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Enclosed liquid

  1. Mar 2, 2012 #1
    in a container there are some pistons of different areas. If we apply a force at one of the pistons and move it a distance. How can we determine the distance moved by the other pistons?
     
  2. jcsd
  3. Mar 2, 2012 #2

    DaveC426913

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    You must show your attempts at an answer before we can help you.
     
  4. Mar 2, 2012 #3
    what if a child asks this question? i thought this rule was for homework section.
     
  5. Mar 2, 2012 #4

    DaveC426913

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    Yes, this belongs in the homework section. Show your attempt.
     
  6. Mar 2, 2012 #5
    it does not depend on the area. Every piston will be moved by an equal distance. Thats my attempt.
     
  7. Mar 2, 2012 #6

    boneh3ad

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    It is just conservation of mass. For water, this equates to conservation of volume. If you know how far down one piston moved, you can use that displaced volume to determine how high the rest moved.
     
  8. Mar 2, 2012 #7
    boneh3ad
    everybody knows volume is conserved. so u mean that increase in volume of every piston is same? in which case, the distance can be determined.
     
  9. Mar 2, 2012 #8

    boneh3ad

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    No, the pistons will rise the same and the total volume "added" by the rising pistons will equal the volume "lost" by the falling piston(s).
     
  10. Mar 3, 2012 #9
    ok. what if the sum of the areas of the other pistons is less than or greater than the area of that piston times the no. of the other pistons?
     
  11. Mar 3, 2012 #10
    somebody comment on this problem.
     
  12. Mar 3, 2012 #11
    if the other pistons move equal distances. Then the sum of the other areas must equal the no. of the other pistons multiplied by the gien piston.
     
  13. Mar 3, 2012 #12
  14. Mar 3, 2012 #13

    DaveC426913

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    Presumably, only one target piston is allowed to move freely at a time, otherwise, I'm not sure how you could calculate several pistons' movement simultaneously.

    This is identical to filling cylindrical glasses of water.

    If you start with a glass that is 10cm in diameter, and it is filled with 1L of water, the height of the water will reach X.

    If you pour that entire 1L of water (i.e. volume does not change) into another glass of 5cm diameter, what level will the water reach and how do you calculate that?
     
  15. Mar 3, 2012 #14
    well imagine a enclosed sphere under no action of gravity. I am guessing there must be some regularity in the movement of the pistons if we push in one of the pistons.
     
  16. Mar 3, 2012 #15

    DaveC426913

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    I'm not sure how that helps or hinders the problem at-hand. It just seems to kind of restate it.

    Did you read my previous post? Do you understand the correlation between the areas and the volume displacements of the pistons?
     
  17. Mar 3, 2012 #16
    ok so volume is conserved.
    we get, for directed distance
    $A_1$ $d_1$ + $A_2$ $d_2$ + ............... = 0
    And also the sum of the work done is also 0.
    Then?
     
  18. Mar 3, 2012 #17

    DaveC426913

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    Perhaps you need to rethink what you want from us. If this is a question from a book, perhaps you should write out the entire question.

    P1Area x P1Displacement = P2Area x P2Displacement

    What more do you want?
     
  19. Mar 4, 2012 #18
    forget it
     
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