Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 162: 30

Answer

(a) The tension in Ellen's rope is 533 N (b) The tension in the cable is 5250 N

Work Step by Step

(a) To keep the piano descending vertically, the horizontal component of Bob's rope tension and the horizontal component of Ellen's rope tension must be equal in magnitude. We can find the tension $T_E$ in Ellen's rope. $T_E~cos(25^{\circ}) = T_B~cos(15^{\circ})$ $T_E = \frac{T_B~cos(15^{\circ})}{cos(25^{\circ})}$ $T_E = \frac{500~N~cos(15^{\circ})}{cos(25^{\circ})}$ $T_E = 533~N$ The tension in Ellen's rope is 533 N. (b) Since the piano is moving at a constant speed, the tension in the cable $T_C$ must be equal in magnitude to the sum of the downward forces. $T_C = T_E~sin(25^{\circ}) + T_B~sin(15^{\circ})+mg$ $T_C = (533~N)~sin(25^{\circ}) + (500~N)~sin(15^{\circ})+(500~kg)(9.80~m/s^2)$ $T_C = 5250~N$ The tension in the cable is 5250 N.
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