Answer
$\begin{aligned} & F_R=5.93 \mathrm{kN} \\ & \theta=77.8^{\circ} \\ & M_B=11.6 \mathrm{kN} \cdot \mathrm{m} \text { (Counterclockwise) }\end{aligned}$
Work Step by Step
$
\begin{aligned}
& \stackrel{\Rightarrow}{\Rightarrow} F_{R_x}=\Sigma F_x ; \quad F_{R_x}=1.5 \sin 30^{\circ}-2.5\left(\frac{4}{5}\right) \\
& =-1.25 \mathrm{kN}=1.25 \mathrm{kN} \leftarrow \\
& +\uparrow F_{R_y}=\Sigma F_y ; \quad F_{R_y}=-1.5 \cos 30^{\circ}-2.5\left(\frac{3}{5}\right)-3 \\
& =-5.799 \mathrm{kN}=5.799 \mathrm{kN} \downarrow \\
&
\end{aligned}
$
Thus,
$
F_R=\sqrt{F_{R_s}^2+F_{R_y}^2}=\sqrt{1.25^2+5.799^2}=5.93 \mathrm{kN}
$
and
$
\theta=\tan ^{-1}\left(\frac{F_{R_y}}{F_{R_x}}\right)=\tan ^{-1}\left(\frac{5.799}{1.25}\right)=77.8^{\circ}
$
$
\begin{aligned}
↺+M_{R_a}=\Sigma M_{R_a} ; \quad & M_B=1.5 \cos 30^{\circ}(2)+2.5\left(\frac{3}{5}\right)(6) \\
& =11.6 \mathrm{kN} \cdot \mathrm{m} \quad \text { (Counterclockwise) }
\end{aligned}
$
