Answer
$M=52lb\cdot ft$
Work Step by Step
We can determine the magnitude of the applied couple moments as follows:
$F_s=k(4)=2(4)=8lb$
virtual displacements are given as
$\delta_{yB}=\delta_{yc}=\frac{d(4cos\theta)}{d\theta}=-4sin\theta$
and $\delta_{ylink}=\frac{d(2cos\theta)}{d\theta}=-2sin\theta$
Now, according to the virtual-work function equation
$\delta U=0$
$\implies M+\frac{W_{block}}{2}\delta_{yB}+W_{link}\delta_{ylink}+F_s\delta_{yB}=0$
We plug in the known values to obtain:
$M-\frac{50}{2}\cdot 4sin20-10\times 2sin20-8(4)sin20=0$
This simplifies to:
$M=52lb\cdot ft$
