Bukhovtsev-et-al-Problems-in-Elementary-Physics

MECHANICS
183
Fig. 294
Fig. 295
(see Problem 34). Hence,
at
a 2
4ml m3- 3m2m3+ ml m2
g
4ml m3+m 2m3+mtm t
mlm2-4mlm3+m2m3
g
4m1m 3+ m2mS+ m 1m2
4mlmS-3mlm2+m2m3
as= 4mlmS+m2mS+mlm2 g
69. The second monkey will be at the same height as the first.
If the mass of the pulley and the weight of the rope are disregarded,
the force T tensioning both ends of the rope will be the same and, therefore,
the forces acting on each monkey will equal F=T-mg. Both monkeys have
the same accelerations in magnitude and direction and will reach the pul..
ley at the same time.
70. Since the mass of the pulleys and the string is negligibly small, the
tension of the thread is the same everywhere.
Therefore,
mlg-T=m1a1
mgg- 2T= m 2a2
2T-T=O
Hence, T=O and a1=a2= g.
Both weights fall freely with an acceleration g. Pulleys Band C rotate
counterclockwise and pulley A clockwise.
71. (1) The forces acting on the table and the weight are shown in Fig. 295.
The equations of horizontal motion have the following form:
for the table with the pulley
and for' {h'c weight
0 1
F-F+F/r=-a t g