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Calculating the force for a differential pulley

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A differential pulley carries a weight W. The chain used has N links per foot. The bigger pulley has n notches, which can hold two chain links each. The smaller pulley contains n - 1 notches, which can also hold 2 chain links. The friction is such that the ratio of the force required to lift the weight up to the force required to lower it is R. Assuming that the friction is the same in both directions, find the force to lift the weight and find the force to lower the weight. Use the principle of virtual work.

    2. Relevant equations
    [tex] ∆E = 0 [/tex] for virtual displacement
    [tex] F_{dn} = \frac {W} {n(R - 1) + 1} [/tex]
    [tex] F_{up} = RF_{dn} [/tex]
    The solutions
    3. The attempt at a solution
    I set up the two following equations:

    Call the downwards force F, the radius of the big pulley X, and the radius of the smaller pulley x. I then imagine a virtual rotation of the pulleys of dø. I yank on the chain with F, and the chain moves a distance Xdø, So I do a work of FXdø. However, at the same time, the smaller pulley moves a distance xdø opposite the force, so I do work of F(Xdø - xdø). Furthermore, the weight moves down a distance Xdø/2, which is a work of WXdø/2. There is friction, f, which acts along both pulleys and opposite the movement each time, so it does a total work of -f(X + x)dø. Virtual work states:

    [tex] \frac{WXdø}{2} + F(X - x)dø = f(X + x)dø. [/tex]

    The situation is the same to move the weight upwards, except that the work done by gravity is negative and the force applied is RF instead of F. Thus:

    [tex] \frac{-WXdø}{2} + RF(X - x)dø = f(X + x)dø [/tex]

    Equating them, and moving the weight to the right, dividing by dø:

    [tex] RF(X - x) = WX + F(X - x) [/tex]

    To determine the radius X and x, I note that they can hold 2n chain links and 2(n - 1) chain links respectively. There are N links per foot, so the circumference is 2n/N. Dividing by 2π to obtain the radius I get: n/πN = X. I get something similar for x, except that the numerator is (n-1). I also note that because everything in my equation contains a term in X or x, I can eliminate any factors in front of n and n - 1.

    Therefore X - x = n - n + 1 = 1. The equation reduces to

    [tex] RF = Wn + F [/tex]

    Solving for F I get:

    [tex] F = \frac{Wn}{(R -1).} [/tex]

    Which is not the answer. I think I might have made a mistake in setting up my equations but I have no idea what it could be. Thank you for helping.
     
  2. jcsd
  3. Jan 28, 2017 #2

    haruspex

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    Where does your "relevant equation" for Fdn come from?
    As I understand differential pulleys (https://en.m.wikipedia.org/wiki/Differential_pulley), the return section of chain from hand to small pulley is slack. It certainly does not have the same tension as the one you pull on.
     
  4. Jan 28, 2017 #3
    That's true. I did not notice that.

    I tried performing the calculations again, and they become worse. Supposing I yank the chain labelled W in the diagram again, with a rotation dø I get the following:

    [tex] \frac{WXdø}{2} + FXdø = fXdø [/tex]

    Because the friction around the small pulley does no work if nothing is moving around.

    I can then cancel the Xdø, to have:

    [tex] \frac W2 + F = f [/tex]

    But if do that, I will have to equate the equation to pull the weight upwards to f before replacing. And the work done in lifting the weight by the force RF is RF(X - x)dø anyway, because pulling on the Z-section of the chain will cause both the small pulley and the large one to rotate, but in opposite directions. If I equate them, I end up with terms not containing X or x, which means that I will have to take into account the constants in front of them, namely 1/πN.

    I got the solution from the same textbook as the question. And it is unlikely that there is a mistake in the solution given because they wrote the solution for the upwards force in full (they rewrote the expression for Fdn multiplied by R)
     
  5. Jan 28, 2017 #4

    haruspex

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    Please explain your reasoning to reach that. (Shouldn't x feature somewhere?)
    The question does not make it clear, but I would consider friction as a frictional torque, so you can just write it as fdφ, without the X.
    You need to make it clear in the naming that this is Fup.
     
  6. Jan 28, 2017 #5
    I yank the chain section W downwards, causing the big pulley to rotate through an angle dø. The length of arc by which the pulley moves, and hence the distance the chain is pulled is Xdø (because the length of an arc is equal to the radius times the angle). Since the chain section moves in the same direction as the force, the work done is positive.
    Because I need both chain sections Y and W to move downwards in order to allow the weight to move, they both must increase in length. They do so equally, each increasing by half the increase in the length of chain, namely Xdø. Therefore the weight falls distance Xdø/2. Since the displacement is in the same direction as gravity, there is work done of WXdø.
    However, in pulling the chain across the pulley, there is friction against the big pulley. This friction will act against the section of chain I'm pulling downwards, namely Xdø. Because they are in opposite directions, the work done by friction is -fXdø.
    Equating the total work to zero and then moving the friction term to the right is how I got that equation.

    It does make sense that x would be in there somewhere, but I have no idea of where x would go.

    Trying with the fdø helped remove the problem of the leftover 1/Nπ, but it still does not yield the answer. I did find a combination of the x's and X's that give the answer but I have no idea of how it works. If I could end up with the following equation, I would have my answer:

    [tex] W(X - x) + Fx = RFX [/tex]

    since X - x = 1

    But I have no idea of how to obtain this equation. It seems like I would do work of Fxdø when pulling the weight down with this equation, and work RFXdø when pulling the weight up, but why these quantities I don't know. Furthermore, the signs for the work done by lowering or lifting the weight would have to be the same for this to work, and that should be impossible.


    Not too sure of what you mean by this. I used F to indicate the downwards force in the equations because adding the subscripts becomes messy. RF would then be the force upwards.
     
  7. Jan 28, 2017 #6

    haruspex

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    They cannot both move downwards (nor both upwards). The two coaxial pulleys rotate together, not one to the left and one to the right.
    OK.
     
  8. Jan 28, 2017 #7
    Wouldn't that imply that the total work done by F is then F(X - x)dø, because they smaller pulley rotates in a direction opposite that of the bigger pulley? Then I end up with the equation I put in the first post.
     
  9. Jan 29, 2017 #8

    haruspex

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    You only defined F there as "the downwards force". As you can see from my first post, I thought you meant the pull the hand exerts on the chain.
     
  10. Jan 29, 2017 #9
    You'd have to pull downwards on the Y or W section of the chain to lower the weight. So the force exerted by the hand on the Y or W chain is the downwards force.

    When lowering the weight, you pull downwards with a force F on the Y or W section of the chain. When raising the weight, you pull downwards with a force RF on the Z section of the chain, just like in the diagram.

    I must admit my definitions are a bit messy, and I apologise. I've been working on it for quite a while, so I forget that people aren't in my head. By downwards force, I mean the force that you need to exert to lower the weight, not necessarily an indication of the direction of the force you exert (although you do need to pull downwards anyway). This force is represented by F. The upwards force would then mean the force that has to be exerted to lift the weight. Information from the question states that this force is RF.
     
  11. Jan 29, 2017 #10

    haruspex

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    No. The operator merely pulls less hard on the Z section.
     
  12. Jan 29, 2017 #11
    Less than the force required to lift it?
    Wouldn't it just fall back to its original position?
     
  13. Jan 29, 2017 #12

    haruspex

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    The operator has to exert more than some minimum force to set the weight rising. At a slightly lower force the weight won't move in either direction because of friction. At some force less still, that force and the friction are overcome by the weight and it starts to move down.
     
  14. Jan 29, 2017 #13
    I see. Does that change anything in terms of the calculations though?
    I can see that if I pulled the chain hard enough, the weight would rise by a distance of Xdø. But if I pulled the chain, by how much would the weight fall?
     
  15. Jan 29, 2017 #14

    haruspex

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    There's no connection between how hard you pull and how far the weight moves. Once you are pulling hard enough, it's the distance you pull that determines how far the weight rises. Likewise, when you ease off enough for the weight to descend, the distance the weight falls is determined by how much chain you allow to run back up before pulling a bit harder again.
     
  16. Jan 29, 2017 #15
    So what would be the correlation between the length I pull and the height change of the weight? If I pull the chain by Xdø with a force F (so that the weight falls down), then the weight falls by Xdø. Is that correct?

    Furthermore, if I pull on the chain so as the let the weight fall, when the weight falls, it will drag the chains downwards correct? So the weight does work on the chains equal to WXdø as well, since it is moving the chains?
     
  17. Jan 29, 2017 #16

    haruspex

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    If you pull the chain "by Xdø" then you must be lifting the weight. For the weight to go down you must pay out some chain, i.e. let some be taken back up by the cog. In both cases you are applying a tension, so pulling, but in the first case you pull harder than the cog pulls, and in the second you let the cog win.

    Anyway, if you do pull the chain out by Xdø, how far will the weight rise?
     
  18. Jan 29, 2017 #17
    I see what you mean. So if I pulled the chain with a force F, the weight would fall by Xdø/2 and thus drag the chain by Xdø with it?

    If I pulled the chain out by Xdø, the weight would rise by Xdø/2 because each segment of the chain holding the weight (two of them) would have to decrease by Xdø/2
     
  19. Jan 29, 2017 #18

    haruspex

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    No, you have to think about the smaller cog at the top. If the cogs turn anticlockwise by dø, what effect does the smaller cog have on the vertical sections of chain?
     
  20. Jan 29, 2017 #19
    The smaller cog rotates by dø, so I'm guessing it would lower xdø of the chain?

    Thus the weight would rise by (X -x)dø/2?
     
  21. Jan 29, 2017 #20
    Ok I tried my hand at it again, and I got really close to the solution. I'm still missing something apparently.
    So suppose the pulleys rotate clockwise by an angle of dø, while I apply a force F to section Z. Z rotates opposite F by Xdø, so F does negative work.
    The W and Y section of the chain each feel a force of W/2. The Y chain rises by xdø, and the W chain lowers by Xdø.
    Friction does work both across the small and large pulley, so as to oppose the motion of the pulleys, so it does negative work. Summing those 3 up should give me 0, thus:

    [tex] \frac{W(X - x)dø}{2} - FXdø - f(X + x)dø = 0 [/tex]

    Now, I yank at Z again, exerting a force of RF. The pulleys rotate anticlockwise by Xdø, so this time RF does positive work.
    The W and Y sections still feel W/2 each. The Y chain lowers by xdø and the W chain rises by Xdø.
    Friction is still opposite the displacements, so the work done by it is negative. Thus:

    [tex] \frac{-W(X - x)dø}{2} + RFXdø - f(X + x)dø = 0 [/tex]

    Equating the two equations to the work done by the friction, and dividing by dø I obtain:

    [tex] \frac{-W(X - x)}{2} + RFX = \frac{W(X - x)}{2} - FX [/tex]

    Then calculating X -x = 1/Nπand X = n/Nπ, and multiplying throughout by 1/Nπ, then moving the F terms to the left and the W terms to the right gives me:

    [tex] Fn(R + 1) = W [/tex]

    I solve for F getting:

    [tex] F = \frac{W}{n(R + 1)} [/tex]

    So I must have missed something. But what?
     
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