∑ fy=0↔Ray+Rby=12T +2T

Ray+Rby=14 T….¿0

∑MA=0↔Rby (6 )−12T (3 )−2T (3 )=0

Rby (6 )−36T−6T=0

Rby (6)−42T=0

Rby=7T….¿1

Reemplazando(¿¿¿1)en (¿¿¿0)¿¿

Ray+7T=14T

Ray=7T

Calculamos diagramsde fuerzacortante V ( x ) y momento flector M (x ) .

∑ fy=0↔7−2 x−V ( x )=0

V ( x )=7−2x

∑M=0↔M ( x )+2 x . x2−7 x=0

M (x )=7 x−x2

∑ fy=0↔7−2x−2−V ( x )=0

5−2x−V ( x )=0

V ( x )=5−2 x

∑M=0↔M ( x )+2 ( x−3 )+2 x . x2−7 x=0

M (x )=−x2+5 x+6