Answer
$\begin{aligned} & F_R=215 \mathrm{kN} \\ & y=3.68 \mathrm{~m} \\ & x=3.54 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force: By equating the sum of the forces along the $z$ axis to the resultant force
,
$
\begin{gathered}
+\uparrow F_R=\Sigma F_5 \quad-F_R=-30-20-90-35-40 \\
F_R=215 \mathrm{kN}
\end{gathered}
$
Point of Application: By equating the moment of the forces and $\mathbf{F}_R$, about the $x$ and $y$ axes,
$
\begin{aligned}
& \left(M_R\right)_x=\Sigma M_x ; \quad-215(y)=-35(0.75)-30(0.75)-90(3.75)-20(6.75)-40(6.75) \\
& y=3.68 \mathrm{~m} \\
& \left(M_R\right)_y=\Sigma M_y ; \\
& 215(x)=30(0.75)+20(0.75)+90(3.25)+35(5.75)+40(5.75) \\
& x=3.54 \mathrm{~m} \\
&
\end{aligned}
$
