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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.
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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
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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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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}
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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?
Express your answer using two significant figures.
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}
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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
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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}
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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.
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. 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.
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}.
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)
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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).
Express your answer in terms of some or all of the variables
\texttip{\mu _{\rm s}}{mu_s},
\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}}}
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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}}}
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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?
Express your answer using two significant figures.
ANSWER:
Correct
Part C
What minimum horizontal force must the monkey apply to start the
box in motion?
Express your answer using two significant figures.
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}
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What minimum horizontal force must the monkey apply to keep the
box moving at constant velocity once it has beenstarted?
Express your answer using two significant figures.
ANSWER:
Correct
Part E
If the monkey applies a horizontal force of 18.0 {\rm N}, what
is the magnitude of the friction force ?
Express your answer using two significant figures.
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}
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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:
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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}}}
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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.
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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
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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.
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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}