Friction in a 3 block and pulley system

[Maximusw47]

Homework Statement

A block of mass m1 is on top of a block of mass m2. Block 2 is connected by an ideal rope passing through a pulley to a block of unknown mass m3 as shown. The pulley is massless and frictionless. There is friction between block 1 and 2 and between the horizontal surface and block 2. Assume that the coefficient of kinetic friction between block 2 and the surface, μ, is equal to the coefficient of static friction between blocks 1 and 2.
What is the minimum value of m3 for which block 1 will start to move relative to block 2?

Homework Equations

F=ma
f= mu*N

The Attempt at a Solution

I've taken a couple of approaches. The one I think is most valid is this: The point just before block 2 slips relative to block 1 is where the static friction between m1 and m2 reaches its maximum value. At this point the magnitude of acceleration between all the blocks is the same. I'm then solving for:

m1*a = mu*m1*g
m2*a = T-mu*m1*g-mu*(m1+m2)*g
m3*a=T-m3*g

Which gives me:
m3=2*mu*(m1+m2)/(1+mu)

Which is wrong. Am I making a math error or taking the wrong approach entirely?

Thanks in advance for any help!

 

[paisiello2]

Can you post the diagram or facsimile thereof?

 

[haruspex]

Can you post the diagram or facsimile thereof?

Looks the same as https://www.physicsforums.com/threads/friction-3-blocks-and-a-pulley.820010/. There's a diagram attached to that thread.

 

[Nathanael]

m1*a = mu*m1*g
m2*a = T-mu*m1*g-mu*(m1+m2)*g
m3*a=T-m3*g

You have it almost right, except your sign-convention is inconsistent:
"a" is supposed to be the same for all three equations, right?
In the first two equations, "a" takes a positive value, but in the third equation "a" takes a negative value (m₃g ≥ T)

 

[Maximusw47]

You have it almost right, except your sign-convention is inconsistent:
"a" is supposed to be the same for all three equations, right?
In the first two equations, "a" takes a positive value, but in the third equation "a" takes a negative value (m₃g ≥ T)

Got it. Thanks.