Answer
$\begin{aligned} & P=0.533 \mathrm{kN} \\ & N_C=-2 \mathrm{kN} \\ & V_C=-0.533 \mathrm{kN} \\ & M_C=0.400 \mathrm{kN} \cdot \mathrm{m}\end{aligned}$
Work Step by Step
$
\begin{array}{ll}
↺+\Sigma M_A=0 ; & -2(0.6)+P(2.25)=0 \\
\pm \Sigma F_x=0 ; & P=0.533 \mathrm{kN} \\\\
+\uparrow \Sigma F_y=0 ; & A_x=2 \mathrm{kN} \\
\pm \Sigma F_x=0 ; & -N_C-2=0.533 \mathrm{kN} \\
& N_C=-2 \mathrm{kN} \\\\
+\uparrow \Sigma F_y=0 ; & V_C-0.533=0 \\
& V_C=-0.533 \mathrm{kN} \\\\
↺+\Sigma M_C=0 ; & -M_C+0.533(0.75)=0 \\
& M_C=0.400 \mathrm{kN} \cdot \mathrm{m}
\end{array}
$
