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Double pulley Atwood machine (with 3 masses)

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known data
    EC21.png
    I need help with number 4.
    Also I am altogether unsure of this problem... if you catch any mistakes, point them out! :)

    2. Relevant equations
    F = ma

    Here are the variables I used in my attempt at the solution
    mx = m subscript 1, 2, or 3
    ±a1 = acceleration of either mass on smaller pulley
    T1 = tension of string on smaller pully
    ±a2 = acceleration of either mass (one is a pulley) on larger pully
    T2 = tension of string on larger pully
    3. The attempt at a solution

    (a)

    m2*g - T1 = m2*a1 ---> from the free body diagram of mass 2
    T1 - m1*g = m1*a1 ---> from the free body diagram of mass 1
    m3*g - T2 = m3*a2 ---> from the free body diagram of mass 3
    T2 - m1*g - m2*g = (m1+m2)*a2 ---> from the free body diagram of small pulley

    substituting T1 from the second equation above in the first equation
    m2*g - m2*a1 = m1*a1 + m1*g
    a1 = (m2*g - m1*g) / (m1+m2)
    so if downward is the positive direction,
    mass 1's acceleration = -a1
    mass 2's acceleration = +a2

    substituting T2 for the last two equations,
    m3*g - m3*a2 = (m2+m1)*a2 + m1*g + m2*g
    a2 = (m3*g - m1*g - m2*g) / (m3+m2+m1)
    mass 3's acceleration is +a2

    (b)

    a1 = 0 now, because the numerator of a1 would end out equal to zero.
    This means the smaller pulley's masses would not experience any acceleration

    (c)

    since we want a2 to be zero, the numerator of a2 must equal 0, so
    m3 = m1 + m2
    also,
    m2>m1

    (d)

    I'm confused, the expression won't change for a1? It will still be ±a1 for either masses.
     
  2. jcsd
  3. Apr 1, 2016 #2
    For part 4, you can simplify the problem - since m3 isn't moving you can forget about it. The problem then is what is the acceleration of two masses on a pulley?
    Say you have m1 and m2, what is the net force? and then the acceleration?
     
  4. Apr 1, 2016 #3
    Fnet = m2*g - m1*g
    Set Fnet = ma ---> total mass m = m1+m2
    So, a = m2g - m1g / m2+m1
    Which is the same answer I got for the original case.
     
  5. Apr 1, 2016 #4

    haruspex

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    By doing that, you are making an incorrect assumption. It would be ok if you change that to be accelerations relative to the small pulley, but then your equation needs to change.
     
  6. Apr 1, 2016 #5
    Do you mean that the pulley is the one accelerating, thus there is only one acceleration? Isn't m1 accelerating in the upwards direction, and m2 accelerating in the downwards direction?
     
  7. Apr 1, 2016 #6

    haruspex

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    The pulley is accelerating, and m1 and m2 are also accelerating relative to the pulley.
    Let the pulley's acceleration be a, and the absolute accelerations of the masses be a1, a2, taking up as positive for each. What is the algebraic relationship between the three accelerations?
     
  8. Apr 1, 2016 #7
    The magnitude of a, a1, a2 will all be the same.
    I'm not quite sure what you mean by the algebraic relationship.
     
  9. Apr 1, 2016 #8

    haruspex

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    Not so. Let's start with a simplification. Suppose the pulley does not accelerate. What is the algebraic relationship between a1and a2 now?
     
  10. Apr 1, 2016 #9
    Ok, a1 = -a2 or the other way around
    I think the tangential acceleration of the pulley should be equal to a1 or a2
     
  11. Apr 1, 2016 #10

    haruspex

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    Yes.
    Now allow the pulley a vertical acceleration a. What is the relationship between a1, a2 and a?
     
  12. Apr 1, 2016 #11
    -a1 = a2 = a ?
     
  13. Apr 1, 2016 #12

    haruspex

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    No.
    It might help to think about the three heights from the ground. Let these be y1, y2 and y. If the length of the string is s, can you write a relationship between y1, y2, y and s?
     
  14. Apr 1, 2016 #13
    Haha, I think I might be onto something.
    I overlooked the free body diagram of the smaller pulley itself.

    It would look something like this right?
    T2 - 2*T1 = 0

    Meaning haruspex, the acceleration of the pulley is the negative of the acceleration of m3?
     
  15. Apr 1, 2016 #14

    haruspex

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    Since the pulley has no mass, that is not going to help in finding the acceleration of the pulley. The relationship between the three accelerations is a purely kinematic one, nothing to do with masses or forces. Please try to answer my question in post #12.
     
  16. Apr 1, 2016 #15
    Sorry, I edited my post, the vertical acceleration of the pulley is equal to the negative of the acceleration of m3.
     
  17. Apr 1, 2016 #16

    haruspex

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    That is true, but it does not give us the relationship between that acceleration and the accelerations a1 and a2. The relationship we need results from the fact that the string connecting m1 and m2 has a fixed length.
     
  18. Apr 1, 2016 #17
    haruspex, I've tried to answer post 12, but it seems like a pretty hard question to answer (to me).
    Here is an attempt,
    For m1, it has both a1 and a2 contributing to its height.
    y1 = y0 + y_due m2 accelerating + y_due to m3 accelerating
    I can substitute by using kinematics equation y = 1/2at^2, not sure where to go from here.

    I think that,
    y1 = y0 + y2 + y3 ---> [y3 = y]
    is that right?
     
  19. Apr 1, 2016 #18

    haruspex

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    No, you're making it a hard question. Forget about accelerations for now. This is just a question about the relationship between four distances that comes straight from the geometry.
    Treat the pulley as a point. If m1 is height y1 above the ground, m2 is height y2 above the ground, the pulley above them is height y above the ground, and a string of length s runs from m1 up to the pulley and back down to m2, there is a very simple algebraic relationship between those four distances. What is it?
     
  20. Apr 1, 2016 #19
    I was confused on which acceleration of the pulley you meant (Tangential due to m2 or Vertical due to m3). This is what you mean right?

    acceleration of m1 = a1+a
    acceleration of m2 = -a1+a
    With respect to the ground
     
  21. Apr 1, 2016 #20
    y1 = y2
    y = y1 + 0.5*s
     
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