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Homework Help: Atwood Machine Proof.

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    It appears that the subscript is not working properly, please take m1 to means mass 1 and m2 to mean mass2

    Atwood's machine consists of two masses connected by a string that passes over a pulley, as show in the figure. Consider the pulley to be massless and frictionless. Show that, if released for rest, m2 takes a time t=[tex]\sqrt{}2h(m2+m1)/g(m2+m1)[/tex] to reach the floor.
    [PLAIN]http://rawrspace.com/atwood.jpg [Broken]

    2. Relevant equations
    I believe,
    T-m1g=m1a
    m2g-T=m2a


    3. The attempt at a solution
    So from the relevant equations I solve for T and set them equal. I get
    m1a+m1g=m2g-m2a

    I solved the equation so that the a's were on one side and the g's were on the other and factored.

    a(m2+m1)=g(m2-m1)

    Then I divided both sides by (m2-m1)

    a(m2+m1)/(m2-m1)=g

    Now I am kind of close I think but I am not sure where to go. I know that acceleration (a) is the distance travelled over time squared. So would I replace a with h/t2 and then solve the equation to get

    t=[tex]\sqrt{}h(m2+m1)/g(m2-m1)[/tex]

    That is where I have hit a brick wall because they have it as 2h , I know that both masses move h distance, how did that get incorporated in however ? I also know that there is an equation 1/2(g)t2 that may be the way it was introduced, but I do not know how to relate them. Any help would be greatly appreciated.
     
    Last edited by a moderator: May 4, 2017
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  3. Sep 24, 2010 #2

    vela

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    If you solve for the acceleration, you get a=g(m2-m1)/(m2+m1), which is constant. That means you can apply the equations you have for constant acceleration.
     
  4. Sep 24, 2010 #3
    Alright so
    We know that for mass 1 , T-m1g=m1a and for mass 2 , m2g-T=m2a
    The reason I set this up is because I was trying to make sure the force is positive and since m1 is moving upward, T is greater and on the otherside, since m2 is moving downward, it is greater than T , is that the correct reasoning for how I came up with the formulas above ?

    Once I have those formulas I rewrote it to

    a(m2+m1)=g(m2-m1)

    then solving for a you get the following
    for a=g(m2-m1)/(m2+m1)

    then using
    x=x_0+v_0t+(1/2)at^2

    you plugin and get
    h=(1/2)[g(m2-m1)/(m2+m1)]t2

    and finally rewriting you get
    t=[tex]
    \sqrt{}2h(m2+m1)/g(m2+m1)
    [/tex]

    Is that the correct reasoning ? I just want to make sure I understand the process.
     
  5. Sep 24, 2010 #4

    vela

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    Yup, that's correct. Good work.
     
  6. Sep 24, 2010 #5
    Thank you very much for the help.
     
  7. Sep 24, 2010 #6
    I had one additional question, It also asks

    Let M be the mass of the pulley in the previous question. If the effect of friction on the pulley is considered , how long does it take m2 to reach the floor? The mass is again released from rest and height h.

    To solve this would I follow the same general format as the last except now in this case I will have a frictional force acting on m2 since it is pulling downward. Or is this where something like torque would come in.

    I believe for the equations I would have

    T=m1a+m1g and for mass 2 , T=m2g-m2a-f

    Then setting them equal and solving I get



    (m2-m1)g-f=(m2+m1)a

    Solving for a I get
    a=[(m2-m1)g-f]/(m2+m1)

    Plugging into the constant acceleration equation, I get

    t=[tex]
    \sqrt{}2h(m2+m1)/g(m2-m1)-f
    [/tex]

    Does that seem correct ? I think it is because it is taking the external forces and subtracting frictional force from it, thus reducing the force that is doing the work.
     
    Last edited: Sep 24, 2010
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