Physics 130 General Physics
October 30, 2012
Fall 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. A labeled picture or diagram, if appropriate.
2. A list of given variables.
3. A list of the unknown quantities (i.e., what you are being asked to find).
4. A force-interaction diagram, if appropriate.
5. One or more free-body diagrams, as appropriate, with labeled 1D or 2D coordinate axes.
6. Algebraic expression for the net force along each dimension, as appropriate.
7. An algebraic solution of the unknown variables in terms of the known variables.
8. A final numerical solution, including units, with a box around it.
9. 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: p F net q y
“
ÿ
F y p F net q x
“
ÿ
F x
“ T
1 y
` T
2 y
` f ky
“ 0 N
“ T
1 x
` T
2 x
` f kx
“ 0 N
Examining the free-body diagram, along the y -axis we have
(1)
(2)
T
T f
1 y
2 y ky
“ ` T
1 sin p θ { 2 q
“ ´ T
2 sin p θ { 2 q
“
Substituting into equation (1), we get
0 .
ñ
T
1 sin p θ { 2 q ´ T
2 sin p θ { 2 q “ 0
T
1
“ T
2
“ T.
(3)
(4)
(5)
(6)
(7)
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Physics 130 General Physics
In other words, the tension force of each rope is the same.
From the free-body diagram along the x -axis we have
Fall 2012
T
1 x
T
2 x
“ T
1 cos p θ { 2 q “ T cos p θ { 2 q
“ T
2 cos p θ { 2 q “ T cos p θ { 2 q .
Substituting into equation (2) and solving for T we get
(8)
(9)
T cos p θ { 2 q ` T cos p θ { 2 q ´ f kx
“ 0
2 T cos p θ { 2 q “ f kx
ñ T “ f kx
2 cos p θ { 2 q
1000 N
T “
2 ˆ cos p 20 ˝ { 2 q
“ 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 General Physics Fall 2012
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 µ s
“ 0 .
60 and µ k
“ 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 General Physics
The free-body diagram of just the box of antiques is:
Fall 2012
(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 p F net q x
“
ÿ
F x
“ ´ f “ p m ` M q a x
“ p m ` M q a p F net q y
“
ÿ
F y
“ N ´ p F
G q
P A
“ 0
“ N ´ p m ` M q g “ 0
ñ N “ p m ` M q g.
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
(1)
(2)
(3)
(4)
´ f “ ´ µN “ ´ µ p m ` M q g “ p m ` M q a
ñ a “ ´ µg.
(5)
(6)
The mass of the pickup-antiques system drops out! Next, from the constant-acceleration kinematics equation with v 2
1
“ 0 and x
0
“ 0, we have
ñ v
2
1
ñ a “ ´ v 2
0
2 x
1
µ min
“ v
2
0
“
2 v
2
0
` 2 a p x
1
` x
0 q gx
1
“ p 25 m { s q 2
2 ˆ p 9 .
8 m { s 2 q ˆ p 55 m q
“ 0 .
58 .
(7)
(8)
(9)
(10)
(11)
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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 of static friction ( µ s
µ min
“ 0 .
58. This value is smaller than the given coefficient
“ 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, F
G
, the normal force, n , and the kinetic friction force, f k
. 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 p F net q x
“
ÿ
F x
“ F
G sin 25 ˝ ´ f k
“ ma p F net q y
“
ÿ
F y
“ n ´ F
G cos 25
ñ n “ F
G cos 25 ˝
˝ “ 0
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 F
G
“ mg , and the magnitude of the kinetic force of friction is
(1)
(2)
(3)
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Physics 130 General Physics Fall 2012 f k
“ µ k n
“ µ k
F
G cos 25 ˝
“ µ k mg cos 25 ˝ .
(4)
(5)
(6)
Substituting equation (6) into equation (1) and solving for the acceleration a we get mg sin 25 ˝ ´ µ k mg cos 25 ˝ “ ma
ñ a “ p sin 25 ˝ ´ µ k cos 25 ˝ q g
“ p sin 25 ˝ ´ 0 .
“ ´ 0 .
743 m { s 2
55 ˆ cos 25 ˝ q ˆ 9 .
8 m { s 2
(7)
(8)
(9)
(10) where the minus sign indicates the acceleration is directed up the incline. To find the initial speed, v i
, necessary to have the box come to rest (i.e., v f
“ 0) after ∆ x “ 5 .
0 m can be found from the kinematic equation linking velocity and acceleration: v
2 f
“ v
?
ñ v i
“
2 i
` 2 a ∆
´ 2 a ∆ x x
“ b
´ 2 p´ 0 .
743 m { s 2 qp 5 .
0 m q
“ 2 .
7 m { s .
(11)
(12)
(13)
(14)
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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 µ s
“ 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 p F net q x
“
ÿ
F x
“ n ´ F push cos θ “ 0
ñ n “ F push cos θ.
From our model for friction, the maximum static friction force is
(1)
(2) p f s q max
“ µ s n “ µ s
F push cos θ
“ 0 .
5 ˆ p 12 N q ˆ cos p 30 ˝ 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, p F net q y
, is greater or smaller than p f s q max
. The net force excluding the force of friction is
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Physics 130 General Physics Fall 2012 p F net q y
“
ÿ
F y
“ F push sin θ ´ F
G
“ F push sin θ ´ mg
“ p 12 N q ˆ sin p 30 ˝ q ´ p 1 .
0 kg q ˆ p 9 .
8 m { s 2 q
“ ´ 3 .
8 N .
In other words, a static friction force f s
“ ` 3 .
8 N would prevent the block from moving. Since this force is smaller than p f s q max the box stays in place and does not move.
(6)
(7)
(8)
(9)
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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
µ k
“ 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
13

Physics 130 General Physics Fall 2012 pairs. Notice how the top block (block 1) both pushes down on the bottom block
(block 2) with force of block 2.
~n 1
1
, and exerts a retarding friction force f
2 , top on the top surface
(a) Block 1 is in static equilibrium ( a
1
“ 0 m { s 2 right. Newton’s second law for block 1 is
), but block 2 is accelerating to the p F net on 1 q x
“ f
1
´ T rope
“ 0 ñ T rope
“ f
1 p F net on 1 q y
“ n
1
´ m
1 g “ 0 ñ n
1
“ m
1 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 f
1
“ µ k n
1
“ µ k m
1 g.
Substituting this result into equation (1) we get the tension of the rope:
(3)
T rope
“ f
1
“ µ k m
1 g
“ p 0 .
40 q ˆ p 1 .
0 kg q ˆ p 9 .
8 m { s 2 q
“ 3 .
9 N .
(b) Newton’s second law for block 2 is
(4)
(5)
(6) a x
” a “ a y
“ p F net on 2 q x m
2 p F net on 2 q y m
2
“
“
T pull
´ f
2 top
´ f
2 bot n
2
´ n 1
1 m
2 m
2
´ m
2 g
“ 0
(7)
(8)
Forces ~n
1 and ~n 1
1 equation (8) gives are an action/reaction pair, so ~n 1
1
“ ~n
1
“ m
1 g . Substituting into n
2
“ p m
1
` m
2 q g.
(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 f
2 bot
“ µ k n
2
“ µ k p m
1
` m
2 q g.
(10)
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Physics 130 General Physics Fall 2012
Next, we recognize that the forces f
1 and f
~
2 top are an action/reaction pair, so f
2 top
“ f
1
“ µ k m
1 g.
Finally inserting these friction results into equation (7) gives
(11) a “
“
T pull
´ µ k m
1 g ´ µ k p m
1
` m
2 q g m
2
20 N ´ p 0 .
40 qp 1 .
0 kg qp 9 .
8 m { s 2 q ´ p 0 .
40 qp 1 .
0 kg ` 2 .
0 kg qp 9 .
8 m { s 2 q
2 .
0 kg
“ 2 .
2 m { s 2 .
(12)
(13)
(14)
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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 a
2
“ a “ ´ a
1
, 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 1
1 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 p F net on 1 q x
“ f
1
´ T
1
“ µ k n
1
´ T
1
“ m
1 a
1
“ ´ m
1 a p F net on 1 q y
“ n
1
´ m
1 g “ 0
ñ n
1
“ m
1 g.
(1)
(2)
(3)
(4)
For block 2 we have p F net on 2 q x
“ T pull
´ f 1
1
“ T pull
´ f 1
1
´ f
2
´ T
2
´ µ k n
2
´ T
2
“ m
2 a
2
“ m
2 a p F net on 2 q y
“ n
2
´ n 1
1
ñ n
2
“ n 1
1
´ m
2 g “ 0
` m
2 g.
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: n 1
1 f 1
1
T
1
“ n
1
“ m
1 g
“ f
1
“ µ k n
1
“ µ k m
1 g
“ T
2
“ T.
(9)
(10)
(11)
(5)
(6)
(7)
(8)
Knowing n 1
1 we can now use the y -equation for block 2 to find n
2
. Substituting all these pieces into the two x -equations, we end up with two equations with two unknowns:
µ k m
1 g ´ T “ ´ m
1 a
T pull
´ T ´ µ k m
1 g ´ µ k p m
1
` m
2 q g “ m
2 a.
(12)
(13)
Subtracting the first equation from the second we get
T pull
´ T ´ µ k m
1 g ´ µ k p m
1
` m
2 q g ´ µ k m
1 g ` T “ m
2 a ` m
1 a
T pull
´ 3 µ k m
1 g ´ µ k m
2 g “ p m
2
` m
1 q a
T pull
´ µ k p 3 m
1
` m
2 q g “ p m
2
` m
1 q a.
And finally solving for a we get
ñ a “
“
T pull
´ µ k p 3 m
1
` m
2 q g m
1
` m
2
20 N ´ p 0 .
30 qp 3 ˆ 1 .
0 kg ` 2 .
0 kg qp 9 .
8 m { s 2 q
1 .
0 kg ` 2 .
0 kg
“ 1 .
8 m { s 2 .
(14)
(15)
(16)
(17)
(18)
(19)
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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 m
1 and m
3 gives
T
1
T
2
´ p F
G q
1
“ T
1
´ m
1 g “ m
1 a
1
´ p F
G q
3
“ T
2
´ m
3 g “ m
3 a
3
.
For m
2
Newton’s second law gives
18
(1)
(2)

Physics 130 General Physics Fall 2012
ÿ p F on m
2 q y
“ n
2
´ p F
G q
2
“ n
2
´ m
2 g “ 0
ñ n
2
“ m
2 g
ÿ p F on m
2 q x
“ T
2
´ f k 2
´ T
1
“ T
2
´ µ k n
2
´ T
1
“ T
2
´ µ k m
2 g ´ T
1
“ m
2 a
2
.
Since all three blocks move together, the acceleration constraint gives a
1
“ a
2
“ ´ a
3
” a.
Therefore, the equations for the three masses are:
(8)
T
2
T
1
´ m
1 g “ m
1 a
´ µ k m
2 g ´ T
1
“ m
2 a
T
2
´ m
3 g “ ´ m
3 a.
Subtracting the third equation from the sum of the first two equations yields:
(9)
(10)
(11) p T
1
´ m
1 g q ` p T
2
´ µ k m
2 g ´ T
1 q ´ p T
2
´ m
3 g q “ m
1 a ` m
2 a ` m
3 a (12) p´ m
1
´ µ k m
2
` m
3 q g “ a p m
1
` m
2
` m
3 q (13)
ñ a “
´ m
1 m
1
´ µ k m
2
` m
3 q g
` m
2
` m
3
´ 1 .
0 kg ´ 0 .
30 ˆ 2 .
0 kg ` 3 .
0 kg
“
1 .
0 kg ` 2 .
0 kg ` 3 .
0 kg
“ 2 .
3 m { s 2 .
ˆ p 9 .
8 m { s 2 q
(14)
(15)
(16)
(3)
(4)
(5)
(6)
(7)
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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 µ s
“ 0 .
50 and µ k
“ 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:
20

Physics 130 General Physics 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 x
1 the book slides we must find its acceleration. Newton’s second law applied to the book gives
ÿ p F on B q y
“ n
B
´ p F
G q
B cos θ “ 0
ñ n
B
“ m
B g cos θ
ÿ p F on B q x
“ ´ T ´ f k
´ p F
G q
B sin θ
“ ´ T ´ µ k n
B
´ m
B g sin θ
“ ´ T ´ µ k m
B g cos θ ´ m
B g sin θ “ m
B a
B
.
Similarly, for the coffee cup we have
ÿ p F on C q y
“ T ´ p F
G q
C
“ T ´ m
C g “ m
C a
C
.
(6)
Equations (5) and (6) can be rewritten using the acceleration constrain a
C as
“ a
B
“ a
(1)
(2)
(3)
(4)
(5)
´ T ´ µ k m
B g cos θ ´ m
B g sin θ “ m
B a
T ´ m
C g “ m
C a.
Adding these two equations and solving for the acceleration gives
(7)
(8)
´ µ k m
B g cos θ ´ m
B g sin θ ´ m
C g “ p m
B
` m
C q a (9)
ñ a “ ´
„ m
B p µ k cos θ ` sin θ q ` m
C m
B
` m
C
 g (10)
“ ´
„ p 1 .
0 kg q ˆ p 0 .
20 cos 20 ˝ ` sin 20
1 .
0 kg ` 0 .
5 kg
˝ q ` 0 .
5 kg 
“ ´ 6 .
73 m { s 2 .
ˆ p 9 .
8 m { s 2 q (11)
(12)
Finally, to solve for the distance x
1 we use the following kinematic equation with v
1 x
“ 0, v
0 x
“ 3 .
0 m { s, and x
0
“ 0:
21

Physics 130 General Physics Fall 2012 v
2
1 x
“ v
2
0 x
` 2 a p x
1
´ x
0 q
0 “ v
2
0 x
` 2 ax
1
ñ x
1
“ ´
“ v
2
0 x
2 a
´p 3 .
0 m { s q 2
2 ˆ p´ 6 .
73 m { s 2 q
“ 0 .
67 m .
(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, f s is larger or smaller than the maximum static friction force p f s q max
“ µ s n
B
“ µ s m
B gcosθ
“ p 0 .
50 q ˆ p 9 .
8 m { s 2 q ˆ cos 20 ˝
“ 4 .
60 N .
(18)
(19)
(20)
When the cup is at rest, the string tension is T “ m
C g . In this case, Newton’s first law for the book becomes
ÿ p F on B q x
“ f s
´ T ´ m
B g sin θ
“ f s
´ m
C g ´ m
B g sin θ “ 0
ñ f s
“ p m
C
` m
B sin θ q g
“ p 0 .
5 kg ` 1 .
0 kg sin 20 ˝ q ˆ p 9 .
8 m { s 2 q
“ 8 .
25 N .
Because f s ą p f s q max
, the book slides back down.
(21)
(22)
(23)
(24)
(25)
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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 { s 2 ?
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
2 F ´ F
G
“ p m
P
` m
C q a
2 F ´ p m
P
` m
C q g “ p m
P
` m
C q a
ñ F “ p m
P
` m
C q a ` p m
P
` m
C q g
“
2 p m
P
` m
C qp a ` g q
“
2 p 70 kg ` 10 kg q ˆ p 0 .
20 m { s 2
2
“ 400 N .
` 9 .
8 m { s 2 q
(1)
(2)
(3)
(4)
(5)
(6)
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Physics 130 General Physics Fall 2012
12. Based on the figure below, find an expression for the acceleration of m
1
. 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 m the acceleration constraint is
2 has to be lengthened by one-half meter. Therefore, a
2
“ ´
1
2 a
1
.
Newton’s second law applied to block 1 and block 2, respectively, gives
(1)
T “ m
1 a
1
2 T ´ p F
G q
2
“ m
2 a
2
“ ´ m
2 a
1
2
.
(2)
(3)
24

Physics 130 General Physics Fall 2012
Combining these two equations gives
4 m
1 a
1 m
2 a
1
2 m
1 a
1
´ m
2 g “ ´
2
´ 2 m
2 g “ ´ m
2 a
1
4 m
1 a
1
` m
2 a
1
“ 2 m
2 g a
1 p 4 m
1
` m
2 q “ 2 m
2 g
2 m
2 g
ñ a
1
“ p 4 m
1
` m
2 q
Note: if m
1 object.
“ 0 then a
2
“ ´ g , which is what we would expect for a free-falling
(4)
(5)
(6)
(7)
(8)
25

Physics 130 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 a
1
“ ´ 2 a
2
.
(1)
Newton’s second law applied to block one gives
2 T “ m
2 a
2
ñ T “ m
2
2 a
2
.
Similarly, applying Newton’s second law to block two gives
T ´ p F
G q
1
“ m
1 a
1
“ ´ 2 m
1 a
2 m
2
2 a
T ´ m
1 g “ ´ 2 m
1 a
2
2
´ m
1 g “ ´ 2 m
1 a
2
(2)
(3)
(4)
(5)
(6)
26

Physics 130 General Physics m
2 a
2
` 4 m
1 a
2
“ 2 m
1 g a
2 p m
2
` 4 m
1 q “ 2 m
1 g
2 m
ñ a
2
“
“ m
2
` 4 m
1
2 ˆ p 1 .
0 kg q ˆ p 9 .
8 m { s 2 p 2 .
1 g
0 kg ` 4 ˆ 1 .
0 kg q q
“ 3 .
3 m { s 2 .
Note that if m
1
“ 0 then a
2
“ 0, which is what we would expect.
Fall 2012
(7)
(8)
(9)
(10)
(11)
27

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 p F net q y
“ F
R
´ mg “ ma
R
1
ñ a
R
“ m p F
R
´ mg q
1
“
0 .
8 kg r 15 N ´ p 0 .
8 kg q ˆ p 9 .
8 m { s 2 qs
“ 8 .
95 m { s 2
.
(1)
(2)
(3)
(4)
Next, we use the kinematic equation for the rocket to find the time, noting that y
0 R
“ p v
0 R q y
“ t
0 R
“ 0:
ñ y
1 R
“ y
0 R
` p v
0 R q y p t
1 R
´ t
0 R q `
1
2 a
R p t
1 R
´ t
0 R q 2
“
1
2 a
R t
2
1 R c 2 y
1 R t
1 R
“
“ a
R d
2 ˆ 30 m
8 .
95 m { s 2
“ 2 .
589 s .
(8)
(9)
(5)
(6)
(7)
28

Physics 130 General Physics Fall 2012
For the target, we use the same kinematic equation along the x -direction, noting that t
1 T
“ t
1 R
(the times are the same for both the rocket and the target) and that x
0 T
“ a
T
“ 0: x
1 T
“ x
0 T
` p v
0 T q x p t
1 T
´ t
0 T q `
1
2 a
T p t
1 T
´ t
0 T q 2
“ p v
0 T q x t
1 T
“ p 15 m { s q ˆ p 2 .
589 s q
“ 39 m .
In other words, the rocket should be launched when the target is 39 m away.
(10)
(11)
(12)
(13)
29

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,
30

Physics 130 General Physics Fall 2012 adjacent cos θ “ hypotenuse
T z
“ T cos θ
“
T z
T
Apply Newton’s second law in the z-direction
ÿ
F z
“ T cos θ ´ mg “ 0 mg
T “ cos θ
To calculate cos θ , adjacent cos θ “ hypotenuse
ñ θ “ 11 .
5 ˝
“
?
L 2
L
´ r 2
“ a p 1m q 2 ´ p 0 .
2m q 2
1 .
0m
“ 0 .
98
Plugging this in for tension, mg
T “ cos θ
T “ 5 .
0 N
“ p 0 .
500 kg q ˆ p 9 .
80 m { s 2 q
0 .
98
(5)
(6)
(7)
(8)
(b) What is the ball’s angular speed, in rpm?
Referring back to the tension triangle, the radial component of tension is
T r opposite sin θ “ hypotenuse
“ T sin θ
“
T r
T
(9)
(10)
31
(1)
(2)
(3)
(4)

Physics 130 General Physics
Referring back to the triangle for the length of the pendulum, opposite sin θ “ hypotenuse r “ L sin θ r
“
L
Apply Newton’s second law in the r-direction
ÿ
F r
“ T r
“ T sin θ “ mω
2 r
ñ ω “
“ c
T sin θ mr d
5 .
0 N ˆ sin 11 .
5 ˝ p 0 .
500 kg q ˆ p 0 .
2 m q
“ 3 .
16 rad { sec rad
“ 3 .
16 sec
ˆ
ˆ 1 rev
2 π rad
˙
ˆ
ˆ 60 sec ˙
1 min
“ 30 rpm
Fall 2012
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
32

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: f s
“ F g
. The minimum angular velocity occurs when static friction reaches its maximum possible value p f s q max
“ µ s n . Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient
( µ s
“ 0 .
60 q 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
F r
“ n “ mω
2 r
ÿ
F z
“ f s
´ F g
“ 0
ñ f s
“ mg
(1)
(2)
The minimum angular speed occurs when f s
“ p f s q max
“ µ s n “ µ s mrω
2 min
(3)
Substituting this expression into the z -equation gives f s
“ µ s mrω
2 min
“ mg (4)
33

Physics 130 General Physics Fall 2012
Solving for ω min
ω min
“ c g
“
µ s r d
9 .
80 m { s 2
0 .
60 ˆ p 2 .
5 m q
“ 2 .
56 rad { s ˆ
ˆ 1 rev
2 π rad
˙
ˆ
ˆ 60 sec ˙
1 min
“ 24 rpm
Note that the angular velocity does not depend on the mass of the individual.
(5)
(6)
(7)
(8)
34

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 r direction in the free body diagram is the component of the tension in the r -direction.
To find the r -component,
T r opposite sin θ “ hypotenuse
“ T sin θ
“
T r
T
35
(1)
(2)

Physics 130 General Physics
Newton’s second law along the r -axis is
ÿ
F r
“ T r
“ ma r
“ m v 2 r
T sin θ “ mω
2 r
“ mω
2 r
The radius, r , is a component of the length of the chain, L .
Fall 2012
(3)
(4) opposite sin θ “ adjacent r “ L sin θ
“ r
L
Plugging this in, we find
T sin θ “ mω
2
L sin θ
T “ mω
2
L
T “ p 150 kg q ˆ
„ 2 π
4 .
0 sec

2
ˆ p 9 .
0 m q
T “ 3300 N
Thus, the 3000 N chain is not strong enough for the ride.
(5)
(6)
(7)
(8)
(9)
(10)
36

Physics 130 General Physics Fall 2012
18. Mass m
1 on the frictionless table shown below is connected by a string through a hole in the table to a hanging mass m
2
. With what speed must m
1 r if m
2 is to remain hanging at rest?
rotate in a circle of radius
Solution:
This is a coupled problem where the two masses are connected by a string.
m
1 is moving in circular motion and m
2
’s motion is linear. Notice in the free body diagrams that the coordinate system for m
1 is only defined for one direction. The coordinate system for m
2 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 m
2 moving. So, the sum of the forces on m
2 equals zero.
is hanging - it’s not m
2
:
ÿ p F on m
2 q y
“ T ´ p F
G q
2
“ T ´ m
2 g
“ 0
T “ m
2 g
(1)
(2)
(3)
(4)
37

Physics 130 General Physics Fall 2012
For m
2
, 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 m
1
.
m
1
:
ÿ p F on m
1 q r
“ T “ m
1 a “ m
1 v
2 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 m
1
ñ v
2 v 2 m
2 g “
“ m
1 m
2 r gr m
1
ñ v “ c m
2 gr m
1
(6)
(7)
(8)
38

Physics 130
Kinematics and Mechanics x f v xf
“ x i
` v xi
∆ t `
1
2 a x p ∆ t q 2
“ v xi
` a x
∆ t v
2 xf
“ v
2 xi
` 2 a x p x f
´ x i q y f v yf
“ y i
` v yi t `
1
2 a y t
2
“ v yi
` a y t v
2 yf
“ v
2 yi
` 2 a y p y f
´ y i q
θ f
ω f
“ θ i
` ω i
∆ t `
1
2
α ∆ t
2
“ ω i
` α ∆ t
ω f
2 “ ω i
2 ` 2 α ∆ θ s “ rθ c “ 2 πr v t
2 πr v “
T
“ ωr a r a t v 2
“ r
“ rα
“ ω
2 r
Forces
F net
F net
F g
“ Σ
~
“ m~a
“ Σ
~ r
“ m~a “ m v
2 r
“ mg
0 ă f s ă“ µ s
F
N f k
“ µ k
F
N
F
A on B
“ ´
~
B on A
Constants m g “ 9 .
8 s 2
General Physics
39
Fall 2012