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Hydraulics equilibrium and the straw in my drink

  1. Apr 21, 2007 #1
    Sorry for the noobie question, but I've got myself in a mental bind trying to figure out static hydraulics and why things happen the way they do.

    Could somebody give me an equation that shows me why the fluid level on the outside of the straw in my drink is equal to the fluid level on the inside of my drink.

    For ease of math and my understanding, let's assume:
    1. The air pressure is 16 psi
    2. The area of water exposed to air on the inside of the straw is 1 sq in (it's a big straw :rolleyes: )
    3. The area of water exposed to air on the outside of the straw is 10 sq in (it's a huge glass :rolleyes: )

    For some strange reason, I keep thinking that with 160 total pounds pushing on the water in the glass, any straw would produce a nice fountain affect at rest. How do I mathmatically show that the backpressure in the straw is 16 psi and is in equilibrium with the pressure of the water outside the straw?

    Thanks in advance.
     
  2. jcsd
  3. Apr 21, 2007 #2

    marcusl

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    Science Advisor
    Gold Member

    Pressure is defined as force per unit area, so the relative areas of straw and glass are irrelevant. The air pressures on the surface of the water in the glass and in the straw are equal, therefore the water columns are in equilibrium.
     
  4. Apr 21, 2007 #3
    Marcusl,

    Thank you for the reply, but can you prove this to be true given equations used to determine force in static hydraulics?
     
  5. Apr 22, 2007 #4

    russ_watters

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    Staff: Mentor

    You already did all the math in your first post! The equation you used is what marcus described in words: p=f/a (or f=p*a).

    The problem you are having comes from the conceptual misunderstanding in your final paragraph. The first sentence talks about fforce and the second sentence talks about pressure. Pressure and force are not the same thing. It is pressure, not force, that moves a fluid. So equal pressure means no motion.

    Think about it this way: the pressure at the surface of the water (with no straw) is 14.7 psi (the actual air pressure) and the total force is 147 lb. That means that if you cut up the surface into 1 cu in squares, the pressure is still 14.7 psi and the force on each square is 14.7 lb. This you know because you used it in your calculation. So what changes when you put a straw into the water? Nothing.
     
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