RECITATION
CHAPTER 5
APPLICATIONS OF NEWTON’S LAWS
5.3. Two 25.0 N weights are suspended at opposite ends of a rope that passes over a light,
frictionless pulley. The pulley is attached to a chain that is fastened to the ceiling. (See the
figure.) Start solving this problem by making a free-body diagram of each weight.
(a) What is the tension in the rope? (b) What is the tension in the chain?
For each object a = 0. Apply
F
y
 ma y to each weight and to the pulley. Take upward
to be the +y direction. The pulley has negligible mass. Let Tr be the tension in the rope
and let Tc be the tension in the chain.
(a) the free-body diagram for each weight is the same and is given here
Tr
w
F
 ma y gives Tr  w  25.0N .
(b) The free-body diagram for the pulley is given here
y
Tc
2Tr
Tc  2Tr  50.0 N.
1
5-12 In a rescue, the 73 kg police officer is suspended by two cables, as shown in the
figure below. (a) Sketch a free-body diagram of him. (b) Find the tension in each
cable.
(a)
Following is the free body diagram.
 cos 35 
(b)  Fx  0 gives T2 cos 48  T1 cos 35  0; T2  
T1
 cos 48 
 Fy  0 gives T1 sin 35  T2 sin 48  w  0.
Solving for T1 we get,


 cos 35 
T1  sin 35  
 sin 48   w and T1  482 N . Then T2  590 N .
 cos 48 


2
5.13. A tetherball leans against the smooth, frictionless post to which it is attached. (See the
figure) The string is attached to the ball such that a line along the string passes through the
center of the ball. The string is 1.40 m long and the ball has a radius of 0.110 m with mass
0.270 kg. (a) Make a free body diagram of the ball. b) What is the tension in the rope?
(c) What is the force the pole exerts on the ball?


Apply  F  ma to the ball. Since the post is frictionless, the force it exerts on the ball
is a horizontal normal force. Use coordinates with +y upward and +x to the right.
y
Tcos θ
T
θ
n
Tsin θ
mg
l is the length of the string and r is the radius of the ball.
0.110m
 r 

1 
  sin 1 
  sin 
  4.18
r l
 1.40m  0.110m 
(a) The free-body diagram of the ball is given above in the figure.
(b)
F
 ma y
T cos   mg  0
y



mg
0.27kg  9.80m / s 2
T

 2.65 N
cos 
cos 4.18
3
x
(c)
F
 max
T sin   n  0
n  T sin   2.65 N sin 4.18  0.193N
x
5-15 Two blocks, each with weight w, are held in place on a frictionless incline as shown in
the figure. In terms of w and the angle  of the incline, calculate the tension in (a) the rope
connecting the blocks and (b) the rope that connects block A to the wall. (c) Calculate the
magnitude of the force that the incline exerts on each block. Interpret your answers for the
cases   0 and   90.
y
x
When choosing a coordinate system for an inclined surface, it is generally best to have the
x and y axis of the system parallel and perpendicular to the surface respectively. One can
imagine the coordinate system to be “bolted down” to the surface, so that when the surface
is tilted the coordinate system tilts along with it.
Consider the +x direction down the incline. The weight vector makes an angle  with the
–y axis. Suppose the tension in the rope connecting the two blocks has magnitude T1 and
the tension in the rope connecting the block A to the wall has a magnitude T2 .
(a) For B,
F
x
 max  0 and hence T1  w sin  .
(b) For block A,  Fx  max gives T1  T2  w sin   0 and T2  2w sin  . Since both
the blocks have the same weight, the x component of the weight vector for both
the blocks is the same.
(c)  Fy  ma y for each block gives n A  nB  w cos  .
(d) For   0, T1  T2  0 and n A  nB  w.
For   90, T1  w, T2  2w and nA  nB  0.
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5.22. A short train (an engine plus four cars) is accelerating at 1.10 m/s2. The mass of each
car is 38,000 kg, and each car has negligible frictional forces acting on it.
(a) What is the force of the engine on the first car? (b) What is the force of the first car on
the second car? (c) What is the force of the second car on the third car? (d) And what is
the force of the third car on the fourth car?
In solving this problem, note the importance of selecting the correct set of cars to isolate as
your object.
Let the acceleration be in the +x direction.
(a) Apply  Fx  max to the four cars taken as a single object. m is the mass of one car
and let F1 be the force the engine exerts on the first car.
F
x
 max gives F1  4m a and F1  438,000kg1.10m / s 2   1.67  10 5 N .
(b) Apply  Fx  max to the last three cars taken as a single object. Let F2 be the
force the first car exerts on the second car.
 Fx  max give F2  3ma  338,000kg 1.10m / s 2  1.25 105 N.

(c) Apply
F
x

 max to the last two cars taken as a single object. F3 is the force that
the second car exerts on the third car.


F
x
 max gives the force F3.
F3  (2m)a  238,000kg  1.10m / s 2  8.36  10 4 N .
 F  ma to the fourth car. F4 is the force that the third car exerts on the
fourth car.  F  ma gives F  ma  38,000kg1.10m / s   4.18  10 N .
(d) Apply
x
x
2
x
x
4
5
4
5.33. At a construction site, a pallet of bricks is to be suspended by attaching a rope to it
and connecting the other end to a couple of heavy crates on the roof of a building, as shown
in the figure. The rope pulls horizontally on the lower crate, and the coefficient of static
friction between the lower crate and the roof is 0.666. (a) What is the weight of the
heaviest pallet of bricks that can be supported this way? Start with appropriate free-body
diagrams. (b) What is the friction force on the upper crate under the conditions of part (A)?
(a) The free-body diagram for the two crates (treated as a single object) and the bricks is
shown below. wc and wb represent the weights of the crates and the bricks
respectively.
y
y
n
T
T
x
fs
x
wb
wc
The system doesn’t move so the friction force exerted by the roof is static friction.
For the heaviest pallet of bricks this force has its maximum possible, f s   s n .The
free-body diagram for the pallet of bricks is given in Figure (right).Solve: (a) For the
crates,  Fy  ma y gives n  wc  0 and n  400lb . Then
f s   s n  0.666400lb   266lb.
F
x
 max gives T  f s  0 and T  f s  266lb .
For the bricks,
F
y
 ma y gives T  wb  0 and wb  T  266lb
(b) For the upper crate the only horizontal force on the crate would be friction. This crate
has a x  0 so  Fx  0 and the friction force is zero.
6
5.35. A hockey puck leaves a player's stick with a speed of 9.9 m/s and slides 32.0 m
before coming to rest. Find the coefficient of friction between the puck and the ice.
Use the information about the motion to find the acceleration of the puck and then use


 F  ma to relate a to the friction force. Take +x to be in the direction the puck is
moving.
v0 x  9.9m / s, x  32.0m and vx  0.
v x2  v02x 0  9.9m / s 2

 1.53m / s 2 .
2x
232.0m 
 Fy  may gives n – mg = 0 and n = mg.
v x2  v02x  2a x x gives a x 
F
x
 max gives   k n  max
. k  
max
a
1.53m / s 2
 x 
 0.16.
mg
g 9.80m / s 2
5-36. Stopping distance of a car. (a) If the coefficient of kinetic friction between tires
and dry pavement is 0.80, what is the shortest distance in which you can stop an
automobile by locking the brakes when traveling at 29.1 m/s (about 65 mi/h)? (b) On wet
pavement, the coefficient of kinetic friction may be only 0.25. How fast should you drive
on wet pavement in order to be able to stop in the same distance as in part (a)? (Note:
Locking the brakes is not the safest way to stop.)
(a)
F
y
F
x
 ma y gives n = mg.
 max gives   k n  max . This equation gives
  k n   k mg

  k g. Then
m
m
v x  0 and v x2  v02x  2a x x gives
ax 
2
v02x
v02x

29.1m / s 
x  


 54.0m.
2a x
2 k g 20.80 9.80m / s 2

(b) Solving the equation x  


v02x
for v0x gives
2 k g

v0 x  2 k gx  20.25 9.80m / s 2 54.0m  16.3m / s.
7
5.48. A person pushes on a stationary 125 N box with 75 N at 30° below the horizontal, as
shown in the figure. The coefficient of static friction between the box and the horizontal
floor is 0.80. (a) Make a free body diagram of the box. (b) What is the normal force on
the box? (c) What is the friction force on the box? (d) What is the largest the friction force
could be? (e) The person now replaces his push with a 75 N pull at 30° above the
horizontal. Find the normal force on the box in this case.
Use coordinates where +y is upward and +x is horizontal to the right. The applied force
pushes to the left so the friction force is to the right.
(a) The free-body diagram is given here.
+y
n
a=0
Fcos30°
θ
f
+x
~
Fsin30°
F
w
`
The applied force F has been replaced by its x and y components.
(b)  Fy  ma y gives n  F sin   w  0 and
n  w  F sin   125N  75N sin 30  163N.
(c)
F
x
 max gives f  F cos 30  0 and f  F cos 30  65 N .
(d) The maximum possible static friction force is f s   s n  0.8163N   130 N .
(e) The person now replaces his push with a 75 N pull at 30° above the horizontal. This
results in the vertical component of the pull in the upward direction resulting in less
normal force as calculated below:
.  Fy  ma y gives n  F sin   w  0 and n  w  F sin 30  88N .
8
5-73. A student attaches a series of weights to a tendon and measures the total length of the
tendon for each weight. He then uses the data he has gathered to construct the graph shown
in the figure, giving the weight as a function of the length of the tendon. (a) Does this
tendon obey Hooke’s law? How do you know? (b) What is the force constant (in N/m) for
the tendon? (c) What weight should you hang from the tendon to make it stretch by 8.0
cm from its unstretched length?
The unstretched length L0 of the tendon is 20.0 cm. The weight W suspended from the end
of the tendon equals the force applied by the tendon.
(a) x = L - L0, so Hooke’s law says W = kx = kL - kL0. The graph of W versus L is of this
form.
15.0 N
 50.0 N / m .
(b) k is the slope of W versus x, so k 
0.500m  0.200m
(c) W = k x = (50.0 N/m) (0.080m) = 4.0 N.
9