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Vessel of water under free fall

  1. Mar 21, 2015 #1
    When a Vessel of water is under free fall and there is a hole in the side of the vessel, why does no water come out of the vessel? (its given in my textbook that pressure exerted on sides of vessel is zero. But what does the horizontal force have to do with the vertically acting force due to gravity?)
     
  2. jcsd
  3. Mar 21, 2015 #2

    DaveC426913

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    Water has weight. It pushes down on the water below it, creating horizontal pressure. Under gravitational influence, the water seeks the lowest point of potential energy. It can do this by pouring out the hole.
     
  4. Mar 21, 2015 #3
    The text book says water does not flow out.
     
  5. Mar 21, 2015 #4

    A.T.

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    Why should it? The vessel falls with the same acceleration as the water, so it's not in its way.
     
  6. Mar 21, 2015 #5
    Capture.PNG
    (its in an Indian text book, so there are grammatical errors)
     
  7. Mar 21, 2015 #6

    DaveC426913

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    I was addressing the default state: when the vessel is sitting on a surface.

    You asked: "what does the horizontal force have to do with the vertically acting force due to gravity?"
    I provided the answer. Under gravity, the water tries to squeeze horizontally.

    So, with that part answered, does it now make more sense why the water would not flow out once gravity is neutralized?
     
  8. Mar 21, 2015 #7
    Understood.
     
  9. Mar 21, 2015 #8

    A.T.

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    There is no pressure due to the weight of the water, because the weight is not supported by the vessel (the vessel falls itself). There could still be pressure if the vessel is pressurized.
     
  10. Mar 21, 2015 #9

    DaveC426913

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    This does not answer A.T.'s question.

    Why should the water come out of the hole while the vessel AND water are freely falling?
     
  11. Mar 21, 2015 #10
    It doesn't right?
     
  12. Mar 21, 2015 #11

    DaveC426913

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    Well, you know that, having been told by the textbook. But do you now understand why?
     
  13. Mar 21, 2015 #12
    Under free fall, if you consider an arbitrary cylinder of length y like this:
    Capture.PNG
    From newtons laws, ##\Delta P.\Delta S+W=mg##
    Then ##(P_1-P_2)\Delta S+y\Delta S\sigma g=y\Delta S\sigma g##, sigma is the density of liquid. The force acting are due to weight and due to pressure difference.
    So, ##P_1-P_2=0##
    hence, ##P_1=P_2##
    So there is no pressure difference between 2 levels of the liquid.
     
  14. Mar 21, 2015 #13
    I don't know if what i did is correct, but it does make sense and i have used nothing other than newton's laws. Please verify this.
     
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