#### Answer

$x=16in$

#### Work Step by Step

We can determine the required position $x$ as follows:
The virtual displacements are given as
$\delta_{yF}=\frac{d(2cos\theta)}{d\theta}=2sin\theta$
and $\delta_{yG}=\frac{d(4+xcos\theta)}{d\theta}=(4+x)sin\theta$
Now, according to the virtual-work equation
$\delta U=0$
$\implies F\delta_{yF}+G\delta_{yG}=0$
We plug in the known values to obtain:
$20(2)sin\theta-2(4+x)sin\theta=0$
This simplifies to:
$x=16in$