Answer
The horizontal component of the force from the hinge is $~~\frac{W~x~~cot~\theta}{L}$
Work Step by Step
In part (a), we found that the tension in the wire is $~~T = \frac{W~x}{L~sin~\theta}$
The horizontal component of the force from the hinge is equal and opposite to the horizontal component of the tension.
We can find the magnitude of the horizontal component of the force from the hinge:
$F_x = T~cos~\theta$
$F_x = (\frac{W~x}{L~sin~\theta})~cos~\theta$
$F_x = \frac{W~x~~cot~\theta}{L}$
The horizontal component of the force from the hinge is $~~\frac{W~x~~cot~\theta}{L}$
