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 161: 3


The tension in each cable is 147 N

Work Step by Step

We can find the angle $\theta$ that each cable makes with the vertical. $cos(\theta) = \frac{2.0}{3.0}$ $\theta = arccos(\frac{2.0}{3.0})$ $\theta = 48.2^{\circ}$ The sum of the vertical component of the tension in each cable is equal in magnitude to the weight of the speaker. $2~T_y = mg$ $2~T~cos(\theta) = mg$ $T = \frac{mg}{2~cos(\theta) }$ $T = \frac{(20~kg)(9.80~m/s^2)}{2~cos(48.2^{\circ})}$ $T = 147~N$ The tension in each cable is 147 N.
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