Answer
$M=13.1N\cdot m$
Work Step by Step
We can determine the magnitude of the applied couple moments as follows:
$F_s=k(100)=3(100)=300N$
and the virtual displacements are given as
$\delta_{yC}=\frac{d(300cos\theta)}{d\theta}=-300sin\theta$
$\delta_{xC}=\frac{d(300sin\theta)}{d\theta}=-300cos\theta$
Now, according to the virtual-work equation
$\delta U=0$
$\implies M+\frac{W_{block}}{2}\delta_{xC}+F_s\delta_{yC}=0$
We plug in the known values to obtain:
$M+\frac{25(9.81)}{2}(300)cos30-300(300)sin30=0$
This simplifies to:
$M=13.1N\cdot m$
