Answer
$\begin{aligned} & F_R=12.0 \mathrm{kN} \\ & \theta=48.4^{\circ} \\ & d=3.28 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force. Summing the forces along the $x$ and $y$ axes,
$
\begin{array}{ll}
\stackrel{+}{\rightarrow}\left(F_R\right)_x=\Sigma F_x ; & \left(F_R\right)_x=-2(4)=-8 \mathrm{kN}=8 \mathrm{kN} \leftarrow \\
+\uparrow\left(F_R\right)_y=\Sigma F_y ; & \left(F_R\right)_y=-3(3)=-9 \mathrm{kN}=9 \mathrm{kN} \downarrow
\end{array}
$
Then
$
F_K=\sqrt{\left(F_R\right)_x^2+\left(F_R\right)_y^2}=\sqrt{8^2+9^2}=12.04 \mathrm{kN}=12.0 \mathrm{kN}
$
And
$
\theta=\tan ^{-1}\left[\frac{\left(F_R\right)_y}{\left(F_R\right)_s}\right]=\tan ^{-1}\left(\frac{9}{8}\right)=48.37^{\circ}=48.4^{\circ}
$
Location of the Resultant Force. Summing the moments about point $A$,
$
\begin{aligned}
↺+\left(M_R\right)_A=\Sigma M_A ; \quad-8 x-9 y & =-3(3)(1.5)-2(4)(2) \\
8 x+9 y & =29.5
\end{aligned}
$
Along $A B, x=0$. Then Eq becomes
$
\begin{aligned}
8(0)+9 y & =29.5 \\
y & =3.278 \mathrm{~m}
\end{aligned}
$
Thus, the inter section point of line of action of $\mathbf{F}_R$ on $A B$ measured to the right from point $A$ is
$
d=y=3.28 \mathrm{~m}
$
