College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 2 - Section 2.7 - Contact Forces - Practice Problem - Page 56: 2.15


The tension in the cable is $3125~N$

Work Step by Step

When the cable sags, both ends of the cable make an angle of $\theta$ below the horizontal. Note that $sin~\theta = \frac{0.12~m}{3.00~m}$ We can consider the left side and the right side of the cable. The sum of the vertical component of the tension $T$ in each side of the cable is equal in magnitude to Denisha's weight. We can find the tension $T$ in the cable: $2T~sin~\theta = 250~N$ $T = \frac{250~N}{2~sin~\theta}$ $T = \frac{250~N}{(2)~(\frac{0.12~m}{3.00~m})}$ $T = 3125~N$ The tension in the cable is $3125~N$.
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