Dynamics- Two Body Problems

[hsphysics2]

Homework Statement

Two blocks are stack on each other. The lower block has a mass of 6.8kg and μ_s=0.34 with the floor. The top block has a mass of 3.4kg and μ_s=0.65 between the two blocks. A force of 18N is being applied to the top box to the left. What maximum force can be applied to the lower block without the upper block slipping? What is the acceleration of the system?

Homework Equations

F_x= ma_x

The Attempt at a Solution

I drew free body diagrams for the system, and started the 3 equations as the following;

for m_A,
F_x= m_Aa_x
T- μ_sη-m_Agsinθ= m_Aa_x 1)F_y= m_aa_y
η-m_agcosθ=0for m_B,
F_x= m_Ba_x
m_Bg-T=m_Ba_x 2)I think I should add 1) and 2) to each other so I can get rid of T but I'm not sure about where I should go from here.

 

[haruspex]

Since the OP has chosen an unusual set of variable names and not defined them, I cannot comment on the equations given.

The first thing to check is the state of things with no force applied to the lower block. We find that the 18N is sufficient neither to make the top block slide on the lower block, nor to make both blocks slide on the table.
Next, we must ask what the force applied to the lower block must do in order to make the blocks slide relative to each other. If it also makes nothing move then the 18N still won't. So the extra force must make the lower block accelerate so fast that the combination of the top block's inertia and the 18N force overcomes the static friction.

|m_Agμ_sA|<|F_A+am_A|, where the acceleration a is measured oppositely to the 18N force.
Note that this has two solutions.
To achieve this acceleration, a(m_A+m_B)=F_B+/-(m_A+m_B)gμ_kB, where the +/- sign is opposite to the sign of F_B.
Since we are not told any coefficients of kinetic friction, we cannot go further without making assumptions.