#### Answer

Since the required tension of 15,300 N is greater than the 15,000-N tension that the cable is rated, the worked should be a little worried.

#### 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.
$\sum \tau = 0$
$r~T~sin(150^{\circ}) - r_w~m_w~g-r_b~m_b~g = 0$
$r~T~sin(150^{\circ}) = r_w~m_w~g+r_b~m_b~g$
$T = \frac{r_w~m_w~g+r_b~m_b~g}{r~sin(150^{\circ})}$
$T = \frac{(4.0~m)(80~kg)(9.80~m/s^2)+(3.0~m)(1450~kg)(9.80~m/s^2)}{(6.0~m)~sin(150^{\circ})}$
$T = 15,300~N$
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.