Atwood’s Machine. A 15.0-kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0-kg counterweight is suspended from the other end of the rope (Fig. E5.15). The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?
Solution 15E (a) The free body diagram can be drawn as shown below. (b) Let the upward acceleration for the load of the bricks is a. Now, from the free body diagram of the load of the bricks, we get T 15g = 15a…..(1) From the free body diagram of the counterweight, we get 28g T = 28a…..(2) Now, adding equations (1) and (2), we have 28g 15g = 28a + 15a 13g = 43a a = 13g/43 m/s 2 a = 2.96 m/s 2 …..(3) Therefore, the value of upward acceleration is 2.96 m/s .2 (c) Now, using the value of a from equation (3) in equation (1), we can calculate the tension in the rope. T 15g = 15a T = 15g + 15a T = 15 × 9.8 + 15 × 2.96 N T = 191 N The tension value in the string is 191 N. The weight of the load of the bricks = 15.0 kg × 9.8 m/s 2 = 147 N Therefore, the tension value is greater than the weight of the load of the bricks by = 191 147 N = 44 N The weight of the counterweight = 28.0 × 9.8 N = 274 N This value is greater than the tension value by = 274 191 N = 83 N .