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

$F_A=4.46lb$
We can determine the required force as follows: $z=0.1sin2\theta$ Taking the derivative, we have $z^{\cdot}=0.2\theta^{\cdot}cos 2\theta$ $\implies z^{\cdot \cdot}=-0.4\theta^{\cdot2}sin2\theta+0.2\theta^{\cdot \cdot}cos2\theta$ We plug in the known values to obtain: $z^{\cdot \cdot}=-0.4(6)^2sin90+0=-14.4ft/s^2$ We know that $F_A-F_s=\frac{W}{g}z^{\cdot \cdot}$ $\implies F_A-k(z+s)=\frac{0.75}{32.2}(-14.4)$ $\implies F_A-12(0.1sin90+0.3)=\frac{0.75}{32.2}(-14.4)$ This simplifies to: $F_A=4.46lb$