MOT= 485i - 1000j + 1020k
\(FAX=F\ A\ \cos\ \theta=\frac{4}{5}\ F\ A\)
\(FAY=\ F\ A\ \sin\ \theta\ =\frac{3}{5}F\ A\ F\ B\ X=\ FB\ \cos\ \theta60\)
\(FAY\ =\ FB\ \sin60\ \)
\(\sum_{}^{}\)Fx=0
\(\sum_{}^{}\) Fbx - Fby = 0
\(-30lb\ \cos60\ -\ \frac{4}{5}\ F\ A=\ 0-30lb\ \cos\ 60-\frac{4}{5}\ FA=0\)
\(FA\ =\frac{5}{4}\left(-30lb\ \cos\ 60\right)=18.75\)
\(RESULTADO=\ 18.75\)