its magnitude is 50 sin 20° = 17.1lb. Then,

30- 17.1 - F = 1.553 x 1.25 a= 1.941a

and F = 12.9- 1.941a

When the two expressions for Fare equated, the result is

0.970 a= 12.9 -1.941a

· Hence, 12.9 d a = 2_911 = 4.43 ra per sec per sec

The required values are

~ I ac I = 1.25 x 4.43 = 5.54 fps per sec

F = 0.970 X 4.43 = 4.30 lb

Engineering Mechanics, Part 4 51

Problem 2: The wheel in example problem 1 is rolling downward along the specified inclined plane surface under the action of its own weight and a clockwise moment MR. If the linear acceleration of the center of gravity of the wheel is to be 3 fps per sec at a certain instant, what should be the magnitude of the moment MR?

Solution: The FBD is similar to that in Figure 23B. The values of m and Ic were computed in example problem 1. By formula 1, Article 24,

3 = 1.25a

Hence, a = 2.4 rad per sec per sec

By equation 1, Article 21,

MR -1.25F = 1.213 a= 1.213 x 2.4 = 2.91

From the equation Lfx' = m(ac)x', it follows that

and

From equation A,

-17.1 + F = 1.553 X (-3) =- 4.66

F= 12.44

MR - 1.25 X 12.44 = 2.91

MR - 15.55 = 2.91

MR = 15.55 + 2.91 = 18.46 ft-lb

(A)

52 Engineering Mechanics, Part 4

Practice Problems 5 1. The diameter of a solid wheel is 3.5 ft, and its weight is 60 lb. The wheel is rolled toward

the left along a horizontal surface under the action of a horizontal force whose magnitude is 20 lb. If the friction force is sufficient to prevent slipping, what are a) the magnitude of the linear acceleration of the center of gravity of the wheel and b) the magnitude of the friction force?

2. The wheel in problem 1 is to be rolled upward along a plane surface which slopes upward toward the left and is inclined so that the angle between it and a horizontal reference plane is 10°. The desired linear acceleration of the center of gravity of the wheel along the inclined surface is 4 fps per sec, and this motion is to be produced by a force whose line of action is parallel to the inclined surface. What is the required magnitude of that force?

3. The wheel in problem 1 is to roll down the inclined plane in problem 2 under the action of its own weight and a counterclockwise moment whose magnitude is 40 ft-lb. Deter- mine a) the magnitude of the linear acceleration of the center of gravity of the wheel and b) the magnitude of the friction force.

Check your answers with those on page 93.

Engineering Mechanics, Part 4 53

WORK-ENERGY METHOD

Application of Method for Translation

General Statement of Work-Energy Principle

The basic principle that is applied in the work-energy method may be stated as follows:

The total amount of work done on a body by the resultant force acting on the body while it moves from one position to another is equal to the total change in the kinetic energy of the body during the movement.

The work-energy method is useful when there is rectilinear translation in either of the following types of problems:

1. The magnitude of the velocity of a body changes while the body moves from one position to another.

2. Two connected bodies move in a related manner.

The work-energy method is sometimes useful in a problem pertaining to rotation alone or to general plane motion.

General Equation for Rectilinear Translation of Body

When the motion of a body is rectilinear translation, the resultant force acting on the body must act in the same direction during the entire interval of time under consideration. Also, the direction of the velocity must remain unchanged. Sometimes the magnitude of the resultant force remains constant. In other cases, the magnitude of the resultant force varies, but it is possible to determine the mean, or average, mag- nitude of the force. In such a case, the force may be treated as if it had a constant magnitude equal to this mean value. Unless the body is moving under the action of a balanced system of forces, the magnitude of its velocity will change from instant to instant. The basic equation for rectilinear translation may be derived as follows.

As explained in Article 9, the amount of work done by a force on a body in a certain interval of time may be found by multiplying the magni- tude of the force by the distance through which the body moves during that time interval in the direction of the line of action of the force. If FR denotes the magnitude of the resultant force acting on a body, Lls denotes the displacement of the body in the direction of the line of action of the resultant force, and U denotes the work done during the time interval in which this displacement occurs, then

(1)

54 Engineering Mechanics, Part 4

Also, the kinetic energy of translation of a body at any instant can be computed by applying the formula in Article 11. If v1 denotes the magni- tude of the velocity at the beginning of a selected time interval, and if v2 denotes the magnitude of the velocity at the end of that time interval, then the change t.£1 in kinetic energy of translation during that interval is

w 2 2 M 1 = 2g (v2 - v1 ) (2)

In which W is the weight of the body and g is the acceleration due to gravity.

According to the basic principle for the work-energy method, the fol- lowing equation can be written for rectilinear translation:

w 2 2 FR (&) = 2g (v2 - v1 ) (3)

In this equation, v1 denotes the smaller velocity and v2 denotes the larger velocity. It does not matter whether the velocity increased dur- ing the specified time interval or decreased during that interval. The given conditions will determine which velocity is larger in a particular problem. It is necessary to express the displacement L).s in feet.

The correctness of equation 3 for a simple case of rectilinear translation will be demonstrated here. As indicated in Figure 24, a body whose weight is W has been slid along a horizontal surface under the action of a horizontal force FR. Friction will be neglected. When the body was in position 1, in which its rectangular outline is shown by the solid lines, the magnitude of its velocity was v1• Also, when the body was in position 2, represented by the broken lines, the magnitude of its ve- locity was v2• The distance through which the center of gravity of the body moved from position 1 to position 2 is denoted by&; and the in- terval of time required for this motion is denoted by M. Vectors repre- senting the velocities are included in Figure 24.

POSITION 2 ___ ) BODY 1 1 V2 ~ 1---+

FIGURE 24-Displacement and Change in Velocity of Body during Time Interval

Engineering Mechanics, Part 4

In accordance with explanations in Engineering Mechanics, Part 3, the mean velocity Vm during the time interval M can be expressed in the following two ways:

Hence,

As v=-

m L'1t and

Also, the magnitude of a uniform acceleration a during the time interval b.twould be

v2 -vl a=--

When M is replaced by its equivalent value, it is found that

According to Newton's second law of motion,

w F =-xa

r g

When a is replaced by its equivalent value,

FR ~; ( V, '~v,') The system of forces acting on the body consists of the horizontal force FR, the weight W, and the vertical reaction exerted on the body by the supporting surface. However, since each particle of the body moves along a horizontal path, neither of the vertical forces does any work. The work done by the horizontal force FR is

w 2 2 FR (As)= 2g (v2 - v1 )

This is equation 3. This demonstration also indicates that the work- energy method is really another way of applying Newton's second law of motion.

Action of Coil Spring

A type of device that is often used to apply a force to a body is known as a coil spring. A spring in its normal condition is pictured in Figure 25A. The length of the spring in this condition is the distance L0• A spring is so constructed that either its length can be increased by pulling its ends further apart, as indicated in Figure 25B, or its length can be reduced by pushing its ends closer together, as indicated in Figure 25C.

56 Engineering Mechanics, Part 4

A. Spring in natural condition B. Stretched spring C. Shortened spring

FIGURE 25-Deformation of Spring

In order to stretch or shorten a spring, a force must be applied to the spring, and work must be done on the spring by such a force. When a spring has been either stretched or shortened, it is said that the spring is deformed. The difference between the normal length L0 of a spring and its length when stretched or shortened is called the deformation of the spring for the particular condition. In Figure 25B, the deformation is d1, and in Figure 25C, the deformation is d2•

,:M-----Lo = dl-----+1

B. Stretched spring

C. Shortened spring

FIGURE 26-Forces Produced by Deformed Spring

Engineering Mechanics, Part 4 57

A deformed spring tends to return to its original length L0• The deformed spring therefore exerts a force on any body to which it is attached. In Figure 26 is shown a spring (which is represented schematically by a zigzag line) that is connected to a stationary support at the left-hand end and to a movable block at the right-hand end. It is assumed that the block must move along a horizontal path. In Figure 26A, the spring has its normal length L0, and it does not exert any force on either the sta- tionary support or the movable block. In Figure 26B, the block is moved to the position shown and the spring is stretched to the length L0 + d1• Because of its tendency to return to its normal length, the spring exerts a pull toward the right on the stationary support and also exerts an equal pull toward the left on the movable block. The magnitude of each force is denoted by F1. In Figure 26C, the block is moved to the position shown and the spring is compressed to the length L0 - d2• In this condition, the spring exerts a push toward the right against the block. The magnitude of each force is denoted by F2•

If the force that keeps a spring deformed is no longer applied and the spring is permitted to return to its normal length, the spring will cause translation of a movable body to which it is attached and will do work on that body. For instance, if the spring in Figures 26B or 26C is per- mitted to return to its normal length, the block will move back to the position it occupied in Figure 26A because of the action of the force exerted on the block by the spring.

Spring Constant

For an ideal spring, there is a constant ratio between the force applied to or by a spring and the deformation of the spring. For a particular spring, the magnitude of this ratio would be the same regardless of whether the spring were stretched or shortened, and the following relationship may be applied:

in which k = constant ratio F = force exerted on or by a spring d =deformation of the spring

The constant ratio k is usually called the spring constant. The unit for a spring constant must be a unit of force divided by a unit of length. In the English system, the common unit for a spring constant is pounds per inch. For instance, if the constant for a certain spring is 50 lb per in. (pounds per inch), and it is desired to increase or decrease the length of the spring by 1.5 in., the required magnitude F of the force would be F = kd = 50 x 1.5 = 75lb. Also, if a force equal to 25lb is applied to the spring, the deformation d will bed = % = 25/so = 0.50 in. If the spring is

58 Engineering Mechanics, Part 4

deformed by 2 in., the magnitude of the force exerted by the spring on a body to which it is attached will be F = kd = 50 x 2 = 100 lb.

Work Done on or by Deformed Spring

As mdicated in Figure 27, a spring is attached to a stationary support and a movable block, which can only move along a horizontal path. It is assumed that a horizontal force, which acts toward the right and has an increasing magnitude, is applied to the block so as to move the block and stretch the spring. Since the distance through which the block moves is in the same direction as the line of action of the force, the work done by the force during a specified interval of time is equal to the product of the mean magnitude of the force and the distance through which the block moves .

._---------------Lo+d2------------------~

FIGURE 27-Positions of Movable Block Attached to Stretched Spring

The block is shown in three positions. The normal length of the spring is denoted by La, and the spring has this length when the block is in the left-hand position in Figure 27. When the block is in the middle position, the length of the spring is La+ d1• When the block is in the right-hand position, the length of the spring is La + d2• In this case, it is assumed that the spring was stretched, but similar reasoning could be applied if the spring were shortened. Also, the spring constant will be denoted by k. At the instant at which the length of the spring was La, no force was applied to the block. At the instant at which the length was La + d11 the magnitude of the force was kd1• At the instant at which the length was La + d2, the magnitude of the force was kd2• While the spring was being stretched, it was necessary to apply a force to the spring through the block. At the same time, the spring also exerted a force on the block. At any instant these two forces necessarily had the same magnitude. During a certain specified interval of time, the deformation of the spring in Figure 27 increased from d1 to d2, and the block moved through a horizontal distance equal to d2 - d1• It will be assumed that while the deformation increased from d1 to d2, the magnitude of the force that was applied to or by the spring increased from F1 to F2 at a uniform

Engineering Mechanics, Part 4

Example Problem

rate. The mean force F m that was applied to or by the spring during the specified time interval was

(1)

While the spring was being stretched, the applied force did work on the spring. The amount of this work U during the specified time interval was

1 u = 2 k (d2 + dl) (d2 - dl)

or u = _! k (d2 2 - dl 2) 2

(2)

This amount of work was stored in the spring because it was stretched.

If the force that caused the spring to stretch is no longer applied, the tendency of the spring to return to its normal length will cause the block to move toward the left. As a result of this movement, the deformation of the spring will decrease and the magnitude of the force applied to or by the spring will be decreased proportionately. During the interval of time during which the deformation decreased from d2 to d11 the hori- zontal distance through which the block moved will be d2 - d11 and the magnitude of the mean force applied by the spring to the block will be Y2(F2 + F1). Therefore, the amount of work U done by the spring on the block during the specified time interval will again be h k (d2 2 - dl 2).

Equation 2 can be used to compute either the amount of work done on a spring or the amount of work done by a spring while the defor- mation of the spring changes from d1 to d2 or from d2 to d1• In generaC d1 should be taken as the smallest deformation during the specified time interval, and d2 the largest deformation. The unit used for the deformations must correspond to the unit of length specified in the spring constant.

Problem: The normal length of a spring is 16 in. and the constant for it is 3lb per in. One end of the spring is attached to a stationary sup- port, and the other end is attached to a movable block which can only move along a straight horizontal path. If the block was moved from a point 20 in. from the support to a point 28 in. from the support, how much work was done on the spring?

Solution: The conditions are represented in Figure 28. The block moved from position 1 to position 2. Since the natural length L0 of the spring is 16 in., the deformation of the spring when the block was 20 in. from the support was d1 = 20- 16 = 4 in. When the block was 28 in. from the support, the deformation was d2 = 28 - 16 = 12 in. By equation 2,

60 Engineering Mechanics, Part 4

in which the deformations must be expressed in inches, the amount of work done on the spring was

U = ~ x 3 x (122 - 42) = 192 in-lb (inch-pounds)

STATIONARY SUPPORT

~--------- 28 in .. -------~

FIGURE 28-conditions in Example Proble, Article 30

In some devices, a spring is deformed in a certain direction but a movable body, to which the spring is attached, must move in a different direction. For instance, as indicated in Figure 29, one end of a spring is attached to a stationary support, and the other end of the spring is attached to a movable block, which can only move along a horizontal path. The normal length of the spring will be denoted by L0, and the spring constant will be denoted by k. When the spring is in the vertical position, as indicated in Figure 29A, the length of the spring is L0 + d11 the deformation of the spring is d11 and the force applied to the spring is kd1. When the spring is in the inclined position indicated in Figure 29B, the length of the spring is Lo + d21 the deformation is d21 and the force is kd2• In this case, the work done on the spring by the force applied through the block is equal to % k (d2 + d1) (d2 - d1), or

u = l k (d/- dl 2) 2

This equation is the same as equation 2, Article 30. For the conditions in Figure 29, d2 and d1 are the deformations of the spring. The actual distance moved by the block need not be considered.

Engineering Mechanics, Part 4 61

If it is desired to determine the work done by the spring on the block when the spring and the block return to their original positions, it is convenient to consider the change in the length of the spring rather than the movement of the block The preceding equation would apply also for this condition.

BLOCK

A. Spring vertical B. Spring inclined

FIGURE 29-Movement of Block and Deformation of Spring in Different Directions

Example Problem

Problem: A block that is attached to a spring was moved from the position indicated in Figure 29A to the position in Figure 29B. The normal length of the spring is L0 = 4 in. and the spring constant is 6 lb per in. If the actual length L0 + d1 of the spring in the original vertical position was 5 in. and the block was moved horizontally toward the right for a distance equal to 8 in., how much work was required to deform the spring?

Solution: The actual length L0 + d2 of the spring in the inclined position was ...;52 + 82 = 9.43 in. Hence, d1 =5-4= 1 in. and d2 = 9.43-4 = 5.43 in. The required amount of work was

U =~X 6 X (5.432 -12) = 85.5 in-lb

62 Engineering Mechanics, Part 4

Practice Problems 6 1. The normal length of a spring is 6 in., and the spring constant is 20 lb per in. How much

work would have to be done on the spring to reduce its length from 6 in. to 4.5 in.?

2. How much work would have to be done on the spring in problem 1 to increase its length from 8 in. to 10 in.?

3. One end of the spring in problem 1 was connected to a stationary support, and the other end of the spring was connected to a movable block which could only move in a vertical direction. A force was applied to the block until the block was moved to a point 14 in. from the support. Then the block was released and the spring was permitted to bring the block back to a point 8 in. from the support. How much work was done by the spring while it was being shortened?

4. A spring is attached to a stationary support and a movable block in the manner indicated in Figure 29A. The normal length of the spring is 10 in., and the spring constant is 4lb per in. When the spring is vertical, its actual length is 12 in. The block was moved along a hori- zontal surface from an original position with the spring vertical to a new position 9 in. to the right of the original position, as indicated in Figure 29B. How much work was done on the spring during this movement of the block?

Check your answers with those on page 93.

Engineering Mechanics, Part 4

Example Problems

Application of Work-Energy Method for Translation of Single Body

When equation 3, Article 27, is to be applied in the solution of a practical problem, the weight W of the body will usually be known and the acceleration g is a constant. The other four quantities in the equation are the magnitude of the resultant force acting on the body, the distance moved by the body, and the initial and final velocities of the body. When any three of these quantities are known, the fourth one can be computed. When a force is applied to or by a spring, the term FR represents the mean force that is exerted on or by the spring and & represents the change in deformation of the spring. However, v1 and v2 represent the magnitudes of the velocities of the body at the beginning and end of the specified time interval.

Problem 1: As indicated in Figure 30A, a block is being slid along a horizontal surface under the action of a horizontal force P that has a constant magnitude. The block weighs 40 lb, and the coefficient of friction f..L (Greek mu) for the materials used for the block and the sur- face is 0.1. If the magnitude of the force Pis 20 lb and the velocity of the block is 2 fps when the block is in the position shown, what will be the velocity of the block when it reaches a position 8 ft to the right of the position shown?

w = 40 lb BLOCK

p = 20 lb

A. Force acting on block B. FBD for block

FIGURE 3D-Conditions in Example Problem 1, Article 32

Solution: The free-body diagram for the block is shown in Figure 30B. The forces to be considered are the weight W, the vertical force N ex- erted on the block by the supporting surface, the applied force P, and the friction force F m, which must act in the direction indicated. To de- termine the magnitude of the resultant force FR that causes move- ment of the block, it is necessary to compute the maximum friction force F m, which depends on the magnitude of N. In this case, N = W = 40 lb andFm = f..LN = 0.1 x 40 =4lb. Therefore, FR = P-Fm = 20-4 = 16 lb. Other known values in equation 3, Article 27, are & =8ft, W = 40 lb,

64 Engineering Mechanics, Part 4

g = 322 fps per sec, and v1 = 2 fps. The calculations for computing v2 follow:

16 X 8 = 40 X (v2 2 - 22) 2x32.2

v2 2 -4=206 v2 = 14.5 fps

Problem 2: The block in the example problem in Article 30 weighs 25 lb. After the block had been moved to a point somewhat more than 28 in. from the stationary support, the spring was permitted to return to its natural length. When the block was 28 in. from the support, the magnitude of its velocity was 3.5 fps. toward the left. If the coefficient of friction is 0.2, what was the velocity of the block when it reached the point 20 in. from the support?

Solution: The FBD for the block is shown in Figure 31. The magni- tude of the weight W of the block is constant. The magnitude of the up- ward vertical force exerted on the block by the horizontal surface must be equal to Wand constant. The magnitude of the horizontal force P exerted on the block by the spring is variable, but the force must act toward the left during the specified interval of time. Since the magni- tude of the friction force F m depends only on the magnitude of Nand the coefficient of friction, it remains constant. The direction of this fric~ tion force must be opposite to the direction of P, or toward the right.

FIGURE 31-FBD for Block in Example Problem 2, Arficle32

w = 25lb

During the interval of time that was required for the specified movement, the displacement & was 28- 20 = 8 in., or 0.667 ft. Also, the deforma- tion of the spring changed from d2 = 28- 16 = 12 in. to d1 = 20 -16 = 4 in. Since the spring constant is 3lb per in., the mean magnitude of the force exerted by the spring on the block was 1J2 k (d2 + d1) = 1J2 x 3 x (12 + 4) = 24lb. Since the magnitude of the force N is 25 lb and the magnitude of the force F m is 0.2 x 25 = Sib, the magnitude of the resultant horizon- tal force acting on the block was 24- 5 = 19lb. Therefore, equation 3, Article 27, becomes

25 2 2 19 X 0.667 = 2 X 32.2 X (V2 - 3.5 )

Engineering Mechanics, Part 4 65

The value of v2 can be found as follows:

v2 2 - 12.25 = 32.6 v2 = 6.70 fps

Problem 3: The block in the example problem in Article 31 weighs 12 lb. After the block had been moved slightly more than 8 in. to the right of its original position with the spring vertical, the spring was permit- ted to return the block to that original position. If the velocity of the block was 2 fps toward the left when the block was 8 in. from the original position, and if friction can be neglected, what was the veloc- ity of the block when it reached that original position?

Solution: When the block was 8 in. from its original position, the length

of the spring was ...J52 + 82 = 9.43 in. and the deformation of the spring was d2 = 9.43-4 = 5.43 in. When the block returned to its original po- sition, the deformation of the spring was d1 = 5 - 4 = 1 in. The change in deformation of the spring was 5.43 -1 = 4.43 in., or 0.369 ft. Also, the mean magnitude of the force exerted by the spring during the specified movement was Yz k (d2 + d1) = 1h x 6 x (5.43 + 1) = 19.29 lb. According to equation 3, Article 27,

12 2 2 19.29 X 0.369 = 2 X 32.2 X (v2 -2)

Hence, v2 = 6.50 fps

Problem 4: As indicated in Figure 32A, a block is sliding down an inclined plane surface. The block weighs 20 lb, and the angle between the inclined surface and a horizontal reference plane is 30°. Also, the coefficient of friction for the materials for the block and the surface is 0.6. If the magnitude of the velocity of the block in the position indicated is 2 fps, how far will the block slide from that position before it comes to rest?

Solution: The free-body diagram for the block is shown in Figure 32B. The three forces acting on the block are W which represents the weight of the block and acts vertically downward; N, which represents the force exerted on the block by the inclined surface and acts perpendicular to that surface in the direction indicated; and F m, which represents the maximum friction force and acts parallel to the inclined surface in the direction indicated.

66 Engineering Mechanics, Part 4

w = 20 lb

A. Block on inclined surface B. FBD for block

FIGURE 32-conditions in Example Problem 4, Article 32

The first step is to compute the magnitude of Nby applying the prin- ciple that the vector sum of the components of the forces at right angles to the inclined surface must be zero. Thus,

- 20 cos 30° + N = 0

and N = 17.32lb

The next step is to compute the magnitude ofF m by multiplying N by the coefficient of friction. Thus,

F m = 0.6 X 17.32 = 10.39lb

In order that the block will come to rest and will not continue sliding down the plane, the magnitude of the maximum friction force must be greater than the magnitude of the component of the weight of the block that acts parallel to the inclined surface. Since 20 sin 30° = 10.00 lb and Fm = 10.39lb, the block will come to rest.

The magnitude of the resultant force that acts parallel to the inclined surface and tends to change the linear velocity of the block along the surface is

FR = 10.39 -10.00 = 0.39lb

The line of action of this force and the required distance will be parallel to the inclined surface, but the block will move downward while the force acts upward. As a result, the magnitude of the linear velocity of the block decreases.

The velocity of the block at the beginning of the specified interval of time is 2 fps. Since the block will be at rest at the end of that time inter- val, the velocity at that instant will be zero.

Engineering Mechanics, Part 4 67

The required distance & through which the block will move can now be computed by applying equation 3, Article 27. When the known values are substituted for FR, W, g, v2, and v11 the resulting equation is

0.39 (&) = 2~ X (22 - 0Z) 2x 2.2

Hence, &=3.2 ft

It may be of interest to know that the same result would be obtained for any value of the weight W. In other words, the distance & would not be affected by the weight of the block if other conditions remained the same.

Work-Energy Method for Translation of Two Connected Bodies

When two bodies are connected, they must move in such a manner that the magnitudes of the velocities of both bodies are the same at any instant. However, the bodies may not be moving in the same direction. Two simple cases are illustrated in Figure 33A and Figure 34A.

In Figure 33A, blocks 1 and 2 are connected by a flexible cable, which passes around a rotating sheave or pulley. When the blocks are in the positions shown, the weight of block 2 tends to cause that block to move vertically downward. The block exerts a vertical force on the cable, and a horizontal force having the same magnitude is exerted by the cable on block 1. (It may be assumed that the sheave rotates without friction.) Block 1, which is supported on a horizontal surface, tends to slide along that surface under the action of the horizontal force ap- plied to the block by the cable. At any instant, the magnitude of the velocity of block 1 must be the same as the magnitude of the velocity of block2.

W= lOib

BLOCK l

CABLE

T = l51b

A. Relative positions of blocks B. FBD for block 1

FIGURE 33-System Composed of Two Connected Bodies

68 Engineering Mechanics, Part 4

In Figure 34A, two blocks 1 and 2 having different weights are connected by a flexible cable, which passes around two rotating sheaves.

Each block exerts a force on the cable. Both forces act vertically down- ward and the magnitude of each force is equal to the weight of the block producing it. Since these weights are not the same, the heavier block tends to move downward while the lighter block is compelled to move upward. At any instant, the magnitudes of the velocities of the two blocks must be the same. It may be assumed that the magni- tude of the force acting upward on either block is equal to the weight of the other block. The work-energy method is used in the following example problems in which two connected bodies are considered.

CABLE

T =50 lb

CABLE

w = 80 lb

BLOCK 2

A. Relative positions of blocks B. FBD for block 1

FIGURE 34-System Composed of Two Connected Bodies

Example Problems

Problem 1: At a certain instant the two blocks 1 and 2 in Figure 33A are in the indicated positions. Block 1 weighs 10 lb, and block 2 weighs 15lb. Block 1 slides along a horizontal surface, for which the coefficient of friction is 0.2, under the action of a horizontal force exerted by the cable. If the friction in the sheave is neglected and the magnitude of the velocity of each block at the specified instant is 3 fps, what would be the magnitude of each velocity after block 2 has moved 2 ft downward from the position shown?

Solution: Since the velocities of both blocks must have the same mag- nitude, it is sufficient to determine the velocity of block 1. The FBD for block 1 is shown in Figure 33B. The weight W of the block acts ver- tically downward; the force N exerted on the block by the horizontal sur- face acts vertically upward; the force T exerted on the block by the

Engineering Mechanics, Part 4 69

cable acts horizontally toward the right; and the friction force F m acts horizontally toward the left. The magnitude of W is given as 10 lb; the magnitude of N is also 10 lb; the magnitude of Tis equal to the weight of block 2, or 15lb; and the magnitude ofF m is 0.2 N = 0.2 x 10 = 2lb. Since the block 1 must move horizontally and the lines of action of Wand N are vertical, the magnitude of the resultant force FR that affects the movement of the block 1 is T- Fm = 15- 2 = 13lb. In equation 3, Arti- cle 27, As is equal to the distance moved by the block 2 in the speci- fied interval of time, or 2 ft, and v1 = 3 fps. Since both blocks must be moved, the value to be substituted for W is the sum of their weights, or 10 + 15 = 25lb. Hence,

and

25 13 x2= 2 2 x(v/-Y) x3 .2

v2 = 8.72 fps

Problem 2: The weight of block 1 in Figure 34A is 50 lb, and the weight of block 2 is 80 lb. At a certain instant, downward movement of block 2 is prevented, and the two blocks are stationary. If the support for block 2 is removed and friction in the sheaves is neglected, how far would each block move in a vertical direction before the magnitude of its velocity became 5 fps?

Solution: The movement of either block may be considered. The FBD for the heavier block is shown in Figure 34B. Only two vertical forces need be considered in this case. The force W representing the weight of block 2 acts downward, while the force T representing the pull ex- erted on block 2 by the cable acts upward. The magnitude of Tis equal to the weight of block 1.

The magnitude of the resultant force FR that acts on block 2 is 80 - 50 = 30 lb. Also, in equation 3, Article 27, v1 = 0 and v2 = 5 fps. Since both blocks must be moved, the value to be substituted for Win the equation is the sum of their weights, or 50 + 80 = 130 lb. Hence,

and

30 Lls) = 130 X (52 - 02) 2 x32.2

As= 1.68ft

70 Engineering Mechanics, Part 4

Practice Problems 7 1. The block in Figure 30A weighs 30 lb, and it is being pulled toward the right along a

horizontal surface by a horizontal force P that has a constant magnitude equal to 12lb. If the coefficient of friction is 0.15 and the velocity of the block is 4 fps toward the right when it is in the position shown, what will be the magnitude of its velocity when it is 6 ft further to the right?

2. The normal length of a spring is 5 in. and the spring constant is 10 lb per in. As indicated in Figure 35, one end of the spring is connected to a stationary support, and the other end is connected to a movable block which weighs 60 lb and can only travel along a horizontal surface in a straight line. In the position shown, the block is 15 in. to the left of the stationary support and has a velocity toward the right whose magnitude is 2.5 fps. If the coefficient of friction is 0.2, what would be the velocity of the block at the instant at which it reached a point 9 in. to the left of the support?

~----15i1.-----~

STATIONARY SUPPORT

FIGURE 35-lnitial Position of Block in Practice Problem 2

3. The spring in problem 2 is connected to a stationary support and to a movable block which weighs 18 lb and can only travel along a straight horizontal path. When the spring is vertical, as indicated in Figure 36A, its actual length is 6 in. After the block had been moved toward the left for a certain distance, the spring was permitted to return the block to the position in Figure 36A. At the instant at which the block was 10 in. from the position in Figure 36A, as indicated in Figure 36B, the magnitude of its velocity was 3.5 fps and it was moving toward the right. If friction is neglected, what was the magni- tude of the velocity of the block when it was 4 in. to the left of the position in Figure 36A?

(Continued)

Engineering Mechanics, Part 4

Practice Problems 7

BLOCK l 0 in.---~

A. Spring vertical B. Spring inclined

FIGURE 36-Conditions in Practice Problem 3

4. As indicated in Figure 37, a block is sliding down an inclined plane surface. The angle between the inclined surface and a horizontal reference plane is 20°, the weight of the block is 25lb, the coefficient of friction is 0.4, and the magnitude of the velocity of the block in the position shown is 2.8 fps. How much further will the block slide before it comes to rest?

FIGURE 37-conditions in Practice Problem 4

5. As indicated in Figure 33A, two blocks are connected by a cable. The block 1, which weighs 36 lb, slides along a horizontal surface under the action of the force exerted on the cable by the block 2, which weighs 20 lb. The coefficient of friction is 0.25, and the magnitude of the velocity of each block in the position shown is 4 fps. If friction in the sheave is neglected, what would be the magnitude of the velocity of each block after the block 2 has moved downward 1 ft from the position shown?

6. As indicated in Figure 34A two blocks are connected by a cable. The weight of block 1 is 40 lb, and the weight of block 2 is 65lb. Mter the blocks have been held stationary, block 2 is permitted to move downward. If friction in the sheaves is neglected, how far would each block move in a vertical direction before the magnitude of its velocity be- came 6fps?

Check your answers with those on page 93.

72 Engineering Mechanics, Part 4

Other Applications of Work-Energy Method

Rotation of Disk

A typical case in which the work-energy method is useful is illustrated in Figure 38. A solid cylindrical disk is caused to rotate about a sta- tionary axis, which passes through the center of gravity of the disk. The force that produces the rotation is applied to the disk by means of a cable, which is wound around the disk and is attached to a body. The weight of this body exerts a pull on the cable. If the disk is permitted to rotate, the body will move downward .

FIGURE 38-Disk Rotated by Falling Body

.a AXIS

BODY

As the body moves downward under the action of its own weight, it does work. The amount of work done in a specified interval of time is equal to the product of the weight of the body and the vertical distance through which the body moves during that time interval. Since the center of gravity of the disk remains stationary, the disk does not do any work. Both the disk and the body possess kinetic energy. The kinetic energy of the disk is due to its rotation, while the kinetic energy of the body is due to its translation. According to the work-energy princi- ple, the amount of work done by the body in a specified interval of time must be equal to the sum of the change in kinetic energy of the disk and the change in kinetic energy of the body.

Engineering Mechanics, Part 4

Example Problem

Problem: The radius of the disk in Figure 38 is 3 ft, the weight of the disk is 500 lb, and the weight of the body is 200 lb. Both the disk and the body are at rest when the body is supported in the position shown. If the support is taken away and downward movement of the body is permitted, what will be the angular velocity of the disk after the body has moved 2 ft? Friction in the disk can be neglected.

Solution: The amount of work done by the body in the specified interval of time is

U =200 X 2 = 400 ft-lb

Before movement began, both the kinetic energy of the disk and the kinetic energy of the body were zero. At the end of the specified time interval, the kinetic energy of the body is, by the formula in Article 11,

1 200 2 Et = 2 X 32.2 X v

where v is the linear velocity of the body. In this case, v is related to the angular velocity ro of the disk by the equation v = 3ro. Hence,

1 200 2 2 Et = 2 x 32.2 x (3ro) = 28.0ro ft-lb

The kinetic energy of the disk is computed by formula 1, Article 12, in which the mass moment of inertia Ic is determined by formula 2, Article 7. Thus,

1 500 2 .2 Ic = 2 x 32.2 x 3 = 69.9 slug-fr

and 1 2 2 E, = 2 x 69.9 x ro = 35.0 ro ft-lb

The total change in kinetic energy of the disk and the body is

28.0 (J) + 35.0 ro2 = 63.0 ro2

When this change is equated to the work done by the block, the result is

63.0 ro2 = 400

and ro = 2.52 rad per sec

Example Problem

Engineering Mechanics, Part 4

Wheel Rolling Along Horizontal Surface

The work-energy method can also be applied to advantage in the case of a wheel that rolls along a horizontal surface, without slipping, under the action of a horizontal force. The conditions are represented in Figure 21A. In such a case, the horizontal force P does work when the wheel moves, and the amount of work done by this force is equal to the product of the magnitude of the force and the horizontal distance moved by the center of gravity of the wheel. No work is done by the weight W of the wheel or by the vertical force N exerted on the wheel by the supporting surface, because there is no displacement of the center of gravity of the wheel in the direction of the line of action of either of these forces. Also, no work is done by the friction force F because there is no displacement of that force if the wheel does not slip.

The wheel possesses two kinds of kinetic energy; namely, that due to translation and that due to rotation. The change in the total kinetic energy of the wheel during a specified interval of time must be equal to the amount of work done by the force P during that time interval.

Problem: A solid wheel is rolled along a horizontal plane, without slipping, by a horizontal force whose magnitude is Slb. The radius of the wheel is 2ft, and its weight is 30 lb. If the linear velocity of the center of gravity of the wheel at a certain instant is 4 fps, what would be the linear velocity at the instant at which the center of gravity of the wheel is 10ft from its first position?

Solution: The amount of work done by the applied horizontal force during the specified interval of time is

U = 5 X 10 = 50 ft-lb

The mass moment of inertia of the wheel is

1 30 2 2 Ic = 2 X 32.2 X 2 = 1.863 sl ug-ft

The angular velocity of the wheel at the beginning of the specified time interval is found by dividing the linear velocity of the center of gravity of the wheel by the radius. Thus, o/z = 2 rad per sec. Therefore, the total kinetic energy of the wheel at that instant is

1 30 2 1 2 2 X 32.2 X 4 + 2 X 1.863 X 2 = 11.18 ft-lb

Engineering Mechanics, Part 4 75

At the end of the specified time interval, when the linear velocity of the center of gravity of the wheel is denoted by v2, the total kinetic energy of the wheel is

1 30 2 1 (V2J 2 2 X 32_2 X V2 + 2 X 1.863 X 2 = 0.699 v2 ft-lb

When the total change in the kinetic energy of the wheel is equated to the work done by the horizontal force, the result is

0.699v2 2 -11.18 =50

Hence, V2 =9.36 fps

76 Engineering Mechanics, Part 4

Practice Problems 8 1. The disk in Figure 38 has a radius equal to 2.5 ft and weighs 480 lb. The weight of the

body is 150 lb. For a certain position of the body, the angular velocity of the disk is 2 rad per sec. If friction in the disk is neglected, what would be the angular velocity of the disk at the instant at which the body is 3ft below its first position?

2. A solid wheel that has a radius equal to 2.25 ft and weighs 40 lb is rolled along a horizontal plane, without slipping, by a horizontal force whose magnitude is 6lb. If the wheel started from rest, what would be the linear velocity of the center of gravity of the wheel when the center of gravity had moved 8 ft horizontally from its starting position?

Check your answers with those on page 93.

Engineering Mechanics, Part 4 77

IMPULSE ... MOMENTUM METHOD

Rectilinear Translation of Single Body

General Statement of Impulse-Momentum Principle

The basic principle that is applied in the impulse-momentum method may be stated as follows:

The linear ~pulse of a force which acts upon a body during a specified interval of time is equal to the change in the linear momentum of the body during that time interval.

Since impulse is the product of the magnitude of a force and an interval of time, and since momentum is the product of a mass and the magni- tude of a velocity, the impulse-momentum method is useful for deter- mining the change in the linear velocity of a body that is subjected to a system of forces for a definite interval of time. The impulse-momentum method is also applied in the analysis of problems involving the collision of two bodies. Such problems are commonly called impact problems.

Basic Equation for Rectilinear Translation

If FR denotes the magnitude of the resultant force acting on a body and .6..t denotes the interval of time during which this force acts, the linear impulse I of the resultant force is

(1)

Also, if v1 denotes the magnitude of the velocity of the body at the beginning of the time interval and v2 denotes the magnitude of the velocity at the end of the time interval, then the change ~in the momentum of the body during that time interval is

(2)

where W denotes the weight of the body, and g the acceleration due to gravity.

You should remember that impulse and momentum are really vector quantities. However, for rectilinear translation, which is the only type of motion that will be considered in this text, the direction of each velocity must be along the line of action of the resultant force FR. The resultant force FR may act either to increase the velocity of the body or to decrease that velocity.

Example Problems

Engineering Mechanics, Part 4

According to the impulse-momentum principle, the following equa- tion can be written for rectilinear translation of a body:

(3)

In this equation, v1 denotes the smaller velocity, and v2 the larger velocity. It does not matter whether the velocity increased during the specified time interval or decreased during that interval. The given conditions will determine which velocity is larger in a particular prob- lem. The unit used for expressing the time interval flt must be the same as the unit of time used in each velocity and in the acceleration g. In a particular problem, it will usually be assumed that the magnitude of the resultant force will remain constant during the specified interval of time.

Application of Method for Translation of Single Body

The impulse-momentum method will be applied in the following example problems involving rectilinear translation of a single body. Each of these problems can also be solved by applying the force- mass-acceleration method.

Problem 1: A block that weighs 20 lb is being slid along a horizontal surface by a horizontal force whose magnitude is 5 lb. The coefficient of friction is 0.1. If the velocity of the block at a certain instant is 4 fps, how many seconds will be required to increase the velocity to 30 fps?

Solution: The conditions are represented in Figure 30A, and the FBD for the block will be similar to that in Figure 30B. However, in this problem, W = 20 lb and P = 5 lb.

Since the magnitude of the vertical force N exerted on the block by the supporting surface is 20 lb, the magnitude of the friction force F m is 0.1 x 20 = 2lb. Hence, the magnitude of the resultant horizontal force is P- F m = 5-2 = 3lb. Other known values in equation 3 in Article 37 are W = 20 lb, g = 32.2 fps per sec, v1 = 4 fps, and v2 = 30 fps. The relation obtained by substituting these values in the general equation is

20 3(flt) = 32.2 X (30 - 4)

Therefore, flt=5.38sec

Engineering Mechanics, Part 4 79

Problem Z: A block that weighs 100 lb is being slid along a horizontal surface by an inclined force P which acts as indicated in Figure 39A. The magnitude of Pis 30 lb and the angle His zoo. Also, the coefficient of fric- tion is 0.15. If the block will not overturn and its velocity at a certain instant was Z fps, what will be the velocity at an instant 4 sec later?

FIGURE 39-conditions in Example Problem 2, Article38

A. Position of Applied Force

B. FBD for Block

Solution: The horizontal and vertical components of the force P are P x = 30 cos zoo = 28.2 lb toward the left

and Py = 30 sin 20° = 10.3lb downward

The magnitude of the force N exerted on the block by the supporting surface is W + Py = 100 + 10.3 = 110.3lb, and the magnitude of the friction force F m is 0.15 X 110.3 = 16.5lb. In equation 3, Article 37, FR = 28.2-16.5 = 11.7lb, L1t = 4 sec, v1 = 2 fps, and v2 denotes the required velocity. Therefore,

and

100 11.7x4= 32.2 x(v2 -2)

v2 - 2 = 15.07

v2 = 17.07 fps

80 Engineering Mechanics, Part 4

Problem 3: As indicated in Figure 32A, a block is sliding down an in- clined plane surface that makes an angle equal to 30° with a horizontal reference plane. The block weighs 20 lb, the coefficient of friction is 0.2, and the magnitude of the velocity of the block in the position shown is 2 fps. What would be the magnitude of the velocity 3 sec after the block passes the position shown?

Solution: The FBD for the block is shown in Figure 32B. The magni- tude of the force N exerted on the block by the supporting surface is 20 cos 30° = 17.32lb, and the magnitude of the friction force Fm is 0.2 x 17.32 = 3.46lb. The component of the weight of the block that acts parallel to the inclined surface is 20 sin 30° = 10 lb. Since this com- ponent is greater than F mt the block will slide down the surface with increasing velocity. If forces parallel and perpendicular to the inclined surface are considered, the line of action of the resultant force will be parallel to that surface and its magnitude will be 10- 3.46 = 6.54lb. By equation 3, Article 37, in which ilt = 3 sec, v1 = 2 fps, and v2 denotes the required velocity,

20 6.54 X 3 = 32.2 X (V2 - 2)

Hence, v2 =33.6 fps

Engineering Mechanics, Part 4

Practice Problems 9 1. As indicated in Figure 30A, a block is being slid along a horizontal surface by a

horizontal force P. The block weighs 40 lb, the magnitude of the force Pis 12lb, and the coefficient of friction is 0.25. If the velocity of the block at a certain instant was 4 fps, what would be the velocity at an instant 6 sec later?

2. The block in Figure 39A weighs 50 lb, the magnitude of the force Pis 25lb, and the angle H is 30°. If the block will not overturn, the coefficient of friction is 0.1, and the velocity of the block at a certain instant was 3 fps, how many seconds would elapse from that instant before the velocity increased to 20 fps?

3. As indicated in Figure 40, a block is being pushed up an inclined plane surface by a force P whose line of action is parallel to the inclined surface. The block weighs 60 lb, the angle between a horizontal reference plane and the inclined surface is 15°, the coef- ficient of friction is 0.3, and the magnitude of the force Pis 35.5lb. If the velocity of the block at a certain instant was 5 fps upward, what would be the velocity at an instant 4 sec later?

BLOCK

FIGURE 4D-Conditions in Practice Problem3

Check your answers with those on page 94.

82 Engineering Mechanics, Part 4

Collision of Two Bodies

- A

Conditions before Collision

In this text we shall consider only impact problems in which there is a collision of two bodies that are moving along a common straight path. The bodies may be moving either in the same direction along that path or in opposite directions. If they are moving in the same direction, the magnitude of the velocity of the body in front must be less than the magnitude of the velocity of the other body. In Figure 41A, it is as- sumed that both bodies A and B are moving toward the right and that the velocity of A is greater than the velocity of B. In Figure 41B, it is assumed that the body A is moving toward the right while the body B is moving toward the left. In such a case, the velocities may have any magnitudes; in other words, the magnitudes may be the same or either velocity may be greater than the other. It is possible that one of the bodies involved in a collision will be at rest before the collision occurs.

- - - B A B

A. Bodies moving in same direction B. Bodies moving in opposite directions

FIGURE 41-Velocities of Two Bodies before They Collide

In problems involving the collision of two bodies, we shall use the --7

following notation: (vA)1 denotes the velocity of a body A just before --7

collision occurs, and ( vB)1 denotes the velocity of a body Bat the same instant. This notation is used in Figure 41. Also, if a body is moving toward the right, its velocity will be considered to be a positive quantity; and if a body is moving toward the left, its velocity will be considered negative.

Engineering Mechanics, Part 4

- A

Types of Impact

When two bodies collide, it is said that there is impact. What happens at the instant of impact depends on the nature of the materials in the bodies. One type of impact is called plastic impact. When there is plastic impact, the two bodies move together after the collision occurs. Two common examples of plastic impact are the following: 1) A bullet that is fired from a gun strikes a block of wood and enters the wood. 2) Two railroad cars collide and become connected by a coupling device.

If the two colliding bodies do not move together after impact, the im- pact is called elastic impact. When there is elastic impact, the colliding bodies must have different velocities after the collision occurs. After elastic impact, the two bodies may move either in the same direction or in opposite directions. These two possible conditions after there has been elastic impact are illustrated in Figure 42. In Figure 42A, it is as- sumed that the bodies A and B move in the same direction after im- pact. In Figure 42B, it is assumed that the bodies move in opposite directions. The notation used here and later is as follows: ( v Ah denotes the velocity of a body A just after a collision, and (v8 )z denotes the ve- locity of a body B at the same instant. When the bodies move in the same direction after the collision has occurred, the body that had the greater velocity before the collision will have the smaller velocity af- ter the collision. One of the bodies may be at rest after a collision has occurred.

- - - B A B

B. Bodies moving in opposite directions

FIGURE 42-Ve/ocities of Two Bodies after They Have Collided

Impact Forces

Whenever there is impact, each body that is involved in the collision must exert a force on the other body. In accordance with Newton's third law of motion, one of these forces may be treated as an action while the other force is the equal and opposite reaction. Therefore, the magnitudes of the two forces produced by impact must be equal, and the forces must act in opposite directions. This means that the magnitude of the resultant of the two forces must be zero.

84 Engineering Mechanics, Part 4

When there is elastic impact, the two bodies that collide must remain in contact for a certain very short interval of time while the impact forces act. During this time interval, the magnitude of each impact force first increases from zero to some maximum value and then decreases to zero again. Neither the length of the time interval nor the nature of the variation of the impact forces during that interval can be determined accurately. For this reason, the force-mass-acceleration method cannot be readily applied in the analysis of the motion of colliding bodies. Furthermore, while the bodies are in contact, some of the kinetic en- ergy of each body is converted into heat energy, sound energy, light energy, and energy needed to deform the body. Since the amount of this "lost" energy cannot be determined accurately, the work-energy method cannot be applied in a problem involving impact. However, when the impulse momentum method is used to solve a problem in which two bodies collide, it is not necessary to consider the actions that take place while the bodies are in contact.

In a free-body diagram for a body involved in a collision, we shall use the notation P* to indicate the impact force. For example, the free- body diagrams for the bodies A and Bin Figure 42A are shown in Fig- ures 43A and 43B. Here W denotes the weight of a body; N, the vertical force exerted on the body by the supporting surface; F m the friction force; and P*, the impact force. Similar free-body diagrams can be drawn for the bodies A and B in Figure 42B. In the diagram for body A in this case, the force F m would act toward the right. The actual magnitude of an impact force is not important when the impulse-mo- mentum method is used.

r P* o~

r P* ·D

·r (FmJA NA r (Fmls NB A. FBD for body A B. FBD for body B

FIGURE 43-FBD'S for Bodies in Figure 42A

Engineering Mechanics, Part 4 85

Conservation of linear Momentum

When the impulse-momentum method is applied in a problem in- volving the collision of two bodies, it is convenient to consider the total impulse and the change in the total momentum of both bodies. Before the collision occurs or after impact has been completed, there are no impact forces. Also, while the bodies are in contact during elas- tic impact, the magnitude of the resultant of the impact forces is zero. Therefore, in the case of a collision of two bodies that move along a common path, where the only forces that change the momentum of the bodies are the impulse forces, the linear impulse of the resultant force must be zero. When equation 3, Article 37, is applied, the result is

WA -7 -7 WB -7 -7 0 =- [(vAh- (vAhl -+7- [(vBh- (vBh] g g (1)

When one velocity is subtracted from another, the velocities must be treated as vector quantities. For equation 1,

This reduces to

(2)

You should note that the weights of the bodies can be used instead of their masses in equation 2.

Instead of saying that there is no change in the total linear momen- tum of the two bodies, it may be said that "the linear momentum of the system of bodies is conserved."

Even though there is no change in the total linear momentum of two bodies that collide, there will always be a reduction in the total ki- netic energy of the bodies. Such a reduction is to be expected because some energy must be "lost" whenever there is impact.

In equation 2, there are six variable quantities, namely, the weights of the two bodies, the velocities of the two bodies before the collision, and the velocities of the bodies after the collision. In a particular problem, each of the first four quantities will be known. Hence, it will only be necessary to determine the two velocities after the collision. Since it is not possible to determine two unknown quantities with only one equation, it is necessary to obtain another relation between the two velocities after the collision. This can be done by considering the nature of the impact.

Example Problems

Engineering Mechanics, Part 4

Application of Impulse-Momentum Method for Plastic Impact

When there is plastic impact, the two colliding bodies must move together after the collision. Hence, in such a case, the second relation that is used with equation 2, Article 42, is

The application of the impulse-momentum method for plastic impact is illustrated in the following example problems. When there is plastic impact, the total kinetic energy possessed by the two bodies after the collision is only a relatively small percentage of the energy available before the collision.

Problem 1: Two railroad cars were moving toward each other, as indicated in Figure 41B. Each car weighed 30,000 lb, the magnitude of the velocity of car A was 4 fps, and the magnitude of the velocity of car B was 8 fps. When the cars collided, they became coupled together and both moved with the same velocity. a) If friction between the cars and the rails is neglected, what was the velocity of the cars after the collision? b) How much kinetic energy was lost by the two cars because of the impact?

Solution: a) In equation 2, Article 42, the known quantities are WA = 30,000 Ib, WB = 30,000 lb, (vAh = +4 fps, and (vBh =- 8 fps. An analysis of the specified conditions will indicate that both cars will move toward the left after the collision has occurred, because both cars have the same weight and car B was moving toward the left faster than car A was moving to- ward the right before the collision. However, the algebra in the solution of the problem will be simplified if it is assumed at first that both cars will move toward the right after the collision. For this assumption,

30,000 x [(vAh- (+4)] = 30,000x [-8- (vBh]

or ( V A)z + ( VB)2 = - 8 + 4 = - 4 (A)

Since there is plastic impact, it is known that the magnitude of (vBh is the same as the magnitude of (vAh· Also, since it is assumed that the cars will be moving toward the right after the collision, their velocities will be positive quantities. If we substitute +( v Ah for both ( v A)2 and (vBh, equation (A) becomes

( V Ah + ( V Ah = - 4

Engineering Mechanics, Part 4 87

from which (vAh =- 2 fps. Since this result is negative, the assumed direction of movement of the cars is incorrect. Hence, (vAh = (Vsh = 2 fps toward the left.

b) The total kinetic energy of the two cars before the collision was

1 30,000 2 1 30,000 2 (Eth = 2 X 32.2 X 4 + 2 X· 32.2 X 8 = 37,300 ft-lb

After the collision, the total kinetic energy was

(E) = 2 30,000 22 2 30,000 22 = 3730 ft-lb 1 2 2 X 32.2 X + 2 X 32.2 X

The loss of kinetic energy was 37,300 - 3730 = 33,570 = 33,600 ft-lb.

In this problem, the kinetic energy remaining in the bodies after the collision was only 10% (percent) of the energy they possessed before the collision.

Problem 2: As indicated in Figure 44, a block of wood that weighs 10 lb was at rest on a smooth horizontal surface when it was struck by a bullet that weighs 0.1lb and was traveling horizontally through the air toward the right with a velocity of 1800 fps. It may be assumed that the bullet became embedded in the block and that friction between the block and the supporting surface can be neglected. a) Determine

. the magnitude of the velocity of the block containing the bullet after the bullet stopped moving into the block. b) What was the impulse of the impact force that acted on the bullet while it was entering the block?

FIGURE 44-conditions in Example Problem 2, Article43

BULLET

'-c:>

WOOD BLOCK

Solution: a) In equation 2 in Article 42, we shall let the subscript A refer to the bullet and shall let the subscript B refer to the block. Then, W A = 0.1lb, W8 = 10 lb, (vAh = + 1800 fps, and (vBh = 0. An analysis will indicate that the block will move in the direction in which the bullet was traveling before the collision, or toward the right. Hence,

0.1 X [(vA)z- (+1800)] = 10 X [0- (V8)2]

For plastic impact, which occurred in this problem, ( v8 )z = ( v Ah· If we substitute +(vA)z for each velocity,

88 Engineering Mechanics, Part 4

0.1 (vAh + 10(vAh = 180

and (v A)z = + 17.82 fps

Since this result is positive, the assumed direction of movement is correct and the desired velocity is 17.82 fps toward the right.

b) To determine the impulse of the force that acted on the bullet, it is only necessary to compute the change in the momentum of the bullet. The magnitude of this change is

0.1 lb 32.2 x (1800 - 18) = 5.53 -sec

The value 5.53lb-sec is the magnitude of the impulse of the impact force that acted on the bullet. Since the velocity of the bullet was reduced, the direction of this force was opposite to the direction of the movement of the bullet before it struck the block, or toward the left. An impact force having the same magnitude acted on the block, and its direction was toward the right. You should note that the magnitude of the impact force at a particular instant cannot be determined from the magni- tude of the impulse of the force, because the nature of the variation of the impact force is not known.

Application of Impulse-Momentum Method for Elastic Impact

When there is elastic impact, the second relation that is used with equation 2, Article 42, is the following:

In this relation, e represents a ratio called the coefficient of restitution. As indicated by the equation just given, this coefficient may be defined as the ratio of the following two quantities: 1) the difference between the vectors representing the velocities of the two bodies after the colli- sion and 2) the difference between the vectors representing the velocities of the bodies before the collision. The value of e depends on the nature of the materials in the bodies. For two specific materials, the coefficient of restitution is constant and its value can be determined only by suitable tests. The value of e will always be less than 1.

The application of the impulse-momentum method for elastic impact is illustrated in the following example problems. The percentage of the original kinetic energy of the bodies that will be retained after the collision will be greater for a high value of e than for a low value of e.

Engineering Mechanics, Part 4

Example Problems

Problem 1: As indicated in Figure 41B, two blocks A and B were moving toward each other along a horizontal surface. When they collided, there was elastic impact. The block A weighs 4 lb and its velocity before the collision was 5 fps toward the right. The block B weighs 6lb and its original velocity was 8 fps toward the left. If the coef- ficient of restitution for the blocks is 0.5 and friction between the blocks and the supporting surface can be neglected, what were the magnitude and the direction of the velocity of each block after the collision?

Solution: In equation 2, Article 42, WA = 4lb, W8 = 6lb, (vAh = +5 fps, and (v8)1 =- 8 fps. To simplify the algebra, it will be assumed that both blocks will move toward the right after the collision has occurred. The equation then becomes

4 X [(VA)z- (+ 5)] = 6 X[- 8- (v8)2]

or 4( v A)z + 6( v8)z = - 48 + 20 = - 28 (A)

By the definition of the coefficient of restitution,

or (vB)z- (vA)z = 6.5

and (vBh = (vA)z + 6.5 (B)

When this value of (v8)z is substituted in equation (A), the result is

Therefore, ---7

(vAh =- 6.7 fps

Since this result is negative, (vA)z is 6.7 fps toward the left.

By equation (B),

(vB)z =- 6.7 + 6.5 =- 0.2 fps

---7 Since this result is negative, (vB)z is 0.2 fps toward the left.

Problem 2: A tennis ball was held stationary at a point 5 ft above the surface of a tennis court and was then dropped. When the ball struck that surface, the surface did not move and the coefficient of restitution was 0.8. To what height above the surface did the ball bounce?

90 Engineering Mechanics, Part 4

Solution: The ball is shown in the important positions in Figure 45. At the beginning of its motion, the ball was 5 ft above the surface of the court and its velocity v1 was zero, as indicated in Figure 45A. Just before the ball struck the surface of the court, it was moving downward and the magnitude of its velocity was v2, as indicated in Figure 45B. Just after the ball struck the surface of the court, it was moving upward and the magnitude of its velocity was v3, as indicated in Figure 45C. After the ball bounced, it rose to a maximum height H2 and its velocity at that height was zero.

BALL-o 1 v, 'COURT".

A. At beginning of motion B. Just before striking court

rOBAlL I"' v4 = 0

~·.COURT BALLD L

\COURT

D. Just after striking court C. At top of bounce

FIGURE 45-Positions of Ball in Example Problem 2, Article 44

In order to solve the problem, it is necessary to proceed as follows: First, the magnitude of the velocity v2 should be determined either by apply- ing the method described in Engineering Mechanics, Part 3, for the motion of a body falling under the action of its own weight alone, or by applying the work-energy method. Then the magnitude of the velocity v3 should be determined by applying the definition of the coefficient of restitu- tion. Finally, the required height after the bounce should be determined by applying either the method for the motion of a body moving upward against the action of its weight or the work-energy method.

Engineering Mechanics, Part 4 91

In this problem, the actual weight of the ball does not affect the result, as will be shown. So the weight of the ball will be denoted by W. The work-energy method will be used for determining v2 and Hz. In the calculations for computing Vz, the amount of work done by the force representing the weight of the ball while the ball fellS ft was SW, the kinetic energy possessed by the ball before it was dropped was 0, and the kinetic energy possessed by the ball when it struck the surface of

1 w the court was 2 x g x Vz z. Hence,

and Vz = 17.9 fps

You should note that the factor W appeared in each member of the equation and was eliminated by cancellation.

In the equation in Article 44, the ball will be called body A and the surface of the court will be called body B. The velocity of the ball before the collision was v21 and its velocity after the collision was v3• The veloc- ity of the surface of the court was zero both before and after the colli- sion. Therefore, if upward velocities are considered positive quantities,

and

O-v3 0'8 = - 17.9 - 0

v3 = 14.3 fps

The amount of work done against the force representing the weight of the ball while the ball moved upward to the height Hz was WHz, the kinetic energy possessed by the ball when it started its upward

movement was~ x; x v3 z = Yz x 3~2 x 14.3z, and the kinetic energy possessed by the ball when it reached the height Hz was zero. Hence,

1 w z WHz= 2 x 32.2 x14.3

and Hz= 3.18 ft

Again, the factor W was eliminated by cancellation.

92 Engineering Mechanics, Part 4

Practice Problems 10 1. In Figure 41A, body A represents a railroad car that weighs 20,000 lb and was moving

. toward the right with a velocity whose magnitude was 6 fps; body B represents another car that weighs 40,000 lb and was moving toward the right with a velocity whose mag- nitude was 2 fps. When the cars collided, they became coupled together and then both cars moved with the same velocity. If friction between the cars and the rails is neglected, what were the magnitude and the direction of the velocity of the cars after the collision?

2. A block of wood weighing 20 lb was sliding toward the right on a smooth horizontal surface with a velocity of 10 fps when it was struck by a bullet weighing 0.16lb. The bullet was traveling horizontally through the air toward the left with a velocity whose magnitude was 2000 fps. If the bullet became embedded in the block and friction be- tween the block and the supporting surface can be neglected, what were the magnitude and the direction of the velocity of the block after the bullet stopped moving inside the block?

3. In Figure 46, body A represents a block that weighs 80 lb and was sliding along a smooth horizontal surface toward the left with a velocity whose magnitude was 5 fps. Body B represents a block that weighs 60 lb and was sliding on the same surface toward the left with a velocity whose magnitude was 15 fps. The coefficient of restitution when the bodies collided was 0.6, and friction between the bodies and the supporting surface can be neglected. After the collision, what were a) the magnitude and direction of the velocity of block A and b) the magnitude and direction of the velocity of block B?

5 fps 15 fps

A B

FIGURE 46-lnitial Velocities of Bodies in Practice Problem 3, Article44

4. A tennis ball was held 4 ft above the surface of a tennis court and was thrown downward vertically with an initial velocity whose magnitude was 10 fps. The coefficient of resti- tution for the impact was 0.85. Determine a) the magnitude of the velocity of the ball just before it struck the surface of the court, b) the magnitude of the velocity of the ball just after it started moving upward, and c) the greatest height reached by the ball after it bounced.

Check your answers with those on page 94.

Practice Problems Answers

• 1. 6.21 slugs

2. 0.437 slug-ft2

3. 32.6 slug-ft2

4. 61.6 slug-ft2

1. 564 ft-lb

2. 15.53 ft-lb

3. 1000 lb-sec

4. 15.53lb-sec

II 1. 3810 lb

2. 6.44 fps per sec

3. 3010lb

II 1. 2.00 rad per sec per sec

2. a) 64.3lb toward the left b) 1533 lb upward

3. a) 2.58 rad per sec per sec, clockwise b) 3.23 fps per sec upward and

toward the left c) 5.00 fps per sec upward and toward

the right

4. a) 20.1lb toward the right b) 43.9lb upward

B 1. a) 7.16 fps per sec

b) 6.66lb

2. 21.6lb

3. a) 13.39 fps per sec b) 35.4lb

1. 22.5 in-lb

2. 120 in-lb

3. 600 in-lb

4. 42 in-lb

1. 10.56 fps

2. 6.12 fps

3. 8.44 fps

4. 3.58 ft

5. 5.35 fps

6. 2.35 ft

1. 3.98 rad per sec

2. 7.18 fps

1. 13.66 fps

2. 1.71 sec

3. 10.58 fps upward

1. 3.33 fps toward the right

2. 5.95 fps toward the left

3. a) 11.86 fps toward the left b) 5.86 fps toward the left

4. a) 18.91 fps b) 16.1 fps c) 4.03 ft

Practice Problems Answers

Examination

EXA INATI N 28603900

Engineering Mechanics Part 4

Whichever testing option you choose for your answers, you must use this

EXAMINATION NUMBER:

28603900

When you feel confident that you have mastered the material in this study unit, complete the following examination. Then submit only your answers to school headquarters for grading, using one of the examination answer options described in your first shipment. Send your answers for this examination as soon as you complete it Do not wait until another examination is ready.

Questions 1-20: Select the one best answer to each question.

i. A solid sphere whose diameter is 3.5 ft weighs 600 lb. The mass moment of inertia of the sphere with respect to a horizontal line through its center of gravity is

A. 22.8 slug-fr'. B. 28.5 slug-ff.

2 C. 36.0 slug-ft . 2 D. 45.6 slug-ft .

2. As indicated in Examination Figure 1, a steel ball which weighs 800 lb is suspended from a long cable so that it can swing in a vertical plane. When the cable was vertical during a certain swing, the magnitude of the linear velocity of the ball was 18 fps. The kinetic energy of translation of the ball at that instant was

A. 445 ft-lb. B. 4025 ft-lb.

c. 8040 ft-lb. D. 14,400 ft-lb.

POINT OF SUPPORT

Examination Figure 1

Examination

3. A truck with a velocity of 35 mph is traveling along a horizontal stretch of highway. If the total weight of the truck and its contents is 8400 lb, the magnitude of the linear momentum of the truck is

A. 9150 lb-sec. C. 13,390 lb-sec. B. 11 ,250 lb-sec. D. 18,300 lb-sec.

4. A spring has a normal length equal to 15 in. and the constant for it is 4 lb per in. One end of the spring is attached to a stationary support, and the other end of the spring is attached to a movable block which can only travel along a straight horizontal path. After the block has been moved so that the length of the spring was reduced to 10 in., the force that was applied to the block to cause this movement was taken away and the spring was permitted to return to its original length. The amount of work done by the spring in bringing the block back to its original starting position was

A. 50 in-lb. C. 175 in-lb. B. 125 in-lb. D. 250 in-lb.

5. An automobile weighing 3220 lb and having a uniform linear velocity of 55 mph was traveling along a curved portion of a highway. If the radius of the curve was 600 ft, the magnitude of the centrifugal force was

A. 5031b. B. 650 lb.

c. 8721b. D. 10851b.

6. When a body is rotating under the action of a system of forces and the resultant moment of all forces does not change, the body is subjected to alan

A. axis of rotation. C. mass moment of inertia. B. centripetal force. D. uniform angular acceleration.

Examination 97

7. The diameter of the right circular cylinder represented in front elevation in Examination Figure 2 is 9 in., the length of the cylinder is 8 ft, and the weight of the cylinder is 150 lb. The specified reference line for the mass moment of inertia of the cylinder is a vertical line perpendicular to the axis of the cylinder and passing through the centroid of a cross section of the cylinder. The section is located 1 ft from the right-hand end of the cylinder. The required mass moment of inertia is

A. 25.0 slug-f~. C. 66.9 slug-ft2. B. 41.9 slug-f~. D. 83.8 slug-ft2.

Examination Figure 2

8. A solid wheel is being rolled toward the right along a horizontal surface by a force whose line of action is horizontal. The diameter of the wheel is 2.5 ft, the wheel weighs 80 lb, and the magnitude of the force is 15 lb. If the friction force is sufficient to prevent slipping of the wheel on the supporting surface, the magnitude of the linear acceleration of the center of gravity would be

A. 3.22 fps per sec. C. 5.15 fps per sec. B. 4.03 fps per sec. D. 6.44 fps per sec.

9. The characteristics of a force are indicated in Examination Figure 3. If this force acts on a certain body for 4 sec, the magnitude of the vertical component of the linear impulse of the force during that interval of time would be

A. 138 lb-sec. C. 240 lb-sec. B. 190 lb-sec. D. 276 lb-sec.

Examination Figure 3

10. The radius of a cylindrical flywheel is 2 ft and the weight of the flywheel is 350 lb. A force whose constant magnitude is 40 lb is applied to the rim of the wheel along a line that is tangent to the rim. The magnitude of the angular acceleration of the wheel would be

A. 1 .84 rad per sec per sec. C. 2.90 rad per sec per sec. B. 2.25 rad per sec per sec. D. 3.69 rad per sec per sec.

98 Examination

i 1. As indicated in Examination Figure 4, one end of a spring is attached to a stationary fixture and the other end of the spring is attached to a block that can only move on a horizontal supporting surface

along a certain straight line. The natural length of the spring is 8 in., and the constant for it is 10 lb per in. When the spring is vertical, as indicated in the illustration, its actual length is g1;2 in. If the block is

moved toward the left for a horizontal distance equal to 15 in., the amount of work required to deform

the spring during this movement of the block is

A. 83 in-lb. B. 232 in-lb.

C. 465 in-lb. D. 830 in-lb.

STATIONARY FIXTURE

Examination Figure 4

12. As indicated in Examination Figure 5, a solid cylindrical disk is caused to rotate about a stationary axis, which passes through the center of gravity of the disk. The force that produces rotation is applied to the

rim of the disk by means of a cable, which is wound around the disk and is attached to a body. The

radius of the disk is 2.5 ft, the weight of the disk is 800 lb, and the weight of the body is 300 lb. When the body is supported in the position shown, the body and the disk are at rest. If the support tor the body

is taken away, the disk will rotate without friction while the body moves downward under the action of

its weight. The angular velocity of the disk after the body has moved 5 tt from the position shown would be

A. 3. 70 rad per sec. C. 5.70 rad per sec. B. 4.70 rad per sec. D. 6.70 rad per sec.

ffAXIS

CABLE

Examination Figure 5

Examination

13. An elevator car in a building is expected to move upward with an acceleration whose magnitude is 6 fps per sec when the magnitude of the force that tends to lift the car is 4000 lb. If the effect of friction is neglected, the greatest total weight of the car and its contents can be

A. 3000 lb. B. 3160 lb.

c. 3290 lb. D. 3370 lb.

14. The force that acts toward the center of the circle along a radial line is known as

A. centrifugal force. C. liner acceleration. B. centripetal force. D. linear velocity.

15. The usual equation of motion for translation along a straight path is

A. A=mz. C. Fx= ma. B. F= ma. D. ilt =F.

16. Blocks 1 and 2 in Examination Figure 6 are connected by a cable, which passes over a sheave. Block 1 slides along a horizontal surface under the action of a horizontal force applied by the cable when block 2 is permitted to fall vertically through the air. Block 1 weighs 120 lb, block 2 weighs 50 lb, the coefficient of friction for the materials of block 1 and the supporting surface is 0.3, and friction in the sheave can be neglected. At the instant at which the blocks were in the positions shown, the magnitude of the velocity of each block was 5 fps. After block 2 has moved 3 ft downward from the position shown, the magnitude of the velocity of each block would be

A. 6.40 fps. C. 7.04 fps. B. 6.75 fps. D. 7.30 fps.

Examination Figure 6

17. The normal length of a spring is 6 in., and the spring constant is 20 lb per inch. How much work would have to be done on the spring to reduce its length from 6 in. to 4.5 in.?

A. 12.2 in.-lb C. 32.5 in.-lb B. 22.5 in.-lb D. 80 in.-lb

100 Examination

18. The diameter of a solid wheel is 3.5 ft, and its weight is 60 lb. The wheel is rolled toward the left along

a horizontal surface under the action of a horizontal force whose magnitude is 20 lb. If the friction

force is sufficient to prevent slipping, what is the magnitude of the linear acceleration of the center of

gravity of the wheel?

A.. 3.21 fps per sec B. 5.32 tps per sec

C. 7.16 fps per sec D. 8.15 fps per sec

19. A body that is under the action of a system of forces moves along a straight path. One force of the sys-

tem has a constant magnitude equal to 200 lb, and the angle between the line of action of this force

and the path of the moving body is 20°. If the body moved 3ft along its path during 5 sec, how much

work was done on the body by the specified force during that time interval?

A.. 269 ft-lb B. 329 ft-lb

c. 435 ft-lb D. 564 tt-lb

20. An automobile weighing 3600 lb and having a uniform linear velocity of 50 mph was traveling along a

curved portion of highway. If the radius of the curve was 720ft, the magnitude of the centrifugal force was

A. 5031b. B. 650 lb.

c. 8351b. D. 8721b.

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