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Static friction 2 boxes and force question

  1. Feb 22, 2017 #1
    Question:
    A large box of mass M is pulled across a horizontal, friction-less surface by a horizontal rope with tension T. A small box mass m sits on top of the box. The coefficients of static and kinetic frciton between the two boxes are Us and Uk. find an expreeion for the maximum tension Tmax for which the small box rides on top of the large box without slipping.


    Relevant equations
    None given

    3. The attempt at a solution
    Know formulas:
    FFriction (or FF) = m2 x a (the friction of the top on lower box)
    FTension (or FT) - FFriction = m1 x a
    the two forces must have the same acceleration in the second formula, will get both acc. alone on both formulasand equal them.

    Top block acc. is:
    FT = m2 x a (get a alone)
    a = FF/m2

    Bottom block acc. is:
    FT - FF = = m1 x a
    FT – (m2 x a) = m1 x a (substituting Ff for the m2 x a value in order to have one force value only)
    FT = m1 x a + m2 x a
    FT = a(m1 + m2)
    a = FT/(m1 + m2)

    a = a, getting Tension or FT alone:
    FF/m2 = FT/(m1 + m2)
    FT = FF(m1 + m2)/m2
    FMax = m1 x g x Us (m1 + m2)/m2 (substituting FT with max as it will represent max).

    Is this correct, thank
     
  2. jcsd
  3. Feb 22, 2017 #2

    BvU

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    The idea is that here you list the equations you're using in the solution attempt. Like ##F = ma ## and ##F_{\rm fric, max} = \mu_s N##....
    In particular the subscript max is important: that's the friction force that determines the maximum T. But you understand that already.

    And don't change notation on the way: if your problem statement uses Tmax, stay with it and don't change to Fmax.

    You mean FF

    Other than that, it looks flawless to me :smile:
     
  4. Feb 22, 2017 #3
    What's m1 and m2?
     
  5. Feb 22, 2017 #4
    Thanks! Late nights for me, i sometimes makes mistakes lol!
     
  6. Feb 22, 2017 #5
    M1 is top box M2 is bottom box
     
  7. Feb 22, 2017 #6

    BvU

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    Go to bed ! :sleep:
     
  8. Feb 22, 2017 #7
    Your answers wrong. You isolated the acceleration incorrectly for the top block. This is what happens if you change notation.
     
  9. Feb 22, 2017 #8

    BvU

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    I rest my case
    in this case M and m

    How about m1 = bottom box and m2 = top box ? :smile:
     
  10. Feb 22, 2017 #9
    I think it's still wrong, if I remember the general solution correctly.
     
  11. Feb 22, 2017 #10
    Yep, I worked it out, it's wrong. Though I can't find out where he went wrong through all the m1 and m2's. :sorry:

    EDIT: Found it, you messed up the last step.
     
  12. Feb 22, 2017 #11
    Sorry lol, m1 is bottom, m2 is top! Spot on
     
  13. Feb 22, 2017 #12
    sorry m2 is top box, m1 is bottom box.
     
  14. Feb 22, 2017 #13
    I get it, you still messed up the last step
     
  15. Feb 22, 2017 #14

    BvU

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    You are absolutely correct, MM, through all the m1 and m2 he missed FF = m2 μ g and filled in m1 in the last step.

    When wbh wakes up he'll see it more clearly. I should go :sleep: too for overlooking that !
     
  16. Feb 22, 2017 #15
    Where did i mess up last step, sorry been spending a week on this question lol
     
  17. Feb 22, 2017 #16
    Bvu is correct, that's where you messed up. Go to sleep now!
     
  18. Feb 22, 2017 #17
    Ah ok lol i see, thanks for your help!
     
  19. Feb 22, 2017 #18

    BvU

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    Then it's really time to go :sleep: !

    And: you're welcome. Well done, MM, wbh and me :smile:
     
  20. Feb 22, 2017 #19
    Thanks! Yes off to bed!:wink:
     
  21. Feb 22, 2017 #20
    :biggrin:
     
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