Answer
$$
\begin{aligned}
& \mathbf{F}_R=\{-28.3 \mathbf{j}-68.3 \mathbf{k}\} \mathrm{N} \\
& \mathbf{M}_{R A}=\{-20.5 \mathbf{j}+8.49 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \mathbf{F}_{\boldsymbol{n}}=\mathbf{F}_4+\mathbf{F}_2 \\
& --40 \cos 45 \mathrm{j}+\left(-40-40 \sin 45^{\circ}\right) \mathbf{k} \\
& \mathbf{F}_h=|-28 . \mathbf{j}-68.7 \mathbf{k}| \mathrm{N} \\
&
\end{aligned}
$$
$\left.\mathbf{r}_{A P}=\mid-0.3 \mathbf{i}+0.0 \mathrm{j} \mathbf{j}\right]=$
$$
\begin{aligned}
& \mathbf{r}_{A P 2}=-0.3 \mathbf{i}-0.08 \sin 45^{\prime \prime} \mathbf{j}+0.08 \cos 45 \mathrm{k} \\
& -|-0.3-0.0666 j+0.006 f i k| m \\
& \mathbf{M}_{K A}=\left\langle\mathbf{r}_{A F 1} \times \mathbf{F}_1\right\rangle+\left\langle\mathbf{r}_{A F} \times \mathbf{F}_2\right\rangle \\
& -\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-0.3 & 0.06 & 0 \\
0 & 0 & -40
\end{array}\right|+\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-0.3 & -0.0566 & 0.0666 \\
0 & -40 \cos 45^* & -40 \sin 45^*
\end{array}\right| \\
& \mathbf{M}_{A A}=|-20.5 \mathrm{j}+8.49 \mathrm{k}| \mathrm{N} \cdot \mathrm{m} \\
&
\end{aligned}
$$
Also.
$$
\begin{aligned}
& M_{\text {AA, }}=\mathrm{Y} \cdot \mathrm{M}_{\mathrm{A}} \\
& M_{\text {A. }}-2828(0.0566)+2828(0.0566)-40(0.08) \\
& M_{\text {AA, }}=0 \\
& M_{A_A}=\mathbf{\Sigma} \cdot M_A \\
& M_{\text {KA, }}=-28.28(0.3)-40(0.3) \\
& M_{\hbar A}=-20.5 \mathrm{~N} \cdot \mathrm{m} \\
& M_{B A,}=\Sigma \cdot M_A \\
& \left.M_{E A}=282 \times 0.3\right) \\
& M_{\text {EA, }}=8.49 \mathrm{~N}-\mathrm{m} \\
& \mathbf{M}_{K A}=|-20.5+8.49 k| \mathrm{N} \cdot \mathrm{m} \\
&
\end{aligned}
$$
