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Three Blocks Contact force

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Three blocks on a frictionless horizontal surface are in contact with each other. A force F is applied to block A (mass mA ). If mA=mB=mC=10.0 kg and F = 96.0 N, determine the force of contact that each block exerts on its neighbor.


    2. Relevant equations



    3. The attempt at a solution

    a=96/(10+10+10)=3.2N

    contact force between mA and mB: 96-3.2*10=64
    contact force between mB and mC: 3.2*10=32 or 64-3.2*10=32

    did i solve this right?

    thank you
     
  2. jcsd
  3. Sep 27, 2008 #2

    tiny-tim

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    Hi robvba! :smile:

    Correct result … and it looks like you got it the right way, but I'm not sure.

    This is a Newton's second law problem … you really shouldn't talk about the "force between" two bodies … each equation should only concern the forces on a particular body. :smile:
     
  4. Sep 27, 2008 #3
    thank you. you're right. it's force of contact that each block exerts on it's neighbor. which would mean that block mA exerts a force of 64N on block mB while block mB exerts an equal and opposite force back; so mA = -mB. is that a fair statement?
     
  5. Sep 27, 2008 #4

    tiny-tim

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    erm … I don't think you mean mA = -mB, do you?

    I was thinking more of mass times acceleration = sum of forces …

    you know the acceleration of block A, so the sum of forces on block A is … , and so the reaction force from block B on block A is … ? :wink:
     
  6. Sep 27, 2008 #5
    right. i meant Fba = - Fab (Force of A on B = negative Force of B on A)...?
     
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