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 7 - Newton's Third Law - Exercises and Problems: 48

Answer

The tension in the rope is 3590 N.

Work Step by Step

The vertical component of the tension in each side of the rope provides the force to accelerate the person up in the air. Let $T$ be the tension in each side of the rope. Note that $T_y = T~sin(\theta)$. We can set up a force equation to find $T$: $\sum F = ma$ $2T_y - mg = ma$ $2T~sin(\theta) = m(g+a)$ $T = \frac{m(g+a)}{2~sin(\theta)}$ $T = \frac{(70~kg)(9.80~m/s^2+8.0~m/s^2)}{2~sin(10^{\circ})}$ $T = 3590~N$ The tension in the rope is 3590 N
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