Download Midterm Exam 2

Physics 130
General Physics
Fall 2012
Midterm Exam 2
October 30, 2012
Name:
Instructions
1. This examination is closed book and closed notes. All your belongings
except a pen or pencil and a calculator should be put away and your
bookbag should be placed on the floor.
2. You will find one page of useful formulae on the last page of the exam.
3. Please show all your work in the space provided on each page. If you
need more space, feel free to use the back side of each page.
4. Academic dishonesty (i.e., copying or cheating in any way) will
result in a zero for the exam, and may cause you to fail the
class.
In order to receive maximum credit,
each solution should have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
A labeled picture or diagram, if appropriate.
A list of given variables.
A list of the unknown quantities (i.e., what you are being asked to find).
A force-interaction diagram, if appropriate.
One or more free-body diagrams, as appropriate, with labeled 1D or 2D
coordinate axes.
Algebraic expression for the net force along each dimension, as appropriate.
An algebraic solution of the unknown variables in terms of the
known variables.
A final numerical solution, including units, with a box around it.
An answer to additional questions posed in the problem, if any.
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Physics 130
General Physics
Fall 2012
1. A heavy box is in the back of a truck. The truck is accelerating to the right. Draw a
motion diagram, a force-interaction diagram, and a free-body diagram for the box.
Solution:
You can see from the motion diagram that the box accelerates to the right along with
the truck. According to Newton’s second law, F~ “ m~a, there must be a force to the
right acting on the box. This is friction, but not kinetic friction, because the box
is not sliding against the truck. Instead, it is static friction, the force that prevents
slipping. Were it not for static friction, the box would slip off the back of the truck.
Static friction acts in the direction needed to prevent slipping. In this case, friction
must act in the forward (toward the right) direction.
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Physics 130
General Physics
Fall 2012
2. A bag of groceries is on the seat of your car as you stop at a stop light. The bag does
not slide. Draw a motion diagram, a force-interaction diagram, and a free-body diagram
for the bag.
Solution:
You can see from the motion diagram that the bag accelerates to the left along with
the car as the car slows down. According to Newton’s second law, F~ “ m~a, there
must be a force to the left acting on the bag. This force must be static friction, the
force that prevents slipping, because the bag is not sliding across the seat. Were it
not for static friction, the bag would slide off the seat as the car stops. Static friction
acts in the direction needed to prevent slipping, and in this case, friction must act
in the backward (toward the left) direction.
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Physics 130
General Physics
Fall 2012
3. A football coach sits on a sled while two of his players build their strength by dragging
the sled across the field with ropes. Assume that the two players pull with equal strength.
The friction force on the sled is 1000 N and the angle between the two ropes is 20˝ . How
hard must each player pull to drag the coach at a steady 2.0 m{s?
Solution:
The pictorial representation and free-body diagram are shown below:
This is a 1D dynamics problem. The relevant forces are the force of kinetic friction
on the sled and the tension of each rope. Although there is also gravity, we are given
the friction force, and therefore we do not need to know the weight of the sled.
Since the sled is not accelerating, it is in dynamic equilibrium along both axes, and
Newton’s first law applies:
pFnet qy “
pFnet qx “
ÿ
ÿ
Fy “ T1y ` T2y ` fky “ 0 N
(1)
Fx “ T1x ` T2x ` fkx “ 0 N
(2)
Examining the free-body diagram, along the y-axis we have
T1y “ `T1 sinpθ{2q
T2y “ ´T2 sinpθ{2q
fky “ 0.
(3)
(4)
(5)
Substituting into equation (1), we get
T1 sinpθ{2q ´ T2 sinpθ{2q “ 0
ñ
T1 “ T2 “ T.
4
(6)
(7)
Physics 130
General Physics
Fall 2012
In other words, the tension force of each rope is the same.
From the free-body diagram along the x-axis we have
T1x “ T1 cospθ{2q “ T cospθ{2q
T2x “ T2 cospθ{2q “ T cospθ{2q.
(8)
(9)
Substituting into equation (2) and solving for T we get
T cospθ{2q ` T cospθ{2q ´ fkx “ 0
2T cospθ{2q “ fkx
fkx
ñT “
2 cospθ{2q
1000 N
T “
2 ˆ cosp20˝ {2q
“ 508 N « 510 N.
(10)
(11)
(12)
(13)
(14)
Note that in the last line we simplified our result to have just two significant figures.
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Physics 130
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4. You’re driving along at 25 m{s with your aunt’s valuable antiques in the back of your
pickup truck when suddenly you see a giant hole in the road 55 m ahead of you. Fortunately, your foot is right beside the brake and your reaction time is zero! Will the
antiques be as fortunate? Assume that the coefficient of kinetic friction for rubber (i.e.,
a tire) on concrete is 0.80.
(a) Can you stop the truck before it falls in the hole?
(b) If your answer to part (a) is yes, can you stop without the antiques sliding and
being damaged? Their coefficients of friction are µs “ 0.60 and µk “ 0.30.
Solution:
To solve this problem we will treat the antiques (mass “ m) in the back of the pickup
(mass “ M ) both as particles. The antiques touch the truck’s steel bed, so only the
steel bed can exert contact forces on the antiques. The pickup-antiques system will
also be treated as a particle, and the contact force on this particle will be due to the
road.
The pictorial representation, motion diagram, force-interaction diagram, and freebody diagram for the combined pickup-antiques system are shown below:
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Physics 130
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Fall 2012
The free-body diagram of just the box of antiques is:
(a) Our strategy for solving this problem is to find the smallest coefficient of friction
that will allow the truck to stop in 55 m, then compare that to the known coefficients
for rubber on concrete. For the pickup-antiques system, with mass m ` M , Newton’s
second law is
pFnet qx “
pFnet qy “
ÿ
ÿ
Fx “ ´f “ pm ` M qax “ pm ` M qa
(1)
Fy “ N ´ pFG qP A “ 0
(2)
“ N ´ pm ` M qg “ 0
ñ N “ pm ` M qg.
(3)
(4)
The model of static friction is f “ µN , where µ is the coefficient of friction between
the tires and the road. Substituting into the equations above, we get
´f “ ´µN “ ´µpm ` M qg “ pm ` M qa
ñ a “ ´µg.
(5)
(6)
The mass of the pickup-antiques system drops out! Next, from the constant-acceleration
kinematics equation with v12 “ 0 and x0 “ 0, we have
v12 “ v02 ` 2apx1 ` x0 q
v2
ña “ ´ 0
2x1
v02
ñ µmin “
2gx1
p25 m{sq2
“
2 ˆ p9.8 m{s2 q ˆ p55 mq
“ 0.58.
7
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
Since this is smaller than the coefficient of static friction of rubber on concrete, 0.80,
the truck can stop.
(b) The analysis of the pickup-antiques system applies to the antiques as well, and
gives the same value of µmin “ 0.58. This value is smaller than the given coefficient
of static friction (µs “ 0.60) between the antiques and the truck bed. Therefore, the
antiques will not slide as the truck stops over a distance of 55 m.
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Physics 130
General Physics
Fall 2012
5. You and your friend Peter are putting new shingles on a roof pitched at 25˝ . You’re
sitting on the very top of the roof when Peter, who is at the edge of the roof directly
below you, 5.0 m away, asks you for the box of nails. Rather than carry the 2.5 kg box
of nails down to Peter, you decide to give the box a push and have it slide down to him.
If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed
should you push the box to have it gently come to rest right at the edge of the roof?
Solution:
The pictorial representation and free-body diagram are shown below:
This is a 1D dynamics problem. The relevant forces are gravity, FG , the normal
force, n, and the kinetic friction force, fk . Note that we do not include the initial
force that was applied to the box of nails to get it moving, but we do include the
fact that the box has some initial velocity.
The most natural coordinate system is one that is rotated by 25˝ and therefore
aligned with the roof. The interaction and free-body diagrams are shown above.
The net force along the x- and y-axis is
pFnet qx “
pFnet qy “
ÿ
ÿ
Fx “ FG sin 25˝ ´ fk “ ma
(1)
Fy “ n ´ FG cos 25˝ “ 0
(2)
ñ n “ FG cos 25˝
(3)
In the second line we used the fact that the shingles are not leaping off the roof to
set the acceleration in the y-direction equal to zero. The magnitude of the force of
gravity is FG “ mg, and the magnitude of the kinetic force of friction is
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Physics 130
General Physics
fk “ µk n
“ µk FG cos 25˝
“ µk mg cos 25˝ .
Fall 2012
(4)
(5)
(6)
Substituting equation (6) into equation (1) and solving for the acceleration a we get
mg sin 25˝ ´ µk mg cos 25˝ “ ma
ñ a “ psin 25˝ ´ µk cos 25˝ qg
“ psin 25˝ ´ 0.55 ˆ cos 25˝ q ˆ 9.8 m{s2
“ ´0.743 m{s2
(7)
(8)
(9)
(10)
where the minus sign indicates the acceleration is directed up the incline. To find the
initial speed, vi , necessary to have the box come to rest (i.e., vf “ 0) after ∆x “ 5.0 m
can be found from the kinematic equation linking velocity and acceleration:
vf2 “ vi2 ` 2a∆x
?
´2a∆x
ñ vi “
b
“
´2p´0.743 m{s2 qp5.0 mq
“ 2.7 m{s.
10
(11)
(12)
(13)
(14)
Physics 130
General Physics
Fall 2012
6. A 1.0 kg wood block is pressed against a vertical wood wall by the 12 N force shown in
the figure below. If the block is initially at rest, will it move upward, move downward, or
stay at rest? Assume that the coefficient of static friction for wood on wood is µs “ 0.5.
Solution:
The block is initially at rest, so initially the friction force is static friction. If the
force of the push is too strong, the box will begin to move up the wall. If it is too
weak, the box will begin to slide down the wall. And if the pushing force is within
the proper range, the box will remain stuck in place.
From the free-body diagram, the block is in static equilibrium along the x-axis.
Therefore, according to Newton’s first law the net force is
pFnet qx “
ÿ
Fx “ n ´ Fpush cos θ “ 0
ñ n “ Fpush cos θ.
(1)
(2)
From our model for friction, the maximum static friction force is
pfs qmax “ µs n “ µs Fpush cos θ
“ 0.5 ˆ p12 Nq ˆ cosp30˝ q
“ 5.2 N.
(3)
(4)
(5)
To determine if the block stays in place or moves up or down, we need to determine
if the net force along the y-axis, pFnet qy , is greater or smaller than pfs qmax . The net
force excluding the force of friction is
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Physics 130
General Physics
pFnet qy “
ÿ
Fy “ Fpush sin θ ´ FG
“ Fpush sin θ ´ mg
“ p12 Nq ˆ sinp30˝ q ´ p1.0 kgq ˆ p9.8 m{s2 q
“ ´3.8 N.
Fall 2012
(6)
(7)
(8)
(9)
In other words, a static friction force fs “ `3.8 N would prevent the block from
moving. Since this force is smaller than pfs qmax the box stays in place and does not
move.
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Physics 130
General Physics
Fall 2012
7. The 1.0 kg block in the figure below is tied to the wall with a rope. It sits on top of the
2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The
coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is
µk “ 0.40.
(a) What is the tension in the rope holding the 1.0 kg block to the wall?
(b) What is the acceleration of the 2.0 kg block?
Solution:
The free-body diagram for the problem is shown below:
To solve this problem we need to use both Newton’s third and second laws. The separate free-body diagrams for the two blocks show that there are two action/reaction
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Fall 2012
pairs. Notice how the top block (block 1) both pushes down on the bottom block
(block 2) with force ~n11 , and exerts a retarding friction force f~2,top on the top surface
of block 2.
(a) Block 1 is in static equilibrium (a1 “ 0 m{s2 ), but block 2 is accelerating to the
right. Newton’s second law for block 1 is
pFnet
pFnet
on 1 qx
on 1 qy
“ f1 ´ Trope “ 0 ñ Trope “ f1
“ n1 ´ m1 g “ 0 ñ n1 “ m1 g.
(1)
(2)
Although block 1 is stationary, there is a kinetic force of friction because there is
motion between block 1 and block 2. The friction model means
f1 “ µk n1 “ µk m1 g.
(3)
Substituting this result into equation (1) we get the tension of the rope:
Trope “ f1 “ µk m1 g
“ p0.40q ˆ p1.0 kgq ˆ p9.8 m{s2 q
“ 3.9 N.
(4)
(5)
(6)
(b) Newton’s second law for block 2 is
pFnet on 2 qx
Tpull ´ f2 top ´ f2 bot
“
m2
m2
1
n2 ´ n1 ´ m 2 g
pFnet on 2 qy
“
“0
“
m2
m2
ax ” a “
(7)
ay
(8)
Forces ~n1 and ~n11 are an action/reaction pair, so ~n11 “ ~n1 “ m1 g. Substituting into
equation (8) gives
n2 “ pm1 ` m2 qg.
(9)
This result is not surprising because the combined weight of both objects presses
down on the surface. The kinetic friction on the bottom surface of block 2 is then
f2
bot
“ µk n2 “ µk pm1 ` m2 qg.
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(10)
Physics 130
General Physics
Next, we recognize that the forces f~1 and f~2
f2
top
top
Fall 2012
are an action/reaction pair, so
“ f1 “ µk m1 g.
(11)
Finally inserting these friction results into equation (7) gives
Tpull ´ µk m1 g ´ µk pm1 ` m2 qg
(12)
m2
20 N ´ p0.40qp1.0 kgqp9.8 m{s2 q ´ p0.40qp1.0 kg ` 2.0 kgqp9.8 m{s2 q
“
(13)
2.0 kg
“ 2.2 m{s2 .
(14)
a “
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Physics 130
General Physics
Fall 2012
8. The lower block in the figure below is pulled on by a rope with a tension force of 20 N.
The coefficient of kinetic friction between the lower block and the surface is 0.30. The
coefficient of kinetic friction between the lower block and the upper block is also 0.30.
What is the acceleration of the 2.0 kg block?
Solution:
The pictorial representation and free-body diagrams are shown below:
The blocks accelerate with the same magnitude but in opposite directions. Thus the
acceleration constraint is a2 “ a “ ´a1 , where a will have a positive value because
of how we have defined the `x direction. There are two real action/reaction pairs.
The two tension forces will act as if they are action/reaction pairs because we are
assuming a massless rope and a frictionless pulley.
Make sure you understand why the friction forces point in the directions shown in
the free-body diagrams, especially force f~11 exerted on the bottom block (block 2) by
the top block (block 1).
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Physics 130
General Physics
Fall 2012
We have quite a few pieces of information to include. First, Newton’s second law
applied to block 1 gives
pFnet
on 1 qx
pFnet
on 1 qy
“
“
“
ñ
f1 ´ T1
µk n1 ´ T1 “ m1 a1 “ ´m1 a
n1 ´ m 1 g “ 0
n1 “ m1 g.
(1)
(2)
(3)
(4)
Tpull ´ f11 ´ f2 ´ T2
Tpull ´ f11 ´ µk n2 ´ T2 “ m2 a2 “ m2 a
n2 ´ n11 ´ m2 g “ 0
n2 “ n11 ` m2 g.
(5)
(6)
(7)
(8)
For block 2 we have
pFnet
on 2 qx
pFnet
on 2 qy
“
“
“
ñ
Note that to simplify the two x-equations above we have already used the kinetic
friction model. Next, from Newton’s third law we have three additional constraints:
n11 “ n1 “ m1 g
f11 “ f1 “ µk n1 “ µk m1 g
T1 “ T2 “ T.
(9)
(10)
(11)
Knowing n11 we can now use the y-equation for block 2 to find n2 . Substituting
all these pieces into the two x-equations, we end up with two equations with two
unknowns:
µk m1 g ´ T “ ´m1 a
Tpull ´ T ´ µk m1 g ´ µk pm1 ` m2 qg “ m2 a.
(12)
(13)
Subtracting the first equation from the second we get
Tpull ´ T ´ µk m1 g ´ µk pm1 ` m2 qg ´ µk m1 g ` T “ m2 a ` m1 a
Tpull ´ 3µk m1 g ´ µk m2 g “ pm2 ` m1 qa
Tpull ´ µk p3m1 ` m2 qg “ pm2 ` m1 qa.
(14)
(15)
(16)
And finally solving for a we get
Tpull ´ µk p3m1 ` m2 qg
m1 ` m2
20 N ´ p0.30qp3 ˆ 1.0 kg ` 2.0 kgqp9.8 m{s2 q
“
1.0 kg ` 2.0 kg
2
“ 1.8 m{s .
ña “
17
(17)
(18)
(19)
Physics 130
General Physics
Fall 2012
9. The coefficient of kinetic friction between the 2.0 kg block and the table in the figure
below is 0.30. What is the acceleration of the 2.0 kg block?
Solution:
The pictorial representation and free-body diagrams are shown below:
Newton’s second law for m1 and m3 gives
T1 ´ pFG q1 “ T1 ´ m1 g “ m1 a1
T2 ´ pFG q3 “ T2 ´ m3 g “ m3 a3 .
For m2 Newton’s second law gives
18
(1)
(2)
Physics 130
General Physics
ÿ
pFon
m2 qy
ÿ
pFon
m2 qx
Fall 2012
“ n2 ´ pFG q2 “ n2 ´ m2 g “ 0
(3)
“ T2 ´ fk2 ´ T1
(5)
ñ n2 “ m 2 g
(4)
“ T 2 ´ µ k n2 ´ T 1
“ T 2 ´ µ k m 2 g ´ T 1 “ m 2 a2 .
(6)
(7)
Since all three blocks move together, the acceleration constraint gives
a1 “ a2 “ ´a3 ” a.
(8)
Therefore, the equations for the three masses are:
T1 ´ m1 g “ m1 a
T 2 ´ µk m 2 g ´ T 1 “ m 2 a
T2 ´ m3 g “ ´m3 a.
(9)
(10)
(11)
Subtracting the third equation from the sum of the first two equations yields:
pT1 ´ m1 gq ` pT2 ´ µk m2 g ´ T1 q ´ pT2 ´ m3 gq “ m1 a ` m2 a ` m3 a
p´m1 ´ µk m2 ` m3 qg “ apm1 ` m2 ` m3 q
´m1 ´ µk m2 ` m3 q
g
m1 ` m2 ` m3
´1.0 kg ´ 0.30 ˆ 2.0 kg ` 3.0 kg
ˆ p9.8 m{s2 q
“
1.0 kg ` 2.0 kg ` 3.0 kg
“ 2.3 m{s2 .
ña “
19
(12)
(13)
(14)
(15)
(16)
Physics 130
General Physics
Fall 2012
10. The 1.0 kg physics book in the figure below is connected by a string to a 500 g coffee
cup. The book is given a push up the slope and released with a speed of 3.0 m{s. The
coefficients of friction are µs “ 0.50 and µk “ 0.20.
(a) How far does the book slide?
(b) At the highest point, does the book stick to the slope, or does it slide back down?
Solution:
The pictorial representation and free-body diagrams are shown below:
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Physics 130
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Fall 2012
To solve this problem we use the particle model for the book (B) and the coffee cup
(C), the models of kinetic and static friction, and the constant-acceleration kinematic
equations.
(a) To find the distance x1 the book slides we must find its acceleration. Newton’s
second law applied to the book gives
ÿ
ÿ
pFon B qy
“ nB ´ pFG qB cos θ “ 0
(1)
ñ nB “ mB g cos θ
(2)
“ ´T ´ µk nB ´ mB g sin θ
“ ´T ´ µk mB g cos θ ´ mB g sin θ “ mB aB .
(4)
(5)
pFon B qx “ ´T ´ fk ´ pFG qB sin θ
(3)
Similarly, for the coffee cup we have
ÿ
pFon C qy “ T ´ pFG qC “ T ´ mC g “ mC aC .
(6)
Equations (5) and (6) can be rewritten using the acceleration constrain aC “ aB “ a
as
´T ´ µk mB g cos θ ´ mB g sin θ “ mB a
T ´ mC g “ mC a.
(7)
(8)
Adding these two equations and solving for the acceleration gives
´µk mB g cos θ ´ mB g sin θ ´ mC g “ pmB ` mC qa
„

mB pµk cos θ ` sin θq ` mC
ña “ ´
g
mB ` mC

„
p1.0 kgq ˆ p0.20 cos 20˝ ` sin 20˝ q ` 0.5 kg
ˆ p9.8 m{s2 q
“ ´
1.0 kg ` 0.5 kg
“ ´6.73 m{s2 .
(9)
(10)
(11)
(12)
Finally, to solve for the distance x1 we use the following kinematic equation with
v1x “ 0, v0x “ 3.0 m{s, and x0 “ 0:
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Physics 130
General Physics
2
2
v1x
“ v0x
` 2apx1 ´ x0 q
2
0 “ v0x ` 2ax1
v2
ñ x1 “ ´ 0x
2a
´p3.0 m{sq2
“
2 ˆ p´6.73 m{s2 q
“ 0.67 m.
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(13)
(14)
(15)
(16)
(17)
(b) In order to figure out if the book sticks to the slope or slides back down we have
to determine if the static friction force needed to keep the book in place, fs is larger
or smaller than the maximum static friction force
pfs qmax “ µs nB “ µs mB gcosθ
“ p0.50q ˆ p9.8 m{s2 q ˆ cos 20˝
“ 4.60 N.
(18)
(19)
(20)
When the cup is at rest, the string tension is T “ mC g. In this case, Newton’s first
law for the book becomes
ÿ
pFon B qx “ fs ´ T ´ mB g sin θ
ñ fs
“
“
“
“
fs ´ mC g ´ mB g sin θ “ 0
pmC ` mB sin θqg
p0.5 kg ` 1.0 kg sin 20˝ q ˆ p9.8 m{s2 q
8.25 N.
Because fs ą pfs qmax , the book slides back down.
22
(21)
(22)
(23)
(24)
(25)
Physics 130
General Physics
Fall 2012
11. A house painter uses the chair and pully arrangement shown below to lift himself up the
side of a house. The painter’s mass is 70 kg and the chair’s mass is 10 kg. With what
force must he pull down on the rope in order to accelerate upward at 0.20 m{s2 ?
Solution:
The free-body diagram for the problem is shown below:
If the painter pulls down on the rope with force F , Newton’s third law requires the
rope to pull up on the painter with force F , which is just the tension in the rope.
Based on the picture, the same tension force F also pulls up on the painter’s chair.
Newton’s second law for the painter and the chair gives
2F ´ FG “ pmP ` mC qa
2F ´ pmP ` mC qg “ pmP ` mC qa
pmP ` mC qa ` pmP ` mC qg
ñF “
2
pmP ` mC qpa ` gq
“
2
p70 kg ` 10 kgq ˆ p0.20 m{s2 ` 9.8 m{s2 q
“
2
“ 400 N.
23
(1)
(2)
(3)
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
12. Based on the figure below, find an expression for the acceleration of m1 . Assume the
table is frictionless. Hint: Think carefully about the acceleration constraint.
Solution:
The pictorial representation and free-body diagram for the problem are shown below:
For every meter that block 1 moves forward, one meter is provided to block 2. In
other words, each rope on m2 has to be lengthened by one-half meter. Therefore,
the acceleration constraint is
1
a2 “ ´ a1 .
2
(1)
Newton’s second law applied to block 1 and block 2, respectively, gives
T “ m 1 a1
2T ´ pFG q2
m 2 a1
“ m 2 a2 “ ´
.
2
24
(2)
(3)
Physics 130
General Physics
Fall 2012
Combining these two equations gives
m 2 a1
2
´m2 a1
2m2 g
2m2 g
2m2 g
p4m1 ` m2 q
2m1 a1 ´ m2 g “ ´
4m1 a1 ´ 2m2 g “
4m1 a1 ` m2 a1 “
a1 p4m1 ` m2 q “
ñ a1 “
(4)
(5)
(6)
(7)
(8)
Note: if m1 “ 0 then a2 “ ´g, which is what we would expect for a free-falling
object.
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General Physics
Fall 2012
13. What is the acceleration of the 2.0 kg block shown in the figure below across the frictionless table? Hint: Think carefully about the acceleration constraint.
Solution:
The pictorial representation and free-body diagram for the problem are shown below:
For every one meter that block 1 goes down, each rope on block 2 will be shortened
by one-half meter. Therefore, the acceleration constraint is
a1 “ ´2a2 .
(1)
Newton’s second law applied to block one gives
2T “ m2 a2
m 2 a2
.
ñT “
2
(2)
(3)
Similarly, applying Newton’s second law to block two gives
T ´ pFG q1 “ m1 a1 “ ´2m1 a2
T ´ m1 g “ ´2m1 a2
m 2 a2
´ m1 g “ ´2m1 a2
2
26
(4)
(5)
(6)
Physics 130
General Physics
m2 a2 ` 4m1 a2 “ 2m1 g
a2 pm2 ` 4m1 q “ 2m1 g
2m1 g
ñ a2 “
m2 ` 4m1
2 ˆ p1.0 kgq ˆ p9.8 m{s2 q
“
p2.0 kg ` 4 ˆ 1.0 kgq
“ 3.3 m{s2 .
Note that if m1 “ 0 then a2 “ 0, which is what we would expect.
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Fall 2012
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
14. As a science fair project, you want to launch an 800 g model rocket straight up and hit a
horizontally moving target as it passes 30 m above the launch point. The rocket engine
provides a constant thrust of 15 N. The target is approaching at a speed of 15 m/s. At
what horizontal distance between the target and the rocket should you launch?
Solution:
The pictorial representation and motion diagram for the problem are shown below:
This is a 2D dynamics/kinematics problem. For the rocket, Newton’s second law
along the y-direction is
pFnet qy “ FR ´ mg “ maR
1
ñ aR “
pFR ´ mgq
m
1
r15 N ´ p0.8 kgq ˆ p9.8 m{s2 qs
“
0.8 kg
“ 8.95 m{s2 .
(1)
(2)
(3)
(4)
Next, we use the kinematic equation for the rocket to find the time, noting that
y0R “ pv0R qy “ t0R “ 0:
1
y1R “ y0R ` pv0R qy pt1R ´ t0R q ` aR pt1R ´ t0R q2
2
1
“
aR t21R
2
c
2y1R
ñ t1R “
a
d R
2 ˆ 30 m
“
8.95 m{s2
“ 2.589 s.
(5)
(6)
(7)
(8)
(9)
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Physics 130
General Physics
Fall 2012
For the target, we use the same kinematic equation along the x-direction, noting
that t1T “ t1R (the times are the same for both the rocket and the target) and that
x0T “ aT “ 0:
1
x1T “ x0T ` pv0T qx pt1T ´ t0T q ` aT pt1T ´ t0T q2
2
“ pv0T qx t1T
“ p15 m{sq ˆ p2.589 sq
“ 39 m.
In other words, the rocket should be launched when the target is 39 m away.
29
(10)
(11)
(12)
(13)
Physics 130
General Physics
Fall 2012
15. A conical pendulum is formed by attaching a 500 g ball to a 1.0 m-long string, then
allowing the mass to move in a horizontal circle of radius 20 cm. The figure shows that
the string traces out the surface of a cone, hence the name.
(a) What is the tension in the string?
(b) What is the ball’s angular speed, in rpm?
Solution:
(a) What is the tension in the string?
The forces in the z-direction in the free body diagram are the component of the
tension in the z-direction and the force of gravity. To find the z-component of
the tension,
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Physics 130
General Physics
Fall 2012
Tz
adjacent
“
hypotenuse
T
“ T cos θ
cos θ “
Tz
Apply Newton’s second law in the z-direction
ÿ
Fz “ T cos θ ´ mg “ 0
mg
T “
cos θ
(1)
(2)
(3)
(4)
To calculate cos θ,
adjacent
“
cos θ “
hypotenuse
ñ θ “ 11.5˝
?
L2 ´ r 2
“
L
a
p1mq2 ´ p0.2mq2
“ 0.98
1.0m
(5)
(6)
Plugging this in for tension,
mg
p0.500 kgq ˆ p9.80 m{s2 q
“
cos θ
0.98
T “ 5.0 N
T “
(7)
(8)
(b) What is the ball’s angular speed, in rpm?
Referring back to the tension triangle, the radial component of tension is
opposite
Tr
“
hypotenuse
T
“ T sin θ
sin θ “
Tr
31
(9)
(10)
Physics 130
General Physics
Fall 2012
Referring back to the triangle for the length of the pendulum,
r
opposite
“
hypotenuse
L
r “ L sin θ
sin θ “
Apply Newton’s second law in the r-direction
ÿ
Fr “ Tr “ T sin θ “ mω 2 r
c
T sin θ
ñω “
d mr
“
5.0 N ˆ sin 11.5˝
p0.500 kgq ˆ p0.2 mq
“ 3.16 rad{sec
ˆ
˙ ˆ
˙
rad
1 rev
60 sec
“ 3.16
ˆ
ˆ
sec
2π rad
1 min
“ 30 rpm
32
(11)
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(13)
(14)
(15)
(16)
(17)
(18)
Physics 130
General Physics
Fall 2012
16. In an old-fashioned amusement park ride, passengers stand inside a 5.0 m diameter hollow
steel cylinder with their backs against the wall. The cylinder begins to rotate about a
vertical axis. Then the floor on which the passengers are standing suddenly drops away!
If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a
static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient
in the range 0.40 to 0.70. A sign next to the entrance says, “No children under 30 kg
allowed.” What is the minimum angular speed, in rpm, for which the ride is safe?
Solution:
The passengers stick to the wall if the static friction force is sufficient to support the
gravitational force on them: fs “ Fg . The minimum angular velocity occurs when
static friction reaches its maximum possible value pfs qmax “ µs n. Although clothing
has a range of coefficients of friction, it is the clothing with the smallest coefficient
(µs “ 0.60q that will slip first, so this is the case we need to examine. Assuming that
the person is stuck to the wall, Newton’s second law gives
ÿ
ÿ
Fr “ n “ mω 2 r
F z “ fs ´ F g “ 0
(1)
ñ fs “ mg
(2)
The minimum angular speed occurs when
2
fs “ pfs qmax “ µs n “ µs mrωmin
(3)
Substituting this expression into the z-equation gives
2
fs “ µs mrωmin
“ mg
33
(4)
Physics 130
General Physics
Fall 2012
Solving for ωmin
ωmin “
c
g
µs r
(5)
d
9.80 m{s2
0.60 ˆ p2.5 mq
˙ ˆ
˙
ˆ
60 sec
1 rev
ˆ
“ 2.56 rad{s ˆ
2π rad
1 min
“ 24 rpm
“
Note that the angular velocity does not depend on the mass of the individual.
34
(6)
(7)
(8)
Physics 130
General Physics
Fall 2012
17. You have a new job designing rides for an amusement park. In one ride, the rider’s chair
is attached by a 9.0 m long chain to the top of a tall rotating tower. The tower spins the
chair and rider around at the rate of 1.0 rev every 4.0 s. In your design, you’ve assumed
that the maximum possible combined weight of the chair and rider is 150 kg. You’ve
found a great price for chain at the local discount store, but your supervisor wonders if
the chain is strong enough. You contact the manufacturer and learn that the chain is
rated to withstand a tension of 3000 N. Will this chain be strong enough for the ride?
Solution:
For circular motion, the force is to the center of the circle. The only force in the rdirection in the free body diagram is the component of the tension in the r-direction.
To find the r-component,
opposite
Tr
“
hypotenuse
T
“ T sin θ
sin θ “
Tr
35
(1)
(2)
Physics 130
General Physics
Fall 2012
Newton’s second law along the r-axis is
ÿ
Fr “ Tr “ mar “ m
T sin θ “ mω 2 r
v2
“ mω 2 r
r
(3)
(4)
The radius, r, is a component of the length of the chain, L.
r
opposite
“
adjacent
L
r “ L sin θ
sin θ “
(5)
(6)
Plugging this in, we find
T sin θ “ mω 2 L sin θ
T “ mω 2 L
(7)
(8)
„
2π
T “ p150 kgq ˆ
4.0 sec
T “ 3300 N
2
ˆ p9.0 mq
Thus, the 3000 N chain is not strong enough for the ride.
36
(9)
(10)
Physics 130
General Physics
Fall 2012
18. Mass m1 on the frictionless table shown below is connected by a string through a hole in
the table to a hanging mass m2 . With what speed must m1 rotate in a circle of radius
r if m2 is to remain hanging at rest?
Solution:
This is a coupled problem where the two masses are connected by a string. m1 is
moving in circular motion and m2 ’s motion is linear. Notice in the free body diagrams
that the coordinate system for m1 is only defined for one direction. The coordinate
system for m2 is the rtz-coordinate system. The tangential coordinate isn’t shown
because it isn’t needed for this problem.
In a coupled two body problem, a free body diagram is needed for each mass.
Next we use Newton’s second law for each mass. Notice that m2 is hanging - it’s not
moving. So, the sum of the forces on m2 equals zero.
m2 :
ÿ
pFon
m2 qy
“ T ´ pFG q2
“ T ´ m2 g
“ 0
T “ m2 g
37
(1)
(2)
(3)
(4)
Physics 130
General Physics
Fall 2012
For m2 , object isn’t moving in the z-direction and the normal force and the gravitational force are balanced. The two objects are connected by the string. The tension
of the string is in the radial direction and it’s providing the force for the centripetal
force that’s creating the circular motion of m1 .
m1 :
ÿ
pFon
m1 qr
“ T “ m1 a “ m1
v2
r
(5)
The tension is the same for both masses and it’s what connects the objects in the
equations. We can set the tensions equal and solve for the velocity of m1
v2
m2 g “ m1
r
m2 gr
2
ñv “
m
c 1
m2 gr
ñv “
m1
38
(6)
(7)
(8)
Physics 130
General Physics
Kinematics and Mechanics
1
xf “ xi ` vxi ∆t ` ax p∆tq2
2
vxf “ vxi ` ax ∆t
2
2
vxf
“ vxi
` 2ax pxf ´ xi q
1
yf “ yi ` vyi t ` ay t2
2
vyf “ vyi ` ay t
2
2
vyf
“ vyi
` 2ay pyf ´ yi q
1
θf “ θi ` ωi ∆t ` α∆t2
2
ωf “ ωi ` α∆t
ωf 2 “ ωi 2 ` 2α∆θ
s “ rθ
c “ 2πr
2πr
v“
T
vt “ ωr
v2
“ ω2r
ar “
r
at “ rα
Forces
F~net “ ΣF~ “ m~a
2
v
F~net “ ΣF~r “ m~a “ m
r
Fg “ mg
0 ă fs ă“ µs FN
fk “ µk FN
F~A on B “ ´F~B
on A
Constants
m
g “ 9.8 2
s
39
Fall 2012