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Physics 111
Solution to Exam 2
Gary Morris
Chapters 5 – 6, 9.1 – 9.4
Staple Notecard Here
Tuesday, October 19, 2004
Do not open this exam until you are instructed to do so. Carefully read all the
instructions listed below.
•
•
•
•
•
•
•
Look at the seat to your left. Look at the seat to your right. If they are not BOTH unoccupied,
you must change seats until the seats on either side of you contain no persons.
Clear your desktop of all items except a pencil/pen, a (scientific) calculator, and a single 3”×5”
note card (hand-written). You may not use your book or any additional notes during this exam.
You must staple your note card to this exam before turning in the exam.
You will have 75 minutes to complete the exam. When you are instructed to stop working on the
exam, you must put down your pencil/pen immediately. I will notify you when there are 30
minutes, 15 minutes, and 5 minutes remaining in the exam period.
Read through the entire exam before you start. The last page contains useful constants and
mathematical formulas.
Generous partial credit is awarded. However, you must show your work to receive credit. Do
not simply write down the answers. No partial credit will be given on the multiple-choice
section.
The Multiple Choice and Essay sections are mandatory.
You need only work 2 out of the 3 “Sections” on this exam. NOTE: you must circle on this
page the numbers corresponding to the two sections you would like us to grade. We will only
grade the problems that you circle. If you do not indicate which problems you would like us to
grade, we will grade the first two sections you attempt.
Please sign the following statement:
On my honor, I have not given, received, or tolerated the use of any unauthorized aid during this
examination.
_____________________________________
“Education is what survives when what has been learned has been
forgotten.” – B. F. Skinner
Use this page if you need extra space to work any of the following problems.
Section 1 (work two from Sect. 1–3)
You are participating in some bizarre reality TV/game
show/physical challenge. Your goal is to lower a box
containing fragile supplies of mass m = 15.0 kg down a
wall using only a pole. Since the contents of the box
are so fragile, you must try to lower it at a small,
constant velocity, otherwise it will hit the ground
moving too fast and smash into useless bits. The pole
has been greased and so there is a maximum force you
can exert on it before your hands start to slip. Assume
that the force applied to the box by the pole is parallel
to the pole. For parts a) and b) assume the wall is frictionless.
m
θ
a) [ 5 points ] Draw a free-body diagram for the box when the pole makes an angle of θ with the
vertical. Describe each force present in your free-body diagram. Describe the 3rd Law Pair forces
for each force in your free-body diagram.
b) [ 4 points ] If the maximum force you can apply to the pole is 500 N, at what angle will you lose
control of the box. (Hint: Use Newton’s 2nd Law and find the angle if the force is equal to 500 N.)
For parts c) through e), assume the coefficient of kinetic friction between the box and the wall is µk =
0.30.
c) [ 4 points ] Draw a free-body diagram for the box when the pole makes an angle of θ with the
vertical. Describe each force present in your free-body diagram.
d) [ 7 points ] Use Newton’s 2nd Law to find an expression for the force of the pole on the box
required to maintain constant speed as a function of µk, g, m, and θ.
e) [ 5 points ] If the maximum force you can apply to the pole is 500 N, will you ever lose control of
the box? (Hint: Consider the limiting case of a horizontal pole.) Justify your answer with physics.
a)
+y
θ
+x
Fp
force of the pole on box
reaction: force of box on
N
Normal force of wall on box
Reaction: force of box on
W = mg
force of gravity of Earth on box (weight)
reaction: force of gravity of box on
“It takes most men five years to recover from a college education,
and to learn that poetry is as vital to thinking as knowledge.”
– Brooks Atkinson, Once Around the Sun, 1951
b) Let’s apply Newton’s second law in component form. First, in the x-direction:
∑ Fx = max
1
FP sin θ − N = 0
where we have used the fact that ax=0 since it is not burrowing into the wall or hopping off of it. In
the y-direction, we have
∑ Fx = ma y
FP cos θ − mg = 0
2
where ay=0 since the box is supposed to move down at a constant velocity.
Our known are FP = 500 N, m = 15.0 kg, and g = 9.8 m/s2. We are trying to find θ. Using equation 2,
we find
mg
cos θ =
FP
 mg 
D
θ = cos 
 = 73
 FP 
−1
c)
+y
Force of kinetic
friction opposing the
slide down the wall
fk
+x
θ
Fp
force of the pole on
N
Normal force of wall on
W = mg
force of gravity of Earth on box
3
d) Our knowns are m, µk and g. Applying N’s 2nd with zero acceleration, we get:
∑ Fx = max
4
FP sin θ − N = 0
∑F
x
= ma y
5
FP cos θ + f k − mg = 0
Since f k = µk N , we can write this as two equations with two unknowns, N and FP:
FP sin θ = N
6
FP cos θ + µk N = mg
Substituting for N, we can solve for FP:
FP cos θ + µk FP sin θ = mg
FP ( cos θ + µk sin θ ) = mg
FP =
7
mg
cos θ + µk sin θ
e) Well, if I substitue in for m, µk and g and use a horizontal pole so θ=90°, then I find that
FP = 490 N.
So, you will just barely be able to keep the box under control until it dips a little below the horizontal.
Section 2 (work two from Sect. 1–3)
Two buckets of sand hang from opposite ends of a rope that
hangs over a pulley. One bucket is full and weights 170 N; the
other is only partly filled and weighs 97 N. The heavier bucket
is initially h = 2.7 m above the ground. Assume that the pulley
has negligible mass and friction and that the rope is very light.
Initially, you hold onto the lighter bucket to keep it from
moving.
H
a) [4 points] Draw free-body diagrams for the two buckets.
The magnitudes of the forces should be to scale.
b) [3 points] Determine the magnitude of the tension in the
rope.
You release the lighter bucket and the heavier one descends.
c) [4 points] Draw free-body diagrams for the two buckets.
The magnitudes of the forces should be to scale.
d) [7 points] Determine the magnitude of the tension in the
rope.
e) [7 points] Determine the magnitude of the momentum of
the heavier bucket the instant before it hits the ground.
L
a)
T
FH
L
T
H
WL
WH
“How much time he gains who does not look to see what his neighbor
says or does or thinks, but only at what he does himself, to make it just
and holy.” – Marcus Aurelius Antoninus (121 - 180), Meditations
h
b) From the free-body diagram of the second block and the fact that a=0, we have
T − wH = 0
1
T = wH = 170 N
T
c)
L
T
H
WL
WH
d) Use N’s 2nd law for both objects in the y-direction (+y taken to be up):
T − wL = mL aL
2
T − wH = mH aH
Then because they are tied by a string and because when one goes up, the other goes down
3
aL = − aH
Using this fact, we see we have two equations with two unknowns
T − wL = − mL aH
4
T − wH = mH aH
and so all we have to do is algebra. If I solve for aH in the second equation and plug that into the first
equation, I get
 T − wH 
T − wL = − mL 

 mH 
5
mH (T − wL ) = − mL (T − wH )
T (mH + mL ) = mH wL + mL wH
T=
mH wL + mL wH
mH + mL
Also, since we were not given the masses, we must remember:
wL
g
w
mH = H
g
mL =
6
Plugging this all in, we get T = 124 N.
e) To solve the, we have three steps. First, find acceleration. Then, find final velocity. Then find
momentum.
If we subtract the second equation in (4) from the first equation in (4), we find
wH − wL = −(mH + mL )aH
aH = −
wH − wL
= −2.7 m/s 2
mH + mL
We then can find the final velocity using the kinematics equation:
v 2fy = viy2 + 2a y ∆y
v fy = − 2aH (0 − h) = −3.8 m/s
Finally, the final momentum is p fy = mH v fy = −66 kg ⋅ m/s and no x-component.
Section 3 (work two from Sect. 1 – 3)
A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a
pendulum). You are to investigate the motion in which the string moves in a cone with half-angle θ.
In other words, the bob is moving around in a horizontal circle with constant speed.
a) [4 points] Draw a free-body diagram of the bob.
Describe each force present in your free-body
diagram. Describe the 3rd Law Pair force for each
force in your free-body diagrams.
b) [3 points] Is the bob in equilibrium? If not, in what
direction does it accelerate?
c) [6 points] Call v the tangential speed of the bob.
Write down Newton’s 2nd Law in component form for
the vertical and the radial directions.
d) [4 points] Find the tension T in the string. Express
your answer in terms of some or all of θ, L, m, and g.
e) [4 points] Find the tangential speed v in the string.
Express your answer in terms of some or all of θ, L, m,
and g.
f) [4 points] Find the period of time it takes for the mass to make a complete circle. Express your
answer in terms of some or all of θ, L, m, and g. Note: If you have not found an expression for v
in the previous part, you may include it in your answer for reduced credit.
a)
Tension, i.e. force of string on
bob
θ
T
+y
+x
force of gravity of Earth on box (weight)
reaction: force of gravity of box on
W = mg
b) No. It is accelerating toward the center of the circle in which it is traveling.
c) First, note that the distance from the bob to the center of the circle is L sin θ = R .
∑ Fx = max
T sin θ = mac =
mv 2
R
mv 2
T sin θ =
L sin θ
∑ Fy = ma y
T cos θ − mg = 0
d) Based on the y-component equation
mg
cos θ
e) Substituing this into the x-component equation, we get
mg
mv 2
sin θ =
cos θ
L sin θ
2
gL sin θ
v2 =
cos θ
T=
v=
gL sin 2 θ
cos θ
f) The circumference of the circle is ∆s = 2π R = 2π L sin θ . Call the period ∆t . We know
∆s
v=
∆t
so therefore,
∆s
2π L sin θ
L cos θ
∆t =
=
= 2π
v
g
gL sin 2 θ
cos θ
Essay Section (Required)
[ 25 points ] A Bulgarian Olympic weightlifter of mass M
stands on a scale. In his event, he must lift a barbell of mass
m over his head. The barbell begins at rest on the floor. The
weightlifter squats to pick it up, lifts the barbell, and then
stands up straight with the barbell waist high. After a pause,
he then lifts the barbell overhead and holds it steady for ∆t
seconds before dropping the barbell to the floor. Make a
graph of the reading on the scale versus time, marking specific times during the lifting process.
Defend and justify this graph. Draw separate free-body-diagrams for the weightlifter and the barbell
at different times to help justify your argument. Make sure you communicate the assumptions you
make about the speed/acceleration of the barbell in the lifting process and how they affect your
answers.
Normal force
of scale on
weightlifter
The Squat:
As the weightlifter squats down, the scale fluctuates. While standing still, the
scale reads Mg. When he begins to accelerate downward during the squat, the
scale briefly ticks to a number less than Mg since the net force on the
weightlifter is downward, the normal force (and hence, the reading on the
scale) will be less than Mg. As he stops in a squat, the scale will briefly tick
upward since to stop his downward motion, he needs a net force upward,
meaning the normal force (and hence, the reading on the scale) will be greater
than Mg. When he comes to rest, the reading will again be Mg.
Weightlifter
Weight of
weightlifter
Contact force
of weightlifter
on barbell
Grabbing the barbell, standing up, barbell waist high:
If the weightlifter grabs the barbell with it resting on the floor, the reading on
Barbell
the scale will not change. As he begins to stand, the reading will tick upward
to a number greater than (M+m)g since the net force must be upward to
Weight of
change the velocity from 0 to some value upward. If he moves the barbell at
barbell
constant speed most of the way to waist high, the reading on the scale will
Normal force
equal (M+m)g since the net force must be zero. Finally, as the barbell comes
of scale on
to rest waist high, the net force on the barbell must be downward, meaning the
weightlifter
contact force of the man on the barbell will be less that its weight. Similarly,
the man must have a net downward force as well to stop his upward motion,
Weightlifter
so the normal force upward will be less than the contact force of the barbell
plus his weight. The scale will read something less than (M+m)g briefly. Contact force
Weight of
of barbell on
Once all motion has stopped, the reading will be (M+m)g.
weightlifter
weightlifter
Raising the barbell over his head:
The process is exactly the same as described in the process above for getting
the barbell waist high off the floor.
Dropping the barbell:
The weightlifter experiences no acceleration during this process, but the
contact force of the barbell on him disappears. Therefore, the reading on the
scale reverts to just Mg. If the scale is an analog spring scale, there may be
some brief oscillation about Mg.
Normal force
of scale on
weightlifter
Weightlifter
“America's present need is not heroics, but healing; not nostrums
but normalcy; not revolution, but restoration.”
– Warren G. Harding (1865 - 1923), Speech in Boston, 1920
Weight of
weightlifter
Weightlifter (con’t)
Here’s a sketch of the graph:
(M+m)g
Mg
t1
t2
t3
t4
t5
t6
t7
From time 0 until t1, the weightlifter stands still. At t1, he begins to squat, causing the
scale to tick to a slightly lower number.
At t2, he slows his downward motion, causing the scale to briefly tick upward to a slightly
higher number. He grabs the barbell, and the scale reads simply his weight.
At t3, he picks up the barbell and begins to stand. The scale briefly registers a number
greater than the sum of the weights of the barbell and him since they must accelerate
upward from rest. As he proceeds at constant velocity, the scale reads just the sum of the
weights.
At t4, the upward motion ceases, indicating a downward acceleration, resulting in a slight
downward tick on scale to a number less than the sum of their weights.
At t5, the barbell accelerates upward, resulting in a slight upward tick on the scale. If the
barbell then proceeds at constant speed, the scale returns to the reading of the sum of the
weights of the barbell and weightlifter, until at time t6 the barbell must slow down,
meaning a slight downward tick on the scale.
The weightlifter holds the barbell steady for Dt, which must be the time between t6 and t7.
At time t7, he drops the barbell, resulting in the scale returning to a reading of just his own
weight.
Multiple Choice Section (Required)
Each question in this section is worth 5 points. Please record your answers in the blanks on this
page. There will be NO partial credit on this section of the test. In each case, choose the BEST
possible answer from the choices provided.
1) ___C____ 2) ___E____ 3) ___D____ 4) ___C____ 5) ___D____
1) A constant force is exerted for a short time interval on a cart that is initially at rest (v = 0, a = 0) on
an air track. This force gives the cart a certain final speed. The same force is exerted for the same
length of time on another cart, also initially at rest (v = 0, a = 0) that has twice the mass of the first
one. The final momentum of the heavier cart is________________ that of the lighter cart.
a)
b)
c)
d)
e)
one-fourth
half
the same as
double
four times
2) A horizontal force of 15 N acts on a block A (mass = 3 kg) which sits next to and is in contact with
block B (mass = 2 kg). The blocks can slide freely on a horizontal frictionless table. What are the
magnitude of the acceleration (a) of the two blocks and the magnitude of the contact force (F) that A
exerts on B?
a)
b)
c)
d)
e)
a = 5 m/s/s; F = 2 N
a = 5 m/s/s; F = 15 N
a = 3 m/s/s; F = 15 N
a = 3 m/s/s; F = 9 N
a = 3 m/s/s; F = 6 N
A
B
3) The four situations below show before and after “snapshots” of a car’s velocity. The time interval
between the snapshots is the same in each situation. All cars have the same mass. “Before” is
depicted on the left and “after” on the right. Rank these four situations in terms of the impulse on
these cars from most positive to most negative.
I
II
+ 10 m/s
+20 m/s
III
+ 10 m/s
–10 m/s
IV
+ 30 m/s
a)
b)
c)
d)
e)
+10 m/s
–30 m/s
0
I > III > IV > II
II = III > I > IV
IV > III > I > II
IV > I > II = III
IV > II = III > I
“It is the mark of an educated mind to be able to entertain a
thought without accepting it.” – Aristotle
4) A piano mover raises a 400-kg piano at a constant rate
using the frictionless pulley system shown here. With how
much force is he pulling on the rope? Ignore friction. Use
g = 9.8 m/s2.
a)
b)
c)
d)
e)
7840 N
3920 N
1960 N
400 N
not enough information to determine
400 kg
5) A constant force F acts on a mass m which can move without friction. The x-component of F is
not zero. The mass is at rest at the origin of our coordinate system (x = 0, y = 0) at time t = 0. Which
of the graphs below is a possible graph of x-coordinate of the mass versus time for this experiment?
x
x
t
x
x
t
t
x
t
t
“Destiny is no matter of chance. It is a matter of choice. It is not a thing
to be waited for, it is a thing to be achieved.” – William Jennings Bryan