Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 29

Answer

(a) $\tau = 14~m\cdot N$ (counterclockwise) (b) $\tau = 13~m\cdot N$ (clockwise) Since the net torque is clockwise, we can express this as $\tau = -13~m\cdot N$

Work Step by Step

(a) $\tau = \sum r\cdot F~sin(\theta)$ $\tau = (1.0~m)(52~N)~sin(58^{\circ}) - (1.0~m)(56~N)~sin(32^{\circ})$ $\tau = 14~m\cdot N$ (counterclockwise) (a) $\tau = \sum r\cdot F~sin(\theta)$ $\tau = (2.0~m)(56~N)~sin(32^{\circ}) - (1.0~m)(65~N)~sin(45^{\circ})$ $\tau = 13~m\cdot N$ (clockwise)
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