Answer
$\begin{aligned} & F_R=294 \mathrm{~N} \\ & \theta=40.1^{\circ} \\ & M_{R O}=39.6 \mathrm{~N} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
$\begin{aligned} & \pm F_{R x}=\Sigma F_x ; \quad F_{R x}=450 \sin 30^{\circ}=225.0 \\ & +\downarrow F_{R y}=\Sigma F_y ; \quad F_{R y}=450 \cos 30^{\circ}-200=189.7 \\ & F_R=\sqrt{(225)^2+(189.7)^2}=294 \mathrm{~N} \\ & \theta=\tan ^{-1}\left(\frac{189.7}{225}\right)=40.1^{\circ} \\ & ↺+M_{R O}=\Sigma M_O ; \quad M_{R O}=450 \cos 30^{\circ}(1.5)-450\left(\sin 30^{\circ}\right)(0.2)-200(3.5) \\ & \left.M_{R O}=39.6 \mathrm{~N} \cdot \mathrm{m}\right)\end{aligned}$
