Answer
$\begin{aligned} & V_A=0 \\ & N_A=-39 \mathrm{kN} \\ & M_A=-2.425 \mathrm{kN} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
$$
\begin{aligned}
& \Sigma F_x=0 ; \quad 6-6-V_A=0 \\
& V_A=0 \\\\
& +\uparrow \Sigma F_y=0 ; \quad-N_A-16-23=0 \\
& N_A=-39 \mathrm{kN} \\\\
& ↺
+\Sigma M_A=0 ; \quad-M_A+16(0.155)-23(0.165)-6(0.185)=0 \\
& M_A=-2.42 \mathrm{kN} \cdot \mathrm{m} \\
&
\end{aligned}
$$
