Answer
$\begin{aligned} & F_R=27.0 \mathrm{kN} \\ & \left(M_R\right)_A=81.0 \mathrm{kN} \cdot \mathrm{m} \text { (clockwise) }\end{aligned}$
Work Step by Step
Equivalent Resultant Force And Couple Moment About Point A. 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)(3)-3(6)-\frac{1}{2}(3)(3) \\
& =-27.0 \mathrm{kN}=27.0 \mathrm{kN} \downarrow
\end{aligned}
$
Summing the moments about point $A$,
$
\begin{aligned}
C_{+}\left(M_R\right)_A=\Sigma M_A ; \quad\left(M_R\right)_A & \\=-\frac{1}{2}(3)(3)(1)-3(6)(3)-\frac{1}{2}(3)(3)(5) \\
=-81.0 \mathrm{kN} \cdot \mathrm{m}=81.0 \mathrm{kN} \cdot \mathrm{m} \text { (clockwise) }
\end{aligned}
$
