Answer
$\begin{aligned} & N_C=2.20 \mathrm{kip} \\ & V_C=0.336 \mathrm{kip} \\ & M_C=1.76 \mathrm{kip} \cdot \mathrm{ft}\end{aligned}$
Work Step by Step
Entire beam:
$
\begin{array}{ll}
↺+\Sigma M_A=0 ; & -2.8(3)+B_x(8)=0 \\
& B_x=1.05 \mathrm{kip} \\
\pm \Sigma F_x=0 ; & A_x=1.05=0 \\
& A_x=1.05 \mathrm{kip} \\
+\uparrow \Sigma F_y=0 ; & A_y=2.8=0 \\
& A_y=2.80 \mathrm{kip}
\end{array}
$
Segment $A C$ :
$
\begin{aligned}
& +\nearrow \Sigma F_s=0 ; \quad-N_C+1.05 \cos 53.13^{\circ}+2.80 \sin 53.13^{\circ}-0.84 \sin 53.13^{\circ}=0 \\
& N_C=2.20 \mathrm{kip} \\
& \text { } \\
& \Sigma+\Sigma F_y=0 ; \quad-V_C-0.84 \cos 53.13^{\circ}+2.80 \cos 53.13^{\circ}-1.05 \sin 53.13^{\circ}=0 \\
& V_C=0.336 \mathrm{kip} \\
& ↺+\Sigma M_C=0, \quad-2.80 \cos 53.13^{\circ}(3)+1.05 \sin 53.13^{\circ}(3)+0.84 \cos 53.13^{\circ}(1.5) \\
& +M_C=0 \\
& M_C=1.76 \mathrm{kip} \cdot \mathrm{ft} \\
&
\end{aligned}
$
