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Two boxes sliding down incline

  1. Oct 6, 2011 #1
    Hello,

    I have two boxes with the same mass sitting on an incline:
    http://img31.imageshack.us/img31/97/boxesy.png [Broken]

    Uploaded with ImageShack.us

    The coefficient of kinetic friction between A and the incline is greater than the coefficient of kinetic friction between B and the incline. When I simulate this in Algodoo, I observe that A and B move as one. I assume they have the same acceleration.

    I want to determine this acceleration. Firstly, here are my free body diagrams for both boxes:
    http://img402.imageshack.us/img402/1545/98782524.png [Broken]

    Uploaded with ImageShack.us

    http://img543.imageshack.us/img543/4619/85772392.png [Broken]

    Uploaded with ImageShack.us

    Are these correct?

    Thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 7, 2011 #2
    they look ok to me
     
  4. Oct 7, 2011 #3
    This is actually not a homework/coursework question. I am doing this for a personal project. But if that's the best place to put it, that's fine with me.
    I'll post my equations soon.
     
  5. Oct 8, 2011 #4
    I got an email saying someone had replied to my question, but I see that the reply is not shown on this thread. I'm not sure what happened.
    Anyway, here is the equation I had some up with for resultant force on A:
    [PLAIN]http://www.texify.com/img/%5CLARGE%5C%21ma%20%3D%20mg%20%5Csin%20%5Ctheta%20%2B%20T%20-%20%5Cmu_a%20mg%20%5Ccos%20%5Ctheta.gif [Broken]

    and for B:
    [PLAIN]http://www.texify.com/img/%5CLARGE%5C%21ma%20%3D%20mg%20%5Csin%20%5Ctheta%20-%20T%20-%20%5Cmu_b%20mg%20%5Ccos%20%5Ctheta.gif [Broken]

    The problem was that I was assuming T was equal to the 'x' portion of weight, and I was getting a weird result. In the email I had gotten, the person said I have to solve for T and a separately.
    By setting the equations equal to each other, I have T:
    [PLAIN]http://www.texify.com/img/%5CLARGE%5C%21%5Cfrac%7B%5Cmu_amg%5Ccos%20%5Ctheta%20-%20%5Cmu_bmg%5Ccos%20%5Ctheta%7D%7B2%7D.gif [Broken]

    and the acceleration as:
    [PLAIN]http://www.texify.com/img/%5CLARGE%5C%21g%5Csin%5Ctheta%20-%20%5Cmu_bg%5Ccos%5Ctheta%20-%20%5Cfrac%7B%5Cmu_ag%5Ccos%5Ctheta%20-%20%5Cmu_bg%5Ccos%5Ctheta%7D%7B2%7D.gif [Broken]

    Thanks for the person who replied.
     
    Last edited by a moderator: May 5, 2017
  6. Oct 8, 2011 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You can simplify the formula for acceleration to a=gsinθ-0.5gcosθ(μab).

    When you have similar problems with interacting bodies having the same acceleration, you can add all equations to cancel the internal forces, and then you get the common acceleration as the sum of all external forces divided by the total mass.

    ehild
     
  7. Oct 8, 2011 #6
    I replied, but I misread the OP and wrote the equation too soon, so it was deleted.
     
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