IDENTIFY: Use = Fl = rF sin to calculate the magnitude of each torque and use the right-hand rule to
determine the direction of each torque. Add the torques to find the net torque.
SET UP: Let counterclockwise torques be positive. For the 11.9 N forc
IDENTIFY: Use the equations in Table 9.2. I for the rod is the sum of I for each segment. The parallel-axis
theorem says I p = I cm + Md 2 .
SET UP: The bent rod and axes a and b are shown in Figure 9.59. Each segment has length L / 2 and mass M / 2
z1 + z 2
. 0 = av-z t . When z is linear in t, av-z for the time interval t1 to t2 is
SET UP: From the information given, z (t ) = 6.00 rad/s + (2.00 rad/s 2 )t
EXECUTE: (a) The angular acceleration is positive, sin
IDENTIFY: Eqs. 8.24 and 8.25 apply, with object A being the neutron.
SET UP: Let +x be the direction of the initial momentum of the neutron. The mass of a neutron is mn = 1.0 u .
1.0 u 2.0 u
EXECUTE: (a) v A 2 x = A
v A1x = v A1x / 3.0 . The s
IDENTIFY: Apply Eq. 8.9 to relate the change in momentum of the momentum to the components of the average
force on it.
SET UP: Let +x be to the right and +y be upward.
EXECUTE: (a) J x = px = mv2 x mv1x = (0.145 kg)( [65.0 m/s]cos30 50.0 m/s) = 15.4
IDENTIFY: For the system of two blocks, only gravity does work. Apply Eq.(7.5).
SET UP: Call the blocks A and B, where A is the more massive one. v A1 = vB1 = 0 . Let y = 0 for each block to be
at the initial height of that block, so y A1 = yB1 = 0 .
IDENTIFY: All forces are constant and each block moves in a straight line. so W = Fs cos . The only direction the
system can move at constant speed is for the 12.0 N block to descend and the 20.0 N block to move to the right.
SET UP: Since the 12.0 N
F = ma to the point where the three wires join and also to one of the balls. By symmetry
the tension in each of the 35.0 cm wires is the same.
SET UP: The geometry of the situation is sketched in Figure 5.81a. The angle that each wir
IDENTIFY: If the box moves in the + x-direction it must have a y = 0, so
The smallest force the child can exert and still
produce such motion is a force that makes the
y-components of all three forces sum to zero,
but that doesnt have any x-
IDENTIFY: Apply Eq. (3.30).
SET UP: T = 24 h .
4 2 (6.38 106 m)
= 0.034 m/s 2 = 3.4 103 g .
EXECUTE: (a) arad =
(24 h)(3600 s/h) 2
4 2 (6.38 106 m)
= 5070 s =1.4 h.
9.80 m/s 2
EVALUATE: arad is proportional to 1/ T 2 , so to increase arad by a fact
IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment
plus the distance the ship moves while the equipment is in the air.
SET UP: For the motion of the equipment take + x to be to the right and +
IDENTIFY: Apply x x0 = v0 xt + 1 axt 2 to the motion of each train. A collision means the front of the passenger
train is at the same location as the caboose of the freight train at some common time.
SET UP: Let P be the passenger train and F be th