Answer
$\begin{aligned} & F_R=6.75 \mathrm{kN} \downarrow \\ & \left(M_R\right)_D=4.05 \mathrm{kN} \cdot \mathrm{m} \text { }\end{aligned}$
Work Step by Step
Equivalent Resultant Force And Couple Moment About Point $\boldsymbol{O}$. The differential force indicated in Fig. $a$ is $d F_R=w d x=\frac{1}{3} x^3 d x$. Thus, summing the forces along the $y$ axis,
$
\begin{aligned}
+\uparrow\left(F_R\right)_y=\Sigma F_y ; \quad F_R & =-\int d F_R=-\int_0^{3 \mathrm{~m}} \frac{1}{3} x^3 d x \\
=-\frac{1}{12} x^4 \int_0^{3 \mathrm{~m}} \\
=-6.75 \mathrm{kN}=6.75 \mathrm{kN} \downarrow
\end{aligned}
$
Summing the moments about point $O$,
$
\begin{aligned}
↺+\left(M_R\right)_o=\Sigma M_O ; \quad\left(M_R\right)_O & =\int(3-x) d F_R \\
=\int_0^{3 \mathrm{~m}}(3-x)\left(\frac{1}{3} x^3 d x\right) \\
=\int_0^{3 \mathrm{~m}}\left(x^3-\frac{1}{3} x^4\right) d x \\
=\left.\left(\frac{x^4}{4}-\frac{1}{15} x^5\right)\right|_0 ^{3 \mathrm{~m}} \\
=4.05 \mathrm{kN} \cdot \mathrm{m} \text { (counterclockwise) }
\end{aligned}
$
