f1=(100-120j+75k)lb
f2=(-200i+250j+100k)lb
rA=(0I+0J+0K) rB=(4j+5j+3k)
Mo=M1+M2=rAxF1+rBxF2
\(rA\ X\ F1=\left|I\ \ \ J\ \ \ \ K\right|\)
\(\left|0\ \ \ 0\ \ \ 0\right|=0i+oj+ok\)
\(\left|4\ \ \ \ s\ \ \ \ 3\right|=\left(875-390\right)i-\left(700-\left(-300\right)\right)j+\left(520-\left(-500\right)\right)k\)
\(\left|-100\ \ \ 130\ \ \ 175\right|\)
\(MOT=485i-1000j+1020k\)
\(F\ AX=\ FA\ \cos\theta=\frac{4}{5}\ FA\)
\(F\ AY\ =\ FA\sin\theta=\frac{3}{5}\ FAFBX\ =\ FB\ \cos\theta\ 60\ \)
\(FAY=FB\sin\theta60\)\(-301b\ \cos\theta60-\ \frac{1}{3}FA\ =\ 0\ \)
\(FA=\ \frac{5}{4}\ \left(-30lb\ \cos60\right)=18.75\)
Este valor solo seria valido si las fuerzas estuvieran actuando en el mismo brazo
Para B
rbx=6ft
ray=0
para A rbx=9
ft rby=0
Fax=30 lb \(\cos60\)°
Fay=30lb sin 60°
MA= rax X Fay-Ray X Fax = (9 ft)
(\(\left(\frac{3}{5}FA\right)-\left(0\right)\left(\frac{4}{5}\right)=\frac{27}{5}\ FA\ lb\ ft\ \)
rbx X Fby-rby X Fb=\(\left(6\right)\left(30\sin60\right)-\left(0\right)\left(30\cos60\right)=155.88\ \frac{lb}{ft}\)
\(\Sigma M=0\)
\(Mb-Ma=0\)
\(155.88lb\ ft\ -\ \frac{27}{5}\ FA\ lb\ ft\ \)\(FA=\left(\frac{27}{5}\right)\left(155.88\ lb\ ft\right)\)
\(\frac{27}{5}FA\ =155.88\)
\(FA\ =28.9\ lb\ ft\ \)