Answer
$\begin{aligned} & F_{H G}=1125 \mathrm{lb}(\mathrm{T}) \\ & F_{D E}=3375 \mathrm{lb}(\mathrm{C}) \\ & F_{E H}=3750 \mathrm{lb}(\mathrm{T})\end{aligned}$
Work Step by Step
Joint $D$ : From the free-body diagram in Fig.
$
\begin{array}{ll}
↺+\Sigma M_E=0 ; & F_{H G}(4)-1500(3)=0 \\
& F_{H G}=1125 \mathrm{lb}(\mathrm{T}) \\
↺+\Sigma M_H=0 ; & F_{D E}(4)-1500(6)-1500(3)=0 \\
& F_{D E}=3375 \mathrm{lb}(\mathrm{C}) \\
+\uparrow \Sigma F_y=0 ; & F_{H E}\left(\frac{4}{5}\right)-1500-1500=0 \\
& F_{E H}=3750 \mathrm{lb}(\mathrm{T})
\end{array}
$
