Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 154: 4

Answer

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; $cos(\theta) = \frac{2}{3}$ $\theta = arccos(\frac{2}{3})$ $\theta = 48.2^{\circ}$ 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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