University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 164: 5.60

Answer

(a) $T = 2540~N$ (b) The minimum angle is $\theta = 1.01^{\circ}$.

Work Step by Step

(a) The tension $T$ is equal in both sides of the rope. The sum of the vertical components in each side of the rope will be equal in magnitude to the person's weight. $2T~sin(\theta) = mg$ $T = \frac{mg}{2~sin(\theta)}=\frac{(90.0~kg)(9.80~m/s^2)}{2~sin(10.0^{\circ})}$ $T = 2540~N$ (b) Let's suppose the tension in the rope is $2.50\times 10^4~N$. $2T~sin(\theta) = mg$ $sin(\theta) = \frac{mg}{2T}$ $\theta =arcsin(\frac{(90.0~kg)(9.80~m/s^2)}{(2)(2.50\times 10^4~N)})$ $\theta = 1.01^{\circ}$ The minimum angle is $\theta = 1.01^{\circ}$.
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