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Homework Help: Water pressure question

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Not exactly sure how to post this but i will give it my best.

    You have a tube and for simplicity sake lets say it is 1 ft by 1ft square. The tube is 42 inches tall. This tube is full of water and has a water tight piston at the top that is 1 ft by 1 ft. At the bottom is a 1 inch hole coming out of the side that is connected to a tube that takes a right angle back up to the top (42inches). Ok on top of the piston is 100 pds that presses the piston down thus forcing water up the 1 inch gap at the bottom of the tube.

    2. Relevant equations

    The question i am asked is how far down the piston goes before equilibrium is reached and how much water comes out the 1 inch tube that leads back up.

    3. The attempt at a solution

    I have been sitting here thinking the obvious that all the water in the 12 inch tube evacuates up the 1 inch tube and the piston comes to rest at the bottom. But i know this cant be that easy. I thank anyones responce. :)
  2. jcsd
  3. Sep 1, 2009 #2


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    Staff: Mentor

    I think you're close, but you do need to do at least one calculation to be sure.

    With no weight on the piston, the water will establish equilibrium with the top surface of the water in the external tube at the same height as the top surface of the water in the big column, right? Then what force does it take to raise the external water above the surface of the internal water? A force equal to the weight of how much water?

    So you need to calculate something about the weight of water somewhere in this setup, to see if the 100 pounds of force is enough to evacuate the inner column...
  4. Sep 1, 2009 #3
    Thx for the responce, i suppose you mean how much force is needed to push water up the external tube? Doesnt psi play a part in this discussion? I figured the push on the piston would be 100pds/144 sq inches = .69 psi plus the the psi for the column of water still in the tube which would diminish as the piston falls??? But the psi in the external tube would remain the same. I thought this would be a no brainer but now im more confused than ever lol. :) pls comment.
  5. Sep 1, 2009 #4
    How much is the weight of one cubic inch of water in pounds?

    How many times does that number go into 100/144?
  6. Sep 1, 2009 #5
    ok thx for the response, the column of water would wiegh 1.26 pds and that number divided by 100/144 equals .547 but im still having trouble finding what that really means. could u elaborate?

    edit: no wait .03 is the weight of one cubic inch water and it goes into .69(100/144) 23 times but does that mean the piston falls to the bottom?
  7. Sep 2, 2009 #6
    We differ on some of the details. I calculated the weight of a cubic inch of water from knowing that a cubic cm of water weighs 1 gram and there are 2.54^3 = 16.387 cc in a cubic inch.

    16.387 g = 16.387/28.35 = 0.578 oz = 0.0361 lbs.

    42 inches of water weighs ? This is the psi it creates.

    How much psi does the 100 lbs create? 100/144

    How many inches of water is that equivalent to?
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