Answer
$F_s=15lb$
Work Step by Step
We can determine the required force as follows:
$y_A=2(3)sin\theta=6sin\theta$
and $y_D=6(3)sin\theta=18sin\theta$
The virtual displacements are given as
$\delta _{yA}=\frac{d6\space sin\theta}{d\theta}=6\cdot cos\theta$
and $\delta_{yD}=\frac{d(18\space sin\theta)}{d\theta}=18\cdot cos\theta$
Now, the virtual work equation is given as
$\delta U=0$
$\implies P\cdot \delta_{yD}+F_s\cdot \delta _{yA}=0$
We plug in the known values to obtain:
$5\cdot 18cos30^{\circ}-F_s\cdot 6cos30^{\circ}=0$
This simplifies to:
$F_s=15lb$
