## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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. $\sum \tau = 0$ $r~T~sin(150^{\circ}) - r_w~m_w~g-r_b~m_b~g = 0$ $r~T~sin(150^{\circ}) = r_w~m_w~g+r_b~m_b~g$ $T = \frac{r_w~m_w~g+r_b~m_b~g}{r~sin(150^{\circ})}$ $T = \frac{(4.0~m)(80~kg)(9.80~m/s^2)+(3.0~m)(1450~kg)(9.80~m/s^2)}{(6.0~m)~sin(150^{\circ})}$ $T = 15,300~N$ The required tension in the cable to maintain equilibrium is 15,300 N. Since the required tension is greater than the 15,000-N tension that the cable is rated, the worked should be a little worried.