Answer
$\begin{aligned} & F_R=1000 \mathrm{~N} \\ & \theta=53.1^{\circ} \\ & d=2.17 \mathrm{~m}\end{aligned}$
Work Step by Step
Equivalent Resultant Force.
$
\begin{array}{ll}
\pm\left(F_R\right)_x=\Sigma F_x ; & \left(F_R\right)_x=600 \mathrm{~N} \rightarrow \\
+\uparrow\left(F_R\right)_y=\Sigma F_y ; & \left(F_R\right)_y=-200-400-200=-800 \mathrm{~N}\\=800 \mathrm{~N} \downarrow
\end{array}
$
As indicated in Fig,
$
F_R=\sqrt{\left(F_R\right)_x^2+\left(F_R\right)_y^2}=\sqrt{600^2+800^2}=1000 \mathrm{~N}
$
And
$
\theta=\tan ^{-1}\left[\frac{\left(F_R\right)_y}{\left(F_R\right)_x}\right]=\tan ^{-1}\left(\frac{800}{600}\right)=53.13^{\circ}=53.1^{\circ} \square
$
Location of Resultant Force. Along $A B$,
$
\begin{gathered}
↺+\left(M_R\right)_B=\Sigma M_B ; \quad 600(1.5-d)=-400(0.5)-200(1) \\
d=2.1667 \mathrm{~m}=2.17 \mathrm{~m}
\end{gathered}
$
