Acceleration of 3 Blocks w/ Pulley

[Coop]

Homework Statement

There's a picture with the problem attached

A block with mass m = 1.5 kg hangs to the right of two blocks with masses m2 = 2.2 kg and m1 = 1.2 kg which sit on a table. All three blocks are connected with a string and pulley. Neglecting the mass of the pulley, string and any friction, find the acceleration.

Homework Equations

F = ma

The Attempt at a Solution

The answer key says that m*g = (m + m1 + m2)*a

Is my reasoning below correct in coming to that conclusion? I was at first confused at why the tension forces are not shown.

I ignore the weight and normal force of the boxes on the table, since they cancel

\sum F_{x} = (T_{m_{2}})_{x} + (T_{m_{1}})_{left} + (T_{m_{1}})_{right}

But (T_{m_{2}})_{x} + (T_{m_{1}})_{left} cancel, right? Because they are equal but opposite?

So, \sum F_{x} = (T_{m_{1}})_{right}

And,

\sum F_{y} = (T_{m})_{y} + w_{y}

So,

F_{net} = \sum F_{x} + \sum F_{y} = (T_{m_{1}})_{right} + (T_{m})_{y} + w_{y}

But since,

(T_{m_{1}})_{right} + (T_{m})_{y} are equal but opposite, they too cancel, right?

So you are left with F_{net} = m_{net}*a = w_{y} = m*g

Then, you can solve

F_{net} = 4.9 kg*a = 1.5 kg*9.81\frac{m}{s^2}

a = 3.00 \frac{m}{s^2}

Does my reasoning make sense of how the tensions cancel and so are not factored into the equation?

Thanks,
Coop

 

[Simon Bridge]

I cannot tell how you've done it because I don't know your labelling.
However, you appear to be trying to do the problem all in one go. This is not the "correct" approach.
You should be drawing a free-body diagram for each mass, apply ƩF=ma to each one (using appropriate choice of + or - signs to show direction), and then solve the system of simultaneous equations. If you do that, questions about what cancels what will resolve themselves.

If you know how long the string between masses is, you will have a time-dependent solution - otherwise you should state "as long as m1 and m2 remain on the table" or something like that.