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Two bars and a spring

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Two bars of masses ##m_1## and ##m_2## connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to ##k##. What minimum constant force has to be applied in the horizontal direction to the bar of mass ##m_1## in order to shift the other bar.


    2. Relevant equations



    3. The attempt at a solution
    The second bar, ##m_2## starts to move when the force due to spring overcomes the force due to friction on ##m_2## i.e ##F_{spring}=km_2g##. When ##m_2## starts moving, ##m_1## is still moving. Applying Newton's second law on ##m_1##, ##F-km_1g-F_{spring}=m_1a## where F is the minimum constant force asked in the question. I can substitute ##F_{spring}## but that gives me an equation with two variable. How am I supposed to find F from that? :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jun 9, 2013 #2

    Doc Al

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    Staff: Mentor

    Good.

    Maybe.

    Force analysis will only get you so far. (The spring force and acceleration of ##m_1## are not constant.)

    Hint: Consider the work done by that constant force.
     
  4. Jun 9, 2013 #3
    Why maybe? :confused:

    From energy conservation,
    [tex]Fx=\frac{1}{2}cx^2+\frac{1}{2}m_1v^2+km_1gx[/tex]
    where ##x## is the displacement of ##m_1## and ##c## is the spring constant.

    I don't know v but if I put v=0, I do get the right answer but why is v=0? :confused:
     
  5. Jun 9, 2013 #4

    Doc Al

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    Staff: Mentor

    Because you want the minimum force F that will just move ##m_2##.
     
  6. Jun 9, 2013 #5
    Thanks Doc Al! :smile:
     
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