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Drinking straw question

  1. Jul 17, 2006 #1
    Hi, firstly allow me to say that this is not a homework question but the question fits mostly in the introductory physics section. I’m sorry if this is truly the wrong section for my question. Please direct me to right section and I will delete this thread from this section immediately. The question is just something I’ve been pondering and I would like for someone to help me figure out where I’m going wrong.

    It’s a fluid static’s question I guess. I’ve been thinking about the old dipping a straw into a liquid and then holding your finger over the top, and then you extract the straw and the fluid remains inside. Of course the pressure at the top is lesser than the pressure at the bottom where the pressure at the bottom is atmospheric pressure. Using some equations I get:
    Where Pb = Pressure at the bottom,
    Pt = Pressure at the top,
    Po= atmospheric pressure,
    D = density of the fluid,
    g = gravitational acceleration,
    h = length of the colunm of fluid in the straw

    Pb = Pt + Dgh
    Pb =Po
    Po= Pt +Dgh
    Pt = Po - Dgh
    The pressure at the top is lesser than the pressure at the bottom by Dgh, which makes sense.
    But this is where I become confused and the math doesn’t add up.
    First I’m thinking that since the water is at rest that means it is at equilibrium so the net force is zero, gravitational force = the buoyant force (I assume the force acting upward is the buoyant force)
    Where A= cross sectional area of the straw (equal everywhere; meaning the straw is just a straight tube)
    Ft =force acting downwards on the top of the straw
    Fb = force acting upwards on the bottom of the straw
    Fg = gravitational force
    Fbo = buoyant force
    Ft = Fg
    Fb = Fbo
    Since they are in equilibrium:
    Fbo = Fg
    So Pb = Fbo/A
    And Pt = Fg/A
    Since Fg = Fbo
    Then this must mean Pb = Pt
    Which is incorrect!
    To put it in plain words, the pressures differ by Dgh and since the areas are the same that means that the forces must differ (where Fb – Ft = ADgh = mg where m is the mass of the fluid causing a downward acceleration) but how can they differ if the net force = 0 which means the forces are equal. Can someone please help me out with this?
  2. jcsd
  3. Jul 18, 2006 #2


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    at this stage an error kreeps in ...

    I would prefer if you would rather call the "bouyant force" the upwards force say Fu, since bouyancy has another meaning in physics. This force is caused by the atmospheric pressure on the bottom of the water column in the straw. It can be found by multiplying your eqaution

    Pt = Po - Dgh

    with A giving

    APo = APt + ADgh - which can be written as APo = APt + mg via the density giving

    Fu = APt + mg or Fu = Ft + mg

    the first term gives Fu (or your Fbo). So you have missed a term in your reasoning - the force due to the air pressure on top of the liquid in the straw (I chose Ft for the force caused by this air pressure).
    Last edited: Jul 18, 2006
  4. Jul 18, 2006 #3

    Thanks for responding to my thread. I see where I went wrong! This says there are three forces not two as I first believed, those being the forces brought on by the pressure at the bottom and the pressure at the top and the one that I can't believe I missed (although I stated it) the force caused by the mass of the water! ha ha :rofl: so these three forces are in equilibrium, from your equation:
    Fu = Ft + mg
    Fu - Ft - mg =0
    perfect sense. Thanks.
  5. Jul 18, 2006 #4


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    It's a pleasure.
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