## Homework Statement

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.56. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.6 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley. With what acceleration does the mass m3 fall?

F=ma
Torque=I*alpha

## The Attempt at a Solution

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For a problem like this its always helpful to draw a free body diagram.

You have a weight force acting on m3 equal to (m3)g, and a tension in the rope -T.

m3(g)-T=(m3)a(net)

On m2, you have a force of friction -uFn and the same tension force T, such that -uFn+T=(m2)a(net)

If you add the two together, you will have a net tangential force (Ftan)acting on the pulley. And you know that t=RFtang=I*alpha=R(m3*g-um2g), and you have hte numbers to find I for the pulley, so you can find alpha which is a/r where a is the linear acceleration.

I'm pretty sure I did that right.

thanks a lot, that helped. i figured it out.