\(FAX=FA\cos\theta=\frac{4}{5}FA\)
\(FAY=FA\sin\theta=\frac{3}{5}FA\)\(FBX=FB\cos\ \theta60\)°
\(FAY=FB\sin\theta60\)°
\(\Sigma Fx=0\)
\(\Sigma Fbx-Fby=0\)
\(-30lb\cos60°-\frac{4}{5}FA=0\)
FA=\(\frac{5}{4}\)(-30lb cos60°) = 18.75
Este valor solo solo seria valido si las fuerzas estuvieran actuando con el mismo brazo.
para B Fax =30lb cos 60°
rbx= 6 ft Fay=30lb sin 60°
ray=0 Fbx=\(\frac{4}{5}FA\)
Fby=\(\frac{3}{5}FA\)
para A
rbx=9 ft
rby=0
MA=rax X Fay-ray X Fax
=\(\left(9ft\right)\left(\frac{3}{5}FA\right)-\left(0\right)\left(\frac{4}{5}\right)=\frac{27}{5}FA\ lb\ ft\)
rbx X Fby-rby X Fb=(6)(30sen60°)-(0)(30cos60°)=155.88lb ft
\(\Sigma M=0\)
\(Mb-Ma=0\)
\(155.88\ lb\ ft\ -\frac{27}{5}\ FA\ lb\ ft\)
\(\frac{27}{5}FA=155.88\)
\(FA=\left(\frac{27}{5}\right)\left(155.88\ lb\ ft\right)\)
\(FA=28.9\ lb\ ft\)