#### Answer

$F_T = (W+\frac{mg}{2})~cot~\theta$
When $\theta = 0^{\circ}$, $F_T$ is undefined because in this situation, the tension would be directed at the axis which would make the torque exerted by the tension zero even if the tension in the cable was very large.
When $\theta = 90^{\circ}$, $F_T = 0$ because in this situation, the weight of the girder and the weight of the boom would both be directed at the axis. Then their torques would be zero. In this case, a tension in the cable would not be required to keep the system in equilibrium.

#### Work Step by Step

We can consider the torques about the rotation axis at the hinge.
The weight of the boom exerts a counterclockwise torque about this axis.
The weight of the girder exerts a counterclockwise torque about this axis.
The tension in the cable $F_T$ exerts a clockwise torque about this axis.
Since the system is in equilibrium, the clockwise torque is equal in magnitude to the sum of the counterclockwise torques. Let $L$ be the length of the boom. We can find the tension $F_T$:
$\tau_{cw} = \tau_{ccw}$
$L~(F_T)~sin~\theta = L~W~cos~\theta+\frac{L}{2}mg~cos~\theta$
$F_T = (W+\frac{mg}{2})~cot~\theta$
When $\theta = 0^{\circ}$, $F_T$ is undefined because in this situation, the tension would be directed at the axis which would make the torque exerted by the tension zero even if the tension in the cable was very large.
When $\theta = 90^{\circ}$, $F_T = 0$ because in this situation, the weight of the girder and the weight of the boom would both be directed at the axis. Then their torques would be zero. In this case, a tension in the cable would not be required to keep the system in equilibrium.