Answer
$$ M_{y^{\prime}}=464 \mathrm{lb} \cdot \mathrm{ft}
$$
Work Step by Step
Scalar analysis :
$$
M_y=80\left(6 \cos 15^{\circ}\right)=464 \mathrm{lb} \cdot \mathrm{ft}
$$
Vector analysis :
$$
\mathbf{u}_{A B}=\cos 60^{\circ} \mathbf{i}+\cos 30^{\circ} \mathbf{j}
$$
Coordinates of point $C$ :
$$
\begin{aligned}
& x=3 \sin 30^{\circ}-6 \cos 15^{\circ} \cos 30^{\circ}=-3.52 \mathrm{ft} \\
& y=3 \cos 30^{\circ}+6 \cos 15^{\circ} \sin 30^{\circ}=5.50 \mathrm{ft} \\
& z=6 \sin 15^{\circ}=1.55 \mathrm{ft} \\
& \mathbf{r}_{A C}=-3.52 \mathbf{i}+5.50 \mathbf{j}+1.55 \mathbf{k} \\
& \mathbf{F}=80 \mathrm{k} \\
& M_{y^{\prime}}=\left|\begin{array}{ccc}
\sin 30^{\circ} & \cos 30^{\circ} & 0 \\
-3.52 & 5.50 & 1.55 \\
0 & 0 & 80
\end{array}\right| \\
& M_{y^{\prime}}=464 \mathrm{lb} \cdot \mathrm{ft}
\end{aligned}
$$
