f1=(100-120j+75k)lb f2=(-200i+250j+100k)lb
rA=(0i+0j+0k) rB=(4i+5j+3k)
Mo= M1+M2=rAxF1+rBxF2
rA X F1=
|I J K|
|0 0 0| = 0i+0j+0k
|100 120 75|
rB X F2=
|I J K|
|4 300))j+(520-(-500))k 5 3|=
(875-390)i-(700-(- |−100 130 175|
MOT=485i-1000j+1020k
\(FAX=FA\cos\theta=\frac{4}{5\ }FA\)
\(FAY=FA\sin\theta=\frac{3}{5}FAFBX\ =\ FB\cos\ \theta60\)
\(FAY=FB\sin60\)
\(\Sigma Fx=0\)
\(\Sigma Fbx-Fby=0\)
\(-30lb\cos60-\frac{4}{5}FA=0\)\(-30lb\cos60-\frac{4}{5}FA=0\)
\(FA=\frac{5}{4}\left(-30lb\ \cos\ 60\right)=18.75\) Resultado
Ejercicio #2