## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F_s=15lb$
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: $F_s=15lb$