Answer
$W_G=2.5lb$
Work Step by Step
We can determine the required weight as follows:
The displacement by block $G$ is given as
$\delta_{y_1}=(4+x)\delta\theta$
$\implies \delta_{y_1}=(4+12)\delta_{\theta}=16\delta_{\theta}$
and the displacement by weight $F$ is given as
$\delta_{y_2}=2\delta_{\theta}$
Now, according to the virtual-work equation
$W_G\times \delta_{y_1}-20\times \delta_{y_2}=0$
We plug in the known values to obtain:
$W_G\times 16\delta_{\theta}-20(2\delta_{\theta})=0$
This simplifies to:
$W_G=2.5lb$
