Chapter 04 Homework

3/3/2014 Chapter 4 Homework

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Chapter 4 Homework

Due: 10:00pm on Friday, February 21, 2014

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

Exercise 4.1

Two forces have the same magnitude .

Part A

What is the angle between the two vectors if their sum has a magnitude of ?

ANSWER:

Correct

Part B

What is the angle between the two vectors if their sum has a magnitude of ?

ANSWER:

Correct

Part C

What is the angle between the two vectors if their sum has a magnitude of zero?

ANSWER:

Correct

Two Hanging Masses

Two blocks with masses and hang one under the other. For this problem, take the positive direction to beupward, and use for the magnitude of the acceleration due to gravity.

F

2F

= 0 θ ∘

F2√

= 90 θ ∘

= 180 θ ∘

M1 M2g

Typesetting math: 80%



Case 1: Blocks at rest

For Parts A and B assume the blocks are at rest.

Part A

Find , the tension in the lower rope.

Express your answer in terms of some or all of the variables , , and .

Hint 1. Free-body diagram

Isolate the lower block (mass ) by considering just the forces that act on it. Use Newton's 2nd law while

noting that the acceleration of this block is zero.

Hint 2. Sum of forces

Write down the sum of the y components of all the forces acting on the lower block.


ANSWER:

ANSWER:

Correct

T2

M1 M2 g

M2

M1 M2 T2 g

= ∑ = = 0F2y M2a2 − gT2 M2

= T2 gM2



Part B

Find , the tension in the upper rope.



Write down the sum of the y components of all the forces acting on the upper block.

Express your answer in terms of some or all of the variables , , , , and .

ANSWER:

ANSWER:

Correct

Case 2: Accelerating blocks

For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration ofmagnitude .

Part C

Find , the tension in the lower rope.

Express your answer in terms of some or all of the variables , , , and .


Apply Newton's 2nd law, , to the lower block. Write the sum of the y components of the

forces.


ANSWER:

ANSWER:

T1

M1 M2 g

M1 M2 T1 T2 g

= ∑ = 0F1y − − gT1 T2 M1

= T1 ( + )gM1 M2

a

T2

M1 M2 a g

∑ = aFy M2

M1 M2 T2 g

= ∑ = aF2y M2 − gT2 M2



Correct

Good! Now notice that acceleration acts in the tension equation the same way that gravity does. If we look at alimiting case in which the block of mass is in free fall, accelerating downward with , then the

tension goes to zero.

Part D

Find , the tension in the upper rope.

Express your answer in terms of some or all of the variables , , , and .


Consider the block of mass as an isolated system and apply Newton's 2nd law, , with

the value of found in the previous part.

ANSWER:

Correct

The force exerted by a rope is a result of the rope's tension and points along the rope.

Exercise 4.3

Due to a jaw injury, a patient must wear a strap (see the figure) that produces a net upward force of 5.00 on his chin.The tension is the same throughout the strap.

= T2 (a + g)M2

M2 a = −g

T1

M1 M2 a g

M1 ∑ = aFy M1

T2

= T1 ( + )(a + g)M1 M2

N



Part A

To what tension must the strap be adjusted to provide the necessary upward force?

ANSWER:

Correct

Newton's 1st Law

Learning Goal:

To understand Newton's 1st law.

Newton's Principia states this first law of motion:An object subject to no net force maintains its state of motion, either at rest or at constant speed in a right line.This law may be stated as follows: If the sum of all forces acting on an object is zero, then the acceleration of that

object is zero. Mathematically this is just a special case of the 2nd law of motion, when , promptingscholars to advance the following reasons (among others) for Newton's spelling it out separately:

1. This expression only holds in an inertial coordinate system--one that is not accelerating--and this lawreally says you have to use this type of coordinate system (i.e., Newton's laws won't work inside anaccelerating rocket ship.)

2. This was a direct challenge to the Impetus theory of motion, described as follows:A mover, while moving a body, impresses on it a certain impetus, a certain power capable of moving thisbody in the direction in which the mover set it going, whether upwards, downwards, sideways or in a circle.By the same amount that the mover moves the same body swiftly, by that amount is the impetus that isimpressed on it powerful. It is by this impetus that the stone is moved after the thrower ceases to move it;but because of the resistance of the air and the gravity of the stone, which inclines it to move in adirection opposite to that towards which the impetus tends to move it, this impetus is continuallyweakened. Therefore the movement of the stone will become continually slower, and at length, theimpetus is so diminished or destroyed that the gravity of the stone prevails over it and moves the stonedown towards its natural place.

A. C. Crombie, Medieval and Early Modern Science</>This theory is sometimes called the Animistic theory of motion since it envisions a "life force" beingassociated with motion.

Newton's 1st law is often very difficult to grasp because it contradicts various common-sense ideas of motion that theyhave acquired from experience in everyday life. For example, unaccounted for forces like friction might cause a ballrolling on the playground to eventually stop, even though no obvious forces seem to be acting.When studying Newtonian mechanics, it is best to remember this as two laws:

1. If the net force (i.e., sum of all forces) acting on an object is zero, the object will keep moving withconstant velocity (which may be zero).

2. If an object is moving with constant velocity (not speed), that is, with zero acceleration, then the net forceacting on that object must be zero.

Complete the following sentences to see if you can apply these ideas.

Part A

= 3.15 F N

= mF ⃗ a⃗ = 0F ⃗



If a car is moving to the left with constant velocity, one can conclude that

ANSWER:

Correct

Part B

An object cannot remain at rest unless

ANSWER:

Correct

Understanding Newton's Laws

Part A

An object cannot remain at rest unless which of the following holds?

Hint 1. How to approach the problem

This problem describes a situation of static equilibrium (i.e., a body that remains at rest). Hence, it isappropriate to apply Newton's 1st law.

Hint 2. Newton's 1st law: a body at rest

According to Newton's 1st law, a body at rest remains at rest if the net force acting on it is zero.

ANSWER:

there must be no forces applied to the car.

the net force applied to the car is directed to the left.

the net force applied to the car is zero.

there is exactly one force applied to the car.

there are no forces at all acting on it.

the net force acting on it is zero.

the net force acting on it is constant.

there is only one force acting on it.



Correct

If there is a net force acting on a body, regardless of whether it is a constant force, the body accelerates. If thebody is at rest and the net force acting on it is zero, then it will remain at rest. The net force could be zeroeither because there are no forces acting on the body at all or because several forces are acting on the bodybut they all cancel out.

Part B

If a block is moving to the left at a constant velocity, what can one conclude?


This problem describes a situation of dynamic equilibrium (i.e., a body that moves at a constant velocity).Hence, it is appropriate to apply Newton's 1st law.

Hint 2. Newton's 1st law: a body in motion

According to Newton's 1st law, a body initially in motion continues to move with constant velocity if the netforce acting on it is zero.

ANSWER:

Correct

If there is a net force acting on a body, regardless of whether the body is already moving, the bodyaccelerates. If a body is moving with constant velocity, then it is not accelerating and the net force acting on itis zero. The net force could be zero either because there are no forces acting on the body at all or becauseseveral forces are acting on the body but they all cancel out.

Part C

A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What

can you say about the block's motion?

The net force acting on it is zero.

The net force acting on it is constant and nonzero.

There are no forces at all acting on it.

There is only one force acting on it.

There is exactly one force applied to the block.

The net force applied to the block is directed to the left.

The net force applied to the block is zero.

There must be no forces at all applied to the block.

2 kg 3 N 4 N




This problem describes a situation of dynamic motion (i.e., a body that is acted on by a net force). Hence, itis appropriate to apply Newton's 2nd law, which allows you to relate the net force acting on a body to theacceleration of the body.

Hint 2. Newton's 2nd law

Newton's 2nd law states that a body accelerates if a net force acts on it. The net force is proportional to theacceleration of the body and the constant of proportionality is equal to the mass of the body. In other words,

,

where is the net force acting on the body, and and are the mass and the acceleration of the body,

respectively.

Hint 3. Relating acceleration to velocity

Acceleration is defined as the change in velocity per unit time. Keep in mind that both acceleration andvelocity are vector quantities.

ANSWER:

Correct

The acceleration of an object tells you nothing about its velocity--the direction and speed at which it is moving.In this case, the net force on (and therefore the acceleration of) the block is to the right, but the block could bemoving left, right, or in any other direction.

Part D

A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The blockmust be __________.


This problem describes a situation of dynamic motion (i.e., a body that is acted on by a net force). Hence, itis appropriate to apply Newton's 2nd law, which allows you to relate the net force acting on a body to theacceleration of the body.

Hint 2. Newton's 2nd law

Newton's 2nd law states that a body accelerates if a net force acts on it. The net force is proportional to theacceleration of the body and the constant of proportionality is equal to the mass of the body. In other words,

F = ma

F m a

It must be moving to the left.

It must be moving to the right.

It must be at rest.

It could be moving to the left, moving to the right, or be instantaneously at rest.

F = ma



,

where is the net force acting on the body, and and are the mass and the acceleration of the body,

respectively.

ANSWER:

Correct

Since there is a net force acting, the body does not move at a constant velocity, but it accelerates instead.However, the force acting on the body is constant. Hence, according to Newton's 2nd law of motion, theacceleration of the body is also constant.

Part E

Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown.

The net force acting on the object __________.

Check all that apply.


By definition, the net force is the vector sum of all forces acting on the object. To find the magnitude of thenet force you need to add the components of the two forces acting. Try adding the two forces graphically (byconnecting the head of one force to the tail of the other). The directions of the two forces are arbitrary, but bytrying different possibilities you should be able to determine the maximum and minimum net forces thatcould act on the object.

Hint 2. Find the net force when the two forces act on the object in opposite directions

Find the magnitude of the net force if both the forces acting on the object are horizontal and the 10-N force isdirected to the right, while the 4-N force is directed to the left.

Express your answer in newtons.

Hint 1. Vector addition

The magnitude of the vector sum of two parallel forces is the sum of the magnitudes of the forces.The magnitude of the vector sum of two antiparallel forces is the absolute value of the difference inmagnitudes of the forces.

ANSWER:

F = ma

F m a

continuously changing direction

moving at constant velocity

moving with a constant nonzero acceleration

moving with continuously increasing acceleration

4 N 10 N



Correct

Is there any other orientation of the two forces that would lead to a net force with a smaller magnitudethan what you just calculated?

Hint 3. Find the direction of the net force when the two forces act in opposite directions

If both the forces acting on the object are horizontal and the 10-N force is directed to the right, while the 4-Nforce is directed to the left, the net force is horizontal and directed __________.

ANSWER:

Correct

ANSWER:

Correct

Exercise 4.10

A dockworker applies a constant horizontal force of 80.0 to a block of ice on a smooth horizontal floor. The frictionalforce is negligible. The block starts from rest and moves a distance 10.5 in a time 5.30 .

Part A

What is the mass of the block of ice?

ANSWER:

Correct

6.0 N

in the same direction as the 10-N force

in the opposite direction to the 10-N force

cannot have a magnitude equal to

cannot have a magnitude equal to

cannot have the same direction as the force with magnitude

must have a magnitude greater than

5 N10 N

10 N10 N

Nm s

= 107 m kg



Part B

If the worker stops pushing at the end of 5.30 , how far does the block move in the next 4.80 ?

ANSWER:

Correct

Exercise 4.14

A 3.45- cat moves in a straight line (the x-axis). the figure

shows a graph of the x-component of this cat's velocity as afunction of time.

Part A

When does the maximum net force on his cat occur?

ANSWER:

Correct

Part B

Find the magnitude of the maximum net force on this cat.

ANSWER:

s s

= 19.0 x m

kg

in the interval to .



t = 0 t = 2.00 st = 2.00 t = 6.00 st = 6.00 t = 10.00 s



Correct

Part C

When is the net force on the cat equal to zero?

ANSWER:

Correct

Part D

What is the magnitude of the net force at time 8.5 ?

ANSWER:

Correct

Exercise 4.19

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 . A watermelon has a weight of 57.0 at

the surface of the earth. In this problem, use 9.81 for the acceleration due to gravity on earth.

Part A

What is its mass on the earth's surface?

ANSWER:

Correct

Part B

= 13.8 F N




t = 0 t = 2.00 st = 2.00 t = 6.00 st = 6.00 t = 10.00 s

s

= 3.45 F N

m/s2 Nm/s2

= 5.81 m kg



What is its mass on the surface of Io?

ANSWER:

Correct

Part C

What is its weight on the surface of Io?

ANSWER:

Correct

Tension in a Massless Rope

Learning Goal:

To understand the concept of tension and the relationship between tension and force.

This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from bothends. Let and \texttip{F_{\rm d}}{F_d} (with u for up and dfor down) represent the magnitude of the forces acting on thetop and bottom of the rope, respectively. Assume that therope is massless, so that its weight is negligible comparedwith the tension. (This is not a ridiculous approximation--modern rope materials such as Kevlar can carry tensionsthousands of times greater than the weight of tens of metersof such rope.)

Consider the three sections of rope labeled a, b, and c in thefigure.

At point 1, a downward force of magnitude \texttip{F_{\rm ad}}{F_ad} acts on section a.

At point 1, an upward force of magnitude \texttip{F_{\rm bu}}{F_bu} acts on section b.

At point 1, the tension in the rope is \texttip{T_{\rm 1}}{T_1}.

At point 2, a downward force of magnitude \texttip{F_{\rm bd}}{F_bd} acts on section b.

At point 2, an upward force of magnitude \texttip{F_{\rm cu}}{F_cu} acts on section c.

At point 2, the tension in the rope is \texttip{T_{\rm 2}}{T_2}.

Assume, too, that the rope is at equilibrium.

= 5.81 m kg

= 10.5 w N

Fu



Part A

What is the magnitude \texttip{F_{\rm ad}}{F_ad} of the downward force on section a?

Express your answer in terms of the tension \texttip{T_{\rm 1}}{T_1}.

ANSWER:

Correct

Part B

What is the magnitude \texttip{F_{\rm bu}}{F_bu} of the upward force on section b?

Express your answer in terms of the tension \texttip{T_{\rm 1}}{T_1}.

ANSWER:

Correct

Part C

The magnitude of the upward force on c, \texttip{F_{\rm cu}}{F_cu}, and the magnitude of the downward force on b,

\texttip{F_{\rm bd}}{F_bd}, are equal because of which of Newton's laws?

ANSWER:

Correct

Part D

The magnitude of the force \texttip{F_{\rm bu}}{F_bu} is ____ \texttip{F_{\rm bd}}{F_bd}.

ANSWER:

\texttip{F_{\rm ad}}{F_ad} = T_{1}

\texttip{F_{\rm bu}}{F_bu} = T_{1}

1st

2nd

3rd



Correct

It is important to realize that \texttip{F_{\rm bu}}{F_bu} and \texttip{F_{\rm bd}}{F_bd} are not a Newton's third

law pair of forces. Instead, these forces are equal and opposite due to the fact that the rope is stationary(a_{\rm b}=0) and massless(m_{\rm b}=0). By applying Newtons first or second law to this segment of rope

you obtain F_{\rm b \: net} = F_{bu}-F_{bd}=m_{\rm b}a_{\rm b}=0, since m_{\rm b}=0 and a_{\rm b}=0. Note

that if the rope were accelerating, these forces would still be equal and opposite because m_{\rm b}=0.

Part E

Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these twoforces?

Hint 1. Consider the bigger picture

From Part D, we see that the internal forces of the rope are simply equal to each other. This leaves us withonly the external forces to deal with. Since the rope is massless, what must be the relationship betweenthese external forces?

ANSWER:

Correct

The forces on the two ends of an ideal, massless rope are always equal in magnitude. Furthermore, themagnitude of these forces is equal to the tension in the rope.

Part F

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the ropemakes a straight line. For this situation, which of the following statements are true?


ANSWER:

less than

greater than

equal to

F_u > F_d

F_u = F_d

F_u < F_d



Correct

Forces on Blocks in an Elevator Conceptual Question

Two blocks are stacked on top of each other on the floor of an elevator. For each of the following situations, select thecorrect relationship between the magnitudes of the two forces given.

You will be asked two questions about each of threesituations. Each situation is described above the first in thepair of questions. Do not assume anything about a givensituation except for what is given in the description for thatparticular situation.

First situation

The elevator is moving downward at a constant speed.

Part A

Hint 1. Comparing forces that act on the same object

When comparing forces that act on the same object, draw a free-body diagram of the object being acted on.Then, determine the acceleration of the object. By Newton's 2nd law, the net force must be proportional tothe object's acceleration.

Hint 2. Draw a free-body diagram for the top block

Complete the free-body diagram for the top block by drawing the force on the top block due to the earth. Thisforce should act at the center of the block.

ANSWER:

The tension in the rope is everywhere the same.

The magnitudes of the forces exerted on the two objects by the rope are the same.

The forces exerted on the two objects by the rope must be in opposite directions.

The forces exerted on the two objects by the rope must be in the direction of the rope.



ANSWER:

Correct

Part B

Hint 1. Comparing forces that do not act on the same object

If two forces do not act on the same object, they will not appear on the same free-body diagram. Therefore,Newton's 2nd law cannot be used to determine the relative sizes of these forces. Certain forces can,however, be compared using Newton's 3rd law.

Hint 2. Newton's 3rd law

Newton's 3rd law states that when two objects exert forces on each other, these forces are always equal inmagnitude and opposite in direction. Thus, if you are sitting in a chair, the force the chair exerts upward onyou is exactly the same as the force you exert downward on the chair, regardless of whether you are at rest

The magnitude of the force of the bottom block on the top block is

greater than

equal to

less than

unknown compared to

the magnitude

of the force of the earth on the top block.



in the chair, or have you feet up on your desk, or are in the process of getting up out of the chair, or in the

process of landing in the chair after jumping from a great height, ... it does not matter!

ANSWER:

Correct

Second situation

The elevator is moving downward at an increasing speed.

Part C

Hint 1. Determining acceleration

If the elevator is moving downward at an increasing speed, what is the direction of the elevator'sacceleration?

ANSWER:

ANSWER:

Correct

The magnitude of the force of the bottom block on top block is

greater than

equal to

less than

unknown compared to

the magnitude of

the force of the top block on bottom block.

upward

zero

downward

unknown


greater than

equal to

less than

unknown compared to

the magnitude




Part D

Hint 1. Newton's 3rd law in accelerating elevator

Newton's 3rd law holds that forces come in pairs of equal magnitude and opposite direction in all cases.Thus, the acceleration of the elevator should not affect the relative magnitude of two forces that form a 3rdlaw pair.

ANSWER:

Correct

Third situation

The elevator is moving upward.

Part E

Hint 1. Determining acceleration

If the elevator is moving upward, what is the direction of the elevator's acceleration?

ANSWER:

ANSWER:


greater than

equal to

less than

unknown compared to

the magnitude

of the force of the top block on the bottom block.

upward

zero

downward

unknown



Correct

Even though the elevator is moving upwards, you do not know in which direction it is accelerating, or indeedwhether the elevator is accelerating at all!

Part F

ANSWER:

Correct

Tension in a Hanging Massive Rope

Consider a rope with length \texttip{l}{l}, mass per unit length \texttip{\lambda }{lambda}, experiencing a gravitationalacceleration \texttip{g}{g} and hanging vertically as shown. Let\texttip{y}{y} refer to the height of a point P above the bottomof the rope.


greater than

equal to

less than

unknown compared to

the magnitude



greater than

equal to

less than

unknown compared to

the magnitude

of the force of the top block on the bottom block.



Part A

The force exerted on the rope by the ceiling is in the _____ direction.

Hint 1. Consider the weight of the rope

To support the rope, the ceiling must exert a force opposite to the weight of the rope.

ANSWER:

Correct

Part B

Find \texttip{F}{F}, the magnitude of the force exerted on the rope by the ceiling.

Express \texttip{F}{F} in terms of quantities given in the problem introduction.

Hint 1. Use Newton's 2nd law

Take \texttip{W}{W} to be the weight of the rope. Apply Newton's 2nd law to find an expression for W-F.

ANSWER:

Hint 2. Find the weight of the rope

What is the weight of the rope, \texttip{W}{W}?

Express your answer in terms of \texttip{\lambda }{lambda}, \texttip{l}{l}, and \texttip{g}{g}.

Hint 1. Find the mass of the rope

What is the mass \texttip{M}{M} of the rope?

ANSWER:

ANSWER:

upward

downward

W-F = 0

\texttip{M}{M} = {\lambda} l



ANSWER:

Correct

Part C

What is the tension \texttip{T_{\mit P}}{T_P} at point P in the rope?

Express \texttip{T_{\mit P}}{T_P} in terms of quantitites given in the problem introduction.

Hint 1. How to approach this problem

The tension at point P is the magnitude of the force exerted by the part of the rope above P on the part of therope below P. Remember that point P is located a distance \texttip{y}{y} above the bottom of the rope.

Hint 2. Find the weight of the rope below point P

What is the weight \texttip{W_{\mit P}}{W_P} of the part of the rope below point P?

Express your answer in terms of quantities given in the problem introduction.

ANSWER:

ANSWER:

Correct

Exercise 4.23

Boxes A and B are in contact on a horizontal, frictionless surface, as shown in the following figure. Box A has mass24.0{\rm kg} and box B has mass 6.0{\rm kg} . A horizontal force of 100{\rm N} is exerted on box A.

\texttip{W}{W} = {\lambda} l g

\texttip{F}{F} = l {\lambda} g

\texttip{W_{\mit P}}{W_P} = {\lambda} y g

\texttip{T_{\mit P}}{T_P} = y {\lambda} g



Part A

What is the magnitude of the force that box A exerts on box B?

Express your answer to two significant figures and includethe appropriate units.

ANSWER:

Correct

Free-Body Diagrams: Introduction

Learning Goal:

To learn to draw free-body diagrams for various real-life situations.

Imagine that you are given a description of a real-life situation and are asked to analyze the motion of the objectsinvolved. Frequently, that analysis involves finding the acceleration of the objects, which, in turn, requires that you findthe net force.

To find the net force, you must first identify all of the forces acting on the object and then add them as vectors. Such aprocedure is not always trivial. It is helpful to replace the sketch of the situation by a drawing of the object (representedas a particle) and all the forces applied to it. Such a drawing is called a free-body diagram. This problem will walk youthrough several examples of free-body diagrams and will demonstrate some of the possible pitfalls.Here is the general strategy for drawing free-body diagrams:

Identify the object of interest. This may not always be easy: A sketch of the situation may contain manyobjects, each of which has a different set of forces acting on it. Including forces acting on different objectsin the same diagram will lead to confusion and a wrong solution.Draw the object as a dot. Draw and clearly label all the forces acting on the object of interest. The forcesshould be shown as vectors originating from the dot representing the object of interest. There are twopossible difficulties here: omitting some forces and drawing the forces that either don't exist at all or areapplied to other objects. To avoid these two pitfalls, remember that every force must be applied to the

F = 20 {\rm N}



object of interest by some other object.

Once all of the forces are drawn, draw the coordinate system. The origin should coincide with the dotrepresenting the object of interest and the axes should be chosen so that the subsequent calculations ofvector components of the forces will be relatively simple. That is, as many forces as possible must beeither parallel or perpendicular to one of the axes.

Even though real life can present us with a wide variety of situations, we will be mostly dealing with a very small numberof forces. Here are the principal ones of interest:

Weight, or the force due to gravity. Weight acts on every object and is directed straight down unless weare considering a problem involving the nonflat earth (e.g., satellites).Normal force. The normal force exists between two surfaces that are pressed against each other; it isalways perpendicular to the surfaces.Force of tension. Tension exists in strings, springs, and other objects of finite length. It is directed alongthe string or a spring. Keep in mind that a spring can be either compressed or stretched whereas a stringcan only be stretched.Force of friction. A friction force exists between two surfaces that either move or have a tendency to moverelative to each other. Sometimes, the force of air drag, similar in some ways to the force of friction, maycome into play. These forces are directed so that they resist the relative motion of the surfaces. Tosimplify problems you often assume that friction is negligible on smooth surfaces and can be ignored. Inaddition, the word friction commonly refers to resistive forces other than air drag that are caused bycontact between surfaces, so you can ignore air drag in problems unless you are explicitly told to considerits effects.

The following examples should help you learn to draw free-body diagrams. We will start with relatively simple situationsin which the object of interest is either explicitly suggested or fairly obvious.

Part A

A hockey puck slides along a horizontal, smooth icy surface at a constant velocity as shown. Which of thefollowing forces act on the puck?


ANSWER:



Correct

There is no such thing as "the force of velocity." If the puck is not being pushed, there are no horizontal forcesacting on it. Of course, some horizontal force must have acted on it before, to impart the velocity--however, inthe situation described, no such "force of push" exists. Also, the air drag in such cases is assumed to benegligible. Finally, the word "smooth" usually implies negligible surface friction. Your free-body diagram shouldlook like the one shown here.

Part B

Consider a block pulled by a horizontal rope along a horizontal surface at a constant velocity as shown. There istension in the rope. Which of the following forces act onthe block?


ANSWER:

friction

weight

force of velocity

normal force

force of push

air drag

acceleration



Correct

Because the velocity is constant, there must be a force of friction opposing the force of tension. Since theblock is moving, it is k inetic friction. Your free-body diagram should look like that shown here.

Part C

A block is resting on an slope. Which of the following forces act on theblock?


ANSWER:

weight

air drag

friction

acceleration

force of velocity

normal force

force of tension



Correct

Part D

Draw the free-body diagram for the block resting on a slope.

Draw the force vectors such that their tails align with the center of the block (indicated by the black dot).The orientations of your vectors will be graded but not the lengths.

ANSWER:

Correct

Part E

Now consider a block sliding up a rough slope after having been given a quick push as shown . Which of thefollowing forces act on the block?


static friction

force of push

kinetic friction

weight

normal force



ANSWER:

Correct

The word "rough" implies the presence of friction. Since the block is in motion, it is k inetic friction. Once again,there is no such thing as "the force of velocity." However, it seems a tempting choice to some students sincethe block is going up.

Part F

Draw the free-body diagram for the block sliding up a rough slope after having been given a quick push.

Draw the force vectors such that their tails align with the center of the block (indicated by the black dot).The orientations of your vectors will be graded but not the lengths.

ANSWER:

weight

kinetic friction

static friction

force of push

normal force

the force of velocity



Correct

Part G

Now consider a block being pushed up a smooth slope. The force pushing the block is parallel to the slope. Whichof the following forces are acting on the block?


ANSWER:



Correct

Your free-body diagram should look like the one shown here.

The force of push is the normal force exerted, possibly, by the palm of the hand of the person pushing theblock.

In all the previous situations just described, the object of interest was explicitly given. In the remaining parts of theproblem, consider a situation where choosing the objects for which to draw the free-body diagrams is up to you.Two blocks of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} are connected by a light string that goesover a light frictionless pulley. The block of mass \texttip{m_{\rm 1}}{m_1} is sliding to the right on a rough horizontalsurface of a lab table.

Part H

To solve for the acceleration of the blocks, you will have to draw the free-body diagrams for which objects?


weight

kinetic friction

static friction

force of push

normal force



ANSWER:

Correct

Part I

Draw the free-body diagram for the block of mass \texttip{m_{\rm 1}}{m_1} and draw a free-body diagram for the

block of mass \texttip{m_{\rm 2}}{m_2}.

Draw the force vectors acting on \texttip{m_{\rm 1}}{m_1} such that their tails align with the center of the

block labeled \texttip{m_{\rm 1}}{m_1} (indicated by the black dot). Draw the force vectors acting on

\texttip{m_{\rm 2}}{m_2} with their tails aligned with the center of the block labeled

\texttip{m_{\rm 2}}{m_2}. The orientations of your vectors will be graded but not the lengths.

ANSWER:

the block of mass \texttip{m_{\rm 1}}{m_1}

the block of mass \texttip{m_{\rm 2}}{m_2}

the connecting string

the pulley

the table

the earth



Correct

Exercise 4.32

A skier of mass 65.0 {\rm kg} is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to theground. The ground slopes upward at a constant angle of 26.0 \̂circ above the horizontal and you can ignore friction.

Part A

Calculate the tension in the tow rope.

ANSWER:

Correct

Score Summary:

Your score on this assignment is 95.2%.You received 13.33 out of a possible total of 14 points.

T = 279 {\rm N}

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Chapter 04 Homework

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