PHYS 231 - UNIVERSITY PHYSICS I Equilibrium and Elasticity Question Bank Set 1

PHYS 231 - UNIVERSITY PHYSICS I

- Equilibrium and Elasticity

Question Bank - Set 1

Liberty University

Question 1

Question

A uniform rod of length Land mass Mis suspended horizontally by two vertical

wires attached to its ends. A block of mass mhangs from the rod at a distance

xfrom one end. If the tension in one wire is twice the tension in the other,

determine the distance xin terms of L.

Solution

Step 1: Draw a free body diagram of the system. Label the forces acting on the

rod and block.

Step 2: Write down the equations for translational equilibrium of the system.

Equate the sum of forces in the vertical direction and the sum of forces in the

horizontal direction to zero.

For vertical equilibrium: Forces upward: Tension in wire 1, T1Forces down-

ward: Tension in wire 2, T2, weight of block, mg Setting the forces equal:

T1= 2T2+mg (1)

For horizontal equilibrium: The rod is balanced, so the net force acting on

it must be zero.

T1+T2= 0 (2)

Step 3: Express the tension in terms of forces acting on the system. The

tension in a wire is equal to the force it applies in the direction opposite to

gravity.

T1=Mg +mg and T2=Mg

Step 4: Substitute the expressions for T1and T2back into equations (1) and

(2) and solve for x.

From equation (1):

Mg +mg = 2Mg + 2mg

mg =Mg

x=mL

Therefore, the distance xin terms of Lis mL

2M.

Question 2

Question

A vertical spring with a spring constant of 400 N/m is hung from the ceiling.

A block of mass 2 kg is attached to the end of the spring, causing the spring to

stretch by 0.1 m. If the block is then pulled down an additional distance of d,

what is the value of dsuch that the spring is again in equilibrium?

Solution

Step 1: Calculate the force exerted by gravity on the block to find the equilib-

rium position. The force exerted by gravity is given by Fg=mg, where m= 2

kg and g= 9.8 m/s2. Thus, Fg= (2 kg)(9.8 m/s2) = 19.6 N.

Step 2: Calculate the force exerted by the stretched spring to find the equi-

librium position. The force exerted by the spring is given by Hooke’s Law:

Fs=kx, where k= 400 N/m is the spring constant and x= 0.1 m is the

stretch of the spring. Thus, Fs= (400 N/m)(0.1 m) = 40 N.

Step 3: Set up the equilibrium condition. At equilibrium, the force exerted

by the spring should balance the force exerted by gravity. Therefore, Fs=Fg.

Step 4: Solve for the equilibrium position. From Step 3, we have 40 =

19.6. This implies that the spring is stretched by 0.1 m when the block is in

equilibrium.

Step 5: Determine the total distance the block was pulled down. The desired

position is another equilibrium position below the initial equilibrium position.

Let dbe the distance below the initial equilibrium position. The total distance

the block was pulled down is 0.1 + d

Step 6: Set up the new equilibrium condition. At the new equilibrium

position, the force exerted by the spring should balance the force exerted by

gravity. Therefore, k(d+ 0.1) = mg.

Step 7: Solve for d. Substitute the known values into the equation from Step

6: 400(d+ 0.1) = 19.6. Solve for d: 400d+ 40 = 19.6 400d=−20.4d=−20.4

400

d=−0.051 m.

Therefore, the block is pulled down an additional distance of 0.051 m to

reach a new equilibrium position.

Question 3

Question

A uniform beam of length Land mass Mis supported at its ends by two scales.

A person of mass mstands at a distance xfrom one end of the beam, as shown

in the diagram below. The beam is in equilibrium. Determine the reading on

each scale in terms of M,m,L, and x.

A B

xL−x

Solution

Step 1: Draw the free-body diagram for the beam. We have forces acting on

the beam as follows: - The weight of the beam itself, Mg, acting downward at

the center of mass (L

2from each end). - The reaction forces NAand NBfrom

the scales at each end. - The weight of the person, mg, acting downward at

distance xfrom A.

Step 2: Write out the force equilibrium equations in the vertical (y−) direc-

tion:

NA+NB=Mg +mg

Step 3: Write out the torque equilibrium equation about point A:

XτA= 0

NB·L=mg ·x

Step 4: We now have two equations and two unknowns (NAand NB). We

can solve these equations simultaneously to find the readings on each scale:

From equation (1): NA=Mg +mg −NB

Substitute NAinto equation (2):

(Mg +mg −NB)·L=mg ·x

Solve for NB:

NB=Mg +mg −mgx

Step 5: Now, substitute NBback into equation (1) to find NA:

NA=Mg +mg −(Mg +mg −mgx

NA=mgx

Therefore, the readings on each scale are:

NA=mgx

Land NB=Mg +mg −mgx

Question 4

Question

A uniform beam of length Land mass Mis pivoted at one end. A block of

mass mis hanging from the other end of the beam, a distance xfrom the pivot

point. If the beam is in equilibrium, find the tension in the pivot point and the

reaction force at the pivot point.

Solution

Let’s consider the forces acting on the beam. There are three forces acting on

the beam: the tension force Tat the pivot point, the weight of the beam Mg

(acting at the center of mass of the beam), and the weight of the block mg.

Step 1: Set up coordinate system

Let’s set up our coordinate system with the origin at the pivot point. We will

define the positive direction as counterclockwise.

Step 2: Write the torque equation

The torque equation about the pivot point is given by:

Xτ= 0

where τ=r×Fis the torque produced by a force Facting at a distance rfrom

the pivot point.

The torque contributions are: 1. Tension force Tproduces no torque since its

line of action goes through the pivot point. 2. Weight of the beam Mg produces

a torque in the clockwise direction about the pivot point. The distance of the

center of mass of the beam from the pivot point is L/2. 3. Weight of the block

mg produces a torque in the counterclockwise direction about the pivot point.

The distance of the block from the pivot point is x.

So, the torque equation becomes:

−MgL

2+mgx = 0

Step 3: Solve for tension T

Solving the torque equation, we get:

T=MgL

Step 4: Write the force equation

The force equation in the vertical direction is given by:

XFy= 0

The forces in the vertical direction are: 1. Tension force Tacts upward.

2. Weight of the beam M g acts downward. 3. Weight of the block mg acts

downward.

So, the force equation becomes:

T−Mg −mg = 0

Step 5: Solve for reaction force at the pivot point

Substitute the expression for tension Tinto the force equation:

MgL

2x=Mg +mg

Solving for Rwe get:

R=Mg +mg −MgL

Therefore, the tension at the pivot point is MgL

2xand the reaction force at

the pivot point is M g +mg −MgL

2x.

Question 5

Question

A uniform rod of length Land mass Mis suspended horizontally from two

vertical walls by two identical ropes attached to its ends. If each rope makes an

angle θwith the vertical wall, find the tension in each rope.

Solution

Let’s denote the tension in each rope as T. We will analyze the forces acting on

the rod in the horizontal and vertical directions.

Step 1: Free-body diagram

Consider the forces acting on the rod. In the horizontal direction, the only

force is the tension Tin each rope pulling towards the center. In the vertical

direction, we have the weight of the rod acting downwards and the vertical

components of the tensions in the ropes.

Step 2: Equations of equilibrium

In the vertical direction, the sum of the vertical forces must be zero:

Tcos θ+Tcos θ=Mg

2Tcos θ=Mg

In the horizontal direction, the sum of the horizontal forces must be zero

since the rod is in equilibrium:

Tsin θ= 0

T= 0

Step 3: Solve for tension

From the equation 2Tcos θ=Mg, we can solve for the tension in each rope:

T=Mg

2 cos θ

Therefore, the tension in each rope is Mg

2 cos θ.

Question 6

Question

A uniform beam of length Land mass Mis supported by a pivot at one end.

A rope is attached to the other end of the beam and pulled horizontally with a

force F. Assuming the beam is in equilibrium, determine the tension in the rope

and the reaction force at the pivot in terms of L,M,F, and the acceleration

due to gravity g.

Solution

Step 1: Draw a free-body diagram of the beam to identify the forces acting on

it. Force Direction

Tension in the rope, TUp and to the right

Gravitational force, Mg Downward

Reaction force at pivot, RUpward

Applied force, FTo the right

Step 2: Write the equilibrium equations for the beam in both the horizontal

and vertical directions.

(Sum of forces in the vertical direction: R−Mg = 0

Sum of forces in the horizontal direction: T−F= 0

Step 3: Solve the equilibrium equations to find the tension in the rope and

the reaction force at the pivot. From the vertical equilibrium equation: R=M g

From the horizontal equilibrium equation: T=F

Therefore, the tension in the rope is Fand the reaction force at

the pivot is M g.

Question 7

Question

A 2-meter long steel rod with a radius of 2 cm is placed horizontally on two

supports. A 50 kg mass is hung from the rod at a distance of 1 meter from one

end. If the Young’s modulus of steel is 2.0×1011 N/m2, determine the stress

and strain in the rod at the point where the mass is hung.

Solution

Step 1: Calculate the force applied by the mass. The force applied by the mass

is given by F=mg, where mis the mass and gis the acceleration due to gravity.

Substituting m= 50 kg and g= 9.81 m/s2:

F= 50 kg ×9.81 m/s2= 490.5 N

Step 2: Calculate the moment created by the mass. The moment created

by the mass is given by M=F d, where dis the distance of the mass from one

end of the rod. Substituting F= 490.5 N and d= 1 m:

M= 490.5 N ×1 m = 490.5 Nm

Step 3: Determine the stress in the rod. The stress in the rod can be

calculated using the formula σ=M y

I, where σis the stress, Mis the moment,

yis the perpendicular distance from the neutral axis, and Iis the moment of

inertia of the rod. For a solid cylinder, I=πr4

I=π(0.02 m)4

4= 2.01 ×10−8m4

Given that the rod is 2 m long and the mass is placed 1 m from one end,

y= 1 m. Substituting M= 490.5 Nm and y= 1 m:

σ=490.5 Nm ×1 m

2.01 ×10−8m4≈2.44 ×107N/m2

Step 4: Determine the strain in the rod. The strain in the rod is given by

ε=σ

Y, where εis the strain and Yis the Young’s modulus of the material.

Substituting σ= 2.44 ×107N/m2and Y= 2.0×1011 N/m2:

ε=2.44 ×107N/m2

2.0×1011 N/m2= 0.000122

Therefore, the stress in the rod where the mass is hung is approximately

2.44 ×107N/m2and the strain is 0.000122.

Question 8

Question

A uniform beam of length Land mass Mis supported by a pivot at one end and

a cord attached 3L/4 from the pivot at the other end, as shown in the figure

below. If the beam makes an angle θwith the horizontal and the tension in the

cord is T, determine the tension in the pivot and the angle θin terms of M,L,

and g.

L/4

Solution

Step 1: Set up the free body diagram and write down the torque equilibrium

equation. The forces acting on the beam are the tension Tat the end and

the weight M g, which acts at the center of the beam. The torque equilibrium

equation about the pivot point is:

Xτ= 0

−L

4Mg sin θ+3L

4Tcos θ= 0

Step 2: Solve for the tension Tin terms of M,L,g, and θ.

4Mg sin θ=3L

4Tcos θ

T=Mg sin θ

3 cos θ

Step 3: Sum the forces in the vertical direction to solve for θ.

Tsin θ=Mg

Mg sin2θ

3 cos θ=Mg

sin2θ= 3 cos θ

tan2θ= 3

tan θ=√3

θ= tan−1(√3)

Question 9

Question

A uniform beam of length Land mass Mis supported by two strings, as shown

in the diagram below. The angles θ1and θ2are measured from the vertical. If

the tension in the string at angle θ1is T, what is the tension in the other string

at angle θ2?

θ1

θ2

Solution

Let’s consider the forces acting on the beam in the horizontal and vertical di-

rections to find the tension in the other string at angle θ2.

Step 1: Vertical Forces The forces acting in the vertical direction are the

weight of the beam and the vertical components of tension Tand T2. Summing

up forces in the vertical direction:

T1cos θ1=T2+Mg

Step 2: Horizontal Forces The only horizontal force acting is the horizontal

component of tension T1. Summing up forces in the horizontal direction:

T1sin θ1=T2sin θ2

Step 3: Finding T1and T2From Step 1, we get:

T1=T2+Mg

cos θ1

Substitute this into Step 2:

T2+Mg

cos θ1

sin θ1=T2sin θ2

T2+Mg =T2

sin θ1

cos θ1

tan θ1sin θ2

T2(1 −sin θ1tan θ1sin θ2) = Mg

T2=Mg

1−sin θ1tan θ1sin θ2

Question 10

Question

A uniform beam of length Land mass Mis pivoted at one end. An object of

mass mis hung a distance xfrom the pivot point. If the beam is in equilibrium

at an angle θ, determine the tension in the wire supporting the object.

Solution

1. Draw a free-body diagram of the beam and the hanging mass. Label the forces

acting on the beam and hanging mass. The forces acting on the beam are the

tension in the wire (T) and the gravitational force acting at the center of mass.

The forces acting on the hanging mass are its weight (mg) and the tension in the

wire (T). Note that the angle between the beam and the vertical is θ. 2. Write

out the equilibrium equations for the system. For the beam, the torque (τ) about

the pivot point is given by τ= (L/2)M g sin θ−xT sin θ= 0, since the beam

is in rotational equilibrium. For the hanging mass, the forces in the vertical

direction must balance, giving T=mg. 3. Solving the equation for Tfrom the

hanging mass, we find T=mg. Substitute this result into the torque equation

for the beam to find the value of the tension: (L/2)Mg sin θ−x(mg) sin θ= 0.

4. Simplifying the torque equation gives (L/2)M−xm = 0, which can be

rearranged to find x=L/2. Thus, the tension in the wire supporting the object

is T=mg.

Question 11

Question

A steel cable of length 10 m and cross-sectional area 0.001 m2is suspended

vertically from a ceiling. A 100 kg block is attached to the lower end of the

cable. Calculate the stress in the cable and the elongation of the cable when

supporting the block. Assume Young’s modulus for steel is 2 ×1011 N/m2.

Solution

Step 1: Calculate the weight of the block. The weight of the block can be

calculated using the formula: W=mg, where mis the mass of the block and

gis the acceleration due to gravity. Given that the mass of the block is 100 kg,

and g= 9.81 m/s2, we have:

W= (100 kg)(9.81 m/s2) = 981 N

Step 2: Calculate the stress in the cable. The stress (σ) in the cable can

be calculated using the formula: σ=F

A, where Fis the force applied and Ais

the cross-sectional area of the cable. The force on the cable is the weight of the

block, so σ=W

A. Substitute W= 981 N and A= 0.001 m2:

σ=981 N

0.001 m2= 981 ×103Pa

Step 3: Calculate the elongation of the cable. The elongation (∆L) of the

cable can be calculated using Hooke’s Law: ∆L=F L

AE , where Lis the original

length of the cable and Eis the Young’s modulus. Substitute F=W= 981 N,

L= 10 m, A= 0.001 m2, and E= 2 ×1011 N/m2:

∆L=(981 N)(10 m)

(0.001 m2)(2 ×1011 N/m2)

∆L=9810

2×108= 0.04905 m

Therefore, the stress in the cable is 981,000 Pa and the elongation of the

cable when supporting the block is 0.04905 m.

Question 12

Question

A uniform rod of length Land mass Mis supported by a pivot at one end and

a string attached to the other end, making an angle θwith the horizontal. The

tension in the string is T. Calculate the force exerted by the pivot on the rod.

Solution

Step 1: Draw a free-body diagram of the rod and analyze the forces acting on

it.

The forces acting on the rod are the gravitational force Mg acting at the

center of mass, the tension force Tacting at an angle θwith the horizontal, and

the pivot force Fpacting at the pivot point.

Step 2: Write down the equations for the forces in the horizontal and vertical

directions.

In the vertical direction:

Fnet,y =Tsin(θ)−Mg = 0

In the horizontal direction:

Fnet,x =Tcos(θ)−Fp= 0

Step 3: Solve the equation in the vertical direction for T.

Tsin(θ)−Mg = 0

T=Mg/ sin(θ)

Step 4: Substitute Tback into the equation in the horizontal direction and

solve for Fp.

Tcos(θ)−Fp= 0

Mg cot(θ)−Fp= 0

Fp=Mg cot(θ)

Therefore, the force exerted by the pivot on the rod is Mg cot(θ).

Question 13

Question

A steel wire of length 2.0 m and radius 1.0 mm is stretched between two fixed

supports. If a 100 N weight is hung from the middle of the wire, what is the

elongation of the wire assuming it stretches elastically and the Young’s modulus

for steel is 2.0×1011 N/m2?

Solution

Step 1: Calculate the area of the cross-section of the wire. Given that the radius

of the wire is 1.0 mm, we can calculate the area of the cross-section using the

formula for the area of a circle: A=πr2.

A=π×(1.0×10−3)2=π×1.0×10−6m2

Step 2: Calculate the force of tension in the wire. Since the weight is hung

from the middle of the wire, each half of the wire will support half the weight,

which is 50 N. The force of tension in the wire is equal to the weight supported

by one half of the wire, which is 50 N.

F= 50 N

Step 3: Calculate the elongation of the wire. The force of tension can be

related to the Young’s modulus, cross-sectional area, and the elongation of the

wire using Hooke’s Law: F=Y·A·∆L

L, where ∆Lis the change in length and L

is the original length of the wire.

∆L=F·L

Y·A=50 ×2.0

2.0×1011 ×π×1.0×10−6

∆L=100

π×105=100

3.14 ×105≈3.18 ×10−4m

Therefore, the elongation of the wire is approximately 3.18 ×10−4m.

Question 14

Question

A uniform beam of length Land mass Mis supported by a cable attached 1/3

of the way from the left end. A 2Mmass is suspended from the right end of

the beam. Determine the tension Tin the cable.

Solution

Step 1: We will start by drawing a free body diagram of the beam. Let Mbe

the mass of the beam, Lbe the length of the beam, and Tbe the tension in the

cable. Step 2: The forces acting on the beam are the tension Tin the cable,

the weight of the beam M g acting at the center of mass, and the weight of the

2Mmass at the right end. Step 3: Taking moments about the left end of the

beam, we have:

Xτ= 0

T·L

3−Mg ·L

2+ (2Mg)·L= 0

Step 4: Simplifying the equation, we get:

3−Mg

2+ 2Mg = 0

Step 5: Rearranging the terms, we find:

T=5

6Mg

Step 6: Therefore, the tension Tin the cable is 5

6Mg.

Question 15

Question

A uniform rod of length Land mass mis supported by a pivot at one end, as

shown in the diagram below. A force Fis applied at a distance 2L

3from the

pivot in a direction perpendicular to the rod.

Determine the tension in the rod at the pivot when the rod is in equilibrium.

Solution

Step 1: Identify the forces acting on the rod.

The forces acting on the rod are: - The weight of the rod (mg) acting at the

center of mass of the rod. - The tension (T) at the pivot point. - The applied

force F.

Step 2: Set up equilibrium equations.

In the vertical direction:

T+Fy−mg = 0

In the rotational direction about the pivot point:

F·2L

3−T·L= 0

Step 3: Solve the equations.

From the first equation, we have T=mg −Fy.

Substitute Fy=F·2L

3L=2F

3into the expression for T:

T=mg −2F

Therefore, the tension in the rod at the pivot when the rod is in equilibrium

is mg −2F

Question 16

Question

A block of mass mis suspended by a rope of length Lfrom the ceiling. The

block is in equilibrium and the angle the rope makes with the vertical is θ. Find

the tension in the rope.

Solution

Step 1: Draw a free body diagram of the block. The forces acting on the block

are the tension (T) in the rope and the gravitational force (mg). Resolve the

forces into components.

Step 2: The vertical component of the tension force balances the gravita-

tional force, so Tcos θ=mg.

Step 3: The horizontal component of the tension force is responsible for

providing the centripetal force to keep the block in a circular orbit, so Tsin θ=

mv2

rwhere vis the speed of the block and ris the radius of the circle.

Step 4: Since the block is in equilibrium, the net force acting on it is zero in

both the vertical and horizontal directions.

Step 5: To find the tension in the rope, we first solve for the speed of the

block by realizing that the vertical component of the acceleration is zero. Thus,

r=gtan θ.

Step 6: Substituting the expression for v2/r into Tsin θ=mv2

r, we get

T=mg cot θas the tension in the rope.

Question 17

Question

A uniform beam of length Land mass Mis supported by two strings, as shown

in the figure below. The angle between the left string and the horizontal is θ.

Find the tension in each string.

x1x2

Solution

Step 1: We will begin by setting up the equilibrium conditions for the beam.

The sum of the forces in the vertical direction must be zero:

T1cos θ=Mg

The sum of the torques about any point must also be zero. We will choose the

left end of the beam:

T2Lcos θ=T1x1sin θ

Step 2: We can solve the first equation for T1to get:

T1=Mg

cos θ

Step 3: Substitute T1into the second equation and solve for T2:

T2=Mgx1

Step 4: Both x1and x2are related to each other and to Lthrough the

equation x1+x2=L. Using this, we can express x1as:

x1=L−x2

Step 5: Substitute x1=L−x2back into the equation for T2:

T2=Mg(L−x2)

Step 6: Therefore, the tension in the left string T1is T1=M g

cos θand the

tension in the right string T2is T2=Mg(L−x2)

Question 18

Question

A uniform rod of length Land mass Mis hinged at one end and supported in

a horizontal position by a thin wire attached to the other end. The wire makes

an angle θwith the vertical. Find the tension in the wire and the force at the

hinge in terms of M,L, and θ.

Solution

Step 1: Draw a free body diagram of the rod. Step 2: Resolve the forces acting

on the rod into components. Step 3: Write the force balance equations in the

vertical and horizontal directions to find the unknowns.

Question 19

Question

A steel cable with a diameter of 1.5 cm and a length of 10 m hangs vertically

from a ceiling. If the Young’s modulus for steel is 2.10 ×1011 N/m2, calculate

the elongation of the cable when a 300 kg mass is attached to its end.

Solution

Step 1: Calculate the cross-sectional area of the cable. Given that the diameter

of the cable is 1.5 cm, the radius is r= 0.75 cm = 0.0075 m. The cross-sectional

area, A, of the cable is given by:

A=πr2

A=π(0.0075)2

A= 1.77 ×10−4m2

Step 2: Calculate the weight of the attached mass. The weight, W, is given

by:

W=mg

W= (300 kg)(9.81 m/s2)

W= 2943 N

Step 3: Calculate the tensile stress in the cable. The tensile stress, σ, is

given by:

σ=F

where Fis the force applied and Ais the cross-sectional area. Since the force

applied is equal to the weight of the mass, we have:

σ=W

σ=2943

1.77 ×10−4

σ= 1.66 ×107N/m2

Step 4: Calculate the elongation of the cable. The elongation, ∆L, of the

cable is given by:

∆L=F L

A·Y

where Lis the original length of the cable and Yis the Young’s modulus of

steel. Substitute the given values:

∆L=(2943)(10)

(1.77 ×10−4)(2.10 ×1011)

∆L= 0.00738 m or 7.38 mm

Therefore, the elongation of the cable when the 300 kg mass is attached to

its end is 7.38 mm.

Question 20

Question

A uniform beam of length Land mass mis supported by a hinge at one end

and a rope attached 3/5 of the way along the beam. A weight of mass 2mis

hung from the free end of the beam. The beam makes an angle of 37◦with the

horizontal. Calculate the tension in the rope.

Solution

Step 1: Draw a free body diagram of the beam. Label the forces acting on the

beam.

Step 2: Resolve the forces into horizontal and vertical components. Let T

be the tension in the rope, Wbe the weight of the beam, Fbe the force exerted

by the hinge, and 2Wbe the weight at the free end.

Step 3: Write the equilibrium equations for the beam in the horizontal and

vertical directions. In the vertical direction, the sum of the forces is zero:

Fy=F−W−2W= 0

Step 4: Calculate the weight of the beam W=mg and the weight at the

free end 2W= 2mg.

Step 5: Substitute the known values into the equilibrium equations: For the

vertical direction:

F−mg −2mg = 0

F= 3mg

Step 6: Now consider the torque equation. The torque about the hinge must

be zero for rotational equilibrium. Choose the hinge as the pivot point and sum

the torques:

τ= 0 = −F(L)sin(37◦)+2W(3L/5)sin(37◦) + T(L/5)sin(37◦)

Step 7: Substitute the earlier calculated values of F= 3mg,W=mg, and

2W= 2mg into the torque equation:

0 = −3mgL sin(37◦) + 2(2mg)(3L/5) sin(37◦) + T(L/5) sin(37◦)

Step 8: Solve for the tension T:

3mgL sin(37◦)=4mgL sin(37◦) + T(L/5) sin(37◦)

T(L/5) sin(37◦) = −mgL sin(37◦)

T=−5mg

Therefore, the tension in the rope is −5mg. Note that the negative sign

indicates the direction of the tension force.

Question 21

Question

A heavy object of mass 5 kg is attached to a vertical spring. When the object

is attached, the spring stretches by 0.2 m. If the spring constant is 200 N/m,

determine the magnitude of the force exerted by the object on the spring when

it is in equilibrium.

Solution

Step 1: We first determine the weight of the object using the formula Fweight =

mg, where mis the mass and gis the acceleration due to gravity.

Fweight = 5 kg ×9.8 m/s2= 49 N

Step 2: The magnitude of the force exerted by the object on the spring when

it is in equilibrium is equal to the sum of the weight of the object and the force

exerted by the spring.

Fequilibrium =Fweight +kx

where kis the spring constant and xis the displacement from the equilibrium

position.

Step 3: Since the object is in equilibrium, the net force acting on it is zero.

Fequilibrium = 0

Fweight +kx = 0

Step 4: Substituting the known values:

49 N + 200 N/m ×0.2 m = 0

Step 5: Solving for Fequilibrium:

Fequilibrium =−49 N

Therefore, the magnitude of the force exerted by the object on the spring

when it is in equilibrium is 49 N downward.

Question 22

Question

A wooden block of mass mhangs from the ceiling by three wires as shown in

the figure below. Each wire has a tension T. The block is in equilibrium and

the angle between the wires is θ. Calculate the tension Tin each wire in terms

of mand g.

Solution

Step 1: Draw a free-body diagram for the block m. Resolving forces in the

vertical direction, we have:

Tcos(θ)−Tcos(θ) = 0

Tcos(θ) = mg

T=mg

cos(θ)

Step 2: The tension Tin each wire is given by:

T=mg

cos(θ)

Question 23

Question

A uniform ladder of length Land mass Mrests against a smooth vertical wall.

The ladder makes an angle θwith the horizontal floor. Determine the forces

at point Aand point Bwhere the ladder makes contact with the wall and the

floor respectively.

Solution

Step 1: Draw a free-body diagram of the ladder showing all the forces acting

on it.

Step 2: Resolve the forces into their components. Let NAand NBbe the

normal forces at points Aand Brespectively, fAand fBbe the friction forces

at points Aand Brespectively, and Wbe the weight of the ladder acting at its

center of mass.

Step 3: Write the force equations for equilibrium in the x and y directions:

(Sum of forces in the x-direction: NA−fA= 0

Sum of forces in the y-direction: NB−W= 0

Step 4: Write the torque equations for equilibrium about point A:

τA= 0 =⇒fA·Lsin(θ)−W·L

2cos(θ)=0

Step 5: Solve for the normal force at point A,NA:

NA=fA=W

2 sin(θ) cos(θ)=Mg

2 sin(θ) cos(θ)

Step 6: Solve for the normal force at point B,NB:

NB=W=Mg

Step 7: Therefore, the forces at point Aand point Bare:

At point A: NA=M g

2 sin(θ) cos(θ)and At point B: NB=Mg

Question 24

Question

A uniform plank of length Land mass Mis supported horizontally by two ropes,

with one rope located at each end. The plank is suspended by the ropes from a

ceiling. If a person of mass mstands xmeters from one end of the plank, find

the tension in each rope.

Solution

Step 1: Draw a free-body diagram of the plank and the person. Step 2: Apply

the rotational equilibrium condition about the point where the left rope attaches

to the plank. Step 3: Express the tension in each rope in terms of the given

variables.

Let’s proceed with the detailed solution.

Question 25

Question

A thin uniform rod of mass Mand length Lis suspended horizontally by two

vertical wires attached to its ends. A heavy object of mass mis then hung from

the midpoint of the rod, as shown. Find the tension in each wire.

A BC

Solution

Step 1: We begin by setting up the equations of equilibrium for the system. We

have the following forces acting on the system: - The weight of the rod, acting

at its center of mass at point C, which we’ll call Mg downward. - The tension

in the wire attached to point A, which we’ll call TAupward. - The tension in

the wire attached to point B, which we’ll call TBupward. - The weight of the

object m, acting downward at point D, which is mg.

Since the system is in equilibrium, the sum of the forces in the vertical

direction and the sum of torques about any point must be zero.

Step 2: Summing forces in the vertical direction, we have:

TA+TB=Mg +mg

Step 3: To find the torques about point A, we choose point A as the pivot.

The torque due to the weight of the rod is zero (because it acts at the pivot),

the torque due to TBis TB·L/2, and the torque due to the weight of the object

m is mg ·L/4. Setting the sum of torques about A equal to zero:

TB·L

2−mg ·L

4= 0

Step 4: Solving the system of equations from Step 2 and Step 3, we find:

TA=3M+ 4m

TB=M−m

Therefore, the tension in wire A is 3M+4m

4gand the tension in wire B is

M−m

4g.

Question 26

Question

A uniform steel beam of length 5.0 m and weight 9800 N is to be supported by

a cable at each end. If the cable at one end is attached to the end of the beam

and the other cable is attached 2.0 m from the other end, what should be the

tension in each cable?

Solution

Step 1: Draw a diagram of the system to visualize the forces acting on the beam.

Step 2: Identify all the forces acting on the beam. The forces acting on the

beam are its weight, the tension force at the end of the beam, and the tension

force 2.0 m away from the other end.

Step 3: Write the equilibrium equation for the beam in the vertical direction.

The sum of the forces in the vertical direction must be zero for the beam to be

in equilibrium.

Step 4: Use the equilibrium equation to find the tension in the cable at each

end of the beam.

Step 5: Calculate the tension in each cable, making sure to consider the

weight of the beam as well.

Step 6: Write the final answer with the calculated tensions in each cable.

Step 7: Check the units and ensure they are correct for the tension values.

Therefore, the tension in the cable at the end of the beam should be 5880

N, and the tension in the cable 2.0 m away from the other end should be 2950

Question 27

Question

A uniform ladder of length 4.00 m and weight 400 N rests against a smooth,

vertical wall. The coefficient of static friction between the ladder and the ground

is 0.400. A 70.0 kg painter stands on a rung of the ladder 1.50 m from the

bottom. What minimum angle should the ladder make with the ground so that

it doesn’t slip?

Solution

Step 1: Draw a free-body diagram of the ladder. Step 2: Write the equations of

equilibrium for the ladder in the xand ydirections. Step 3: Solve the equations

to find the minimum angle. Let’s begin by drawing the free-body diagram of

the ladder.

Step 1: The free-body diagram of the ladder includes the following forces:

1. Weight of the ladder acting at its center. 2. Normal force exerted by the

ground on the ladder at the bottom. 3. Normal force exerted by the wall on

the ladder at the top. 4. Friction force exerted by the ground on the ladder at

the bottom. 5. Force due to the painter standing on the ladder.

Step 2: Let’s denote the following: Wl= Weight of the ladder N1= Normal

force at the bottom of the ladder N2= Normal force at the top of the ladder f

= Friction force at the bottom of the ladder Wp= Weight of the painter θ=

Angle the ladder makes with the ground

The equations of equilibrium in the xdirection are:

N2=N1

The equations of equilibrium in the ydirection are:

N1+N2=Wl+Wp

Step 3: Next, we can solve for θusing the following relationships:

N1=Wl·cos(θ)

N2=Wl·sin(θ)

f=µs·N1

Substitute these expressions into the equilibrium equations to solve for the

minimum angle θ.

Question 28

Question

A cylindrical tank with a radius of 2.5 m is filled with water to a height of 4.0

m. The tank has a small hole at the bottom with a radius of 5.0 cm. Determine

the speed at which the water exits the hole.

Solution

Step 1: To find the speed at which the water exits the hole, we can use the

principle of conservation of energy, considering the potential energy at the top

of the tank and the kinetic energy of the water exiting the hole.

Step 2: The potential energy at the top of the tank is given by P E =mgh,

where mis the mass of the water, gis the acceleration due to gravity, and h

is the height of the water column. The mass of the water can be calculated as

m=ρV , where ρis the density of water and Vis the volume of water.

Step 3: The volume of water can be calculated as V=πr2h, where ris the

radius of the tank and his the height of the water column.

Step 4: Substituting the values, we have V=π(2.5 m)2(4.0 m) = 100πcubic

meters.

Step 5: The mass of the water can be calculated as m= 100π×1000 kg/m3=

100000πkg.

Step 6: The potential energy at the top of the tank is P E = 100000πkg ×

9.81 m/s2×4.0 m = 3924000πJ.

Step 7: The kinetic energy of the water exiting the hole can be calculated

as KE =1

2mv2, where vis the speed of the water exiting the hole.

Step 8: Equating the potential energy at the top of the tank to the kinetic

energy of the water exiting the hole, we get 3924000π=1

2×100000π×v2.

Step 9: Solving for v, we find v=q2×3924000π

100000π= 28 m/s.

Step 10: Therefore, the speed at which the water exits the hole is 28 m/s.

Question 29

Question

A uniform, horizontal beam of mass Mand length Lis supported by two vertical

ropes attached to its ends. The ropes make angles θ1and θ2with the horizontal.

Find the tension in each rope.

Solution

Step 1: Draw a free-body diagram of the beam.



The weight of the beam acts downward at its center.



The tension in the ropes acts upward and has components in the horizontal

direction.



There are no forces acting in the horizontal direction.

Step 2: Write out the force equations.



In the vertical direction: PFy=T1cos θ1+T2cos θ2−W= 0



In the horizontal direction: PFx=T1sin θ1−T2sin θ2= 0

Step 3: Express the weight in terms of Mand L. The weight of the beam

is W=Mg, where gis the acceleration due to gravity.

Step 4: Solve the equations.



From the horizontal equation: T1sin θ1=T2sin θ2



From the vertical equation: T1cos θ1+T2cos θ2=Mg

Step 5: Use trigonometric identities to simplify the equations.



Divide the two equations to eliminate T1and T2: tan θ1= tan θ2

Step 6: Solve for the tension in each rope.



Since θ1=θ2, the tensions in the ropes are equal: T1=T2=Mg

2 cos θ1

Therefore, the tension in each rope is Mg

2 cos θ1.

Question 30

Question

A uniform rod of length Land mass Mis attached to a wall by a hinge at

one end and supported by a string of length Lattached to the other end. The

system is in equilibrium when the string makes an angle θwith the vertical.

Calculate the tension in the string.

Solution

Step 1: Draw a free body diagram of the rod. Let Tbe the tension in the string

and Wbe the weight of the rod. The forces acting on the rod are the tension T

in the string, the weight W=M g acting at the center of mass of the rod, and

the normal force Nfrom the hinge.

Step 2: Break the tension force into horizontal and vertical components. The

tension force Tcan be broken down into two components: one acting upwards

along the string and the other acting along the rod.

Step 3: Write down the equations for equilibrium in the x and y directions.

In the x-direction: Tcos θ= 0 In the y-direction: Tsin θ−W= 0

Step 4: Solve for the tension T. From the equation in the y-direction, we

have Tsin θ=M g. Since Mg =Wand the rod is in equilibrium, the tension

in the string is given by T=Mg

sin θ.

Therefore, the tension in the string is T=Mg

sin θ.

Question 31

Question

A uniform beam of length Land mass Mrests horizontally on two vertical

supports, with a box of mass mplaced at a distance 2

3Lfrom one end of the

beam. If the supports exert a force Feach, find the force exerted by the box

on the beam.

Solution

Step 1: Begin by drawing a free-body diagram of the beam. We have four forces

acting on the beam: the weight of the beam Wb, the weight of the box Wbox, and

the normal forces N1and N2from the supports. Let’s assume N1is the force

exerted by the left support and N2is the force exerted by the right support.

Step 2: Write the equations for the forces in the vertical direction. Summing

forces in the vertical direction, we have:

N1+N2−Wb−Wbox = 0

Step 3: Write the torque equation about the point where the left support

exerts a force. The torque equation about the left support is:

N2·L−Wb·L

2−Wbox ·2L

3= 0

Step 4: Express the weights in terms of mass and acceleration due to gravity.

The weights can be expressed as Wb=M g and Wbox =mg.

Step 5: Solve the equations. Substitute the expressions for weights into the

torque equation and vertical force equation, then solve for N2. After solving,

we find:

N2=5

3Mg +4

3mg

Step 6: Determine the force exerted by the box on the beam. Since the box

exerts a force in the opposite direction to the support at the right end, the force

exerted by the box on the beam is 4

3Mg +4

3mg .

Question 32

Question

A uniform beam of mass mand length Lis supported by two ropes attached

to its ends. The beam makes an angle θwith the horizontal. One rope is at

a distance dfrom the center of the beam while the other rope is at a distance

L−dfrom the center of the beam. If the tension in the rope at distance dis T,

determine the tension in the other rope in terms of m,L,g,θ, and d.

Solution

Step 1: Draw a Free-Body Diagram (FBD) for the beam. We will consider the

forces acting on the beam: the weight (mg) acting downward from the center of

mass, the tension (T) acting upward from one end of the beam, and the tension

(T′) acting upward from the other end of the beam.

Step 2: Resolve the forces into components. The weight can be resolved into

two components: one along the beam and one perpendicular to the beam. Let

Txbe the horizontal component of T′and Tybe the vertical component of T′.

Step 3: Set up the equilibrium equations. In the horizontal direction, the

sum of the forces must equal zero:

T=T′cos(θ)

Step 4: In the vertical direction, the sum of the forces must equal zero as

well. The weight (mg) is balanced by the vertical components of the tensions:

mg =T′sin(θ)

Step 5: Now we express T′in terms of m,L,g,θ, and d. From step 3, we

have T=T′cos(θ), so T′=T

cos(θ). Substituting this into the vertical equilibrium

equation from step 4, we get:

mg =T

cos(θ)sin(θ)

Step 6: The only unknown in the equation above is T′, the tension in the

other rope. Simplify the expression and solve for T′:

T′=mg

tan(θ)=mg

L/2

=2mgL

Therefore, the tension in the other rope is 2mgL

Question 33

Question

A uniform beam of length Land mass mrests on a pivot at one end. A block

of mass 2mis placed 2L/3 from the pivot. If the beam is in equilibrium, what

is the mass of the beam?

Solution

Let’s denote the mass of the beam as M. To find the mass of the beam, we can

analyze the torques acting on the beam-block-pivot system.

Step 1: Summing the torques about the pivot point. The torque due to the

block is 2mg(2L/3) and the torque due to the beam is MgL/2. These torques

must balance each other to maintain equilibrium.

Στ= 2mg 2L

3−Mg L

2= 0

Step 2: Solving for the mass of the beam M.

2mg 2L

3=Mg L

2

4mg =ML

M=8m

Thus, the mass of the beam is 8m/L.

Question 34

Question

A steel rod of length 2.0 m is suspended vertically from a ceiling. A 500 N

weight is hung from the bottom of the rod. The rod has a cross-sectional area

of 5.0×10−4m2and Young’s modulus of 2.0×1011 N/m2. What is the elongation

of the rod?

Solution

Step 1: Calculate the stress on the rod. The stress on the rod is given by:

stress = force

area

stress = 500 N

5.0×10−4m2

stress = 1.0×106N/m2

Step 2: Calculate the strain on the rod. The strain on the rod is given by:

strain = stress

Young’s modulus

strain = 1.0×106N/m2

2.0×1011 N/m2

strain = 5.0×10−6

Step 3: Calculate the elongation of the rod. The elongation of the rod is

given by:

elongation = strain ×original length

elongation = 5.0×10−6×2.0 m

elongation = 0.01 mm

Therefore, the elongation of the rod is 0.01 mm.

Question 35

Question

A uniform beam of length Land mass Mis attached to a wall by a hinge at one

end and supported by a cable at angle θfrom the vertical at the other end. The

beam makes an angle αwith the horizontal. The forces acting on the beam are

the gravitational force 

Fg=−Mgˆ

jand the tension force 

Tfrom the cable. If

the beam is in equilibrium, find an expression for the tension force 

Tin terms

of M,L,θ, and α.

Solution

Step 1: Draw a free-body diagram of the beam. We draw the beam with the

gravitational force 

Fgacting downwards at the center of mass and the tension

force 

Tacting at an angle θfrom the vertical. The reaction force from the hinge

at the wall is vertical.

Step 2: Resolve the forces into components. Let Txand Tybe the horizontal

and vertical components of the tension force 

T. Since the beam is in equilibrium,

the sum of the forces in each direction must be zero.

Step 3: Write the equilibrium equations. In the horizontal direction, the

sum of forces must be zero:

Tx= 0

Step 4: Write the equilibrium equations. In the vertical direction, the sum

of forces must be zero:

Ty−Mg = 0

Step 5: Express Txand Tyin terms of known quantities. From the equilib-

rium equations, we know that:

Tx=Tcos θ

Ty=Tsin θ

Step 6: Solve for the tension force T. Substitute the expressions for Txand

Tyinto the equilibrium equations:

Tcos θ= 0

Tsin θ−Mg = 0

Step 7: Solve for T. From the horizontal equilibrium equation, we have

Tcos θ= 0, which implies that T= 0 if cos θ= 0. Since this is not possible, we

ignore this equation. From the vertical equilibrium equation, we have Tsin θ−

Mg = 0, so:

T=Mg

sin θ

From equation (1):

Mg +mg = 2Mg + 2mg

mg =Mg

x=mL

Therefore, the distance xin terms of Lis mL

2M.

Question 2

Question

A vertical spring with a spring constant of 400 N/m is hung from the ceiling.

A block of mass 2 kg is attached to the end of the spring, causing the spring to

stretch by 0.1 m. If the block is then pulled down an additional distance of d,

what is the value of dsuch that the spring is again in equilibrium?

Solution

Step 1: Calculate the force exerted by gravity on the block to find the equilib-

rium position. The force exerted by gravity is given by Fg=mg, where m= 2

kg and g= 9.8 m/s2. Thus, Fg= (2 kg)(9.8 m/s2) = 19.6 N.

Step 2: Calculate the force exerted by the stretched spring to find the equi-

librium position. The force exerted by the spring is given by Hooke’s Law:

Fs=kx, where k= 400 N/m is the spring constant and x= 0.1 m is the

stretch of the spring. Thus, Fs= (400 N/m)(0.1 m) = 40 N.

Step 3: Set up the equilibrium condition. At equilibrium, the force exerted

by the spring should balance the force exerted by gravity. Therefore, Fs=Fg.

Step 4: Solve for the equilibrium position. From Step 3, we have 40 =

19.6. This implies that the spring is stretched by 0.1 m when the block is in

equilibrium.

Step 5: Determine the total distance the block was pulled down. The desired

position is another equilibrium position below the initial equilibrium position.

Let dbe the distance below the initial equilibrium position. The total distance

the block was pulled down is 0.1 + d

Step 6: Set up the new equilibrium condition. At the new equilibrium

position, the force exerted by the spring should balance the force exerted by

gravity. Therefore, k(d+ 0.1) = mg.

Step 7: Solve for d. Substitute the known values into the equation from Step

6: 400(d+ 0.1) = 19.6. Solve for d: 400d+ 40 = 19.6 400d=−20.4d=−20.4

400

d=−0.051 m.

Therefore, the block is pulled down an additional distance of 0.051 m to

reach a new equilibrium position.

Question 3

Question

A uniform beam of length Land mass Mis supported at its ends by two scales.

A person of mass mstands at a distance xfrom one end of the beam, as shown

in the diagram below. The beam is in equilibrium. Determine the reading on

each scale in terms of M,m,L, and x.

A B

xL−x

Solution

Step 1: Draw the free-body diagram for the beam. We have forces acting on

the beam as follows: - The weight of the beam itself, Mg, acting downward at

the center of mass (L

2from each end). - The reaction forces NAand NBfrom

the scales at each end. - The weight of the person, mg, acting downward at

distance xfrom A.

Step 2: Write out the force equilibrium equations in the vertical (y−) direc-

tion:

NA+NB=Mg +mg

Step 3: Write out the torque equilibrium equation about point A:

XτA= 0

NB·L=mg ·x

Step 4: We now have two equations and two unknowns (NAand NB). We

can solve these equations simultaneously to find the readings on each scale:

From equation (1): NA=Mg +mg −NB

Substitute NAinto equation (2):

(Mg +mg −NB)·L=mg ·x

Solve for NB:

NB=Mg +mg −mgx

Step 5: Now, substitute NBback into equation (1) to find NA:

NA=Mg +mg −(Mg +mg −mgx

NA=mgx

Therefore, the readings on each scale are:

NA=mgx

Land NB=Mg +mg −mgx

Question 4

Question

A uniform beam of length Land mass Mis pivoted at one end. A block of

mass mis hanging from the other end of the beam, a distance xfrom the pivot

point. If the beam is in equilibrium, find the tension in the pivot point and the

reaction force at the pivot point.

Solution

Let’s consider the forces acting on the beam. There are three forces acting on

the beam: the tension force Tat the pivot point, the weight of the beam Mg

(acting at the center of mass of the beam), and the weight of the block mg.

Step 1: Set up coordinate system

Let’s set up our coordinate system with the origin at the pivot point. We will

define the positive direction as counterclockwise.

Step 2: Write the torque equation

The torque equation about the pivot point is given by:

Xτ= 0

where τ=r×Fis the torque produced by a force Facting at a distance rfrom

the pivot point.

The torque contributions are: 1. Tension force Tproduces no torque since its

line of action goes through the pivot point. 2. Weight of the beam Mg produces

a torque in the clockwise direction about the pivot point. The distance of the

center of mass of the beam from the pivot point is L/2. 3. Weight of the block

mg produces a torque in the counterclockwise direction about the pivot point.

The distance of the block from the pivot point is x.

So, the torque equation becomes:

−MgL

2+mgx = 0

Step 3: Solve for tension T

Solving the torque equation, we get:

T=MgL

Step 4: Write the force equation

The force equation in the vertical direction is given by:

XFy= 0

The forces in the vertical direction are: 1. Tension force Tacts upward.

2. Weight of the beam M g acts downward. 3. Weight of the block mg acts

downward.

So, the force equation becomes:

T−Mg −mg = 0

Step 5: Solve for reaction force at the pivot point

Substitute the expression for tension Tinto the force equation:

MgL

2x=Mg +mg

Solving for Rwe get:

R=Mg +mg −MgL

Therefore, the tension at the pivot point is MgL

2xand the reaction force at

the pivot point is M g +mg −MgL

2x.

Question 5

Question

A uniform rod of length Land mass Mis suspended horizontally from two

vertical walls by two identical ropes attached to its ends. If each rope makes an

angle θwith the vertical wall, find the tension in each rope.

Solution

Let’s denote the tension in each rope as T. We will analyze the forces acting on

the rod in the horizontal and vertical directions.

Step 1: Free-body diagram

Consider the forces acting on the rod. In the horizontal direction, the only

force is the tension Tin each rope pulling towards the center. In the vertical

direction, we have the weight of the rod acting downwards and the vertical

components of the tensions in the ropes.

Step 2: Equations of equilibrium

In the vertical direction, the sum of the vertical forces must be zero:

Tcos θ+Tcos θ=Mg

2Tcos θ=Mg

In the horizontal direction, the sum of the horizontal forces must be zero

since the rod is in equilibrium:

Tsin θ= 0

T= 0

Step 3: Solve for tension

From the equation 2Tcos θ=Mg, we can solve for the tension in each rope:

T=Mg

2 cos θ

Therefore, the tension in each rope is Mg

2 cos θ.

Question 6

Question

A uniform beam of length Land mass Mis supported by a pivot at one end.

A rope is attached to the other end of the beam and pulled horizontally with a

force F. Assuming the beam is in equilibrium, determine the tension in the rope

and the reaction force at the pivot in terms of L,M,F, and the acceleration

due to gravity g.

Solution

Step 1: Draw a free-body diagram of the beam to identify the forces acting on

it. Force Direction

Tension in the rope, TUp and to the right

Gravitational force, Mg Downward

Reaction force at pivot, RUpward

Applied force, FTo the right

Step 2: Write the equilibrium equations for the beam in both the horizontal

and vertical directions.

(Sum of forces in the vertical direction: R−Mg = 0

Sum of forces in the horizontal direction: T−F= 0

Step 3: Solve the equilibrium equations to find the tension in the rope and

the reaction force at the pivot. From the vertical equilibrium equation: R=M g

From the horizontal equilibrium equation: T=F

Therefore, the tension in the rope is Fand the reaction force at

the pivot is M g.

Question 7

Question

A 2-meter long steel rod with a radius of 2 cm is placed horizontally on two

supports. A 50 kg mass is hung from the rod at a distance of 1 meter from one

end. If the Young’s modulus of steel is 2.0×1011 N/m2, determine the stress

and strain in the rod at the point where the mass is hung.

Solution

Step 1: Calculate the force applied by the mass. The force applied by the mass

is given by F=mg, where mis the mass and gis the acceleration due to gravity.

Substituting m= 50 kg and g= 9.81 m/s2:

F= 50 kg ×9.81 m/s2= 490.5 N

Step 2: Calculate the moment created by the mass. The moment created

by the mass is given by M=F d, where dis the distance of the mass from one

end of the rod. Substituting F= 490.5 N and d= 1 m:

M= 490.5 N ×1 m = 490.5 Nm

Step 3: Determine the stress in the rod. The stress in the rod can be

calculated using the formula σ=M y

I, where σis the stress, Mis the moment,

yis the perpendicular distance from the neutral axis, and Iis the moment of

inertia of the rod. For a solid cylinder, I=πr4

I=π(0.02 m)4

4= 2.01 ×10−8m4

Given that the rod is 2 m long and the mass is placed 1 m from one end,

y= 1 m. Substituting M= 490.5 Nm and y= 1 m:

σ=490.5 Nm ×1 m

2.01 ×10−8m4≈2.44 ×107N/m2

Step 4: Determine the strain in the rod. The strain in the rod is given by

ε=σ

Y, where εis the strain and Yis the Young’s modulus of the material.

Substituting σ= 2.44 ×107N/m2and Y= 2.0×1011 N/m2:

ε=2.44 ×107N/m2

2.0×1011 N/m2= 0.000122

Therefore, the stress in the rod where the mass is hung is approximately

2.44 ×107N/m2and the strain is 0.000122.

Question 8

Question

A uniform beam of length Land mass Mis supported by a pivot at one end and

a cord attached 3L/4 from the pivot at the other end, as shown in the figure

below. If the beam makes an angle θwith the horizontal and the tension in the

cord is T, determine the tension in the pivot and the angle θin terms of M,L,

and g.

L/4

Solution

Step 1: Set up the free body diagram and write down the torque equilibrium

equation. The forces acting on the beam are the tension Tat the end and

the weight M g, which acts at the center of the beam. The torque equilibrium

equation about the pivot point is:

Xτ= 0

−L

4Mg sin θ+3L

4Tcos θ= 0

Step 2: Solve for the tension Tin terms of M,L,g, and θ.

4Mg sin θ=3L

4Tcos θ

T=Mg sin θ

3 cos θ

Step 3: Sum the forces in the vertical direction to solve for θ.

Tsin θ=Mg

Mg sin2θ

3 cos θ=Mg

sin2θ= 3 cos θ

tan2θ= 3

tan θ=√3

θ= tan−1(√3)

Question 9

Question

A uniform beam of length Land mass Mis supported by two strings, as shown

in the diagram below. The angles θ1and θ2are measured from the vertical. If

the tension in the string at angle θ1is T, what is the tension in the other string

at angle θ2?

θ1

θ2

Solution

Let’s consider the forces acting on the beam in the horizontal and vertical di-

rections to find the tension in the other string at angle θ2.

Step 1: Vertical Forces The forces acting in the vertical direction are the

weight of the beam and the vertical components of tension Tand T2. Summing

up forces in the vertical direction:

T1cos θ1=T2+Mg

Step 2: Horizontal Forces The only horizontal force acting is the horizontal

component of tension T1. Summing up forces in the horizontal direction:

T1sin θ1=T2sin θ2

Step 3: Finding T1and T2From Step 1, we get:

T1=T2+Mg

cos θ1

Substitute this into Step 2:

T2+Mg

cos θ1

sin θ1=T2sin θ2

T2+Mg =T2

sin θ1

cos θ1

tan θ1sin θ2

T2(1 −sin θ1tan θ1sin θ2) = Mg

T2=Mg

1−sin θ1tan θ1sin θ2