Solving Tension on Slope w/ 2 Packages

[cjavier]

The question: Two packages have just started sliding down a ramp that forms an angle of 30 degrees with the horizontal. They are connected with a rope. Package A is 6m up along the ramp, with a mass of 4 kg and a coefficient of kinetic friction of 0.3. Package B is directly above 4 m further up the ramp, with mass of 8kg and a coefficient of kinetic friction of 0.2. The packages are placed such that if B began moving and A did not, B would have to go through A to move down, i.e they cannot pass each other on the side.

1) Find the tension in the rope just after they start sliding.
2) Find the velocity of both packages when package A arrives at the bottom of the ramp.

I am not sure how to approach this problem. My teacher wrote on my test that T=0 because the rope is slack. He then wrote that the final velocity of package A was 5.5m/s (although I cannot read his handwriting) and that the final velocity of package B was 7.7m/s.

Please help! My approach was to sum the forces in the direction of the ramp for each box like so:

BOX A: ƩF_x = m_box-aa = mgsin(30) - T - μ_kmgcos(30)
BOX B: ƩF_x = m_box-ba = mgsin(30) + T - μ_kmgcos(30)

But I could not derive T becoming 0. Thank you all.

 

[haruspex]

If T > 0 then the two accelerations are the same.
a = gsin(30) - T/m_a - μ_kagcos(30)
a = gsin(30) + T/m_b - μ_kbgcos(30)
What do you get if you combine those equations, eliminating a, and use the fact that μ_ka > μ_kb?

 

[cjavier]

If T > 0 then the two accelerations are the same.
a = gsin(30) - T/m_a - μ_kagcos(30)
a = gsin(30) + T/m_b - μ_kbgcos(30)
What do you get if you combine those equations, eliminating a, and use the fact that μ_ka > μ_kb?

I'm not quite sure what I am supposed to get. I get T/m_a + T/m_b= -gcos(30)
[I substituted μ_ka with 2 and μ_kb with 1 for simplicity]

 

[haruspex]

I'm not quite sure what I am supposed to get. I get T/m_a + T/m_b= -gcos(30)
[I substituted μ_ka with 2 and μ_kb with 1 for simplicity]

So can T be positive?