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 12 - Rotation of a Rigid Body - Exercises and Problems - Page 333: 59

Answer

The tension in the cable is 15,300 N

Work Step by Step

To find the required tension $T$ in the cable to maintain equilibrium, we can consider the torque about an axis of rotation at the left end of the beam. Let $m_w$ be the mass of the worker and let $m_b$ be the mass of the beam. $\sum \tau = 0$ $r~T~sin(30^{\circ}) - r_w~m_w~g-r_b~m_b~g = 0$ $r~T~sin(30^{\circ}) = r_w~m_w~g+r_b~m_b~g$ $T = \frac{r_w~m_w~g+r_b~m_b~g}{r~sin(30^{\circ})}$ $T = \frac{(4.0~m)(80~kg)(9.80~m/s^2)+(3.0~m)(1450~kg)(9.80~m/s^2)}{(6.0~m)~sin(30^{\circ})}$ $T = 15,300~N$ The tension in the cable is 15,300 N.
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.