Answer
$\begin{aligned} & N_C=-30 \mathrm{kN} \\ & V_C=-8 \mathrm{kN} \\ & M_C=6 \mathrm{kN} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
$
\begin{array}{ll}
↺+\Sigma M_A=0 ; & -T(0.6)+8(2.25)=0 \\\\
\pm \Sigma F_x=0 ; & A_x=30 \mathrm{kN} \\
+\uparrow \Sigma F_y=0 ; & A_y=8 \mathrm{kN} \\\\
\pm \Sigma F_x=0 ; & -N_C-30=0 \\
& N_C=-30 \mathrm{kN} \\\\
+\uparrow \Sigma F_y=0 ; & V_C+8=0 \\
& V_C=-8 \mathrm{kN} \\\\
↺+\Sigma M_C=0 ; & -M_C+8(0.75)=0 \\
& M_C=6 \mathrm{kN} \cdot \mathrm{m}
\end{array}
$
