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Homework Help: Systems force problem

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    three boxes are slid across a frictionless surface, 100N is applied to m1(mass1) which is pushed against m2 and m3 each box weighs 10 kg
    a draw a free body diagram for each box
    b find the acceleration of the system
    c the net force on each block
    d the force of contact that each block exerts on its neighbor

    2. Relevant equations

    3. The attempt at a solution
    Well I have drawn my free body diagrams

    then I set up my equation for part b
    though I am not sure where to go from there I was hoping somone could show me how to do the rest of the problem with different values than the ones I was given, and explain how they did it so I can get a better grasp on what to do here..

    {now since it is a frictionless surface and a normal force pushes back as much as is applied to it would the third objects normal force be 100 newtons? If so then it would be really simple and turn out to bed
    100=10a} <---I understand this is wrong...

    I feel way off track?

    Thanks for your insights

    Attached Files:

    Last edited: Sep 23, 2010
  2. jcsd
  3. Sep 23, 2010 #2


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    Make a picture with a drawing program and upload as attachment.

    Draw the normal forces in your figure. The boxes are in contact, so they move together with the same acceleration. Apply sum(F) =ma for each block.

  4. Sep 23, 2010 #3
  5. Sep 23, 2010 #4


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    The first block is pushed with Fa. Fa can not act between two blocks.

    Two boxes act to each other with forces of equal magnitude and opposite directions.

    The boxes move on a horizontal surface, their weight, mg, cancels with the normal force of the surface. You need to work only with the horizontal forces .
    There is no normal force at the free end of the third box.

  6. Sep 23, 2010 #5
    if there is no normal force at the end of the last box then what is pushing it forward?
  7. Sep 23, 2010 #6


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    The other box at the other side.

  8. Sep 23, 2010 #7
    right right.. I think I got it now thanks ^^
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