acceleration involving newton's 2nd law

[man_in_motion]

1. The problem statement, all variables and given/known data
There is a box on a platform with mass 3.5kg and it has two ropes connected to it. On the end of each of these ropes are 2 other boxes hanging off the table (so only F_y component). One box is 2.5kg and the other is 1.5kg. I need to find the acceleration of all the boxes.

the answer is 1.3m/s^2 for all of them
2. Relevant equations

F_net=ma
F_g=mg

3. The attempt at a solution
a_box1=F_net/m=((m_1)(g)-a_box2)/m

I don't know how to proceed from here because I end up going in a loop

[Delphi51]

It takes a long time to get attachments approved. Can you describe the diagram?
Or you could put it up on a photo site like photobucket and post a link to it.
Is there friction in this question?

[man_in_motion]

[img=http://img44.imageshack.us/img44/4338/questionw.th.png] (http://img44.imageshack.us/i/questionw.png/)

[img=http://img44.imageshack.us/img44/4338/questionw.th.png] (http://img44.imageshack.us/i/questionw.png/)

stupid too short message

[Delphi51]

I see students posting complicated solutions with several tension forces, but I just do it by eye. There is an extra kg hanging on the right, so it will go right and the acceleration is
a = F/M where the force is the mg on the 1 kg and the M is the total mass.

[Doc Al]

Nothing wrong with "doing it by eye" in one step. But I strongly suggest you also do it the long way, by applying Newton's 2nd law to each mass. That's the method that will allow you to solve more complicated problems, so you might as well learn how to use it on this simpler one.

Write a Newton's 2nd law force equation for each mass. Combine those three equations and you'll be able to solve for the acceleration and the two rope tensions.