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Homework Set 4 part 2
4.27
Choose the +x direction to be horizontal and forward with the +y vertical and
upward. The common acceleration of the car and trailer then has components of
a x  2.15 m s2 and a y  0 .
(a) The net force on the car is horizontal and given by
Fx car  F  T  mcar ax  1000 kg   2.15 m s2  
2.15 103 N forward
(b) The net force on the trailer is also horizontal and given by
 Fx trailer  T  mtrailer ax   300 kg   2.15 m s2  
645 N forward
(c) Consider the free-body diagrams of the car and trailer. The only horizontal
force acting on the trailer is T  645 N forward , and this is exerted on the
trailer by the car. Newton’s third law then states that the force the trailer
exerts on the car is 645 N toward the rear
(d) The road exerts two forces on the car. These are F and nc shown in the freebody diagram of the car.
F  T  2.15 103 N   2.80 103 N
From part (a),
Also,  Fy 
car
 nc  wc  mcar a y  0 , so nc  wc  mcar g  9.80 103 N
The resultant force exerted on the car by the road is then
Rcar  F 2  nc2 
 2.80 10 N   9.80 10 N 
3
2
3
2
 1.02  104 N
n 
at   tan 1  c   tan 1  3.51  74.1 above the horizontal and forward.
F
Newton’s third law then states that the resultant force exerted on the road
by the car is
1.02  104 N at 74.1 below the horizontal and rearward
4.29
When the block is on the verge of moving, the static friction force has a
magnitude f s   f s max  s n .
Since equilibrium still exists and the applied force is 75 N, we have
Fx  75 N  f s  0 or  f s max  75 N
In this case, the normal force is just the weight of the crate, or n  mg . Thus, the
coefficient of static friction is
s 
 f s max
n

 f s max
mg

75 N
 0.38
 20 kg   9.80 m s2 
After motion exists, the friction force is that of kinetic friction, f k  k n
Since the crate moves with constant velocity when the applied force is 60 N, we
find that Fx  60 N  f k  0 or f k  60 N . Therefore, the coefficient of kinetic
friction is
k 
fk
f
60 N
 k 
 0.31
n mg  20 kg   9.80 m s2 
4.32
(a)
ax 
vx  v0 x 6.00 m s  12.0 m s

  1.20 m s2
t
5.00 s
(b) From Newton’s second law, Fx   f k  ma x , or f k  ma x .
The normal force exerted on the puck by the ice is n  mg , so the coefficient
of friction is
k 
(c)
4.39



2
f k m 1.20 m s

 0.122
n
m 9.80 m s2

 v  v   6.00 m s  12.0 m s 
x   vx av t   x 0 x  t  
 5.00 s   45.0 m
2
 2  

The acceleration of the system is found from
1
1
2
y  v0 y t  at 2 , or 1.00 m  0  a 1.20 s 
2
2
which gives a  1.39 m s2
Using the free body diagram of m2 , the second law gives
5.00 kg   9.80
or
m s2   T  5.00 kg  1.39 m s2 
T  42.1 N
Then applying the second law to the horizontal motion of m1
42.1 N  f  10.0 kg  1.39 m s2  , or f  28.2 N
Since n  m1 g  98.0 N , we have k 
f 28.2 N

 0.287
n 98.0 N