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=\(\left(6\right)\left(30\sin60\right)-\left(0\right)\left(30\cos60\right)=155.88\ lb\ ft\)
\(\Sigma M=0\)
\(Mb\ -\ Ma\ =0\)
\(155.88lb\ ft\ -\ \frac{27}{5}FA\ lb\ ft\)
\(\frac{27}{5}FA\ =\ 155.88\)
\(FA=\left(\frac{27}{5}\right)\left(155.88\ lb\ ft\right)\)
\(FA\ =29.9\ lb\ ft\) Resultado