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Applied force and the force between boxes?

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 5.0-kg mass, which pushes against the other two masses. What is the contact force between the 3.0-kg mass and the 2.0 kg mass?

    a) 25 N
    b) 10 N
    c) 40 N
    d) 0 N


    2. Relevant equations

    Maybe F = m a?



    3. The attempt at a solution

    [ 5 kg ] [ 3 kg ] [ 2 kg ]

    I would think that gravity would diminish the applied force of 50 N.

    Fg = m g
    Fg = 5 kg * 10 m / s^2
    Fg = 50 N

    So the 50 N force would be subtracted from the applied force of 50 N, leaving 0 N, right?

    I'm confused. Can someone help me?

    Thank you!
     
  2. jcsd
  3. Jan 27, 2010 #2
    50 newtons here is pushing 10 kg. so you know the acceleration of the system. Since each box will have an acceleration, and a different mass, you have what you need to calculate the applied force for each individual box
     
  4. Jan 27, 2010 #3
    Ah, so the acceleration of the system is 5 m/s^2.

    Fmass1 = m * a = 5 kg * 5 m/s^2 = 25 N
    Fmass2 = m * a = 3 kg * 5 m/s^2 = 15 N
    Fmass3 = m * a = 2 kg * 5 m/s^2 = 10 N

    Because the contact forces between mass2 (the 3-kg mass) and mass3 (the 2-kg mass) are 15 N and 10 N respectively, does that mean that I should subtract 15 and 10 and get 5?

    Or is 10 N already correct?

    Thanks again!
     
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