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Forces due to atmospheric pressure won't cancel in an open tank

  1. Apr 26, 2017 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand why books always point as a fact that hydrostatic force on the bottom of a open liquid filled tank doesn't depend on the force due atmospheric pressure because they these forces cancels each other.

    2. Relevant equations

    P=[P][/o]+ρgh

    F=P*A

    Third/Second's Newton Law.

    3. The attempt at a solution

    Please indicate in which step i went wrong instead just pointing the same argument that books provides without formulas/eqs. Like, "Your application of the Newton's second law is wrong, or your substitution from eq. (1) into (2) is wrong". Thanks very much in advance.

    http://docdro.id/SYbWm95
     
  2. jcsd
  3. Apr 26, 2017 #2

    kuruman

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    You don't show any steps, just two equations both of which are correct and the names of two Laws. Steps would be what you make of all these. So I will present you with a picture and ask you a question.

    Suppose you are at sea level and put a glass (tank) filled with water on a scale and it reads 10 N. Then you climb to the top of a 2 km high mountain where the atmospheric pressure is lower. What do you think the scale will read? Neglect the buoyant force of the air.
     
  4. Apr 26, 2017 #3
    I don't see where your analysis in conflict with the statement. The net force on the bottom of the tank from the liquid pressure above and the air pressure below is just equal to the weight of the fluid in the tank: $$F_{liq}-F(atm\ below)=mg$$
     
  5. Apr 26, 2017 #4
    Sorry maybe i have should pointing from the beginning my attempt to solve the problem was in the pdf file, i repost the link http://docdro.id/SYbWm95
     
  6. Apr 26, 2017 #5
    I must be doing something wrong because books point the F_liq=MassLiquid*g. See by example "Serway - Physics for Scientists, 7th Ed, page 409, exercise 11."

    "11. A swimming pool has dimensions 30.0 m 10.0 m and a
    flat bottom. When the pool is filled to a depth of 2.00 m
    with fresh water, what is the force caused by the water on
    the bottom? On each end? On each side?"

    The answer on the bottom is F_liq=M*g=ρVg=5.88x10^6 N ≠ My result = mg+F_atm
     
  7. Apr 26, 2017 #6
    Well, your answer is more correct than the book.
     
  8. Apr 26, 2017 #7
    I got that feeling too, i have been checking this result for weeks, with few sleep and cant find any mistake in my reasoning. Can anyone else confirm if this result is correct/wrong please?
     
  9. Apr 26, 2017 #8
    You don't need anyone else to confirm this. I have lots of experience in fluid mechanics. This same kind of deal comes into play when doing macroscopic momentum balances on fluid flow in ducts of varying cross section, in determining the additional force exerted on the duct wall when the fluid is flowing. See this thread: https://www.physicsforums.com/threads/nozzle-conservation-of-momentum.901349/
     
    Last edited: Apr 26, 2017
  10. Apr 26, 2017 #9

    kuruman

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    I think that Serway is correct if by "hydrostatic force" is meant force exerted by a fluid other than air. In other words the hydrostatic force is "gauge pressure times area". Take an empty container standing on a table. Pump some air in it to pressurize the container to 2p0. What is the total force at the bottom of the container? Answer: 2p0A (neglecting the weight of the air). Is that the hydrostatic force? No, because air does not count as something that exerts a hydrostatic force. Note that 2p0A is the force due to gauge pressure; we did not subtract p0A which is the force exerted on the outside of the bottom by atmospheric pressure.

    Now put fluid of density ρ to height h in the container and vent some of the air so that the pressure above the surface is maintained at 2p0. What is the total force now? Answer: 2p0 + ρghA. OK, but what is the hydrostatic force on the bottom? Answer: ρghA, the increase to the existing force due to the addition of the fluid. That difference is, quite simply, the weight of the fluid.
     
  11. Apr 26, 2017 #10
    I think the OP understands all this.
     
  12. Apr 26, 2017 #11
    Thanks to both for your response, sorry for late reply, there was a blackout earlier. Ill try to read carefully both answers to see if i can clarify my doubt.
     
  13. Apr 27, 2017 #12

    haruspex

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    The trouble is that "caused by" is ambiguous. It could mean the direct force the water applies to the surface, as you have interpreted it, but it could equally be interpreted as meaning the difference made by the presence of the water. Indeed, the second seems more reasonable to me.
     
  14. Apr 27, 2017 #13
    Yes. One has to be very careful to understand specifically what the literature author intends to be saying in these contexts.

    I would add that I prefer the way Miguel analyzes this because there is less chance for error in solving actual problems. Of course, once he gets comfortable with it, he can switch to working in terms of gauge pressures and forces (for incompressible fluids).
     
  15. Apr 27, 2017 #14
    I agree with you, im tired to check over and over my result, it doesnt seems to be nothing wrong with it, nothing wrong with math or reasoning, its a problem about definition. How do books define "Force exerted by a liquid in open liquid filled tank" is the question to answer. If we define the force caused by liquid as direct force that liquid exerts on the bottom surface, then


    [F][/liq]≡[F][/DirectOnSurflByLiq]=mg+[F][/atm]

    and if we define the force caused by liquid as the weight of liquid, then

    [F][/liq]≡[F][/DirectOnSurflByLiq]-[F][/atm]=mg


    I wish authors in books define concepts in a very clearly way, there should not be space for ambiguity in science. If anyone has anything else to add to this post, i would like to conclude it.

    Thank you very much to Chestermiller, kuruman and haruspex for your time and your sharp comments. I appreciate it.
     
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