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Atwood machine

  1. Nov 19, 2009 #1
    There are two blocks of equal mass M that hang from a massless, frictionless pulley. There is a small square block of mass m sitting on top of one of the blocks. When the block is released, it accelerates downward for a distance H until the block of mass m sitting on tip gets caught by a ring. The block of mass M continues to fall at constant speed for a distance D.

    Derive an expression in terms of g using M, m, D, H, and t, where t is the time that M takes to move at constant speed through the distance D.

    I have made drawing describing the situation:

    http://img691.imageshack.us/img691/3714/38860857.png [Broken]

    I actually have this solution, but I don't understand it. Could someone explain how it was derived?

    [tex]a=\frac{mg}{2M+m}[/tex]

    [tex]v^2=2aH[/tex]

    [tex]v=\frac{D}{t}[/tex]

    [tex]v^2=\frac{2mgH}{2M+m}[/tex]

    [tex]\frac{D^2}{t^2}=\frac{2mgh}{2M+m}[/tex]

    [tex]g=\frac{(2m+m)D^2}{2mHt^2}[/tex]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 19, 2009 #2
    What part are you having difficulty with? What did you try?
     
  4. Nov 19, 2009 #3
    Hi there,

    By looking at your problem, and making the calculations myself, I found this error in your solution:

    [tex]a = \frac{mg}{M + m}[/tex] without the 2M at the bottom
     
  5. Nov 19, 2009 #4
    Are you sure? I did it too and I'm getting 2M.
     
  6. Nov 19, 2009 #5
    From where do you find the second M???

    Let's look at the forces acting on the (M+m) system. You have the tension in the rope, and the weight of both M+m.

    [tex]\sum F = (M+m)a[/tex]
    [tex]T - F_M - F_m = (M+m)a[/tex]
    [tex]\cancel{Mg} - \cancel{Mg} - mg = (M+m)a[/tex]
    [tex]a = -\frac{mg}{M+m}[/tex]
     
  7. Nov 19, 2009 #6
    What i did was :

    (M+m)g - T = (M+m)a for the right side
    T - Mg = Ma => T = M(g+a) for the left
    Then

    (M+m)g - M(g+a) = (M+m)a
    mg-Ma = Ma + ma
    mg = 2Ma + ma
    a = mg / (2M+m)
     
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