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Silly Question on Binding for Piston

  1. Sep 1, 2015 #1
    Hi,

    Another fellow engineer told me that for a dual connected piston guided inside a tube (as shown in the diagram below), the distance D1 has to be chosen correctly in order to avoid binding (I'm guessing from friction only) since there may be a side force present. This is ignoring gravity for now. In the example discussed there is a force F applied with some side force component Fn. I couldn't make sense as to why D1 is relevant. Can anyone explain it easily?

    The only thing that comes to mind is that I see a normal force at point A and point B as a result of Fn. I can see how maybe there is a moment created at point B by the normal force at point A times the distance D1 and then a moment at point B from the force F times D2/2. Would the balance of those moments result in the value for D1?

    b1.jpg
     
  2. jcsd
  3. Sep 1, 2015 #2

    Bystander

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    What happens to the moments you've described as D1 goes to zero?
     
  4. Sep 1, 2015 #3
    The moments at point B would be

    F times D2/2

    minus

    Fn/2 times D1

    As D1 goes to zero the moment at point B would be F times D2/2. I'm not sure how this relates to binding unless a positive moment is contributing to a force normal to the tube and friction comes into play. Is there another correct explanation?
     
  5. Sep 2, 2015 #4

    Nidum

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    As stated the question is not very meaningful . Piston stability depends on many different things .
     
  6. Sep 2, 2015 #5
    Thank you, I thought the same and I know there are other factors to consider for piston stability.

    However I was asked what distance D1 would eliminate binding ignoring all other effects. Is this a matter of solving for the moments about point B and making sure they sum to zero?
     
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