Answer
$$
\begin{aligned}
& F_R=12.5 \mathrm{kN} \\
& d=1.54 \mathrm{~m}
\end{aligned}
$$
Work Step by Step
Summing the forces along the $y$ axis
$$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad-F_R & =-4(2)-\frac{1}{2}(6)(1.5) \\
F_R & =12.5 \mathrm{kN}
\end{aligned}
$$
Now Summing the Moment about point $O$,
$$
\begin{aligned}
↺+\left(M_R\right)_o=\Sigma M_O ; \quad-12.5(d) & =-4(2)(1)-\frac{1}{2}(6)(1.5)(2.5) \\
d & =1.54 \mathrm{~m}
\end{aligned}
$$
