# Atwood Machine Help

Tags:
1. Sep 22, 2016

### Ryan95

1. The problem statement, all variables and given/known data
In the Atwood’s machine, what should M be, in terms of m1 and m2, so that it doesn’t move?

2. Relevant equations
F=ma

3. The attempt at a solution
I've set T1=Mg as T1 is the tension of the rope attached to M. m1 and m2 are both connected together by the same rope so I assumed T1=2T2 .
I set up the equations
T2-m1g=m1a
T2-m2g=-m2a (negative due to m1 and m2 moving in opposite directions).
I then added the two together to get
T2=[m1(g+a)+m2(g-a)]/2.
Which, since T1=2T2=Mg, it can be expressed as
M=[m1(g+a)+m2(g-a)]/g (I think).
Which is where I'm stuck now. I'm unsure of how to get rid of g and a in order to have my answer in terms of just m1 and m2. It's probably pretty simple, but I can't quite get it.. It seems as g can be canceled out from the current equation, but I'm not sure about a.
I've tried canceling out the g and then solving for a which gives me
a=M/m1-m2
But upon plugging that in as a, I am unsure of how to proceed.

Last edited: Sep 22, 2016
2. Sep 22, 2016

### haruspex

Please either describe the set-up or post a diagram. Not all Atwood's machines are the same.

3. Sep 22, 2016

### Ryan95

Right, apologies. I've added a photo.

4. Sep 22, 2016

### Staff: Mentor

You have 2 linear algebraic equations in two unknowns, T2 and a. Solve for them separately.