Answer
$\begin{aligned} & F_R=1.80 \mathrm{kN} \\ & d=2.33 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force. Summing the forces along the $y$ axis,
$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ;-F_R & =-400(3)-\frac{1}{2}(400)(3) \\
F_R & =1800 \mathrm{~N}=1.80 \mathrm{kN} \downarrow
\end{aligned}
$
Location of Resultant Force. Summing the moment about point $A$
$
\begin{aligned}
↺+\left(M_R\right)_A=\Sigma M_A ; \quad-1800 d & =-400(3)(1.5)-\frac{1}{2}(400)(3)(4) \\
d & =2.333 \mathrm{~m}=2.33 \mathrm{~m}
\end{aligned}
$
