How to Find Acceleration and Velocity in a Block and Wedge System

[springo]

Homework Statement

http://img187.imageshack.us/img187/9476/dib.th.png

m_A = 22kg
m_A = 10kg
No friction
At t = 0, at rest

a.- Find a_B
b.- Find v_B/A at t = 0.5s

Homework Equations

The Attempt at a Solution

http://img504.imageshack.us/img504/3786/dib3.th.png http://img504.imageshack.us/img504/9979/dib2.th.png

I got to the following equations:
m_A·a_A = m_A·g·cos(30) + N_B·sin(50)
0 = N_A - m_A·g·sin(30) - N_B·cos(50)

m_B·a_A·sin(30) = N_B - m_B·g·cos(20)
m_B·a_B/A - m_B·a_A·cos(50) = m_B·g·sin(20)

There must be something wrong in these because after solving my results are wrong.
I don't know if I chose the wright base for studying each body (maybe the equations would be simpler with some other base?).

Thanks for your help!

 

[Sonolum]

I'm having a little trouble figuring out how you got to those equations - can you tell me a little more of your reasoning?

 

[springo]

I used Newton's 2nd law and then projected the vectors.

 

[Delphi51]

It seems to me that you have oversimplified. For starters, the acceleration of block B is going to be partly vertical and partly horizontal. It will not be in the direction of the ramp surface because that surface is falling and moving to the left. I think you need both horizontal and vertical accelerations for both blocks.

I'm thinking of the free body diagrams for A and B. I can write sum of forces = ma in each direction for each block, and get 4 equations with 6 unknowns including the 4 accelerations and the two normal forces.

N_Bsin(20) = 10a_Bx horizontal
N_Bcos(20) - 10g = 10a_By vertical
-N_Bcos(20) - N_Acos(60) = 22a_Ax horizontal
-N_Bsin20 - 22g + N_Bsin(60) = 22a_Ay

Block A can only accelerate along the ramp so a_Ay = a_Ax*tan(30)

Still missing one equation. Must be something to do with B accelerating along its ramp, but it is itself accelerating so awkward to write!