Answer
$\begin{aligned} & F_R=35 \mathrm{kN} \\ & y=11.3 \mathrm{~m} \\ & x=11.5 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force.
$
+\uparrow\left(F_R\right)_z=\Sigma F_z ; \quad-F_R=-8-6-12-9 \quad F_R=35 \mathrm{kN} \quad \text { Ans. }
$
Location of the Resultant Force. Sum the moments about the $x$ and $y$ axes by referring to Fig
$
\begin{aligned}
\left(M_R\right)_x=\Sigma M_s ; \quad-35 y & =-12(8)-6(20)-9(20) \\
y & =11.31 \mathrm{~m}=11.3 \mathrm{~m} \\
\left(M_R\right)_y=\Sigma M_y ; \quad 35 x & =12(6)+8(22)+6(26) \\
x & =11.54 \mathrm{~m}=11.5 \mathrm{~m}
\end{aligned}
$
