Question related to Newton's second law

[UNknown 2010]

1. The problem statement, all variables and given/known data

Given data:
- The weight of (A) is 3.6 N.
- The weight of (B) is 5.4 N.
- The kinetic friction coefficient between all surfaces is 0.25
http://img369.imageshack.us/img369/2999/figurehg3.png

The problem statement:
Find the required force for dragging (B) with a constant speed.

2. Relevant equations
\sum F = ma


3. The attempt at a solution
Body (A):
http://img266.imageshack.us/img266/2518/78564322dp4.png

N - W2 = 0
N = W2

Fk - P = 0
Fk = P
uk × N = P
0.25 × 3.6 = P
P = 0.9 N.

Body (B):
http://img440.imageshack.us/img440/6740/35855205ln7.png
P + Fk = F
F = ukN + P
F = [ 0.25 × (3.6+5.4) ] + 0.9
F = (9 × 0.25) + 0.9
F = 3.15 N.

[physics girl phd]

One thing I'm slightly concerned about is that you have a friction force between the ground and the block on body B, but not a friction force between the two blocks A & B (which would be present on the top of the block via the way you've been nicely drawing the diagrams).

[UNknown 2010]

The question says "The kinetic friction coefficient between all surfaces is 0.25" so that doesn't mean there is only a friction between the ground and body (B) .

[physics girl phd]

The question says "The kinetic friction coefficient between all surfaces is 0.25" so that doesn't mean there is only a friction between the ground and body (B) .

Yes: and you have the friction between A & B on block A... but not on block B. Remember Newtons THIRD law...

[UNknown 2010]

the final answer is 4.05 N ?

[UNknown 2010]

body (A):

N - W2 = 0
N = W2
Fk(A) - P = 0
Fk(A) = P
uk × N = P
0.25 × 3.6 = P
P = 0.9 N.

Body (B):

P + Fk(A) + Fk(B) = F
F = uk N(A) + 0.9 + P
F = [ 0.25 × (3.6+5.4) ] + 0.9 + 0.9
F = (9 × 0.25) + 1.8
F = 4.05 N.

Right ??

[UNknown 2010]

Yes: and you have the friction between A & B on block A... but not on block B. Remember Newtons THIRD law...

so there is no any friction force on (A) ?!!

I get confused
help plz