# Question related to Newton's second law

## Homework Statement

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 [Broken]

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

## Homework Equations

$$\sum F = ma$$

## The Attempt at a Solution

Body (A):
http://img266.imageshack.us/img266/2518/78564322dp4.png [Broken]

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 [Broken]
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.

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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).

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) .

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...

the final answer is 4.05 N ?

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 ??

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

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