Physics 1 Homework Solutions
4.24
Ch 4: 24, 27, 29
First consider the block moving along the
horizontal. The only force in the direction
of movement is T. Thus,
Fx  ma x  T   5.00 kg  a
(1)
Next consider the block which moves
vertically. The forces on it are the tension
T and its weight, 98.0 N.
Fy  ma y  98.0 N  T  10.0 kg  a
(2)
Note that both blocks must have the same magnitude of acceleration. Equations
(1) and (2) can be solved simultaneously to give.
a  6.53 m s2 , and T  32.7 N
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 