\(fB=30\ lb\)
\(-FAx=FA\ Cos\theta=\frac{4}{5}\ FA\)
\(-FBx=FBCos60°\)
\(FAy=FA\ Sen\theta=\frac{3}{5}\ FA\)
\(FBy=FBSen60°\)
Para B:
\(RBx=6ft;FBx=30lbCos60°\)
\(RBy=0;FBy=30lbSen60°\) \(RAy=0;\ FAy=\frac{3}{5}FA\)
Para A:
\(FAx=9ft;FAx=\frac{4}{5}FA\)
\(\Sigma Fx=0\)
\(-BBx-FAx=0\)
\(-30lbCos60°-\)\(\frac{4}{5}FA=0\) \(FA=\frac{5}{4}\ \ \) (-\(30lbCos60°\))\(=18.716\ \)
Valor Válido si las fuerzas estuvieran en el mismo punto
\(MA=FAx\) \(FAy-FAy\) \(FAx\)
=\(\left(9ft\right)\left(\frac{3}{5}FA\right)-\left(0\right)\left(\frac{4}{5}FA\right)\) \(=\frac{27}{5}FA\ \ lb\cdot ft\)
\(MB=RBxFB-RByFBx\)
\(=\left(6ft\right)\left(30lbSen60\right)\)\(-\left(0\right)\left(30lbCos60\right)=155.88\ \ lb\cdot ft\)
\(MB-MA=0\)
\(155.88\ lb\cdot ft\)\(\frac{27}{5}FA\ lb\cdot ft\)
\(FA\left(\frac{5}{27}\right)\left(155.88\right)lb\cdot ft=28.9\ lb\cdot ft\)