Download Solutions to Assigned Problems Chapter 5

Solutions to Assigned Problems Chapter 5
2.
A free-body diagram for the box is shown. Since the box does not
accelerate vertically, FN  mg .
(a) To start the box moving, the pulling force must just overcome the
force of static friction, and that means the force of static friction
will reach its maximum value of Ffr  s FN . Thus we have for the
starting motion,
 Fx  FP  Ffr  0 
FP  Ffr  s FN  s mg  s 
FP
35.0 N

mg
 6.0 kg   9.80 m
s2

FN
Ffr
FP
mg
 0.60
(b)
The same force diagram applies, but now the friction is kinetic friction, and the
pulling force is
NOT equal to the frictional force, since the box is accelerating to the right.
 F  FP  Ffr  ma  FP  k FN  ma  FP  k mg  ma 
k 
4.
FP  ma
mg


35.0 N   6.0 kg  0.60 m s 2
 6.0 kg   9.80 m
s
2

  0.53
See the included free-body diagram. To find the maximum angle, assume
that the car is just ready to slide, so that the force of static friction is a maximum. Write
Newton’s second law for both directions. Note that for both directions, the net force must
be zero since the car is not accelerating.
 F y  FN  mg cos   0  FN  mg cos 
F
s 
x
 mg sin   Ffr  0  mg sin   Ffr  s FN   s mg cos 
mg sin 
mg cos 
 tan   0.90    tan 1 0.90  42
11.
A free-body diagram for the box is shown, assuming that it is moving
to the
right. The “push” is not shown on the free-body diagram because as
Ffr
soon as the box moves away from the source of the pushing force, the
push is no longer applied to the box. It is apparent from the diagram that
FN  mg for the vertical direction. We write Newton’s second law for the
horizontal direction, with positive to the right, to find the acceleration of the box.
 Fx   Ffr  ma  ma   k FN   k mg 

FN
mg

a   k g  0.15 9.80 m s 2  1.47 m s 2
Eq. 2-12c can be used to find the distance that the box moves before stopping. The initial
speed is 4.0 m/s, and the final speed will be 0.
v 2  v02  2a  x  x0   x  x0 
v 2  v02
2a

0   3.5 m s 

2
2 1.47 m s 2

 4.17 m  4.2 m
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17.
(a)
Since the two blocks are in contact, they can be
FP
treated as a
m1 +
single object as long as no information is needed about
m2
Ffr
internal forces (like the force of one block pushing on the
 m1  m2  g
FN
other block). Since there is no motion in the vertical
direction, it is apparent that
FN   m1  m2  g , and so Ffr  k FN  k  m1  m2  g. Write Newton’s second law for the
horizontal direction.
 Fx  FP  Ffr   m1  m2  a 
a
FP  Ffr
m1  m2

FP  k  m1  m2  g
m1  m2


650 N   0.18190 kg  9.80 m s 2

190 kg
 1.657 m s2  1.7 m s 2
(b) To solve for the contact forces between the blocks, an individual
block
must be analyzed. Look at the free-body diagram for the second
block. F21 is the force of the first block pushing on the second
F21
m2
Ffr2
m2 g
FN2
block. Again, it is apparent that FN2  m2 g and so
Ffr2  k FN2  k m2 g. Write Newton’s second law for the horizontal direction.
F
x
 F21  Ffr2  m2 a 




F21  k m2 g  m2 a   0.18125kg  9.80 m s 2  125kg  1.657 m s 2  430 N
By Newton’s third law, there will also be a 430 N force to the left on block # 1 due to
block # 2.
(c)
If the crates are reversed, the acceleration of the system will
F12
remain
the same – the analysis from part (a) still applies. We can also
repeat the analysis from part (b) to find the force of one block on
the other, if we simply change m1 to m2 in the free-body diagram
and the resulting equations.
a  1.7 m s2 ;
F
x
m1
Ffr1
FN1
m1g
 F12  Ffr1  m1a 




F12  k m1 g  m1a   0.18 65 kg  9.80 m s 2   65 kg  1.657 m s 2  220 N
23. (a) For mB to not move, the tension must be equal to mB g , and so mB g  FT . For mA to not
move, the tension must be equal to the force of static friction, and so FS  FT . Note
that the normal force on mA is equal to its weight. Use these relationships to solve for
mA .
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mB g  FT  Fs  s mA g  mA 
mB
s

2.0 kg
0.40
 5.0 kg  mA  5.0 kg
(b) For mB to move with constant velocity, the tension must be equal to mB g . For mA to
move with constant velocity, the tension must be equal to the force of kinetic friction.
Note that the normal force on mA is equal to its weight. Use these relationships to
solve for mA .
mB g  Fk  k mA g  mA 
mB
k

2.0 kg
0.30
 6.7 kg
24. We define f to be the fraction of the cord
y
that is handing down, between mB and
FN
x
FT
the pulley. Thus the mass of that piece of
FT
Ffr
cord is fmC . We assume that the system
x
is moving to the right as well. We take
mAg
the tension in the cord to be FT at the
1  f  mCg
pulley. We treat the hanging mass and
mBg  fmCg
hanging fraction of the cord as one mass,
and the sliding mass and horizontal part
of the cord as another mass. See the free-body diagrams. We write Newton’s second law
for each object.
 FyA  FN   mA  1  f  mC  g  0
xA
 FT  Ffr  FT  k FN   mA  1  f  mC  a
xB
  mB  fmC  g  FT   mB  fmC  a
F
F
Combine the relationships to solve for the acceleration. In particular, add the two equations
for the x-direction, and then substitute the normal force.
 mB  fmC  k  mA  1  f  mC  
g
mA  mB  mC


a 
33. To find the limiting value, we assume that the blocks are NOT
slipping, but that the force of static friction on the smaller block is
at its maximum value, so that Ffr   FN . For the two-block
system, there is no friction on the system, and so F   M  m a
FN
Ffr
y
x

describes the horizontal motion of the system. Thus the upper
block has a vertical acceleration of 0 and a horizontal
mg
F
acceleration of
. Write Newton’s second law for the
 M  m
upper block, using the force diagram, and solve for the applied force F. Note that the static
friction force will be DOWN the plane, since the block is on the verge of sliding UP the
plane.
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F
 FN cos   Ffr sin   mg  FN  cos    sin    mg  0  FN 
F
 FN sin   Ffr cos   FN  sin    cos    ma  m
y
x
F  FN  sin    cos  

M m

mg
 cos    sin  
 sin    cos  
 M  m g
 cos    sin  
m
F
 cos    sin  

M m
 sin    cos  
mg
M m
m
34. A free-body diagram for the car at one instant of time is shown. In the
FN
diagram, the car is coming out of the paper at the reader, and the center of
Ffr
the circular path is to the right of the car, in the plane of the paper. If the
car has its maximum speed, it would be on the verge of slipping, and the
mg
force of static friction would be at its maximum value. The vertical forces
(gravity and normal force) are of the same magnitude, because the car is
not accelerating vertically. We assume that the force of friction is the force causing the
circular motion.
FR  Ffr  m v 2 r  s FN  s mg 
v
s rg 
 0.6580.0 m   9.80 m

s 2  22.57 m s  23m s
Notice that the result is independent of the car’s mass .
40. At the top of a circle, a free-body diagram for the passengers would be as shown, assuming
the passengers are upside down. Then the car’s normal force would be pushing DOWN on
the passengers, as shown in the diagram. We assume no safety devices are present.
Choose the positive direction to be down, and write Newton’s second law for the
passengers.
 F  FN  mg  ma  m v 2 r  FN  m  v 2 r  g 
We see from this expression that for a high speed, the normal force is positive, meaning the
passengers are in contact with the car. But as the speed decreases, the normal force also
decreases. If the normal force becomes 0, the passengers are no longer in contact with the
car – they are in free fall. The limiting condition is as follows.
2
vmin
r  g  0  vmin  rg 
 9.80 m s  7.6 m   8.6 m s
2
41. A free-body diagram for the car is shown. Write Newton’s second law for
the car in the vertical direction, assuming that up is positive. The normal
force is twice the weight.
 F  FN  mg  ma  2mg  mg  m v 2 r 
v
rg 
 95 m   9.80 m
FN
mg

s 2  30.51m s  31m s
51.
The force of static friction is causing the circular motion – it is the
centripetal
force. The coin slides off when the static frictional force is not large
enough to move the coin in a circle. The maximum static frictional force
is the coefficient of static friction times the normal force, and the normal
FN
Ffr
mg
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force is equal to the weight of the coin as seen in the free-body diagram, since there is no
vertical acceleration. In the free-body diagram, the coin is coming out of the paper and the
center of the circle is to the right of the coin, in the plane of the paper.
The rotational speed must be changed into a linear speed.
rev   1 min   2  0.120 m  

v   35.0


  0.4398 m s
min   60 s  
1 rev


FR  Ffr  m v 2 r  s FN  s mg  s 
66.
(a)
v2
rg

 0.4398 m s 2
 0.120 m   9.80 m
s2

 0.164
The terminal velocity is given by Eq. 5-9. This can be used to find the value of b.
vT 
mg
 b
mg
 3  10

5

kg 9.80 m s2
  3.27  10
5
kg s  3  105 kg s
b
vT
9 m s
(b) From Example 5-17, the time required for the velocity to reach 63% of terminal velocity
is the time constant,   m b .

m
b

3  105 kg
3.27  105 kg s
 0.917 s  1s
84.
The car moves in a horizontal circle, and so there must be a net
horizontal
centripetal force. The car is not accelerating vertically. Write
Newton’s second law for both the x and y directions.
mg
 Fy  FN cos   mg  0  FN  cos 
 Fx   FR  FN sin   max
y

x
FN
mg

The amount of centripetal force needed for the car to round the curve is as follows.
2

 1.0 m s  
85km h   3.6 km h  

   9.679  103 N
FR  m v 2 r  1250 kg  
72 m
The actual horizontal force available from the normal force is as follows.
mg
FN sin  
sin   mg tan   1250 kg   9.80 m s 2  tan14  3.054  103 N
cos 
Thus more force is necessary for the car to round the curve than
y
can be supplied by the normal force. That extra force will have to
x
have a horizontal component to the right in order to provide the
extra centripetal force. Accordingly, we add a frictional force
pointed down the plane. That corresponds to the car not being
able to make the curve without friction.
mg
Again write Newton’s second law for both directions, and again
the y
acceleration is zero.

FN


Ffr
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F
 FN cos   mg  Ffr sin   0  FN 
F
 FN sin   Ffr cos   m v r
y
x
mg  Ffr sin 
cos 
2
Substitute the expression for the normal force from the y equation into the x equation, and
solve for the friction force.
mg  Ffr sin 
v2
sin   Ffr cos   m v 2 r   mg  Ffr sin   sin   Ffr cos2   m cos 
cos 
r
Ffr  m
v2
r




cos   mg sin   9.679  103 N cos14  1250 kg  9.80 m s 2 sin14
 6.428  103 N
So a frictional force of 6.4  103 N down the plane is needed to provide the necessary
centripetal force to round the curve at the specified speed.
92. From Example 5-15 in the textbook, the no-friction banking angle is given by   tan
v02
1
Rg
.
The centripetal force in this case is provided by a component of
y

the normal force. Driving at a higher speed with the same radius
x
FN
requires more centripetal force than that provided by the normal
force alone. The additional centripetal force is supplied by a force
of static friction, downward along the incline. See the free-body

mg
diagram for the car on the incline. The center of the circle of the
Ffr
car’s motion is to the right of the car in the diagram. Write

Newton’s second law in both the x and y directions. The car will
have no acceleration in the y direction, and centripetal acceleration in the x direction.
Assume that the car is on the verge of skidding, so that the static frictional force has its
maximum value of Ffr   s FN .
F
y
FN 
F
x
FN 
 FN cos   mg  Ffr sin   0  FN cos    s FN sin   mg 
mg
 cos   s sin  
 FR  FN sin   Ffr cos   m v 2 R  FN sin    s FN cos   m v 2 R 
mv 2 R
 sin    s cos  
Equate the two expressions for the normal force, and solve for the speed, which is the
maximum speed that the car can have.
mv 2 R
mg


 sin    s cos    cos    s sin  
vmax 
Rg
sin  1   s tan  
cos  1   s tan  
 v0
1  Rg 
1   v
2
s 0
s

Rg 
v02
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Driving at a slower speed with the same radius requires less
centripetal force than that provided by the normal force alone.
The decrease in centripetal force is supplied by a force of static
friction, upward along the incline. See the free-body diagram
for the car on the incline. Write Newton’s second law in both
the x and y directions. The car will have no acceleration in the
y direction, and centripetal acceleration in the x direction.
Assume that the car is on the verge of skidding, so that the
static frictional force is given by Ffr   s FN .
F
FN

y
x
mg

 FN cos   mg  Ffr sin   0 
y
FN cos    s FN sin   mg 
F
FN 
mg
 cos   s sin  
 FR  FN sin   Ffr cos   m v 2 R  FN sin    s FN cos   m v 2 R 
x
FN 

Ffr
mv 2 R
 sin    s cos  
Equate the two expressions for the normal force, and solve for the speed.
mv 2 R
mg


 sin    s cos    cos    s sin  
vmin 
Rg
Thus vmin  v0
sin  1   s tan  
cos  1   s tan  
1   Rg v 
1   v Rg 
s
2
s 0
2
0
 v0
1   Rg v 
1   v Rg 
2
0
s
2
s 0
and vmax  v0
1  Rg 
1   v
2
s 0
s

Rg 
v02
.
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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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