32. REASONING AND SOLUTION According to Equations 4.4 and 4.5, the acceleration due to gravity at the surface of the neutron star is
a=
Gm (6.67 1011 N m 2 / kg 2 )(2.0 1030 kg) = = 5.3 1012 m/s 2 2 3 2 r (5.0 10 m)
Since the gravitational force is assume
47.
SSM
REASONING AND SOLUTION Since the magnitude of the centripetal
2 acceleration is given by Equation 5.2, a C = v / r , we can solve for r and find that
r=
v2 ( 98.8 m / s ) 2 = = 332 m a C 3.00(9.80 m / s 2 )
_ _
48. REASONING The centripetal force is the name given to the net force pointing toward the center of the circular path. At the lowest point the net force consists of the tension in the arm pointing upward toward the center and the weight pointing downward
48. REASONING At first glance there seems to be very little information given. However, it is enough. In part a of the drawing the bucket is hanging stationary and, therefore, is in equilibrium. The forces acting on it are its weight and the two tension f
50. REASONING AND SOLUTION The period of the moon's motion (approximately the length of a month) is given by
4 2 r 3 T= = GM E
4 2 ( 3.85 108 m ) ( 6.67 1011 N m 2 / kg 2 )( 5.98 1024 kg )
3
= 2.38 106 s = 27.5 days
_ _
51. REASONING AND SOLUTION given by T = 2 r/v. The speed is
The sample makes one revolution in time T as so that v = 55.3 m/s
v2 = rac = (5.00 102 m)(6.25 103)(9.80 m/s2) The period is
T = 2 (5.00 102 m)/(55.3 m/s) = 5.68 103 s = 9.47 105 min The number o
52. REASONING AND SOLUTION The free body diagram for the plane is shown below to the left. The figure at the right shows the forces resolved into components parallel to and perpendicular to the line of motion of the plane. L T R W L T
R
W sin
W cos
If t
54. REASONING AND SOLUTION a. The centripetal force the normal force exerted on the rider by the wall .
is
provided
by
b. Newton's second law applied in the horizontal direction gives FN = mv2/r = (55.0 kg)(10.0 m/s)2/(3.30 m) = 1670 N
c. Newton's second
56. CONCEPT QUESTIONS a. The period for the second hand is the time it takes for it to go once around the circle or Tsecond = 60 s. b. The period for the minute hand is the time it takes for it to go once around the circle or Tminute = 1 h = 3600 s. c. Th
46. REASONING AND SOLUTION Newtons second law applied in the vertical and horizontal directions gives L cos 21.0 W = 0 L sin 21.0 R = 0 a. Equation (1) gives
21.0
(1) (2)
L=
W 53 800 N = = 57 600 N cos 21.0 cos 21.0
R W
L
b. Equation (2) gives
R = L sin 2
45. REASONING AND SOLUTION Let s represent the length of the path of the pebble after it is released. From Conceptual Example 2, we know that the pebble will fly off tangentially. Therefore, the path s is perperpendicular to the radius r of the circle. Th
44. REASONING AND SOLUTION The normal force exerted by the wall on each astronaut is the centripetal force needed to keep him in the circular path, i.e., Fc = mv2/r. Rearranging and letting Fc = (1/2)mg yields r = 2v2/g = 2(35.8 m/s)2/(9.80 m/s2) = 262 m
33. REASONING We place the third particle (mass = m3) as shown in the following drawing: L D
m3 m 2m The magnitude of the gravitational force that one particle exerts on another is given by Newtons law of gravitation as F = Gm1m2/r2. Before the third part
34. REASONING Equation 5.2 for the centripetal acceleration applies to both the plane and the satellite, and the centripetal acceleration is the same for each. Thus, we have
ac =
2 v plane
rplane
=
2 v satellite
rsatellite
or
v plane =
F GGH
rplane rsatel
34. REASONING In each case the object is in equilibrium. According to Equation 4.9b, Fy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of the object(s) and the normal force exerted on them. SO
36. REASONING According to Equation 5.3, the magnitude Fc of the centripetal force that acts on each passenger is Fc = mv 2 / r , where m and v are the mass and speed of a passenger and r is the radius of the turn. From this relation we see that the speed
36.
REASONING AND SOLUTION a. The apparent weight of the person is given by Equation 4.6 as FN = mg + ma = (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 103 N b. c. FN = (95.0 kg)(9.80 m/s2) = 931 N FN = (95.0 kg)(9.80 m/s2 1.30 m/s2) = 808 N
_ _
37. REASONING AND SOLUTION The block will move only if the applied force is greater than the maximum static frictional force acting on the block. That is, if F > sFN = smg = (0.650)(45.0 N) = 29.2 N The applied force is given to be F = 36.0 N which is gre
38. REASONING The normal force (magnitude FN) that the pilots seat exerts on him is part of the centripetal force that keeps him on the vertical circular path. However, there is another contribution to the centripetal force, as the drawing at the right sh
42. REASONING AND SOLUTION The deceleration produced by the frictional force is
a= fk m = k mg m = k g
The speed of the automobile after 1.30 s have elapsed is given by Equation 2.4 as
v = v0 + at = v0 + ( k g ) t = 16.1 m/s (.720 ) ( 9.80 m/s 2 ) (1.30 s
56. REASONING Since the wire beneath the limb is at rest, it is in equilibrium and the net force acting on it must be zero. Three forces comprise the net force, the 151-N force from the limb, the 447-N tension force from the left section of the wire, and
57 SSM REASONING There are four forces that act on the chandelier; they are the forces of tension T in each of the three wires, and the downward force of gravity mg. Under the influence of these forces, the chandelier is at rest and, therefore, in equilib
59. REASONING The toboggan has a constant velocity, so it has no acceleration and is in equilibrium. Therefore, the forces acting on the toboggan must balance, that is, the net force acting on the toboggan must be zero. There are three forces present, the
86.
REASONING AND SOLUTION a. The static frictional force is responsible for accelerating the top block so that it does not slip against the bottom one. The maximum force that can be supplied by friction is fsMAX = sFN = sm1g Newton's second law requires
93. REASONING AND SOLUTION The acceleration needed so that the craft touches down with zero velocity is 2 v 2 v0 (18.0 m/s )2 a= = = 0.982 m/s 2 ( 165 m ) 2s 2
Newton's second law applied in the vertical direction gives F mg = ma Then F = m(a + g) = (1.14
94.
REASONING AND SOLUTION The apparent weight is FN = mw(g + a)
We need to find the acceleration a. Let T represent the force applied by the hoisting cable. Newton's second law applied to the elevator gives T (mw + me)g = (mw + me)a Solving for a gives
a
96. REASONING The magnitude F of the net force acting on the kayak is given by Newtons second law as F = ma (Equation 4.1), where m is the combined mass of the person and kayak, and a is their acceleration. Since the initial and final velocities, v0 and v
97. REASONING According to Equation 4.4, the weights of an object of mass m on the surfaces of planet A (mass = MA, radius = R ) and planet B (mass = MB , radius = R ) are WA = GM A m R
2
and
WB =
GM B m R2
The difference between these weights is given in
102. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. Three forces comprise the net force, her weight, and the tension forces from the left and right sides of the rope. We will resolve th
104. REASONING AND SOLUTION The figure to the left below shows the forces that act on the sports car as it accelerates up the hill. The figure to the right below shows these forces resolved into components parallel to and perpendicular to the line of moti