Answer
$\begin{aligned} & F_R=1.30 \mathrm{kN} \\ & \theta=86.7^{\circ} \\ & \left(M_R\right)_A=1.02 \mathrm{kN} \cdot \mathrm{m} \text { (counterclockwise) }\end{aligned}$
Work Step by Step
Equivalent Resultant Force And Couple Moment at Point $\boldsymbol{A}$.
$
\begin{aligned}
\stackrel{ \pm}{\rightarrow}\left(F_R\right)_x=\Sigma F_x ; \quad\left(F_R\right)_x & \\=650 \sin 30^{\circ}-500 \cos 60^{\circ}=75 \mathrm{~N} \rightarrow \\
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad\left(F_R\right)_y & \\=-650 \cos 30^{\circ}-300-500 \sin 60^{\circ} \\
& =-1295.93 \mathrm{~N}\\=1295.93 \mathrm{~N} \downarrow
\end{aligned}
$
As indicated in Fig.
$
F_R=\sqrt{\left(F_R\right)_x^2+\left(F_R\right)_y^2}=\sqrt{75^2+1295.93^2}=1298.10 \mathrm{~N}=1.30 \mathrm{kN} \text { }
$
$
\theta=\tan ^{-1}\left[\frac{\left(F_R\right)_y}{\left(F_R\right)_x}\right]=\tan ^{-1}\left(\frac{1295.93}{75}\right)=86.69^{\circ}=86.7^{\circ} \quad \text { }
$
Also,
$
\begin{aligned}
↺+\left(M_R\right)_A=\Sigma M_A ; \quad\left(M_R\right)_A & \\=650 \cos 30^{\circ}(3)+1500-500 \sin 60^{\circ}(5) \\
=1023.69 \mathrm{~N} \cdot \mathrm{m} \\
=1.02 \mathrm{kN} \cdot \mathrm{m} \text { (counter clockwise) }
\end{aligned}
$
