Answer
$F_A=4.46lb$
Work Step by Step
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:
$F_A=4.46lb$
