Question related to Newton's second law

1. Nov 5, 2008

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 [Broken]

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 [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.

Last edited by a moderator: May 3, 2017
2. Nov 5, 2008

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).

3. Nov 5, 2008

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) .

4. Nov 5, 2008

physics girl phd

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

5. Nov 5, 2008

UNknown 2010

the final answer is 4.05 N ?

6. Nov 6, 2008

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 ??

7. Nov 8, 2008

UNknown 2010

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

I get confused
help plz

Last edited: Nov 8, 2008