# Newton's laws / pulleys question?

1. Oct 20, 2008

I have uploaded the problem here: http://i36.tinypic.com/mps9i.jpg

These are the answers I have gotten so far:

a) L1 = 2*P1 - P2 - x1 + (pi*R)
L2 = 2*P2 + (pi*R) - x2

b) Forces on M1 --> M1*g downward, and T1 upward, which I think is equal to M2*g. Therefore a1 = [(M2 - M1)*g] / M1

Forces on M2--> M2*g downward, and T2 upward, which I think is equal to M1*g. Therefore a2 = [(M1 - M2)*g] / M2

If someone could tell me if I am correct so far, and also how to approach the remaining parts to the problem, I would really appreciate it.

Thank you.

2. Oct 21, 2008

### JoAuSc

Part a is correct. In part b, you assume T1 = M2*g, and T2 = M1*g, but that's false. Consider a force diagram for pulley 2. You have one rope pulling up and essentially two ropes pulling down. Now, a pulley does not change the force between the two rope halves (on either side of the pulley), it just redirects the force, so the tension T1 going up from M1 is also the force going upwards above pulley 2. Also, if one half of the bottom rope for pulley 2 is pulling down on the pulley with T2, then the other half must also be pulling down with T2. Thus, on pulley #2 we have 2*T2 pulling down and T1 pulling up, and these sum to zero since m*a for the pulley is zero since it is massless. So T1 = -2*T2.