Answer
$\begin{aligned} & F_1=27.6 \mathrm{kN} \\ & F_2=24.0 \mathrm{kN}\end{aligned}$
Work Step by Step
Equivalent Resultant Force. Sum the forces along $z$ axis.
$
+\uparrow\left(F_R\right)_z=\Sigma F_z ; \quad-F_R=-F_1-F_2-12-6 \quad F_R=F_1+F_2+18
$
Location of the Resultant Force. Sum the moments about the $x$ and $y$ axes.
$
\begin{gathered}
\left(M_R\right)_s=\Sigma M_x ; \quad-\left(F_1+F_2+18\right)(10)\\=-12(8)-6(20)-F_2(20) \\
10 F_1-10 F_2=36 \\
\left(M_R\right)_y=\Sigma M_y ; \quad\left(F_1+F_2+18\right)(12)=\\12(6)+6(26)+F_1(22) \\
12 F_2-10 F_1=12
\end{gathered}
$
Solving these Eqs,
$
F_1=27.6 \mathrm{kN} \quad F_2=24.0 \mathrm{kN}
$
