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Atwood Machine diagram

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/a2fLm

    2. Relevant equations

    F = ma is it

    3. The attempt at a solution

    The problem I had is that it's attached to the floor all of a sudden. Never before have we had Atwood problems like this, and I can't find any online... I drew out the free body diagram but probably screwed up somewhere because of it. I know my answer is wrong.

    I had T1-mg = ma1
    and T2-Mg=Ma2

    ***I said if m goes down 1, pulley goes up a 1/2, and M goes up 1/4 (this is where I screwed up I think...)***
    And thus a1 = -4a2

    T1 - mg = m(-4a2)

    And the appropriate work ended me with a solution of 2g/3 (which is wrong I'm 99% sure of).

    So I need help knowing what I did wrong in the stars (presumably I think this is my mistake, as I have never done it with a floor and was very lost...)
     
  2. jcsd
  3. Feb 11, 2010 #2
    The real problem here is to find how accelerations of the masses relate to each other. Here's a nice trick.

    First we assume that the positive direction of everything we consider points down. So, we assume that both masses move down (which is obviously wrong, but you'll see why we do so).

    Let

    [tex]x_{m}[/tex] be the distance from center of the top pulley to the mass m,

    [tex]x_{M}[/tex] the distance from the center of the top pulley to the mass M,

    [tex]h[/tex] the distance from the top pulley to the ground and

    [tex]x[/tex] distance from the top pulley to the center of the bottom pulley.

    We know that acceleration is second derivative of position with respect to time.

    The main assumption is that the lengths of the strings are constant. So, for the length of the string attached to mass m we get

    [tex]L_{m}=x_{m}+x[/tex]

    Now, if we apply second derivative we get accelerations.

    [tex]0=a_{m}+a \Rightarrow a_{m}=-a[/tex]

    Notice that the derivative of constant [tex]L_{m}[/tex] is zero and accelerations of the mass m and the bottom pulley are in opposite directions and same magnitude.

    Now, for the other string

    [tex]L_{M}=(h-x)+(x_{M}-x)[/tex]

    [tex]0=0-2a+a_{M} \Rightarrow a_{M}=2a \Rightarrow a_{M}=-2a_{m}[/tex]

    All constants become zero again and finally we see that acceleration of the mass M is twice the acceleration of the mass m and they point in opposite directions.

    Using this and force diagrams should lead you to the right answer.
     
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