Page 72 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 5] LAWS OF MOTION 57

Fig. 5-3

(b) We can use either of the original equations to find the tension T. From the first,
2
T = m B a = (12 kg)(2.9 m/s ) = 35 N
If B were fixed in place, the tension would equal the weight of A, which is m A g = 49 N. Here the tension is
less because B moves in response to the pull of A’s weight, but it is not zero because of B’s inertia.

SOLVED PROBLEM 5.19

Figure 5-4 shows the same two blocks, A and B, from Fig. 5-3 but now suspended by a string on either
side of a frictionless pulley. (a) Find the accelerations of the two blocks. (b) Find the tension in the string.

(a) Here A moves upward and B moves downward, so we let upward be + for A and downward be − for B. Both
blocks have accelerations of the same magnitude a. Applying the second law of motion to the two blocks with
the help of the free-body diagrams in Fig. 5-4 gives
F A = T − m A g = m A a
F B = m B g − T = m B a

From the first equation we have for the tension T

T = m A a + m A g
Substituting this value of T in the second equation gives

m B g − m A a − m A g = m B a
(m B − m A )g = (m A + m B )a
2
(m B − m A )g (12 kg − 5kg)(9.8 m/s )
a = = = 4.0 m/s 2
m A + m B 5kg + 12 kg