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Force on pulley shaft

  1. Aug 24, 2011 #1

    vrc

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    hello,

    I have some trouble understanding the force vector of a loaded pulley shaft:
    the pulley is driven by the shaft with a torque T, the pulley on his turn is driving a belt.
    because of newton 3th law there's tension in the belt as shown,

    ok, now the question: many books state that the green force is the shaft force => how can this be true:

    the shaft load is de resultant force : adding those 2 red force vectores , so the shaft load must be a vector that is oriented toward right instead of that green force.

    the green force is de reaction of the pulley support = reaction of the shaft

    in shaft calculation they use the green force vector as shaft load....

    do not understand it !
    krachten.jpg

    grtz
     
  2. jcsd
  3. Aug 25, 2011 #2
    Put a diagram up of the problem and make some form of attempt at the solution, then I will take a look.
     
  4. Aug 25, 2011 #3

    vrc

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    oké, a better explenation:

    riem.jpg

    we see a pulley with a shaft; it is driving a belt because the shaft has a torque, the power tis transmitted because of a difference in belt force , logic

    now: suppose I want to calculate the diameter of the shaft, first I have to know the shaft load,
    I say it's the green force vector, but official mechanical books say that the shaft laod is a vector that has an opposed direction, it's pointed tot the secand quadrant

    how could the shaft load be a vector opposed to the green one in the picture ?

    thank you
     
  5. Aug 25, 2011 #4

    jack action

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    The shaft load is all the forces acting ON the shaft. So you have the tension from your belt that is pushing on your shaft but there is also the bearing support that is pushing on your shaft. In your particular case, it just happens - from basic statics calculations - that the belt tension and the bearing support reaction are equal and opposite.
     
  6. Aug 26, 2011 #5

    vrc

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    for my example, an electromotor is driving the pulley, so where the shaft is connected to the pulley, it's a fixed connections

    to in this case there's just the shaft reactions that is opposed to the the green force vector,
    but for deflection calulation/shaft diameter, I must use the green force vector...

    I can't see another explenation.

    thank you
     
  7. Aug 28, 2011 #6

    vrc

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    someone who wants to confirm my ideas....

    thank you
     
  8. Aug 29, 2011 #7
    The force ON the shaft is acting in the green-line direction. The force that the shaft is applying is in the opposite direction. To calculate the deflection of the shaft, yes, use the line you drew. The book likely put it in the opposite direction to maintain force-balance convention.

    Though, I don't see how a belt can have two different tensions on it if it is a single belt...
     
  9. Aug 29, 2011 #8

    vrc

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    I assume also that the book might mean that the oppsite force is the one give the belt e preload to there is enough friction with the pulley. Otherwise it wouldn't be able to transmit power.

    In rotation mode there's a different tension in de belt, that''s rather logic, how would the power be transmitted otherwise...

    thank you

    grtz
     
  10. Aug 29, 2011 #9
    I doubt that. If there is tension in the belt, there will be power transmission. If it was preload, they would tell you. It doesn't matter though. They tell you the tensions in the belt, so that is all you need.
     
  11. Aug 30, 2011 #10

    vrc

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    difference in tension I mean: different forces in the belt,
    if those forces would be equal, why would there be motion...
    I have learned and understood that there only wille be motion if there's a difference in forces

    grtz
     
  12. Aug 30, 2011 #11
    Yea. That was a misunderstanding on my part. Its the initial acceleration that you are studying. And the resulting deflection.
     
  13. Aug 30, 2011 #12

    vrc

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    what about initial acceleration ?
    if the shaft is rotating at constant RPM, there also will be difference in belt forces...
    If that isn't true I have a major problem in understanding physics !

    grtz
     
  14. Aug 30, 2011 #13
    If you are talking real world, yea, there will be because nothing is completely inelastic and nothing is frictionless.

    Newton's laws: F=m*a Force is a direct correlate to acceleration. Zero acceleration means zero net force. If you have a force difference, you have an acceleration. Constant RPM means constant velocity, which means zero net force. I.e. The pulley should be pulling the feed side with the same force that it is pushing the other side.

    Someone chime in if I am wrong, but I don't believe I am.
     
  15. Aug 31, 2011 #14

    vrc

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    conclusion: constant RPM doens't excist and therefore will be a difference in belt force ?

    grtz
     
  16. Aug 31, 2011 #15
  17. Sep 7, 2011 #16

    vrc

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    what do you mean with 'hold' ?

    grtz
     
  18. Sep 7, 2011 #17

    Bandit127

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    VTC - sorry, but please could you re-size your picture to 800 wide or smaller. It has forced the text on all the entries in the thread to the same width and it's forcing scrolling to read the replies.

    At least on my Firefox and XP.

    Thanks.
     
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