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Atwood machine acceleration problem

  1. Mar 14, 2007 #1
    A peculiar Atwood’s machine
    The Atwood’s machine shown in the figure consists of masses m,m/2,m/4,........,m/(2^(N-1)). All the pulleys and springs are mass less, as usual
    a) Put a mass m/(2^(N-1)) at the free end of the bottom spring. What are the acceleration s of all the masses
    b) Remove the mass m= (m/(2^(N-1)) [which was arbitrarily small, for very large
    N] that was attached in part (a). What are the accelerations of all the
    Masses, now that you’ve removed this infinitesimal piece?

    Please solve the Question, I have been trying to answer to it but I am not confident of my method. Please answer it


    picture in the attachment
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2007 #2
    Show us what you did. We'll help you out.
     
  4. Mar 15, 2007 #3
    In part (a), there is perfect balance at all points, so each mass will have acceleration = 0.
    Proof: At the last pulley, there are equal masses on each side, so it will not turn. Now on the next pulley above it, the mass on the left equals the sum of the masses on the right, so it will not turn. This continues all the way to the first pulley.

    In part (b), all the string tensions are zero, so each mass will accelerate downward with acceleration = g.
    Proof: There is a mass on one side of the last pulley, but no mass on the other side. Therefore, the string around the last pully has no tension, so the last mass falls at acceleration g. Since there is no tension in the last string, there is no downward force on the last pulley, so there is no force in the string supporting it. Therefore, the mass on the other end of that string falls freely with acceleration g. This continues all the way to the first pulley.
     
  5. Mar 15, 2007 #4
    Ok. What makes you think that there is perfect balance? There is a mass m/2 on one side, and a mass m/2+m/4+m/8... on the other side which is greater than m/2, so there must be a net acceleration.

    In case of the second question, If you look at the first pulley, there is a mass m/2 on one side and masses m/2+m/4... and so on, on the other side. For the second pulley, there is a mass m/4 on one side and a mass m/8+m/16... and so on on the other... In this way, I think you should get a series for you expression of acceleration.
     
  6. Mar 16, 2007 #5
    i wiil study your solution
     
  7. Mar 16, 2007 #6

    Doc Al

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    Staff: Mentor

    Actually, there's a mass m on one side and a mass m/2 + m/4 + m/8 ... = m on the other side.

    Looks to be in balance to me. For any given pulley, the total mass hanging from the right equals the mass hanging from the left.
     
  8. Mar 17, 2007 #7
    Originally Posted by chaoseverlasting
    Ok. What makes you think that there is perfect balance? There is a mass m/2 on one side, and a mass m/2+m/4+m/8... on the other side which is greater than m/2, so there must be a net acceleration.


    I would like to say that this problem is from the book authored by david morin. I feel Mt answer is correct and secong part there is infinitely small mass attached to the bottom string, which can be treated as there is no mass. Therefore the acceleration is g
     
  9. Mar 17, 2007 #8
    Im sorry. I didnt see the diagram carefully enough. I thought there was a mass m/2 on one sid.
     
  10. Mar 17, 2007 #9
    it is ok. thank you for u r insights
     
  11. Oct 22, 2011 #10

    sxr001

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    Gold Member

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