Answer
$\begin{aligned} & x_{A D}=0.4905 \mathrm{~m} \\ & x_{A C}=0.793 \mathrm{~m} \\ & x_{A B}=0.467 \mathrm{~m}\end{aligned}$
Work Step by Step
$
\begin{gathered}
F_{A D}=2(9.81)=x_{A D}(40) \quad x_{A D}=0.4905 \mathrm{~m} \\
\Rightarrow \Sigma F_x=0 ; \quad F_{A B}\left(\frac{4}{5}\right)-F_{A C}\left(\frac{1}{\sqrt{2}}\right)=0 \\
+\uparrow \Sigma F_y=0 ; \quad F_{A C}\left(\frac{1}{\sqrt{2}}\right)+F_{A B}\left(\frac{3}{5}\right)-2(9.81)=0 \\
F_{A C}=15.86 \mathrm{~N} \\
x_{A C}=\frac{15.86}{20}=0.793 \mathrm{~m} \\
F_{A B}=14.01 \mathrm{~N} \\
x_{A B}=\frac{14.01}{30}=0.467 \mathrm{~m}
\end{gathered}
$
