Answer
$$
\begin{aligned}
& F_R=15.4 \mathrm{kN} \\
& \left(M_R\right)_o=18.5 \mathrm{kN} \cdot \mathrm{m} \text { (clockwise) }
\end{aligned}
$$
Work Step by Step
Summing the forces along the $y$ axis
$$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad F_R & =-\frac{1}{2}(3)(1.5)-5(2.25)-\frac{1}{2}(5)(0.75) \\
& =-15.375 \mathrm{kN}=15.4 \mathrm{kN} \downarrow
\end{aligned}
$$
Now Summing the Moment about point $O$,
$$
\begin{aligned}
↺+\left(M_R\right)_o=\Sigma M_O ; \quad & \left(M_R\right)_o=-\frac{1}{2}(3)(1.5)(0.5)-5(2.25)(1.125) \\
& -\frac{1}{2}(5)(0.75)(2.5) \\
& =-18.46875 \mathrm{kN} \cdot \mathrm{m}=18.5 \mathrm{kN} \cdot \mathrm{m} \text { }
\end{aligned}
$$
