2nd and 3rd Newton Law Problem

[Soil]

Consider the figure below where the coefficient of kinetic friction between the m1= 2.0-kg and
the m2 = 3.0-kg blocks is 0.30. The hanging mass is given to be m3 =10.0-kg and the horizontal
surface supporting the blocks along with the pulleys are frictionless.http://img812.imageshack.us/img812/4836/33515317.jpg

Uploaded with ImageShack.us
(a) Draw the free-body diagram of each block.
(b) Identify the action-reaction pairs of forces between the 2.0-kg and the 3.0-kg blocks.
(c) Apply Newtonʹs second law to each block. Use symbols only.
(d) Determine the magnitude of the acceleration of each block.

 

[Rick88]

what part(s) can't you do?

R.

 

[Soil]

I can do the free body diagram for entire system and for individual box. It means I can do (a) and (b) But I can't do the next step to calculate (c) or (d)
Is the acceleration is the same for two boxes and calculated by (T2- total friction forces of two boxes)/(m1+m2) ? How to find T1?
T1 - friction force of smaller box = (mass of smaller box) x (acceleration of system) ?
I think friction of smaller box =( its weight (or its normal force) x coefficient) because the net internal contacted forces is zero, so they don't count for the normal force. is it true?
Thanks

 

[Rick88]

c) doesn't require calculations, as it specifically asks to write the answer in symbols only.

Part d) is entirely down to tensions, weights and coefficient of friction.
For a start, think about what would actually happen when the bodies start moving. it should be easy enough to extrapolate the acceleration of the 10kg body.
From there you can work backwards using forces.R.

 

[Soil]

When the system starts moving ,the 10kg mass goes downwards pull the 3kg mass to the right. In the same time, the 2kg mass is pulled to the left.
But basically I don't know why each box will move with different accelerations.

 

[Rick88]

Because there is friction between the 2kg and 3kg masses. (with a specific coefficient of friction that was given in the question).