Answer
$m=88.5 \mathrm{~kg}$
Work Step by Step
Equations of Equilibrium.
$
\begin{aligned}
& \Sigma F_x=0 ; \quad F_{O A}\left(\frac{2}{\sqrt{14}}\right)-F_{O C}\left(\frac{3}{\sqrt{22}}\right)+F_{O B}\left(\frac{1}{3}\right)=0 \\
& \Sigma F_y=0 ;-F_{O A}\left(\frac{3}{\sqrt{14}}\right)+F_{O C}\left(\frac{2}{\sqrt{22}}\right)+F_{O B}\left(\frac{2}{3}\right)=0 \\
& \Sigma F_z=0 ; \quad F_{O A}\left(\frac{1}{\sqrt{14}}\right)+F_{O C}\left(\frac{3}{\sqrt{22}}\right)-F_{O B}\left(\frac{2}{3}\right)-\mathrm{m}(9.81)=0
\end{aligned}
$
After Solving Eqs
$
F_{O C}=16.95 m \quad F_{O A}=15.46 \mathrm{~m} \quad F_{O B}=7.745 \mathrm{~m}
$
Since $O C$ subjected to the greatest force, it will reach the limiting force first, that is $F_{O C}=1500 \mathrm{~N}$. Then
$
\begin{aligned}
& 1500=16.95 \mathrm{~m} \\
& m=88.48 \mathrm{~kg}=88.5 \mathrm{~kg}
\end{aligned}
$
