DATOS:
ABC: 2(10-3)M②
E: 60 C1.PA
\(\Sigma\):FY=0
\(\Sigma\): M=0
PROCEDIMIENT0:
\(\Sigma\)FY=0
Fba+fbc-60kn=0
-(2m)(60kn)+(6m)Fpc=0
-120kmn+Fbc(6m)=0
Fbc=\(\frac{120knm}{6m}\) =20 M
Fda+20kn=60kn=0
fad=60-20=40
\(\Sigma\)bc:\(\frac{pl}{ab}\) \(\frac{\left(-20x10-3kn\right)\left(3kn\right)}{2x10-3m\left(60x10-4MKN\right)}\) =5X10-3