Answer
$m=26.7 \mathrm{~kg}$
Work Step by Step
At $H$ :
$
+\uparrow \Sigma F_y=0 ; \quad F_{H A}=W
$
At $A$ :
$
\begin{array}{ll}
+\uparrow \Sigma F_y=0 ; & F_{A B} \sin 60^{\circ}-W=0 \\
& F_{A B}=1.1547 \mathrm{~W} \\
\pm \Sigma F_x=0 ; & F_{A E}-(1.1547 \mathrm{~W}) \cos 60^{\circ}=0 \\
& F_{A E}=0.5774 \mathrm{~W}
\end{array}
$
At $B$ :
$
\begin{array}{ll}
+\uparrow \Sigma F_y=0 ; & F_{B D}\left(\frac{3}{5}\right)-\left(1.1547 \cos 30^{\circ}\right) W=0 \\
& F_{B D}=1.667 \mathrm{~W} \\
\pm \Sigma F_x=0 ; & -F_{B C}+1.667 \mathrm{~W}\left(\frac{4}{5}\right)+1.1547 \sin 30^{\circ}=0 \\
& F_{B C}=1.9107 \mathrm{~W}
\end{array}
$
By comparison, cord $B C$ carries the largest load. Thus
$
\begin{aligned}
500 & =1.9107 W \\
W & =261.69 \mathrm{~N} \\
m & =\frac{261.69}{9.81}=26.7 \mathrm{~kg}
\end{aligned}
$
