Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 105: 74

Answer

(a) The rope sags 1.53 meters. (b) The tension in the rope is 11,600 N, which is still less than the maximum tension of 29,000 N. Therefore the rope should not break.

Work Step by Step

Bob's weight will be supported by the vertical force provided by each side of the rope. We can use a force equation to find the angle that the rope makes with the horizontal. Let $F_T$ be the tension in the rope (which we will set at 2900 N): $2F_T~sin(\theta) = mg$ $sin(\theta) = \frac{mg}{2F_T}$ $\theta = sin^{-1}(\frac{(72.0~kg)(9.80~m/s^2)}{(2)(2900~N)})$ $\theta = 7.0^{\circ}$ We can use this angle to find the value of $x$ which is the amount of sag in the middle of the rope: $\frac{x}{12.5~m} = tan(7.0^{\circ})$ $x = (12.5~m)~tan(7.0^{\circ})$ $x = 1.53~m$ (b) If the rope sags only 1/4 of the distance in part (a), then the rope sags only $\frac{1.53~m}{4}$ which is 0.38 m. The angle that the rope makes with the horizontal is: $tan(\theta) = \frac{0.38~m}{12.5~m}$ $\theta = tan^{-1}(0.0304) = 1.74^{\circ}$ We can find the tension in the rope. $2F_T~sin(\theta) = mg$ $F_T = \frac{(72.0~kg)(9.80~m/s^2)}{2~sin(1.74^{\circ})}$ $F_T = 11,600~N$ The tension in the rope is 11,600 N, which is still less than the maximum tension of 29,000 N. Therefore the rope should not break.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.