1. Sep 22, 2010

### mmmboh

In the Atwood machine, what should M be, in terms of m1 and m2 so that it doesn't move? [PLAIN]http://img163.imageshack.us/img163/8288/atwood.jpg [Broken]

My work: It doesn't move, so I said T1=Mg (T1 is the tension in the rope attached to M), I believe the tension in the rope connecting m1 and m2 is the same since its the same rope, so I said T1=2T2. And I said T2-m1g=m1*a, and T2-m2g=-m2*a, negative because the acceleration of those masses is in opposite directions, but should be of the same magnitude. Adding those two equations together, I find T2=[m1(g+a)+m2(g-a)]/2, and since T1=2T2=Mg, I said
M=[m1(g+a)+m2(g-a)]/g...and I can simplify this to M=2m2*(g-a)/g.
I don't believe this is right because it isn't just in terms of m1 and m2....I did it another way before and got M=(m1+m2)/2, but I don't believe the process was right.
Any help? this is due in the morning.

Thanks.

Last edited by a moderator: May 4, 2017
2. Sep 22, 2010

### PhanthomJay

This is very good, and correct....but, you have written M as a function of m2, g, and a...you need to solve for a to find M as a function of m1, m2, and g
Edit....g cancels out...M is a function of m1 and m2 only

Last edited: Sep 22, 2010
3. Sep 22, 2010

### mmmboh

Ok, if I did my algebra right, I get 4m1m2/(m1+m2), is this what you got?

4. Sep 22, 2010

### PhanthomJay

Yes, that is very good work, I must say. It is difficult setting up the equations, then doing the algebra, with letter vs. numerical values..nice job simplifying, also!