PHYS 231 - UNIVERSITY PHYSICS I Equilibrium and Elasticity Question Bank Set 9

PHYS 231 - UNIVERSITY PHYSICS I

- Equilibrium and Elasticity

Question Bank - Set 9

Liberty University

Question 1

Question

A cylindrical steel rod of length Land diameter dhangs vertically from the

ceiling. If the Young’s modulus of steel is Y, determine the elongation of the

rod under its own weight.

Solution

Let’s assume that the density of steel is ρand the acceleration due to gravity is

Step 1: Calculate the mass of the rod. The mass of the rod can be calculated

using the formula:

m=ρ·V=ρ·A·L

where Vis the volume of the rod, Ais the cross-sectional area of the rod, and

Lis the length of the rod.

The cross-sectional area of the rod is given by:

A=πd2

Therefore, the mass of the rod is:

m=ρ·πd2

4·L

Step 2: Calculate the force due to the weight of the rod. The force due to

the weight of the rod is equal to the gravitational force acting on it, which is

given by:

F=m·g=ρ·πd2

4·L·g

Step 3: Calculate the stress in the rod. The stress in the rod, σ, is given

by:

σ=F

A=ρ·πd2

4·L·g

πd2

=ρ·g·L

Step 4: Calculate the strain in the rod. The strain in the rod, ϵ, is given

by:

ϵ=σ

Step 5: Calculate the elongation of the rod. The elongation of the rod, ∆L,

is given by:

∆L=ϵ·L=ρ·g·L2

Therefore, the elongation of the rod under its own weight is ρ·g·L2

Question 2

Question

A solid cylinder of radius Rand mass Mis resting on a rough inclined plane

with angle θ. The coefficient of kinetic friction between the cylinder and the

incline is µk. If the cylinder starts to slide down the incline, determine the

acceleration of the cylinder.

Solution

Step 1: Draw a free-body diagram for the cylinder. The forces acting on the

cylinder include the gravitational force (mg) acting downwards, the normal force

(N) acting perpendicular to the incline, the frictional force (fk) acting opposite

to the direction of motion, and the component of the gravitational force parallel

to the incline. Step 2: Break down the gravitational force into components

parallel and perpendicular to the incline. The component of the gravitational

force parallel to the incline is mg sin(θ), and the component perpendicular to the

incline is mg cos(θ). Step 3: Write down the net force equation in the direction

parallel to the incline. The net force in the direction parallel to the incline

is equal to the component of the gravitational force minus the force of kinetic

friction. Therefore, we have:

Ma =mg sin(θ)−fk

Step 4: Express the force of kinetic friction based on the coefficient of kinetic

friction. The force of kinetic friction can be expressed as fk=µkN. And

the normal force, N, can be found by balancing forces in the perpendicular

direction:

N=mg cos(θ)

Step 5: Substitute the expression for the force of kinetic friction and the normal

force into the net force equation. Substitute fk=µkNand N=mg cos(θ) into

the net force equation:

Ma =mg sin(θ)−µkmg cos(θ)

Step 6: Solve for acceleration, a. Solving for a, we get:

a=g(sin(θ)−µkcos(θ))

Therefore, the acceleration of the cylinder sliding down the incline is a=

g(sin(θ)−µkcos(θ)).

Question 3

Question

A uniform cylindrical beam of radius Rand length Lis hinged at one end and

propped up by a support at an angle θas shown in the figure below. The beam

has a mass Mand is in static equilibrium. Determine the force exerted by the

support on the beam and the force exerted by the hinge.

Solution

Step 1: Draw free body diagrams of the beam and show all forces acting on it.

Step 2: Write down the equilibrium conditions for the forces acting in the

perpendicular and parallel directions.

Step 3: Solve the equilibrium equations to find the unknown forces.

Question 4

Question

A uniform beam of length Land mass Mis suspended horizontally by two ropes

attached at each end of the beam. The beam is supported by a vertical force F

applied at its midpoint. If the tension in each rope is given by T, determine the

magnitude and direction of the supporting force Fneeded to keep the beam in

equilibrium.

Solution

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

are the tension forces Tin the ropes, the vertical force Fat the midpoint, the

gravitational force Mg acting at the center of mass, and the reaction forces R

at the endpoints holding up the beam.

Step 2: Write down the sum of forces in the vertical direction equals to 0

since the beam is in equilibrium:

R+R−Mg −F= 0

Solving for Fgives:

F= 2R−Mg

Step 3: Write down the sum of torques about the midpoint equals to 0 since

the beam is in rotational equilibrium. Torque due to M g and Fwill cause

clockwise rotation while torque due to Twill cause counterclockwise rotation.

The torques can be expressed as:

0 = L

2T−L

Solving for Fgives:

F= 2T

Step 4: Substitute F= 2Tinto the equation obtained in Step 2 to get:

2T= 2R−Mg =⇒R=T+Mg

So the required supporting force Fneeded to keep the beam in equilibrium

is 2Tin the direction required to counteract the gravitational force.

Question 5

Question

A cylindrical steel column supports a load of 5000 N. The column has a diameter

of 10 cm and a height of 3 m. If the Young’s modulus for steel is 2 ×1011 N/m2,

determine the change in length of the column under the load.

Solution

Step 1: First, we need to find the cross-sectional area of the column in order to

calculate the stress applied to it. The cross-sectional area of a cylinder is given

by the formula:

A=πr2

where ris the radius of the cylinder. Given the diameter of the column is 10

cm, the radius ris half of the diameter, so r= 5 cm = 0 .05 m.

Therefore, the cross-sectional area is:

A=π(0.05)2

Step 2: Now, we can calculate the stress (σ) applied to the column using the

formula:

σ=F

where Fis the load applied. Given F= 5000 N, we can substitute the values

of Fand Ainto the formula to find σ.

Step 3: With the stress found, we can calculate the strain (ε) using Hooke’s

Law:

σ=E·ε

where Eis the Young’s modulus. Given E= 2 ×1011 N/m2, we can substitute

the values of σand Einto the formula to find ε.

Step 4: To find the change in length of the column (∆L), we use the formula:

∆L=ε·L

where Lis the original length of the column. Given L= 3 m, we can substitute

the values of εand Linto the formula to find ∆L.

Question 6

Question

A uniform cylindrical steel rod of length Land radius Ris hanging vertically

from its upper end. If the Young’s modulus of steel is Y, find the elongation of

the rod when a weight Wis attached to the lower end. Assume that the density

of steel is ρ.

Solution

Step 1: First, let’s determine the cross-sectional area of the rod. The cross-

sectional area of a cylinder is given by A=πR2.

Step 2: Next, we need to find the mass mof the rod. The volume of a cylinder

is V=πR2L, and the density is ρ=m

V. Solving for m, we get m=ρπR2L.

Step 3: The weight Wis given by W=mg, where gis the acceleration due to

gravity. Substituting the expression for mfrom Step 2, we have W=ρπR2Lg.

Step 4: The stress σin the rod due to the weight is given by σ=F

A, where

Fis the force due to the weight. Since F=W, we get σ=W

Step 5: The strain ϵin the rod is related to the stress by Hooke’s Law:

σ=Y ϵ, where Yis the Young’s modulus. So, ϵ=W

Y A .

Step 6: Substituting the expression for Wfrom Step 3 and Afrom Step 1

into the equation for ϵin Step 5, we find ϵ=ρπR2Lg

Y πR2.

Step 7: Simplifying, we get ϵ=ρgL

Therefore, the elongation of the rod when a weight Wis attached to the

lower end is ρgL

Question 7

Question

A uniform beam of length Land mass Mis supported by a cable as shown in

the figure. The beam makes an angle θwith the horizontal. If the tension in

the cable is T, find the position of the beam’s center of mass (distance from the

left end) in terms of Land θ.

Solution

Step 1: Let’s find the distance xfrom the left end of the beam to its center of

mass. The weight of the beam acts at its center of mass, which is located at

x=L

Step 2: Next, let’s consider the forces acting on the beam in the y-direction.

The vertical component of tension Tis Tsin(θ), acting upwards. The weight

of the beam is Mg, acting downwards. The net force in the y-direction is zero

since the beam is in equilibrium.

Step 3: Writing the force balance equation in the y-direction: Tsin(θ) = M g

Step 4: Now, let’s consider the torques acting on the beam about the left

end. The torque due to the tension about the left end is T·Lsin(θ), acting

clockwise. The torque due to the weight of the beam about the left end is

Mg ·L

2sin(θ), acting counterclockwise. The net torque about the left end is

zero since the beam is in equilibrium.

Step 5: Writing the torque balance equation about the left end: T·Lsin(θ) =

Mg ·L

2sin(θ)

Step 6: Simplifying the torque balance equation: T=M g

Step 7: Substituting the tension into the force balance equation: Mg

2sin(θ) =

Step 8: Solving for sin(θ): sin(θ) = 2

Step 9: Therefore, the position of the beam’s center of mass is x=2L

Question 8

Question

A uniform thin rod of length Land mass Mis hinged at one end and suspended

vertically. A horizontal force Fis applied at the other end of the rod, keeping

it in equilibrium. Determine the minimum magnitude of the force Fneeded to

make the rod move from its equilibrium position.

Solution

1. Draw a free-body diagram of the rod. The forces acting on the rod are: -

Weight (mg) acting downward at the center of mass of the rod. - Tension (T)

acting upward at the hinge. - Applied force (F) acting to the right at the end

of the rod.

2. Since the system is in equilibrium, the sum of the torques about any point

must equal zero. Let’s choose the hinge point as the pivot.

3. The torque due to the weight and the tension is zero about the pivot

point because they pass through the pivot. Therefore, only the torque due to

the applied force Fis nonzero:

τ=F·L

4. The torque must balance to keep the system in equilibrium, so:

τ=F·L= 0

5. Thus, the minimum force Fneeded to make the rod move from its

equilibrium position is F= 0 . This means that the rod will not move from its

equilibrium position unless a force greater than zero is applied.

Question 9

Question

A mass of 2 kg is suspended from a vertical spring with a spring constant of 500

N/m. The mass is then displaced downward by 5 cm from its equilibrium posi-

tion and released. Determine the amplitude, angular frequency, and maximum

speed of the mass during its motion.

Solution

Step 1: Determine the amplitude (A) of the oscillation.

A= maximum displacement from equilibrium position

= displacement when the mass is at its maximum distance from equilibrium.

Given that the mass is displaced downward by 5 cm, A= 5 cm = 0.05 m.

Step 2: Determine the angular frequency (ω) of the oscillation.

ω=rk

=s500 N/m

2 kg

=√250 s−1

= 5√10 s−1.

Step 3: Determine the maximum speed of the mass.

Maximum speed = Aω

= 0.05 m ×5√10 s−1

= 0.25√10 m/s.

Therefore, the amplitude of the motion is 0.05 m, the angular frequency is

5√10 s−1, and the maximum speed of the mass is 0.25√10 m/s.

Question 10

Question

A block of mass mis suspended by a thin wire attached to the ceiling of an

elevator. The elevator moves upward with an acceleration a. Find the tension in

the wire when the elevator moves with constant speed vand when it accelerates

upward with a.

Solution

To solve this problem, we will consider the forces acting on the block in both

situations - when the elevator moves with constant speed and when it accelerates

upward.

Case 1: Constant Speed v



Step 1: Draw the free-body diagram for the block.

Since the elevator is moving with a constant speed v, the net force on the

block is zero.



Step 2: Write the equation of motion in the vertical direction.

The forces acting on the block are the tension in the wire (T) and the

gravitational force (mg), where gis the acceleration due to gravity. The

equation of motion in the vertical direction is:

T−mg = 0



Step 3: Solve for the tension T.

From the equation of motion, we have T=mg.

Case 2: Acceleration Upward a



Step 1: Draw the free-body diagram for the block.

Since the elevator is accelerating upward with a, the net force on the block

is in the upward direction.



Step 2: Write the equation of motion in the vertical direction.

The forces acting on the block are the tension in the wire (T) and the

gravitational force (mg). The equation of motion in the vertical direction

is:

T−mg =ma



Step 3: Solve for the tension T.

From the equation of motion, we have T=m(g+a).

Question 11

Question

A uniform cylindrical pillar is standing upright on a rough horizontal surface.

The radius of the pillar is 0.5 m, and its height is 2 m. If the coefficient of

static friction between the pillar and the surface is 0.6, determine the maximum

horizontal force that can be applied at the top of the pillar without causing it

to tip over. The density of the pillar is 500 kg/m3.

Solution

Step 1: To determine the maximum force that can be applied at the top of the

pillar without causing it to tip over, we need to find the limit where the friction

force just balances the tendency for the cylinder to tip.

Step 2: The equilibrium condition for the pillar not to tip over is when the

net torque is zero. The friction force provides the necessary torque to prevent

the tipping.

Step 3: The friction force (Ffriction) at the base of the pillar must balance

the external force applied at the top (F) to prevent tipping. The maximum

static friction force is fmax =µs·N, where µsis the coefficient of static friction

and Nis the normal force.

Step 4: The normal force Nacting on the pillar is equal to the weight of

the pillar mg where mis the mass of the pillar and gis the acceleration due to

gravity.

Step 5: The mass of the cylinder can be calculated using its density ρand

volume V. The volume of the cylinder is given by V=πr2h.

Step 6: Substituting the expressions for the mass and weight of the pillar

into the expression for the maximum friction force, we have fmax =µsρπr2hg.

Step 7: The torque about the bottom of the pillar due to the applied force

is T=F r. To keep the pillar from tipping, this torque must be balanced by the

torque due to the friction force acting at the base of the pillar, T=fmax(h/2).

Step 8: Substituting the expressions for torque and the maximum friction

force, we have F r =µsρπr2hg ·(h/2).

Step 9: Solving for the maximum force F, we find F=µsρπr2hgh

2r.

Step 10: Now, substituting the given values (µs= 0.6, r= 0.5 m, h= 2 m,

ρ= 500 kg/m3,g= 9.81 m/s2) into the equation, we can find the maximum

horizontal force that can be applied at the top of the pillar without causing it

to tip over.

Step 11: Calculating F, we get:

F=(0.6)(500)(π)(0.52)(2)(9.81)(2)

2(0.5)

F=150π

2×9.81

F= 738.024 N

Therefore, the maximum horizontal force that can be applied at the top of

the pillar without causing it to tip over is 738.024 N.

Question 12

Question

A steel cable of length 10.0 m and diameter 2.0 cm hangs vertically from a

ceiling. If the Young’s modulus of the steel is 2.00 ×1011 N/m2, what is the

elongation of the cable due to its own weight? Assume the density of steel is

7.8 g/cm3.

Solution

Step 1: We need to find the weight of the cable first. The volume of the steel

cable can be calculated using the formula for the volume of a cylinder:

V=πd

22

·L

where dis the diameter and Lis the length of the cable.

Given that the diameter d= 2.0 cm and the length L= 10.0 m, we have:

V=π2.0 cm

22

·10.0 m

V=π(1.0 cm2)·10.0 m = 10πcm2·m

Since the density of steel is 7.8 g/cm3, the mass of the cable can be calculated

as:

m= density ×V= 7.8 g/cm3×10πcm2·m

Converting grams to kilograms, we have:

m= 7.8×103kg/m3×10πm = 78πkg

The weight of the cable can be calculated by multiplying the mass by the

acceleration due to gravity, g= 9.81 m/s2:

Fweight =m·g= 78πkg ×9.81 m/s2

Fweight = 766.86πN

Step 2: The tension in the cable at equilibrium must balance the weight of

the cable. The strain (ϵ) in the cable can be calculated using Hooke’s law:

ϵ=Fweight ·L

A·Y

where Ais the cross-sectional area of the cable and Yis the Young’s modulus.

The cross-sectional area of the cable can be calculated using the formula for

the area of a circle:

A=πd

22

A=π(1.0 cm)2=πcm2

Now, substitute the values into the equation for strain:

ϵ=766.86π·10.0

π·2.00 ×1011

ϵ=7668.6

2.00 ×1011

ϵ≈3.834 ×10−5

Step 3: The elongation (∆L) due to the cable’s weight can be calculated

using the formula:

∆L=ϵ·L

∆L= 3.834 ×10−5×10.0 m

∆L= 3.834 ×10−4m

Therefore, the elongation of the cable due to its own weight is approximately

3.834 ×10−4meters.

Question 13

Question

A thin uniform rod of length Land mass Mis hinged at one end. A block of

mass mis suspended from the other end of the rod. The system is in equilibrium

with the rod at an angle θabove the horizontal. Find the tension in the rod

and the normal force at the hinge.

Solution

Step 1: Draw a free-body diagram for the block and rod system. Step 2: Write

out the force balance equations for the block and rod system. Step 3: Solve for

the tension in the rod and the normal force at the hinge.

Question 14

Question

A uniform beam of length Land mass Mis supported by a pivot a distance x

from one end. A weight Whangs at the other end of the beam. The beam is

in equilibrium at an angle θwith the horizontal. What is the magnitude of the

force at the pivot in terms of M,L,x,W, and θ?

Solution

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

are: - The weight of the beam acting at its center of mass L/2. - The force at

the pivot point. - The weight Wacting at the end of the beam.

Step 2: Write down the torque equation. Since the beam is in equilibrium,

the net torque about any point must be zero. Choose the pivot point as the

point about which to calculate the torques.

The torque due to the weight of the beam is −Mg

2Lsin θ(negative because

it’s trying to rotate the beam counter-clockwise). The torque due to the weight

Wis W(x+L/2) sin θ. The torque due to the pivot force is Fp(x) cos θ.

Setting the sum of torques equal to zero, we have:

−MgL

2sin θ+W(x+L

2) sin θ−Fp(x) cos θ= 0

Step 3: Solve for the force at the pivot point.

Fp(x) = MgL

2sin θ+W(x+L

2) sin θ

Therefore, the magnitude of the force at the pivot is MgL

2sin θ+W(x+

2) sin θ.

Question 15

Question

A uniform plank of length Land mass Mis supported by a horizontal force Fat

one end and a vertical wall at the other end. The plank makes an angle θwith

the horizontal. If the coefficient of static friction between the plank and the

wall is µs, determine the range of values for Fthat allow the plank to remain

in equilibrium.

Solution

Step 1: We start by drawing a free-body diagram of the plank. There are three

forces acting on the plank: the force of gravity (Mg) acting at the center of

mass, the normal force (N) exerted by the wall, and the horizontal force (F)

applied at one end. Step 2: Break down the forces acting on the plank into their

components. The normal force Nhas both horizontal and vertical components.

The vertical component must balance the gravitational force (Mg), and the

horizontal component must balance the frictional force (ffriction =µsN). Step

3: Write the equilibrium conditions for the plank. In the vertical direction:

Ncos θ=Mg

And in the horizontal direction:

F=Nsin θ+µsN

Step 4: Substitute the expression for Nfrom the vertical equilibrium condition

into the horizontal equilibrium condition:

F=Mg sin θ

cos θ−µssin θ

Step 5: To determine the range of values for Fthat allow the plank to remain

in equilibrium, we set up the inequality based on the static friction condition

(ffriction ≤µsN):

µsN≤µsMg

µsMg

cos θ≤µsMg

F≥Mg tan θ

So the range of values for Fthat allow the plank to remain in equilibrium is

F≥Mg tan θ.

Question 16

Question

A uniform rod of length Land mass Mis pivoted about one end and is held in

a horizontal position by a string attached a distance xfrom the pivot point, as

shown in the figure below. The string breaks and the rod swings down until it

comes to rest in a vertical position. Determine the angular speed of the rod as

it swings into a vertical position.

L−x

r′

Solution

Let’s start by analyzing the forces acting on the rod in the horizontal position

before the string breaks. The forces acting on the rod are the tension in the

string (T) and the force of gravity acting at the center of mass of the rod.

Step 1: Set up the torque equation in the horizontal position In

the horizontal position, the torque due to the tension at the pivot point must

balance the torque due to the gravitational force about the pivot point. The

torque equations can be written as:

T·x=1

2Mg ·L

where gis the acceleration due to gravity.

Step 2: Determine the tension in the string before it breaks From

the torque equation, we can solve for the tension Tas:

T=Mg ·L

Step 3: Set up the conservation of energy equation When the string

breaks, the rod no longer has any external forces acting on it, and so we can

use the conservation of energy to determine the angular speed of the rod in the

vertical position. The initial total mechanical energy is equal to the final total

mechanical energy:

2Iω2=Mgh

where I=1

3ML2is the moment of inertia of the rod about the pivot point,

ωis the angular speed of the rod, and h=Lis the height the center of mass

falls.

Step 4: Solve for the angular speed Substitute the moment of inertia

Iand the height hinto the conservation of energy equation. Solving for ω, we

get:

ω=r3g

Therefore, the angular speed of the rod as it swings into a vertical position

is q3g

2L.

Question 17

Question

A uniform bar of length Land mass Mis supported by a scale at a distance a

from one end. A weight of mass mis hung at a distance bfrom the same end

where the support is located. If the scale reads F, what is the distance ain

terms of m,M,L,b, and Ffor the bar to be in equilibrium?

Solution

Step 1: Draw a free body diagram of the bar. We can consider three forces

acting on the bar: the gravitational force Mg acting at the center of mass,

the normal force Ffrom the scale acting at a distance afrom one end, and

the weight mg acting at a distance bfrom the same end. Step 2: Set up the

equilibrium condition for forces in the vertical direction. The sum of the forces

in the vertical direction must be zero for the bar to be in equilibrium. Therefore,

we have:

F−Mg −mg = 0

Step 3: Calculate the torque about the point where the scale is located. The

torque about the point where the scale is located must also be zero for the bar

to be in equilibrium. The torque due to the normal force Fand the weight mg

are:

τF=F·0=0

τmg =mg(L−b)

The torque due to the gravitational force Mg can be calculated as:

τMg =M a

Step 4: Set up the torque equilibrium condition. The sum of the torques must

be zero for the bar to be in equilibrium. Therefore, we have:

τMg +τmg = 0

Ma +mg(L−b)=0

Step 5: Solve for the distance a. Solving for ain the torque equilibrium equation

gives:

a=−mg(L−b)

Thus, the distance ain terms of m,M,L,b, and Ffor the bar to be in

equilibrium is a=−mg(L−b)

Question 18

Question

A 2 kg mass is suspended from the ceiling by two identical strings, each making

an angle of 30 degrees with the ceiling. Find the tension in each string.

Solution

Step 1: Draw a free body diagram of the mass showing the forces acting on it.

Step 2: Resolve the weight of the mass into its vertical and horizontal com-

ponents.

Step 3: Apply the equilibrium condition in the vertical direction to set up an

equation involving the tension forces and the vertical component of the weight.

Step 4: Apply the equilibrium condition in the horizontal direction to set up

an equation involving the tension forces and the horizontal component of the

weight.

Step 5: Solve the two equations simultaneously to find the tensions in the

strings.

Question 19

Question

A uniform horizontal beam of length Land mass Mis supported by a vertical

cable attached at the beam’s midpoint. If a block of mass mis attached to the

right end of the beam, what is the tension in the cable when the system is in

equilibrium? Assume the beam has negligible mass compared to the block and

no friction exists between the beam and the block.

Solution

Step 1: Draw a free-body diagram for the block at the right end of the beam.

Label the forces acting on it. Step 2: Write out the equation for equilibrium in

the vertical direction: PFy= 0. Step 3: Write out the equation for rotational

equilibrium about the point where the cable is attached: Pτ= 0.

Question 20

Question

A spring with a spring constant of 200 N/m is compressed by a distance of 0.1

m. A block of mass 2 kg is placed on top of the spring and released. The block

then compresses the spring by an additional 0.05 m before coming to rest. Find

the coefficient of kinetic friction between the block and the surface.

Note: Neglect air resistance and assume the surface is rough enough that

slipping does not occur between the block and the surface.

Solution

Step 1: Find the total compression of the spring when the block comes to rest.

The work done by the block in compressing the spring is equal to the po-

tential energy stored in the spring when it comes to rest. We can write:

2kx2=mgh

Where: k= spring constant = 200 N/m x= initial compression = 0.1 m m

= mass of the block = 2 kg g= acceleration due to gravity = 9.81 m/s2h=

additional compression = ?? (what we need to find)

Solving for h:

200 ×0.12= 2 ×9.81 ×h

20 = 19.62h

h=20

19.62 ≈1.02 m

So, the total compression of the spring when the block comes to rest is 1.02

Step 2: Calculate the work done by the friction force.

The work done by the friction force is equal to the friction force multiplied

by the distance over which it acts. We know the friction force does negative

work, so the work done by the friction force can be calculated as:

Wfriction =−Ffriction ×d

Where: Ffriction =µkN d = total compression of the spring = 1.02 m N=

normal force = mg

Step 3: Calculate the work done by the friction force.

The work done by the friction force is equal to the decrease in mechanical

energy of the block-spring system. Thus:

Wfriction =−∆U

The change in potential energy of the block-spring system is equal to the

negative of the work done by the spring force. Hence:

Wfriction =−(1

2kx2

1−1

2kx2

Wfriction =−(1

2×200 ×0.12−1

2×200 ×1.022)

Wfriction =−1

2×200 ×(0.12−1.022)

Wfriction =−1

2×200 ×(−0.996) = 99.6 J

Therefore, the work done by the friction force is 99.6 J.

Step 4: Find the coefficient of kinetic friction.

Since the work done by the friction force is equal to the friction force mul-

tiplied by the distance over which it acts, we have

Ffriction ×1.02 = 99.6

µk×mg ×1.02 = 99.6

µk=99.6

2×9.81 ×1.02 ≈0.492

Therefore, the coefficient of kinetic friction between the block and the surface

is approximately 0.492.

Question 21

Question

A beam of length Lis supported by a hinge at one end and a cable attached to

the wall at a distance xfrom the hinge. The beam has a mass mand a uniform

density. Find the tension in the cable required to support the beam.

Solution

Step 1: Consider the forces acting on the beam. Since the beam is in equilibrium,

the sum of the forces in the vertical direction must be zero. Let Tbe the tension

in the cable. The gravitational force acting on the beam can be split into two

components: one parallel to the beam (mg sin θ) and one perpendicular to the

beam (mg cos θ).

Step 2: Calculate the perpendicular force components. The perpendicular

component of the gravitational force (mg cos θ) creates a torque about the hinge.

The torque due to this component is −mg cos θ·L. In order for the beam to be

in equilibrium, the torque created by the tension (T·(L−x)) must balance the

torque from the perpendicular component of gravity.

Step 3: Set up the equilibrium condition. Equating the torques, we have:

T·(L−x) = mg cos θ·L

Step 4: Determine the expression for cos θ. Using the geometry of the situ-

ation, we see that cos θ=L−x

Step 5: Substitute the expression for cos θinto the equilibrium equation.

Substitute cos θ=L−x

Linto the equilibrium equation: T·(L−x) = mg ·L−x

L·L

Step 6: Simplify the equation. After simplifying, we have: T·(L−x) =

mg ·(L−x)

Step 7: Solve for the tension in the cable. Dividing both sides by (L−x),

we get: T=mg

Therefore, the tension in the cable required to support the beam is mg.

Question 22

Question

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

a vertical cable at a distance dfrom the pivot. A weight of mass mis placed at

the far end of the beam. Determine the tension in the cable when the system

is in equilibrium.

Solution

Step 1: Draw a free-body diagram of the beam and weight. Let Tbe the tension

in the cable, Rbe the reaction force at the pivot, and Wbe the weight of the

beam. We also have the weight of the mass macting downwards.

Step 2: Write down the torque balance equation. For the beam to be in

equilibrium, the sum of the torques acting on the beam about the pivot point

must be zero. Xτ= 0

T d −WL

2+mgL = 0

Step 3: Resolve forces vertically and horizontally. In the vertical direction,

we have:

R+T−W−mg = 0

Step 4: Solve the torque equation. Substitute the expression for Winto the

torque equation and solve for T:

T d −Mg

2L+mgL = 0

T d =Mg

2L−mgL

T=MgL

2d−mg

Therefore, the tension Tin the cable when the system is in equilibrium is

MgL

2d−mg.

Question 23

Question

A uniform 5.0 m long rod of mass 3.0 kg is hinged at one end. A 2.0 kg ball

is attached to the other end. The ball is released from rest at an angle of 60

degrees above the horizontal. Find the force exerted by the hinge on the rod at

the instant the ball is released.

Solution

Step 1: We can break the forces acting on the system into horizontal and vertical

components. Let’s define the positive x-direction as to the right and the positive

y-direction as upwards. The force by the hinge will have both horizontal and

vertical components.

Step 2: For equilibrium in the horizontal direction, the net force must be

zero. The only horizontal force is the tension Tin the rod, and it acts to the

right. Therefore,

Tcos(60◦)=0

Step 3: For equilibrium in the vertical direction, the net force must also be

zero. We have the weight of the ball and the tension in the rod as the vertical

forces. The tension Tacts at an angle of 60 degrees from the vertical.

Tsin(60◦)−mballg= 0

Step 4: Now, we can substitute mball = 2.0 kg, mrod = 3.0 kg, g= 9.81

m/s2, and solve for T.

Tsin(60◦)−(2.0)(9.81) = 0

T=2.0(9.81)

sin(60◦)

Step 5: Calculate Tto find the force exerted by the hinge on the rod. Make

sure to convert the angle from degrees to radians before computing the value.

T≈2.0×9.81

sin60 ×π

180 

Question 24

Question

A uniform rod of length Land mass Mis pivoted about an end and is held

horizontally, with the other end pressed against a vertical wall. The rod makes

an angle θwith the horizontal. Determine the magnitude of the force exerted

by the wall on the rod.

Solution

1. Draw a free body diagram of the rod, showing all the forces acting on it. The

forces include the weight of the rod acting at its center of mass, the force of the

pivot, and the force exerted by the wall.

2. Resolve all the forces into components parallel and perpendicular to the

rod. Let Fwbe the force exerted by the wall, Nbe the normal force exerted

by the pivot, Mg be the weight of the rod, and Rbe the reaction force at the

pivot.

3. Write down the equations for the forces in the perpendicular direction:

N=Mg cos θ

R=Fwsin θ

4. Write down the equations for the forces in the parallel direction:

Fw=Mg sin θ

5. Combining the equations from step 3 and step 4, we can solve for the

magnitude of the force exerted by the wall on the rod:

Fw=Mg sin θ

Question 25

Question

A horizontal bar of length Land mass Mis attached at one end to a wall by

a hinge. The bar is held horizontally at the other end by a cable making an

angle θwith the bar. The tension in the cable is T. Determine the tension in

the cable needed just to lift the bar off the ground.

Solution

Let’s consider the forces acting on the bar in equilibrium. Step 1: Draw the

free-body diagram of the bar.

Force Magnitude

Tension from the cable T

Weight of the bar Mg

Normal force at the hinge N

Step 2: Write the force equations in the vertical and horizontal directions. In

the vertical direction, the forces must balance:

N=Mg

In the horizontal direction, the forces must also balance:

Tsin θ=N

Step 3: Compute the tension Tneeded just to lift the bar off the ground.

Substitute N=Mg into Tsin θ=N:

Tsin θ=Mg

Solving for Tgives:

T=Mg

sin θ

Therefore, the tension in the cable needed just to lift the bar off the ground is

sin θ.

Question 26

Question

A uniform beam of mass Mand length Lis supported by a rope attached a

distance xfrom one end, as shown in the diagram below. The beam has a mass

2Mand length 2L. If the system is in equilibrium, determine the tension in the

rope.

M2M

Solution

Step 1: We will first draw the free body diagram for the beam. There are three

forces acting on the beam: the weight of the beam (2Mg) acting downward at

the center of the beam, the tension force from the rope (T) acting upward at a

distance xfrom one end, and the reaction force at the support point (N) acting

upward at the pivot.

Step 2: Let’s sum the forces in the vertical direction to find the expression

for the tension T. Taking upward forces as positive, we have:

N−2Mg +T= 0

N= 2Mg −T

Step 3: Next, let’s sum the torques about the pivot point. The torque due

to the force Nis zero since it acts at the pivot. The torque due to the force

2Mg about the pivot is MgL. The torque due to the tension force Tabout the

pivot is T(x+ 2L). The condition for rotational equilibrium gives:

T(x+ 2L) = MgL

Step 4: Now we can solve the two equations we obtained in Step 2 and Step

3 to find the tension T. Substituting N= 2Mg −Tinto T(x+ 2L) = MgL, we

get:

(2Mg −T)(x+ 2L) = MgL

2Mgx + 4MgL −T x −2T L =M gL

T x + 2T L = 2Mgx + 4MgL

T(x+ 2L)=2Mg(x+ 2L)

T= 2Mg

Therefore, the tension in the rope is 2Mg.

Question 27

Question

A uniform rod of length Land mass Mis pivoted at one end. A force Fis

applied perpendicular to the rod at a distance dfrom the pivot, as shown in the

figure. Find the magnitude and direction of the force needed to hold the rod in

equilibrium.

Pivot

Solution

Step 1: We will begin by drawing a free-body diagram of the rod. Let’s designate

the forces acting on the rod: the force Fat a distance dfrom the pivot, the

weight Mg acting downwards at the center of mass of the rod, and the normal

force Nacting upwards at the pivot point to keep the rod in equilibrium.

Step 2: We will write the torque equation about the pivot point to find the

force Fin terms of known quantities. The torque due to force Fis:

τF=F·d

The torque due to the weight Mg is:

τMg =M g ·L

Since the rod is in equilibrium, the sum of the torques must be zero:

F·d−Mg ·L

2= 0

Step 3: Solve for the force F:

F·d=Mg ·L

F=MgL

Therefore, the magnitude of the force needed to hold the rod in equilibrium is

MgL

2d. To find the direction of the force, we observe that it must be perpendicular

to the rod and directed outward from the pivot point.

Question 28

Question

A uniform, thin rod of length Land mass Mis initially at rest on a frictionless

horizontal surface. A force Fis then applied to the other end of the rod in a

direction perpendicular to the rod. Find the magnitude of the force needed to

make the rod tip lift off the surface.

Solution

1. Draw a Free Body Diagram (FBD):

Let’s consider the forces acting on the rod when the tip is about to lift off.

These forces are: - The force of gravity acting at the center of mass Cof the

rod (mg downwards) - The normal force Nexerted by the surface on the rod -

The external force Fapplied at the other end of the rod

2. Write the Equations for Equilibrium:

For the rod to be in equilibrium, the net force acting on it must be zero and the

net torque about any point must also be zero. We’ll choose the pivot point to

be at the point where the rod is touching the surface.

The net force equation in the vertical direction is:

N=mg

The net torque equation about the pivot point is:

F·L= (MgL/2)

3. Solve for the External Force F:

Substitute N=mg and solve the torque equation for F:

F·L= (MgL/2)

F=MgL

F=Mg

Thus, the magnitude of the force needed to make the rod tip lift off the

surface is F=Mg

Question 29

Question

A uniform rod of mass Mand length Lis pivoted at its midpoint. Two masses,

each with mass m, are attached to the ends of the rod. Find the period of small

oscillations around the equilibrium position.

Solution

Step 1: Identify the force equations for small oscillations.

The forces acting on mass mon the right side of the pivot are the tension in

the rod, gravity, and the force due to motion. The forces acting on the left side

are similar, except the force due to motion is opposite in direction. We start by

writing the force equation for the right side mass m:

T−mg +F=m¨x

where Tis the tension, gis the acceleration due to gravity, Fis the force due

to motion, and ¨xis the acceleration of the mass.

Step 2: Write the force equation for the left side mass m.

The force equation for the left side mass mis:

T−mg −F=m¨x

Step 3: Determine expressions for the tension and force due to motion.

At equilibrium, the tension is merely supporting the weight, so T=mg.

The force due to motion, F, is equal to −kx, where xis the position of the mass

relative to the equilibrium position.

Step 4: Substitute the expressions for tension and force into the force equa-

tions.

For the right side mass m:

mg −mg −kx =m¨x

−kx =m¨x

−k

mx= ¨x

Step 5: Determine the angular frequency ωof oscillations.

The above equation is the differential equation of simple harmonic motion

with angular frequency ω=qk

Step 6: Calculate the period Tof oscillations.

The period T=2π

ω= 2πpm

Therefore, the period of small oscillations around the equilibrium position

for the system is 2πpm

Question 30

Question

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

wires of equal length L, as shown in the figure below. The distance between the

wires is b. What is the tension in each wire?

[Figure: a rod suspended horizontally by two vertical wires of equal length

with distance bbetween them]

Solution

Let’s denote the tension in each wire as T.

Step 1: The free-body diagram of the rod shows that the forces acting on

the rod are the tensions in the two wires and the gravitational force acting at

the center of the rod.

Step 2: In the horizontal direction, the tensions must balance the weight of

the rod:

2T=Mg

Step 3: Solving for T, we have:

T=Mg

Step 4: So, the tension in each wire is Mg

Question 31

Question

A uniform meter stick of mass 0.2 kg is standing vertically with its lower end

resting on a frictionless horizontal surface. A string is attached to the upper

end and pulled horizontally with a force of 12 N. The coefficient of static friction

between the stick and the surface is 0.6. Determine the tension in the string.

Solution

Step 1: First, let’s analyze the forces acting on the meter stick. There are

three forces to consider: the tension in the string (T) pulling to the right, the

weight of the stick acting downward, and the normal force exerted by the ground

perpendicular to the stick.

Step 2: The weight of the stick can be calculated using the formula Fweight =

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

m= 0.2 kg and g= 9.8 m/s2, we have Fweight = (0.2 kg)(9.8 m/s2).

Step 3: The normal force is equal in magnitude and opposite in direction to

the weight of the stick when the stick is in equilibrium. Therefore, the normal

force is also Fweight.

Step 4: Since the stick is on the verge of sliding horizontally, the maximum

static friction force is µsN, where µsis the coefficient of static friction (0.6) and

Nis the normal force. In this case, the static friction force is acting to the left.

Step 5: For the stick to remain in equilibrium, the net force in the horizontal

direction must be zero. This means that the tension in the string must be equal

in magnitude to the friction force. So we have T=µsN.

Step 6: Substituting N=Fweight and µs= 0.6 into the equation T=µsN,

we get T= 0.6·[0.2 kg ·(9.8 m/s2)].

Step 7: Calculating this expression gives us the tension in the string:

T= 0.6·[0.2 kg ·(9.8 m/s2)] = 1.176 N

Question 32

Question

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

strings, as shown in the diagram below. The rod makes an angle θwith the

horizontal. Determine the tension in each string in terms of L,M, and θ.

Solution

Step 1: Setup the following equations: - Sum of forces in the x-direction:

T2cos θ−T1cos θ= 0 - Sum of forces in the y-direction: T1sin θ+T2sin θ−Mg =

Step 2: Solve equation (1) for T2in terms of T1:

T2=T1

Step 3: Substitute the expression for T2from Step 2 into equation (2):

T1sin θ+T1sin θ−Mg = 0

Step 4: Simplify equation from Step 3:

2T1sin θ=Mg

Step 5: Solve for T1:

T1=Mg

2 sin θ

Step 6: The tension in each string is equal, so T2=T1:

T2=Mg

2 sin θ

Therefore, the tension in each string is M g

2 sin θ.

Question 33

Question

A uniform horizontal beam of length Land mass Mis supported at one end by

a pivot and has a weight (W=Mg) hanging from the other end. A block of

mass mis placed a distance dfrom the pivot as shown in the figure below. If

the block is at rest, find the force exerted by the pivot on the beam.

beam_pivot.png

Solution

Step 1: First, let’s draw a free-body diagram for the beam, showing all the

forces acting on it. The forces acting on the beam are its weight acting at the

center of mass, the normal force exerted by the pivot, and the force exerted by

the block.

Step 2: Since the beam is at rest, the sum of torques about any point must

be zero. For simplicity, we will take torques about the pivot point.

The torque due to the weight of the beam is zero since it acts at the pivot

point.

Step 3: The torque due to the block mabout the pivot is m·g·d(clockwise).

The torque due to the normal force exerted by the pivot is zero since it acts

at the pivot point.

Step 4: Therefore, the sum of torques is equal to zero:

m·g·d= 0

Step 5: Solving for the normal force Nexerted by the pivot on the beam:

N=m·g·d

Hence, the force exerted by the pivot on the beam is m·g·d

Question 34

Question

A steel beam of length 5.0 m is supported by a hinge at one end and a vertical

cable attached to the other end. The beam has a mass of 500 kg. If the beam

makes an angle of 30 degrees with the horizontal, what is the tension in the

cable? (Assume g = 9.81 m/s2)

Solution

Step 1: First, let’s draw a diagram of the situation to visualize the forces acting

on the beam. We have the weight of the beam acting downward at the center,

the tension in the cable pulling upward at an angle of 30 degrees with the

vertical, and a normal force acting horizontally at the hinge.

Step 2: We can start by calculating the weight of the beam using the formula:

Fgravity =m·g, where mis the mass of the beam and gis the acceleration due

to gravity. Fgravity = (500 kg) ·(9.81 m/s2) = 4905 N

Step 3: Next, we can resolve the force due to gravity into its horizontal and

vertical components. The vertical component will counteract the tension in the

cable. The vertical component of the weight is Fgravity ·sin(30◦). Fvertical =

4905 N ·sin(30◦) = 2452.5 N

Step 4: Since the beam is in equilibrium, the vertical component of the

weight must be balanced by the tension in the cable. Therefore, the tension

in the cable is equal to the vertical component of the weight. T=Fvertical =

2452.5 N

Step 5: Therefore, the tension in the cable is 2452.5 N when the beam makes

an angle of 30 degrees with the horizontal.

Question 35

Question

A uniform meterstick of mass 0.20 kg is supported horizontally by two vertical

strings, one at the 0.10 m mark and the other at the 0.80 m mark. A mass of

0.40 kg is hung from the 0.50 m mark. Calculate the tension in each string.

Solution

Step 1: Draw free body diagrams for the meterstick and the hanging mass. For

the meterstick, we have two tensions acting vertically upwards at the 0.10 m

and 0.80 m marks, and the weight of the meterstick acting downwards at its

center of mass. For the hanging mass, we have the tension in the string acting

upwards and the weight of the mass acting downwards.

Step 2: Write out the equations of equilibrium for each object. For the

meterstick: XFy=T1+T2−mg = 0

T1+T2=mg

For the hanging mass:

XFy=T2−mg′= 0

T2=mg′

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

g′is the mass of the hanging mass times the acceleration due to gravity.