Answer
$$
\begin{aligned}
& F_R=15.0 \mathrm{kN} \\
& d=3.40 \mathrm{~m}
\end{aligned}
$$
Work Step by Step
Summing the forces along the $y$ axis
$$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad-F_R & =-2(6)-\frac{1}{2}(2)(3) \\
F_R & =15.0 \mathrm{kN} \downarrow
\end{aligned}
$$
Now Summing the Moments about point $A$,
$$
\begin{aligned}
C+\left(M_R\right)_A=\Sigma M_A ; \quad-15.0(d) & =-2(6)(3)-\frac{1}{2}(2)(3)(5) \\
d & =3.40 \mathrm{~m}
\end{aligned}
$$
