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mmmboh

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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

My work: It doesn't move, so I said T

M=[m1(g+a)+m2(g-a)]/g...and I can simplify this to M=2m

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.

My work: It doesn't move, so I said T

_{1}=Mg (T_{1}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 T_{1}=2T_{2}. And I said T_{2}-m1g=m1*a, and T_{2}-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 T_{2}=[m1(g+a)+m2(g-a)]/2, and since T_{1}=2T_{2}=Mg, I saidM=[m1(g+a)+m2(g-a)]/g...and I can simplify this to M=2m

_{2}*(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.

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