What is the acceleration of a falling block attached to a pulley system?

AI Thread Summary
The discussion revolves around calculating the acceleration of a falling block in a pulley system involving friction and a uniform disk pulley. The block on the table has a mass of 2 kg and a friction coefficient of 0.75, while the falling block has a mass of 4 kg. The user initially calculated the acceleration to be 3.9 m/s² but struggles with incorporating the effects of the pulley in their calculations. They express confusion about determining the tension in the string above the table and below the pulley, noting that the net force on the 4 kg mass is 15.6 N. The conversation highlights the complexities of applying Newton's laws in systems with rotational dynamics.
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Homework Statement


A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.75. 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 = 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.
With what acceleration does m3 fall?

Homework Equations


Frictional Force = mew x normal
F = MA
Newtons' second law

The Attempt at a Solution


If I could treat the pulley as a point and didn't have to pay attention to it this would be a cinch. To be honest, I don't know what to do with it at all, and I'm looking for some conceptual help here.
 
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Much time later, I finally got the acceleration to be 3.9. The next question in this set is: what is the tension in the string above the table, and below the pulley. (They will be separate.)
Now, since F=MA, I thought this would be easy. 3.9 x 4 = 15.6. It itsn't that easy, it turns out. Any ideas?
 
Well, 15.6 N is the net force on the 4 kg mass.

This force is due to the string tension and what else?
 
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