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2 blocks on top of each other, applied force, friction, and tension

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    A 4.32 kg block is placed on top of a 9.16 kg
    block. A horizontal force of F = 66.6 N is
    applied to the 9.16 kg block, and the 4.32 kg
    block is tied to the wall. The coefficient of
    kinetic friction between all moving surfaces
    is 0.169. There is friction both between the
    masses and between the 9.16 kg block and the
    ground.
    The acceleration of gravity is 9.8 m/s^2.
    Determine Tension of string in N

    (see image)
    2. Relevant equations
    ƩF = ma
    fs = μk*FN


    3. The attempt at a solution
    (see attachment for free body diagrams)
    ƩF = ma
    m = 4.32 kg
    atop block = 0
    ƩF = Fapplied - fk - T
    0 = (44.274) - (7.155) - T
    T = 44.274 - 7.155
    T = 37.119

    Apparently I did something wrong somewhere, but I'm not sure where.
     

    Attached Files:

    • PF.png
      PF.png
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    • PF2.png
      PF2.png
      File size:
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      Views:
      428
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What forces act on the upper block? (I don't understand your FBDs.)
     
  4. Oct 12, 2011 #3
    Well maybe it isn't supposed to be there, but I was thinking that since the bottom block has a force in that direction that is not completely canceled out by friction, it would cause the top block to have a force as well, which is equal to Fapplied (to the bottom block) - fk
     
  5. Oct 12, 2011 #4

    Doc Al

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    Staff: Mentor

    That applied force acts on the bottom block, not on the top block. So it doesn't belong in your analysis of the top block. The only influence that the bottom block exerts on the top block is through the normal force and friction.
     
  6. Oct 12, 2011 #5
    Oh! so T = f
    and so it's T = 7.155
    Thanks!
     
  7. Oct 12, 2011 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Yep!
     
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