Table and frictionless pulleys

[apiwowar]

Figure 5-58 shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 4.40 kg, mB = 9.80 kg, and mC = 12.0 kg. When the blocks are released, what is the tension in the cord at the right?

all i could do on this one is make the two outside free body diagrams

for block a i got T - 43.1 = 4.40a

and for block c i got T-118 = 12.0a

how would i get the net force for block b since the net force is just the two tensions?

 

[rl.bhat]

Tension in the two segments of the rope are not the same.
All the masses have the same acceleration.

For block A ...> T2 = mA*a...(1)
For block B...> T1 - T2 = mB*a...(2)
For block C...> mC*g - T1 = mC*a...(3)
Solve these equations and find T1 and T2.

 

[apiwowar]

why isn't the weight taken into account for block a?
and
for block b why isn't it the tension minus the weight?

 

[rl.bhat]

why isn't the weight taken into account for block a?
and
for block b why isn't it the tension minus the weight?

for block b why isn't it the tension minus the weight?

Forces acting on block b are T1 and T2 only.