Answer
The vertical component of the force from the hinge is $~~\frac{(W)(L-x)}{L}$
Work Step by Step
In part (a), we found that the tension in the wire is $~~T = \frac{W~x}{L~sin~\theta}$
The sum of the vertical forces on the bar must be zero.
Since $L \geq x$, then $T~sin~\theta \leq W$
Therefore, the sum of $~~T~sin~\theta~~$ and the vertical component of the force from the hinge must be equal in magnitude to $W$
We can find the magnitude of the vertical component of the force from the hinge:
$F_y + T~sin~\theta = W$
$F_y = W-(\frac{W~x}{L~sin~\theta})~sin~\theta$
$F_y = W - \frac{W~x}{L}$
$F_y = \frac{(W)(L-x)}{L}$
The vertical component of the force from the hinge is $~~\frac{(W)(L-x)}{L}$
