Download Solutions to Assigned Problems Chapter 4

Solutions to Assigned Problems Chapter 4
10. (a) The 20.0 kg box resting on the table has the free-body diagram shown. Its
weight
is mg   20.0 kg   9.80 m s 2   196 N . Since the box is at rest, the net force
FN
on the box must be 0, and so the normal force must also be 196 N .
mg
F12 is the force on box 1
(b) Free-body diagrams are shown for both boxes.
(the
top box) due to box 2 (the bottom box), and is the normal force on box
1. F21 is the force on box 2 due to box 1, and has the same magnitude
FN1  F12
Top box (#1)
as F12 by Newton’s third law. FN2 is the force of the table on box 2.
That is the normal force on box 2. Since both boxes are at rest, the
m1g
net force on each box must be 0. Write Newton’s second law in the
vertical direction for each box, taking the upward direction to be positive.
FN2
 F 1  FN1  m1 g  0


FN1  m1 g  10.0 kg  9.80 m s 2  98.0 N  F12  F21
F
2
Bottom
box (#2)
 FN 2  F21  m2 g  0

m2 g

FN 2  F21  m2 g  98.0 N   20.0 kg  9.80 m s 2  294 N
12.
Choose up to be the positive direction. Write Newton’s second law for the
vertical
direction, and solve for the tension force.
 F  FT  mg  ma  FT  m  g  a 

FT  1200 kg  9.80 m s  0.70 m s
2
2
  1.3  10 N
F21
FT
mg
4
16.
In both cases, a free-body diagram for the elevator would look like the
adjacent
diagram. Choose up to be the positive direction. To find the MAXIMUM tension,
assume that the acceleration is up. Write Newton’s second law for the elevator.
 F  ma  FT  mg 

FT  ma  mg  m  a  g   m  0.0680 g  g    4850 kg 1.0680  9.80 m s 2
FT
mg

 5.08  104 N
To find the MINIMUM tension, assume that the acceleration is down. Then Newton’s
second law for the elevator becomes the following.
 F  ma  FT  mg  FT  ma  mg  m  a  g   m  0.0680g  g 


  4850 kg  0.9320  9.80 m s 2  4.43 104 N
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18. There will be two forces on the person – their weight, and the normal force of
the scales pushing up on the person. A free-body diagram for the person is
shown. Choose up to be the positive direction, and use Newton’s second law
to find the acceleration.
 F  FN  mg  ma  0.75mg  mg  ma 

a  0.25 g  0.25 9.8 m s
2
  2.5 m s
mg
FN
2
Due to the sign of the result, the direction of the acceleration is down . Thus the elevator
must have started to move down since it had been motionless.
32. The window washer pulls down on the rope with her hands with a tension
force FT , so the rope pulls up on her hands with a tension force FT . The
tension in the rope is also applied at the other end of the rope, where it
attaches to the bucket. Thus there is another force FT pulling up on the
FT
bucket. The bucket-washer combination thus has a net force of 2FT
upwards. See the adjacent free-body diagram, showing only forces on the
bucket-washer combination, not forces exerted by the combination (the pull
down on the rope by the person) or internal forces (normal force of bucket on
person).
(a) Write Newton’s second law in the vertical direction, with up as positive.
The net
force must be zero if the bucket and washer have a constant speed.
 F  FT  FT  mg  0  2 FT  mg 
FT  12 mg 
1
2
 72 kg   9.80 m
FT
mg

s 2  352.8 N  350 N
(b) Now the force is increased by 15%, so FT  358.2 N 1.15  405.72 N. Again write
Newton’s
second law, but with a non-zero acceleration.
 F  FT  FT  mg  ma 
a
2 FT  mg
m


2  405.72 N    72 kg  9.80 m s 2
72 kg
  1.47 m s
2
 1.5 m s 2
36. Since all forces of interest in this problem are horizontal, draw the free-body diagram
showing only the horizontal forces. FT1 is the tension in the coupling between the
locomotive and the first car, and it pulls to the right on the first car. FT2 is the tension in the
coupling between the first car an the second car. It pulls to the right on car 2, labeled
FT2R and to the left on car 1, labeled FT2L . Both cars have the same mass m and the same
acceleration a. Note that FT2R  FT2L  FT 2 by Newton’s third law.
FT2
FT2
FT1
Write a Newton’s second law expression for each car.
 F1  FT1  FT 2  ma  F2  FT 2  ma
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Substitute the expression for ma from the second expression into the first one.
FT 1  FT 2  ma  FT 2  FT1  2 FT2 
FT1 FT2  2
This can also be discussed in the sense that the tension between the locomotive and the
first car is
pulling 2 cars, while the tension between the cars is only pulling one car.
46. (a) In the free-body diagrams below, FAB = force on block A exerted by block B, FBA = force
on
block B exerted by block A, FBC = force on block B exerted by block C, and FCB = force
on block C exerted by block B. The magnitudes of FBA and FAB are equal, and the
magnitudes of FBC and FCB are equal, by Newton’s third law.
FA N
FB N
FAB
F
FC N
FCB
FBC
FBA
mCg
m Bg
mAg
(b) All of the vertical forces on each block add up to zero, since there is no acceleration in
the
vertical direction. Thus for each block, FN  mg . For the horizontal direction, we have
the following.
F  F  F
AB
 FBA  FBC  FCB  F   mA  mB  mC  a 
a
F
mA  mB  mC
(c)
For each block, the net force must be ma by Newton’s second law. Each block
has the same
acceleration since they are in contact with each other.
FA net  F
mA
mA  mB  mC
FB net  F
mB
mA  mB  mC
F3 net  F
(d) From the free-body diagram, we see that for mC, FCB  FC net  F
Newton’s third law, FBC  FCB  F
mC
mA  mB  mC
mC
mA  mB  mC
mC
mA  mB  mC
. And by
. Of course, F23 and F32 are in
opposite directions. Also from the free-body diagram, we use the net force on mA.
mA
mA
F  FAB  FA net  F
 FAB  F  F

mA  mB  mC
mA  mB  mC
FAB  F
mB  mC
mA  mB  mC
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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By Newton’s third law, FBC  FAB  F
(e)
Using the given values, a 
m2  m3
m1  m2  m3
F
m1  m2  m3

.
96.0 N
30.0 kg
 3.20 m s 2 . Since all three
masses
are the same value, the net force on each mass is
Fnet  ma  10.0 kg  3.20 m s 2  32.0 N .


This is also the value of FCB and FBC . The value of FAB and FBA is found as follows.


FAB  FBA   m2  m3  a   20.0 kg  3.20 m s 2  64.0 N
To summarize:
FA net  FB net  FC net  32.0 N
FAB  FBA  64.0 N
FBC  FCB  32.0 N
The values make sense in that in order of magnitude, we should have F  FBA  FCB , since
F is the net force pushing the entire set of blocks, FAB is the net force pushing the right two
blocks, and FBC is the net force pushing the right block only.
54. We draw a free-body diagram for each mass. We choose UP to be
the positive direction. The tension force in the cord is found from
analyzing the two hanging masses. Notice that the same tension
force is applied to each mass. Write Newton’s second law for each
of the masses.
FT  m1 g  m1a1 FT  m2 g  m2a2
Since the masses are joined together by the cord, their accelerations
will have the same magnitude but opposite directions. Thus
a1  a2 . Substitute this into the force expressions and solve for the
tension force.
m g  FT
FT  m1 g  m1a2  FT  m1 g  m1a2  a2  1
m1
FC
FT
FT
FT
m2
m1
1.2
kg
3.2
kg
m2 g
 m g  FT 
2m1m2 g
FT  m2 g  m2 a2  m2  1
 FT 

m1  m2
 m1 
FT
m1g
Apply Newton’s second law to the stationary pulley.
2
4m1m2 g 4  3.2 kg 1.2 kg  9.80 m s
FC  2 FT  0  FC  2 FT 

 34 N
m1  m2
4.4 kg


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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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55. If m doesn’t move on the incline, it doesn’t move in the vertical
 FN
direction, and so has no vertical component of acceleration. This
suggests that we analyze the forces parallel and perpendicular to the
floor. See the force diagram for the small block, and use Newton’s
second law to find the acceleration of the small block.

mg
mg
 Fy  FN cos   mg  0  FN  cos 
F sin  mg sin 
 Fx  FN sin   ma  a  N m  m cos   g tan 
Since the small block doesn’t move on the incline, the combination of both masses has the
same horizontal acceleration of g tan  . That can be used to find the applied force.
Fapplied   m  M  a   m  M  g tan 
Note that this gives the correct answer for the case of   0, , where it would take no
applied force to keep m stationary. It also gives a reasonable answer for the limiting case
of   90, where no force would be large enough to keep the block from falling, since
there would be no upward force to counteract the force of
gravity.
F
FTC
TC
69. (a) A free-body diagram is shown for each block.
We define the positive x-direction for mA to
be up its incline, and the positive x-direction
y
for mB to be down its incline. With that
definition the masses will both have the
same acceleration. Write Newton’s second
law for each body in the x direction, and
combine those equations to find the
acceleration.
mA :  Fx  FT  mA g sin A  mAa
mB :
F
x
 mB g sin  B  FT  mBa
mC
aC
FN-B
FT
FN-A
FTA
FT
mCg y
B
FTA B
FTA
x
A
A
aC
x
FTA
mA
aA
mAg
mB
aB
m Bg
mAg
m Bg
add these two equations
 FT  mA g sin  A    mB g sin  B  FT   mA a  mBa
 a
mB sin  B  mA sin  A
mA  mB
g
(b) For the system to be at rest, the acceleration must be 0.
m sin  B  mA sin  A
a B
g  0  mB sin  B  mA sin  A 
mA  m B
m B  mA
sin  A
sin  B
  5.0 kg 
sin 32
sin 23
 6.8 kg
The tension can be found from one of the Newton’s second law expression from part (a).
mA : FT  mA g sin  A  0  FT  mA g sin  A   5.0 kg   9.80 m s 2  sin 32  26 N
(c) As in part (b), the acceleration will be 0 for constant velocity in either direction.
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a
mA
mB
mB sin  B  mA sin  A
mA  m B

sin  B
sin  A

sin 23
sin 32
g  0  mB sin  B  mA sin  A 
 0.74
74. Consider the free-body diagram for the watch. Write Newton’s second law
for both the x and y directions. Note that the net force in the y direction is
0 because there is no acceleration in the y direction.
mg
 Fy  FT cos   mg  0  FT  cos 
mg
 Fx  FT sin   ma  cos  sin   ma


FT
y
x
mg

a  g tan   9.80 m s 2 tan 25  4.57 m s 2
Use Eq. 2-12a with v0  0 to find the final velocity (takeoff speed).

v  v0  at  v  v0  at  0  4.57 m s 2
 16s  
73m s
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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