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Question related to Newton's second law

  1. Nov 5, 2008 #1
    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
    [​IMG]

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

    2. Relevant equations
    [tex]\sum F = ma[/tex]


    3. The attempt at a solution
    Body (A):
    [​IMG]

    N - W2 = 0
    N = W2

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

    Body (B):
    [​IMG]
    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.
     
  2. jcsd
  3. Nov 5, 2008 #2
    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).
     
  4. Nov 5, 2008 #3
    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) .
     
  5. Nov 5, 2008 #4
    Yes: and you have the friction between A & B on block A... but not on block B. Remember Newtons THIRD law...
     
  6. Nov 5, 2008 #5
    the final answer is 4.05 N ?
     
  7. Nov 6, 2008 #6
    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 ??
     
  8. Nov 8, 2008 #7
    so there is no any friction force on (A) ?!!

    I get confused
    help plz
     
    Last edited: Nov 8, 2008
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