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Basic Newton's Laws and Applied Force: 3 boxes

  1. Jul 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Three boxes are in contact with each other on a frictionless horizontal surface as shown. The masses of the boxes are m1 = 10 kg, m2 = 20 kg, and m3 = 30 kg. A horizontal force F = 90 N is applied to m1.
    a. the acceleration of the three boxes.
    b. the net force on each box.
    c. the contact forces between the boxes.

    Screen Shot 2015-07-16 at 3.45.49 PM.png

    2. Relevant equations
    Newtons second law

    3. The attempt at a solution
    total mass = 60 kg
    F = ma

    1) a = F/m = 90/60 = 1.5 m/s^2

    2) net force on each box
    F1 = m1a = 10*1.5 = 15N
    F2 = m2a = 20*1.5 = 30 N
    F3 = m3*a = 30 *1.5 = 45 N

    Am I missing forces in calculating the net force?

    c) contact force
    90-15 = 75 N btwn 1st and 2nd box
    75- 30 = 45 N btwn 2nd and 3rd box

    The above is my best guess although I am not confident that this is the way to calculate contact forces. (What are contact forces?) Should I be using something like f=(F*(m2+3))/(m1+(m2+3) or f=(F*m2)/(m1+m2) to find the contact force between boxes 1 and 2?
  2. jcsd
  3. Jul 17, 2015 #2
    Theres a two force which affects first two boxes (m1 and m2).But theres only one force effects third box which is contact between m2 and m3.Contant force means If you push m1 the force will affect the system and in the system m1 will push m2.But every force has a opposite force so m2 applies force on m1.
    For this problem,If you want to answer question quick,start with the last box.Theres one force act it.Find that force.

    Your find the acceleration right.
    After find that force think this way.
    Theres one box in the middle which its m2.Now theres two forces acting on it the contant force between m2 and m3 and contant force between m1 and m2.And their net force will be m2a=F which its 30N.

    To this solution step by step.Youll find the answer
  4. Jul 17, 2015 #3


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    You should be more confident as you've got the solution precisely correct.
  5. Jul 17, 2015 #4
    I did not notice you were right.What a shame for me
  6. Jul 17, 2015 #5
    So it's right as shown? Never mind the other guesses for contact force? Thanks so much!!
  7. Jul 17, 2015 #6
    Thanks for your explanation! :)
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