\(\Sigma Fy=0\)     \(F_{A+F_B}-60KN=0\)  \(\left(1\right)\)
\(\Sigma MO=0\)    \(\left(2M\right)\left(-60KN\right)+\left(6m\right)FB=0\)  \(\left(2\right)\)

Ahora se despeja FB:

−120Nm + 6m FB = 0

\(\left(6m\right)\left(FB\right)=120KNm\)
\(FB=\frac{120KNm}{6m}=20KN\)   \(\left(3\right)\)