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Exact forces (not net forces) involved in pushing a line of boxes?

  1. Jul 21, 2010 #1
    When you have a line of 4 boxes without gaps between adjacent boxes, and all of them are at rest, what exact forces are involved when a rightward force is applied to the leftmost box to cause it to move straight into the other boxes? To make this easier, lets assume no gravity, friction, or air resistance is involved.

    Here is a picture of the situation:
    fbd_line_problem.jpg

    I have been trying various ways to solve this. One thing I did was figure out how a stack of objects sitting on the ground would interact with eachother (https://www.physicsforums.com/showthread.php?t=417332"). I was hoping this would clear things up, but I am still confused as to how this works. Here is what I came up with:
    fbd_line_problem_solution1.jpg

    To me, this looks like the force being applied to A causes D to accelerate leaving A, B, and C at rest. However, as soon as D moves, C is no longer applying F4 to it so it cannot move. I have been trying to solve this "chicken and egg" problem for a while, but I can't quite get my head around it...

    Thinking of ABCD as a single object of mass A_mass + B_mass + C_mass + D_mass works of course, but since they don't actually form a single solid object, I need to somehow figure out how the individual objects are affecting eachother. I was able to figure out that the net force of each object is rightward facing and has a magnitude equal to a fraction of the force pushing object A to the right. The fraction is based on the mass of the object in relation to the total mass (as I type this I'm guessing it is (mass/total_mass)*force), but I can't figure out how that net force comes about. What exact forces were involved to create that net force?
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jul 21, 2010 #2
    make an experiment ;]
    if you push box A with force F, all four boxes will accelarate at the same rate, with acceleration given by:
    a = F/(Ma+Mb+Mc+Md)
    box D will not start to move away from D, unless there is something else involved besides boxes and force you apply (like some bomb between box C and D which explodes ;]).
    Net force on each box will be the force, that makes this box to accelerate at a:
    net_F(on D) = Md*a = Md*F/(Ma+Mb+Mc+Md)
    excact forces on D when added must give net_F(on D). But there is only one force acting on D, that is force exerter by box C.
    On box C there are two forces - one by box B, o ther by box D. According to Newton's third law force by box D is equal in magnitude to the force by C to D, but acting in opposite direction. Using this you can find force by box B on box C. And similarly forces on B and A.
     
    Last edited: Jul 21, 2010
  4. Jul 21, 2010 #3
    Hmmm, well that's the result I described in my post, but it's good to know that it's actually correct. So given the mass of A, B, and C as 10 and the mass of D as 20, and a rightward force of 1N, the forces will look like this?
    (For simplicity, the scale of each arrow is not taken into account)
    fbd_line_problem_solution2.jpg

    Are these all of the forces involved (assuming no gravity, air resistance, or friction)?

    Edit: I mean all of the forces involved with A, B, C, and D. In other words, only the FBDs of A, B, C, and D. There is something applying the force to A, but I'm not interested in the FBD of that "something".
     
    Last edited: Jul 21, 2010
  5. Jul 21, 2010 #4
    well, that force of 1 N must have come from somewhere, if you pushing box A with your hand, then box A will exert force F on your hand.
     
  6. Jul 21, 2010 #5
    Sorry, I meant to ask if those are the only forces involved when considering only A, B, C, and D. There is something applying the force, but I'm only interested in the FBDs of A, B, C, and D.
     
  7. Jul 21, 2010 #6

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  8. Jul 21, 2010 #7
    I guess that then those are the only forces. At least only ones I could think of ;]
     
  9. Jul 21, 2010 #8
    Thanks for the help.

    This raises another question, but I'll create a new thread for it.
     
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