Answer
$l=2.66 \mathrm{ft}$
Work Step by Step
$
\begin{aligned}
& l=\sqrt{4^2+2^2-2(2)(4) \cos 60^{\circ}} \\
& l=\sqrt{12} \\
& \frac{\sqrt{12}}{\sin 60^{\circ}}=\frac{2}{\sin \phi} \\
& \phi=\sin ^{-1}\left(\frac{2 \sin 60^{\circ}}{\sqrt{12}}\right)=30^{\circ} \\
& +\uparrow \Sigma F_y=0 ; \quad T \sin 60^{\circ}+F_s \sin 30^{\circ}-80=0 \\
& \rightarrow \Sigma F_x=0 ; \quad-T \cos 60^{\circ}+F_s \cos 30^{\circ}=0
\end{aligned}
$
Solving for $F_s$.
$
\begin{aligned}
& F_s=40 \mathrm{lb} \\
& F_s=k x \\
& 40=50\left(\sqrt{12}-l^{\prime}\right) \quad l=\sqrt{12}-\frac{40}{50}=2.66 \mathrm{ft}
\end{aligned}
$
