Answer
$\begin{aligned} & a=1.26 \mathrm{~m} \\ & b=2.53 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force And Couple Moment At Point $\boldsymbol{A}$. Summing the forces along the $y$ axis,
$
\begin{gathered}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad 0=2(a+b)-\frac{1}{2}(6)(b) \\
2 a-b=0
\end{gathered}
$
Summing the moments about point $A$, with the requirement that $\left(M_R\right)_A=8 \mathrm{kN} \cdot \mathrm{m}$,
$
\begin{aligned}
↺+\left(M_R\right)_A=\Sigma M_A ;-8 & \\=2(a+b)\left[4-\frac{1}{2}(a+b)\right]-\frac{1}{2}(6)(b)\left(4-\frac{1}{3} b\right) \\
-8 =8 a-4 b-2 a b-a^2
\end{aligned}
$
Solving Eqs
$
\begin{aligned}
& a=1.264 \mathrm{~m}=1.26 \mathrm{~m} \\
& b=2.530 \mathrm{~m}=2.53 \mathrm{~m}
\end{aligned}
$
