Answer
$M_C=126 \mathrm{lb} \cdot \mathrm{ft}$
Work Step by Step
(a)
$
\begin{aligned}
\mathbf{M}_C & =\mathbf{\Sigma}(\mathbf{r} \times \mathbf{F}) \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 0 & 0 \\
-50 \sin 30^{\circ} & -50 \cos 30^{\circ} & 0
\end{array}\right|+\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -4 & 0 \\
\frac{4}{3}(80) & \frac{3}{5}(80) & 0
\end{array}\right| \\
\mathbf{M}_C & =\{126 \mathbf{k}\} \mathrm{lb} \cdot \mathrm{ft}
\end{aligned}
$
(b) $C+M_C=50 \cos 30^{\circ}(2)-50 \cos 30^{\circ}(5)-\frac{4}{5}(80)(1)+\frac{4}{5}(80)(5)$
$
M_C=126 \mathrm{lb} \cdot \mathrm{ft}
$
