\(\Sigma Fy=0\) \(F_{A+F_B}-60KN=0\) \(\left(1\right)\)
\(\Sigma MO=0\) \(\left(2M\right)\left(-60KN\right)+\left(6m\right)FB=0\) \(\left(2\right)\)
Ahora se despeja FB:
−120Nm + 6m FB = 0
\(\left(6m\right)\left(FB\right)=120KNm\)
\(FB=\frac{120KNm}{6m}=20KN\) \(\left(3\right)\)