| Last revised 1999/09/24 |
39.
Let T be the tension in the rope, Ff be the force due to floor, theta be the angle of T, and m be the mass of the crate. Then
a = [T cos(theta) - Ff] / m = {[T cos(theta) - Ff] g } / {m g }
Use the first form for part (a) and the second for part (b).
40.
Call the force exerted by the blocks on each other N. Then
(a) N = F m2 / (m1 + m2)
(b) N = F m1 / (m1 + m2)
In the two cases, the acceleration is the same. However, the force between the blocks is responsible for accelerating m2 in part (a) and m1 in part (b).
54.
(a) W = m g C = 3260 N;
(b) m = (T2 - T1 ) / a2 = 2.7 x 103 kg;
(c) g C = T1 a2 / (T2 - T1 ) = -1.2 m/sec2
All of these agree with the text's answers.
60.
(a) d = v02 / [ 2 g sin (theta) ] = 1.18 m
(b) t = v0 / [ g sin (theta) ] = 0.67 sec
(c) The final velocity is the reverse of the initial, 3.50 m/s down the plane.
73.
(a) Ask in class about this one.
(b) a = F / ( M + m )
(c) Tend = M F / ( M + m )
(d) Tcenter = (1/2) F (m + 2 M ) / (m + M )
VIII.
One boy pushes against the ground harder than the other, thus the ground pushes back on the first harder than the ground pushes on the second. The sum of the forces on the two boys is not zero, so an acceleration does occur.