Answer
$$
\begin{aligned}
& F_R=365 \mathrm{~N} \\
& \theta=70.8^{\circ} \text {. } \\
& \left(M_R\right)_o=2364 \mathrm{~N} \cdot \mathrm{m} \text { (counterclockwise) }
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \stackrel{+}{\rightarrow}\left(F_R\right)_x=\Sigma F_x ; \quad\left(F_R\right)_x=600 \cos 60^{\circ}-455\left(\frac{12}{13}\right)=-120 \mathrm{~N}=120 \mathrm{~N} \leftarrow \\
& +\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad\left(F_R\right)_y=455\left(\frac{5}{13}\right)-600 \sin 60^{\circ}=-344.62 \mathrm{~N}=344.62 \mathrm{~N} \downarrow
\end{aligned}
$$
$$
F_R=\sqrt{\left(F_R\right)_x^2+\left(F_R\right)_y^2}=\sqrt{120^2+344.62^2}=364.91 \mathrm{~N}=365 \mathrm{~N} \quad \text { Ans. }
$$
And
$$
\theta=\tan ^{-1}\left[\frac{\left(F_R\right)_y}{\left(F_R\right)_x}\right]=\tan ^{-1}\left(\frac{344.62}{120}\right)=70.80^{\circ}=70.8^{\circ} \text { マ }
$$
Also,
$$
\begin{aligned}
↺+\left(M_R\right)_O=\Sigma M_O ;\left(M_R\right)_O & \\=455\left(\frac{12}{13}\right)(2)+600 \cos 60^{\circ}(0.75)+600 \sin 60^{\circ} \\
=2364 \mathrm{~N} \cdot \mathrm{m}(\text { counterclockwise })
\end{aligned}
$$
