Answer
$\begin{aligned} & M_B=150 \mathrm{~N} \cdot \mathrm{m} 3 \\ & M_B=600 \mathrm{~N} \cdot \mathrm{m} 3 \\ & M_B=0\end{aligned}$
Work Step by Step
The forces are resolved into horizontal and vertical componen For $\mathbf{F}_1$,
$
\begin{aligned}
↺+M_B & =250 \cos 30^{\circ}(3)-250 \sin 30^{\circ}(4) \\
& =149.51 \mathrm{~N} \cdot \mathrm{m}=150 \mathrm{~N} \cdot \mathrm{m})
\end{aligned}
$
For $\mathbf{F}_2$,
$
\begin{aligned}
↺+M_B & =300 \sin 60^{\circ}(0)+300 \cos 60^{\circ}(4) \\
& =600 \mathrm{~N} \cdot \mathrm{m} \text { ) }
\end{aligned}
$
Since the line of action of $\mathbf{F}_3$ passes through $B$, its moment arm about point $B$ is zero. Thus
$
M_B=0
$
