chapter 05 homework

3/3/2014 Chapter 5 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2788368 1/24

Chapter 5 Homework

Due: 10:00pm on Friday, February 28, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

The Normal Force

When an object rests on a surface, there is always a force perpendicular to the surface; we call this the normal force,denoted by . The two questions to the right will explore the normal force.

Part A

A man attempts to pick up his suitcase of weight by pulling straight up on the handle. However, he is unable to lift the

suitcase from the floor. Which statement about the magnitudeof the normal force acting on the suitcase is true during the

time that the man pulls upward on the suitcase?

Hint 1. How to approach this problem

First, identify the forces that act on the suitcase and draw a free-body diagram. Then use the fact that the

suitcase is in equilibrium, , to examine how the forces acting on the suitcase relate to each other.

Hint 2. Identify the correct free-body diagram

Which of the figures represents the free-body diagram of the suitcase while the man is pulling on the handle with aforce of magnitude ?

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ANSWER:

ANSWER:

Correct

Part B

Now assume that the man of weight is tired and decides to sit on his suitcase. Which statement about the

magnitude of the normal force acting on the suitcase is true

during the time that the man is sitting on the suitcase?

Hint 1. Identify the correct free-body diagram.

Which of the figures represents the free-body diagram while the man is sitting atop the suitcase? Here the vectorlabeled is a force that has the same magnitude as the man's weight.

A

B

C

D

The magnitude of the normal force is equal to the magnitude of the weight of the suitcase.

The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitudeof the force of the pull.

The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and themagnitude of the suitcase's weight.

The magnitude of the normal force is greater than the magnitude of the weight of the suitcase.

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ANSWER:

ANSWER:

Correct

Recognize that the normal force acting on an object is not always equal to the weight of that object. This is animportant point to understand.

Exercise 5.3

A 73.9- wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.0 . (Use 9.80 for the gravitational

acceleration at the earth's surface.)

Part A

Find the maximum tension in the chain.

ANSWER:

A

B

C

D

The magnitude of the normal force is equal to the magnitude of the suitcase's weight.

The magnitude of the normal force is equal to the magnitude of the suitcase's weight minus the magnitude ofthe man's weight.

The magnitude of the normal force is equal to the sum of the magnitude of the man's weight and the magnitudeof the suitcase's weight.

The magnitude of the normal force is less than the magnitude of the suitcase's weight.

LH LH NT



Correct

Part B

Find the minimum tension in the chain.

ANSWER:

Correct

Part C

What is the tension at a point three-fourths of the way up from the bottom of the chain?

ANSWER:

Correct

Block on an Incline Adjacent to a Wall

A wedge with an inclination of angle rests next to a wall. A block of mass is sliding down the plane, as shown. There isno friction between the wedge and the block or between the wedge and the horizontal surface.

Part A

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Find the magnitude, , of the sum of all forces acting on the block.

Express in terms of and , along with any necessary constants.

Hint 1. Direction of the net force on the block

The net force on the block must be the force in the direction of motion, which is down the incline.

Hint 2. Determine the forces acting on the block

What forces act on the block? Keep in mind that there is no friction between the block and the wedge.

ANSWER:

Hint 3. Find the magnitude of the force acting along the direction of motion

Consider a coordinate system with the x direction pointing down the incline and the y direction perpendicular tothe incline. In these coordinates, what is , the component of the block's weight in the x direction?

Express in terms of , , and .

ANSWER:

ANSWER:

Correct

Part B

Find the magnitude, , of the force that the wall exerts on the wedge.

Express in terms of and , along with any necessary constants.

Hint 1. The force between the wall and the wedge

There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the nethorizontal force on the wedge must be zero. If the block exerts a force with a horizontal component on the wedge,some other horizontal force must act on the wedge so that the net force is zero.

Hint 2. Find the normal force between the block and the wedge

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The weight of the block and friction

The weight of the block and the normal (contact) force

The weight of the block and the weight of the wedge

The weight of the block and the force that the wall exerts on the wedge

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What is the magnitude, , of the normal (contact) force between the block and the wedge? (You might have

computed this already while answering Part A.)

Express in terms of , , and .

ANSWER:

Correct

Hint 3. Find the horizontal component of the normal force

In the previous hint you found the magnitude of the normal force between the block and the wedge. What is themagnitude, , of the horizontal component of this normal force?

Express in terms of and .

ANSWER:

Correct

ANSWER:

Correct

Your answer to Part B could be expressed as either or . In either form, we see

that as gets very small or as approaches 90 degrees ( radians), the contact force between the wall and the

wedge goes to zero. This is what we should expect; in the first limit ( small), the block is accelerating very slowly,

and all horizontal forces are small. In the second limit ( about 90 degrees), the block simply falls vertically and

exerts no horizontal force on the wedge.

Pushing Too Hard

A baggage handler at an airport applies a constant horizontal force with magnitude to push a box, of mass , across arough horizontal surface with a very small constant acceleration .

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Part A

The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of . How does the

magnitude of the force that the handler applies to this box compare to the magnitude of the force applied to the

first box?

Hint 1. How to approach the problem

Apply Newton's 2nd law to the first box to obtain an equation relating the force applied to the acceleration of thebox. Then, do the same for the second box. Compare these equations to determine the relationship between

and .

Hint 2. Identify the forces that act on each box

To apply Newton's 2nd law, you must determine which forces contribute to the acceleration of each box. Of thefollowing forces, which act along the direction of the box's acceleration?

Check all that apply.

ANSWER:

Hint 3. Apply Newton's 2nd law to the first box

The first box has mass and acceleration . It is pushed with a force of magnitude . Applying Newton's 2nd

law to this box yields which of the following equations?

Hint 1. Newton's 2nd law

The normal force exerted by the floor on the box.

The weight of the box.

The force of static friction.

The force of kinetic friction.

The force exerted on the box by the baggage handler.

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Newton's 2nd law states that:,

where is the magnitude of the net force on the object, is the mass of the object, and is the

acceleration of the object along the direction of the applied net force.

Hint 2. Equation for the force of kinetic friction

Recall that the force of kinetic friction on a given object moving relative to a surface is,

where is the coefficient of friction and is the normal force.

ANSWER:

Hint 4. Apply Newton's 2nd law to the second box

The second box has mass and acceleration . It is pushed with a force of magnitude . Applying Newton's

2nd law to this box yields which of the following equations?

ANSWER:

Hint 5. Put it all together

In the previous two subparts, you determined that

and

.

Combine these two equations to obtain one equation that contains both and .

ANSWER:

ANSWER:

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Correct

Now see if you can apply this problem-solving technique to answer the next question.

Part B

Now assume that the baggage handler pushes a third box ofmass so that it accelerates at a rate of . How does

the magnitude of the force that the handler applies to this

box compare to the magnitude of the force applied to the

first box?

Hint 1. Apply Newton's 2nd law to the first box

The first box has mass and acceleration . It is pushed with a force of magnitude . Applying Newton's 2nd

law to this box yields which of the following equations?

ANSWER:

Hint 2. Apply Newton's 2nd law to the third box

The second box has mass and acceleration . It is pushed with a force of magnitude . Applying

Newton's 2nd law to this box yields which of the following equations?

Hint 1. Find the force of kinetic friction

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Recall that the force of kinetic friction on a given object moving relative to a surface is,

where is the coefficient of friction and is the normal force.

If represents the force of kinetic friction that acts on the first box, what is the force of kinetic friction

acting on the third box?

ANSWER:

ANSWER:

Hint 3. Put it all together

In the previous two subparts, you determined that

and

\large{F_3 - \frac{1}{2}f_{\rm k} = ma}.

Combine these two equations to obtain one equation that contains both \texttip{F_{\rm 1}}{F_1} and

\texttip{F_{\rm 3}}{F_3}.

ANSWER:

Hint 4. Determine the importance of a small acceleration

You were told in the problem introduction that \texttip{a}{a} is very small. Consider what a very small \texttip{a}{a}

implies about the relative sizes of \texttip{F_{\rm 1}}{F_1} and \texttip{f_{\rm k}}{f_k}. It will help you to consider the

following expression:

F_1 - f_{\rm k} = ma.

Which of the following statements are correct?

ANSWER:

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\large{F_3 = ma - \frac {1}{2}f_{\rm k}}

\large{F_3 = F_1 - \frac{1}{2}f_{\rm k}}

\large{F_3 = F_1 + \frac{3}{2}f_{\rm k} + 2ma}



ANSWER:

Correct

Exercise 5.18

A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is700 {\rm kg}, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to2200{\rm N} . The tension in the towrope between the transport plane and the first glider is not to exceed 12000 {\rm N}.

Part A

If a speed of 40 {\rm m/s} is required for takeoff, what minimum length of runway is needed?

Express your answer using two significant figures.

ANSWER:

Correct

Part B

What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?


ANSWER:

Correct

\texttip{F_{\rm 1}}{F_1} is much larger than \texttip{f_{\rm k}}{f_k}.

\texttip{F_{\rm 1}}{F_1} is slightly larger than \texttip{f_{\rm k}}{f_k}.

\texttip{f_{\rm k}}{f_k} is larger than \texttip{F_{\rm 1}}{F_1}.

\large{0 \leq F_3 < \frac{1}{2}F_1}

\large{\frac{1}{2}F_1 \leq F_3 2F_1

150 {\rm m}

6000 {\rm N}



Exercise 5.24

A 5.50{\rm kg} crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied tothe end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80{\rm m/s} )t +(0.61{\rm m/s 3^})t 3^

Part A

What is the magnitude of the force F when 4.50{\rm s} ?

Express your answer with the appropriate units.

ANSWER:

Correct

Pushing a Block

Learning Goal:

To understand kinetic and static friction.

A block of mass \texttip{m}{m} lies on a horizontal table. The coefficient of static friction between the block and the table is\texttip{\mu _{\rm s}}{mu_s}. The coefficient of kinetic friction is \texttip{\mu _{\rm k}}{mu_k}, with \mu_{\rm k} < \mu_{\rm s}.

Part A

If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from thetable), what is the magnitude of the force due to friction?

Hint 1. Consider the type of friction at rest

What type of friction is acting in this case?

ANSWER:

ANSWER:

Correct

F = 144 {\rm N}

static friction

kinetic friction

neither static nor kinetic

\texttip{F_{\rm friction}}{F_friction} = 0



Part B

Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With whatforce \texttip{F}{F} must you be pushing the block just before the block begins to move?

Express the magnitude of \texttip{F}{F} in terms of some or all the variables

\texttip{\mu _{\rm s}}{mu_s}, \texttip{\mu _{\rm k}}{mu_k}, and \texttip{m}{m}, as well as the acceleration due

to gravity \texttip{g}{g}.

Hint 1. Consider the type of friction to start movement

What type of friction is acting in this case?

ANSWER:

ANSWER:

Correct

Part C

Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of thefriction force?

Express your answer in terms of some or all of the variables \texttip{\mu _{\rm s}}{mu_s},

\texttip{\mu _{\rm k}}{mu_k}, and \texttip{m}{m}, as well as the acceleration due to gravity \texttip{g}{g}.

Hint 1. What level of force is required?

In this situation, the force of static friction prevents the object from moving. Therefore, the magnitude of the staticfriction force must equal the magnitude of the net horizontal applied force acting on the object, up to a certainmaximum value. In this case,

F_{\rm static \;friction} \le \mu_s N,

where \texttip{\mu _{\rm s}}{mu_s} is the coefficient of static friction and \texttip{N}{N} is the normal force that the

surface exerts on the object.

ANSWER:

Correct

static friction

kinetic friction

neither static nor kinetic

\texttip{F}{F} = g m {\mu}_{s}

\texttip{F_{\rm friction}}{F_friction} = .5 m g {\mu}_{s}



Part D

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to applythe same amount of force even after it starts moving. Find the acceleration \texttip{a}{a} of the block after it begins to

move.



Hint 1. Calculate applied force

What is the magnitude \texttip{F}{F} of the force that you are applying to make the block move?

Express your answer in termsof some or all of the variables \texttip{\mu _{\rm s}}{mu_s}, \texttip{\mu _{\rm k}}{mu_k}, and \texttip{m}{m},

as well as the acceleration due to gravity \texttip{g}{g}.

ANSWER:

Correct

Hint 2. Consider applied force and kinetic friction

When the block is moving, there is a force of kinetic friction acting on it, with magnitude|F_{\rm kinetic \;friction}| = \mu_{\rm k}n,

where \texttip{\mu _{\rm k}}{mu_k} is the coefficient of kinetic friction and \texttip{n}{n} is the magnitude of the

normal force.

Hint 3. Calculate net horizontal force

What is the magnitude of the net horizontal force acting on the block? Remember that the friction force is directedopposite to the motion of the object.



ANSWER:

Correct

ANSWER:

\texttip{F}{F} = g m {\mu}_{s}

\texttip{F_{\rm horizontal}}{F_horizontal} = g m \left({\mu}_{s}-{\mu}_{k}\right)

\texttip{a}{a} = g \left({\mu}_{s}-{\mu}_{k}\right)



Correct

Board Pulled Out from under a Box

A small box of mass \texttip{m_{\rm 1}}{m_1} is sitting on a board of mass \texttip{m_{\rm 2}}{m_2} and length \texttip{L}{L} .The board rests on a frictionless horizontal surface. The coefficientof static friction between the board and the box is \texttip{\mu _{\rms}}{mu_s}. The coefficient of kinetic friction between the board andthe box is, as usual, less than \texttip{\mu _{\rm s}}{mu_s}.

Throughout the problem, use \texttip{g}{g} for the magnitude of theacceleration due to gravity. In the hints, use \texttip{F_{\rm f\hspace{1 pt}}}{F_f} for the magnitude of the friction force betweenthe board and the box.

Part A

Find \texttip{F_{\rm min}}{F_min}, the constant force with the least magnitude that must be applied to the board in order

to pull the board out from under the the box (which will then fall off of the opposite end of the board).


\texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{g}{g}, and \texttip{L}{L}. Do not include

\texttip{F_{\rm f \hspace{1 pt}}}{F_f} in your answer.

Hint 1. Condition for the board sliding out from under the box

The board will slide out from under the box if the magnitude of the board's acceleration exceeds the magnitude ofthe maximum acceleration that friction can give to the box.

Hint 2. Find the acceleration of the box in terms of \texttip{F_{\rm f \hspace{1 pt}}}{F_f}

Assume that the coefficient of static friction between the board and the box is not known at this point. What is themagnitude of the acceleration of the box in terms of the friction force \texttip{F_{\rm f \hspace{1 pt}}}{F_f}?

Express your answer in terms of \texttip{F_{\rm f \hspace{1 pt}}}{F_f} and \texttip{m_{\rm 1}}{m_1}.

ANSWER:

Hint 3. Find the largest acceleration of the box

Now take the coefficient of static friction between the board and the box to be \texttip{\mu _{\rm s}}{mu_s}. What

\texttip{a_{\rm box}}{a_box} = \large{\frac{F_{f}}{m_{1}}}



is the largest possible magnitude of the acceleration of the box?

Express your answer in terms of some or all of the variables \texttip{\mu _{\rm s}}{mu_s}, \texttip{g}{g},

and \texttip{m_{\rm 1}}{m_1}.

Hint 1. Maximum force on the box

Friction is the only horizontal force on the box. What is the largest possible value for \texttip{F_{\rm f \hspace{1 pt}}}{F_f}?

ANSWER:

Hint 4. Find the sum of horizontal forces on the board

Write down the sum of all the horizontal forces acting on the board. Take the positive x direction to be to the right.

Give your answer in terms of \texttip{F}{F}, \texttip{F_{\rm f \hspace{1 pt}}}{F_f}, and any constants

necessary.

Hint 1. Friction and Newton's 3rd law

Remember, by Newton's 3rd law, if there is a force of magnitude \texttip{F_{\rm f \hspace{1 pt}}}{F_f} acting

on the box due to the board, there is a force of equal magnitude and opposite direction acting on the boarddue to the box.

ANSWER:

Hint 5. Find the acceleration of the board for large \texttip{F_{\rm f \hspace{1 pt}}}{F_f}

In Hint 4 you found the net horizontal force on the board. Now, find the acceleration of the board when the force ofstatic friction reaches its maximum possible value.

Express your answer in terms of \texttip{F}{F}, \texttip{\mu _{\rm s}}{mu_s}, \texttip{m_{\rm 1}}{m_1},

\texttip{m_{\rm 2}}{m_2}, and \texttip{g}{g}.

ANSWER:

Hint 6. Putting it all together

Reread Hint 1. In Hint 3, you found the largest possible acceleration of the box, \texttip{a_{\rm box}}{a_box}. In

Hint 5, you found the acceleration of the board, \texttip{a_{\rm board}}{a_board}. What is the minimum value of the

constant force, \texttip{F_{\rm min}}{F_min}, so that \left|a_{\rm board}\right| > \left|a_{\rm box}\right|?

\texttip{a_{\rm box}}{a_box} = {\mu}_{s} g

\sum F_{x} = F-F_{f}

\texttip{a_{\rm board}}{a_board} = \large{\frac{F-{\mu}_{s} m_{1} g}{m_{2}}}



ANSWER:

Correct

Exercise 5.28

A box of bananas weighing 40.0 {\rm N} rests on a horizontal surface. The coefficient of static friction between the box andthe surface is 0.40 and the coefficient of kinetic friction is 0.20.

Part A

If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?

ANSWER:

Correct

Part B

What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 {\rm N} to the box and the box is

initially at rest?


ANSWER:

Correct

Part C

What minimum horizontal force must the monkey apply to start the box in motion?


ANSWER:

Correct

Part D

\texttip{F_{\rm min}}{F_min} = \left(m_{1}+m_{2}\right) g {\mu}_{s}

0 {\rm N}

6.0 {\rm N}

16 {\rm N}



What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has beenstarted?


ANSWER:

Correct

Part E

If the monkey applies a horizontal force of 18.0 {\rm N}, what is the magnitude of the friction force ?


ANSWER:

Correct

Part F

If the monkey applies a horizontal force of 18.0 {\rm N}, what is the box's acceleration?

ANSWER:

Correct

Exercise 5.32

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. Thecoefficients of kinetic and static friction between the box and the bed of the truck are 0.310 and 0.550, respectively.

Part A

Starting from rest, what is the shortest time this truck could accelerate uniformly to 35.0{\rm m/s} (\approx 78.3{\rm mph}

) without causing the box to slide. (Hint: First use Newtons second law to find the maximum acceleration that staticfriction can give the box, and then solve for the time required to reach 35.0{\rm m/s} .)

ANSWER:

8 {\rm N}

8 {\rm N}

2.45 {\rm m/s 2^}

t_{min} = 6.49 {\rm s}



Correct

A Mass on a Turntable: Conceptual

A small metal cylinder rests on a circular turntable that is rotatingat a constant rate, as illustrated in the diagram.

Part A

Which of the following sets of vectors best describes the velocity, acceleration, and net force acting on the cylinder at thepoint indicated in the diagram?

Hint 1. The direction of acceleration can be determined from Newton's second law

According to Newton's second law, the acceleration of an object has the same direction as the net force acting onthat object.

ANSWER:



Correct

Part B

Let \texttip{R}{R} be the distance between the cylinder and the center of the turntable. Now assume that the cylinder is

moved to a new location R/2 from the center of the turntable. Which of the following statements accurately describe the

motion of the cylinder at the new location?

Check all that apply.

Hint 1. Find the speed of the cylinder

Find the speed \texttip{v}{v} of the cylinder at the new location. Assume that the cylinder makes one complete

turn in a period of time \texttip{T}{T}.

Express your answer in terms of \texttip{R}{R} and \texttip{T}{T}.

ANSWER:

Hint 2. Find the acceleration of the cylinder

Find the magnitude of the acceleration \texttip{a}{a} of the cylinder at the new location. Assume that the cylinder

makes one complete turn in a period of time \texttip{T}{T}.

Express your answer in terms of \texttip{R}{R} and \texttip{T}{T}.

Hint 1. Centripetal acceleration

Recall that the acceleration of an object that moves in a circular path of radius \texttip{r}{r} with constant

speed \texttip{v}{v} has magnitude given by

\large{a=\frac{v^ 2}{r}}.

Note that both the velocity and radius of the trajectory change when the cylinder is moved.

ANSWER:

a

b

c

d

e

\texttip{v}{v} = \large{\frac{{\pi} R}{T}}

\texttip{a}{a} = \large{\frac{2 {\pi} {^2} R}{T {^2}}}



ANSWER:

Correct

Mass on Turntable

A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram .

The small metal cylinder has a mass of 0.20 \rm kg, the coefficientof static friction between the cylinder and the turntable is 0.080,and the cylinder is located 0.15 \rm m from the center of theturntable.

Take the magnitude of the acceleration due to gravity to be 9.81\rm m/s 2^.

Part A

What is the maximum speed \texttip{v_{\rm max}}{v_max} that the cylinder can move along its circular path without

slipping off the turntable?

Express your answer numerically in meters per second to two significant figures.

Hint 1. Centripetal acceleration

If you know a body is in uniform circular motion, you know what its acceleration must be. If a body of mass \texttip{m}{m} is traveling with speed \texttip{v}{v} in a circle of radius \texttip{R}{R}, what is the magnitude

\texttip{a_{\rm c}}{a_c} of its centripetal acceleration?

ANSWER:

The speed of the cylinder has decreased.

The speed of the cylinder has increased.

The magnitude of the acceleration of the cylinder has decreased.

The magnitude of the acceleration of the cylinder has increased.

The speed and the acceleration of the cylinder have not changed.



Correct

Hint 2. Determine the force causing acceleration

Whenever you see uniform circular motion, there is a real force that causes the associated centripetalacceleration. In this problem, what force causes the centripetal acceleration?

ANSWER:

Correct

Hint 3. Find the maximum possible friction force

The magnitude \texttip{f_{\rm s}}{f_s} of the force due to static friction satisfies f_{\rm s} \leq f_{\rm max}. What is

\texttip{f_{\rm max}}{f_max} in this problem?

Express your answer numerically in newtons to three significant figures.

ANSWER:

Correct

Hint 4. Newton's 2nd law

To solve this problem, relate the answers to the previous two hints using Newton's 2nd law:\vec{F} = m\,\vec{a}.

ANSWER:

Correct

\large{\frac{m v^ 2}{R}}

m v^ {2} R

v^ 2 R

\large{\frac{v^ 2}{R}}

normal force

static friction

weight of cylinder

a force other than those above

\texttip{f_{\rm max}}{f_max} = 0.157 \rm N

\texttip{v_{\rm max}}{v_max} = 0.34 \rm m/s



Exercise 5.46

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upperend. Each arm supports a seat suspended from a cable 5.00 {\rm m} long, the upper end of the cable being fastened to thearm at a point 3.00 {\rm m} from the central shaft.

Part A

Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30.0 \^circ with the vertical.

ANSWER:

Correct

Part B

Does the angle depend on the weight of the passenger for a given rate of revolution?

ANSWER:

Correct

Exercise 5.51

An airplane flies in a loop (a circular path in a vertical plane) of radius 190{\rm {\rm m}} . The pilot's head always points towardthe center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastestat the bottom.

T = 6.19 {\rm s}

Yes.

No.



Part A

At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?

ANSWER:

Correct

Part B

At the bottom of the loop, the speed of the airplane is 200{\rm {\rm km/h}} . What is the apparent weight of the pilot at

this point? His true weight is 700{\rm {\rm N}} .

ANSWER:

Correct

Score Summary:

Your score on this assignment is 98.4%.You received 13.78 out of a possible total of 14 points.

v = 43.2 {\rm m/s}

w = 1860 {\rm N}

chapter 05 homework

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