Page 71 - Schaum's Outline of Theory and Problems of Applied Physics

56 LAWS OF MOTION [CHAP. 5

Figure 5-2 is a free-body diagram of the situation. Because the maximum tension in the cable is T = 14,000 N,
the net force F available to accelerate the elevator upward is a maximum of

F = T − w = 14,000 N − 11,760 N = 2240 N
The elevator’s acceleration when this net force is applied to it is
F 2240 N 2
a = = = 1.87 m/s
m 1200 kg
(b) The cable is flexible and so cannot push down on the elevator. The maximum downward acceleration therefore
2
corresponds to free fall with an acceleration of g = 9.8 m/s .

Fig. 5-2

SOLVED PROBLEM 5.18
Figure 5-3 shows a 5-kg block A which hangs from a string which passes over a frictionless pulley and
is joined at its other end to a 12-kg block B which lies on a frictionless table. (a) Find the accelerations
of the two blocks. (b) Find the tension in the string. What would the tension be if block B were fixed in
place?

(a) The blocks have accelerations of the same magnitude a because they are joined by the string. The net force
F B on B equals the tension T in the string. From the second law of motion, taking the left as the + direction
so that a will come out positive,
F B = T = m B a
The net force F A on A is the difference between its weight m A g, which acts downward, and the tension T in
the string, which acts upward on it. Taking downward as + so that the two accelerations have the same sign,
F A = m A g − T = m A a

We now have two equations in the two unknowns, a and T. The easiest way to solve them is to start by
substituting T = m B a from the first equation into the second. This gives
m A g − m B a = m A a

m A g = (m A + m B )a
2
m A g (5kg)(9.8 m/s )
a = = = 2.9 m/s 2
m A + m B 5kg + 12 kg