Engineering Mechanics: Statics & Dynamics (14th Edition)

$M=52lb\cdot ft$
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: $M=52lb\cdot ft$