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