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Atwood machine

  1. Oct 10, 2009 #1
    1.In the diagram above, the pulley is frictionless and the ropes are massless. Given that m1 = 14.5 kg and m2 = 23.0 kg, calculate the acceleration of m2 downwards. What is the magnitude of the tension in the string?
    http://loncapa2.fsu.edu/enc/59/442f45016df264899bda870a0d495f3be185f0a9c3a069c0974d3402c80bd41f1205766cdaec4ada41f4efc2666f14750f4fc16641ffd2b0af613da7e393c18c701021fb78fb5ffbd3d77c79fd9b210f.gif


    I am not sure what the relevant equations would be.

    N=mg?




    3. The attempt at a solution

    N1=(14.5)*(9.81)
    N2=(23-14.5)*(a)

    Set the equations equal and then solve for "a"?

    I do know that this is an Atwood machine. I have searched this forum for examples and I am still unsure of how to solve this problem. TIA.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Oct 10, 2009 #2
    Ok i solved the problem thanks to some further research....

    For mass1: Mg - T = ma

    For mass2: T-mg = ma

    Solved for T then put back into the equation for mass2.....However, could someone give me the reasoning behind the order of the two equations i.e. Mg - T and T - Mg....TIA
     
  4. Oct 10, 2009 #3

    Delphi51

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    "order of the two equations"?
    I probably don't understand, but there is no special order - either one could be written first. They are just the "sum of forces = ma" for each of the two masses.

    Can't see your diagram.
     
  5. Oct 10, 2009 #4
    Sorry.....why is the equation for mass1 Mg - T = ma.......and why is the equation for mass2 written as T - Mg = ma.....why are they "reversed"? Where one is -T while the other is +T.....the same question could be asked for Mg...
     
  6. Oct 10, 2009 #5

    Delphi51

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    Okay - sorry I'm a bit obtuse!
    I'm guessing that the tension is up on one mass and down on the other so that's why the sign changes on the T. But it is odd to have a positive Mg as if gravity is up. I wish I could see the diagram.
     
  7. Oct 10, 2009 #6
  8. Oct 10, 2009 #7

    Delphi51

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    It will take a few hours for that attachment to be approved.
    If convenient, post it to a free photo site like photobucket.com and put a link here. If you put IMG in square brackets before the link and /IMG in square brackets after it, the pic will show up right in the post.
     
  9. Oct 10, 2009 #8

    Delphi51

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    I found a picture of the Atwood machine and the equations you used here: http://en.wikipedia.org/wiki/Atwood_machine
    They are taking clockwise to be positive so they get mg positive on the one that is accelerating upward and mg negative on the one that is accelerating downward. You are free to choose whatever sign convention you want on each side, and it makes sense to keep all the forces and accelerations positive.

    Maybe better to think of it this way:
    Assume the acceleration is up on the left and down on the right and use "a" for both. Then you have -mg on the left because it opposes the ma. If the assumption is wrong, you'll end up with a negative value for a.
     
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