Tension in cord for three-block pulley system with block on table

[edgarpokemon]

Homework Statement

The figure shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 4.80 kg, mB = 7.10 kg, and mC = 13.0 kg. When the blocks are released, what is the tension in the cord at the right?

Figure A is on the left, figure C on the right, C being the largest mass.

Homework Equations

For part A I got T-m1g=ma, (Because the tension will be more used when it gets pulled up)
For Part B, This is where i am confused, I originally used T1-T2=M2A, because I thought that since the Box B will move to the right, then its tension would be less because Box C is pulling more with M3g, but I checked the answer and it said T2-T1=M2A, but why is that the right answer
For Part C I got, M3g-T=m3a (since gravity will do most of the work

The Attempt at a Solution

I got the answer of -7.22 on my first attempt and 3.22 for the acceleration on the second, which 3.22 is the correct answer. Help! I am just confused in part B! [/B]

 

[Merlin3189]

Your picture doesn't seem to have uploaded and I think it is needed to understand the problem. Perhaps you can try again.

 

[haruspex]

I think I can guess the diagram. Mass 1 hangs from a pulley on the left. The cord passes up over the pulley and extends horizontally to the right to mass 2. The second cord continues from mass 2 horizontally to the right, over a second pulley and straight down to mass 3.
Your working would be easier to follow if you were to use suffixes consistently.

You correctly presumed mass 1 would rise and mass 3 would fall, and got the right equations for those.
And, as you say, mass 2 will move to the right. But I cannot understand your explanation for why you chose T1-T2 there.
If the acceleration is to the right, which tension must be greater?

 

[edgarpokemon]

I see to understand now! thanks sorry for not posting a picture!