1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?






    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

    (M+m)g - M(g+a) = (M+m)a
    mg-Ma = Ma + ma
    mg = 2Ma + ma
    a = mg / (2M+m)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook