Friction and tension between two stacked bricks

[Atx]

Homework Statement

[PLAIN]http://img693.imageshack.us/img693/8219/utennavnu.png

Homework Equations

n= normal force
µ*n=friction
F=ma.
T = tension , mA = mass of A, mB mass of B

The Attempt at a Solution

a) Static friction should be used. Forces on brick A : x: T - µ*mg=0 y dir: n-mg=0
B: x: (mA+MB)g*µ , tension and friction between A and B, y: mg=n , force from A.
b) Kinetic friction is to be used, have tried to find an expression of the acceleration , but so far hasnt come to the right answer.

 

[JaredJames]

Could you elaborate on the question please? What exactly are they asking here?

Jared

 

[Atx]

wow, my bad.

A force F is pulling on B in the right horizontal direction. The mass of A is 5.00 kg, the mass of B is 3.00 kg. Static friction between all surfaces is 0.600 , kinetic friction is 0.400. Assuming the blocks are at rest, for what value of F will the blocks be put into movement?

 

[Atx]

I'd appreciate any hints, I've been stuck on this problem for a while.

 

[hadsed]

Are you sure it's block B that is getting the force? The picture makes it look like A. Just making sure.

Try modeling the two blocks separately in a free-body diagram, labeling all forces acting on them. Don't forget the forces the blocks exert on each other.

 

[Atx]

Always trust picture. A is getting the force. I've modeled the forces as a free body diagram.
The forces excerted on B is x: T(tension)-0.3*mg = m*a (where a is the same accelleration as the one on block A). B in y direction is mg-mg=0 which means there is no accelleration in y-direction for either A or B. A is tricky though because its affected by a force of mg in y direction (from the other block) , Mg and a normal force of equal magnitude. In x direction A is affected by T(tension)-0.3*mg but there is also friction between A and B, so how can I set up an equation to find out when the blocks move?