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PHYS-207 Honors Section LL, HW4 Solutions
5.6. Since the tire remains stationary, by Newton’s second law, the net force must be zero:
Fnet  FA  FB  FC  ma  0.
From the free-body diagram shown on the right, we have
0   Fnet, x  FC cos   FA cos 
0   Fnet, y  FA sin   FC sin   FB
FB , we first compute  . With FA  220 N ,
FC  170 N , and   47, we get
F cos  (220 N) cos 47.0
cos   A

 0.883    28.0
FC
170 N
To solve for
Substituting the value into the second force equation, we find
FB  FA sin   FC sin   (220 N)sin 47.0  (170 N)sin 28.0  241 N.
5.17. The free-body diagram of the problem is shown to
the right. Since the acceleration of the block is zero, the
components of the Newton’s second law equation yield
T – mg sin  = 0
FN – mg cos  = 0,
where T is the tension in the cord, and FN is the normal
force on the block.
(a) Solving the first equation for the tension in the string,
b gc
h
T  mg sin   8.5 kg 9.8 m / s2 sin 30  42 N .
(b) We solve the second equation in part (a) for the normal force FN:
FN  mg cos    8.5 kg   9.8 m/s 2  cos 30  72 N .
(c) When the cord is cut, it no longer exerts a force on the block and the block accelerates. The xcomponent equation of Newton’s second law becomes –mgsin = ma, so the acceleration becomes
a   g sin   (9.8 m/s2 )sin 30   4.9 m/s 2 .
The negative sign indicates the acceleration is down the plane. The magnitude is 4.9 m/s 2.
Note: The normal force FN on the block must be equal to mg cos  so that the block is in contact with
the surface of the incline at all time. When the cord is cut, the block has an acceleration a   g sin  ,
which in the limit
  90 becomes  g .
5.41. The mass of the bundle is m = (449 N)/(9.80 m/s2) = 45.8 kg and we choose +y upward.
(a) Newton’s second law, applied to the bundle, leads to
T  mg  ma  a 
387 N  449 N
45.8 kg
which yields a = –1.4 m/s2 (or |a| = 1.4 m/s2) for the acceleration. The minus sign in the result indicates
the acceleration vector points down. Any downward acceleration of magnitude greater than this is also
acceptable (since that would lead to even smaller values of tension).
(b) We use Eq. 2-16 (with x replaced by y = –6.1 m). We assume 0 = 0.
v  2ay  2  1.35 m/s2   6.1 m   4.1 m/s.
For downward accelerations greater than 1.4 m/s2, the speeds at impact will be larger than 4.1 m/s.
5.51. The free-body diagrams for
m1 and m2 are shown in the figures below. The only forces on the blocks
are the upward tension T and the downward gravitational forces
F1  m1 g and F2  m2 g . Applying
Newton’s second law, we obtain:
T  m1 g  m1a
m2 g  T  m2 a
which can be solved to yield
 m  m1 
a 2
g
m

m
 2
1
 2m1m2 
T 
g
m

m
 1
2 
(a) With m1  1.3 kg and m2  2.8 kg , the acceleration becomes
Substituting the result back, we have:
 2.80 kg  1.30 kg 
2
2
2
a
 (9.80 m/s )  3.59 m/s  3.6 m/s .
 2.80 kg  1.30 kg 
(b) Similarly, the tension in the cord is
T
2(1.30 kg)(2.80 kg)
(9.80 m/s 2 )  17.4 N  17 N.
1.30 kg  2.80 kg
5.87. From the reading when the elevator was at rest, we know the mass of the object is m = (65 N)/(9.8
m/s2) = 6.6 kg. We choose +y upward and note there are two forces on the object: mg downward and T
upward (in the cord that connects it to the balance; T is the reading on the scale by Newton’s third law).
(a) “Upward at constant speed” means constant velocity, which means no acceleration. Thus, the
situation is just as it was at rest: T = 65 N.
(b) The term “deceleration” is used when the acceleration vector points in the direction opposite to the
velocity vector. We’re told the velocity is upward, so the acceleration vector points downward (a = –2.4
m/s2). Newton’s second law gives
T  mg  ma  T  (6.6 kg)(9.8 m/s2  2.4 m/s2 )  49 N.

6.16. (a) In this situation, we take f s to point uphill and to be equal to its maximum value, in which case
fs, max =
 s FN applies, where s = 0.25. Applying Newton’s second law to the block of mass m = W/g = 8.2
kg, in the x and y directions, produces
Fmin 1  mg sin   f s , max  ma  0
FN  mg cos   0
which (with  = 20°) leads to
Fmin 1  mg  sin   s cos   8.6 N.

(b) Now we take f s to point downhill and to be equal to its maximum value, in which case fs, max = sFN
applies, where s = 0.25. Applying Newton’s second law to the block of mass m = W/g = 8.2 kg, in the x
and y directions, produces
Fmin 2  mg sin   f s , max  ma  0
FN  mg cos   0
which (with  = 20°) leads to
Fmin 2  mg  sin   s cos    46 N.
6.48. We will start by assuming that the normal force (on the car from the rail) points up. Note that
gravity points down, and the y axis is chosen positive upward. Also, the direction to the center of the
circle (the direction of centripetal acceleration) is down. Thus, Newton’s second law leads to
 v2 
FN  mg  m    .
 r 
(a) When v = 11 m/s, we obtain FN = 3.7  103 N.
(b)
FN points upward.
(c) When v = 14 m/s, we obtain FN = –1.3  103 N, or | FN | = 1.3  103 N.
(d) The fact that this answer is negative means that
FN points opposite to what we had assumed. Thus,
r
the magnitude of FN is | FN |  1.3 kN and its direction is down.
6.57. For the puck to remain at rest the magnitude of the tension force T of the cord must equal the
gravitational force Mg on the cylinder. The tension force supplies the centripetal force that keeps the puck
in its circular orbit, so T = mv2/r. Thus Mg = mv2/r. We solve for the speed:
v
Mgr
(2.50 kg)(9.80 m/s 2 )(0.200 m)

 1.81 m/s.
m
1.50 kg