Answer
The tension in the cable is $~~408.5~N$
Work Step by Step
We can find the angle of the cable above the horizontal:
$tan~\theta = \frac{4.00~m}{3.00~m}$
$\theta = tan^{-1}~\frac{4.00~m}{3.00~m}$
$\theta = 53.1^{\circ}$
To find the tension $T$ in the cable, we can consider the net torque about the rotation axis at the hinge:
$\sum~\tau_i = 0$
$(3.00~m)~T~sin~126.9^{\circ}-(2.00~m)(50.0~kg)(9.8~m/s^2) = 0$
$(3.00~m)~T~sin~126.9^{\circ} = (2.00~m)(50.0~kg)(9.8~m/s^2)$
$T = \frac{(2.00~m)(50.0~kg)(9.8~m/s^2)}{(3.00~m)~sin~126.9^{\circ}}$
$T = 408.5~N$
The tension in the cable is $~~408.5~N$
