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Homework Help: Where does pressure act upon

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    P=hρg is the pressure difference but where does this pressure act upon? In a standard barometer there is a vacuum on top so 0 Pa, and at the bottom the normal force acts on the glass base. So hρg=Pweight. But theoretically, where would the hρg calculated be acted upon? Or will that pressure just be exerted at all directions?

    If so in this example http://postimage.org/image/53k5i2cqz/full/ [Broken] will the hρg be exerted onto the air as well?

    2. Relevant equations


    3. The attempt at a solution
    I think so because under forces, we learned that by newton's third law besides the normal force balancing out weight, there is also a reaction force acting on the other body that applied the normal force. So now in terms of pressure in that scenario the liquid would exert the pressure on the gas below.

    However, the main part that confuses me is that the normal force balancing the weight will be countered by weight itself. So there is no net force acting on the liquid. So why should there be a pressure being exerted on it?

    Thanks for the help!
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 8, 2012 #2


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    Yes pressure acts in all directions in a fluid. Lets suppose you can make a tiny spherical sensor that can measure the force acting at say 1000 points all over its surface.

    If the sensor was large you might measure a pressure difference between the top and bottom, but the pressure on each side would be the same besause they are at the same depth in the fluid.

    As the diameter of this sensor is reduced towards zero would find that the pressure on all the points approached the same value.

    If the sensor is really small then you would find that the pressure just either side of the liquid air surface was the same. It doesn't suddenly change polarity from positive to negative as you go through the surface.

    Aside: I'm ignoring any effect due to surface tension
    Last edited: Dec 8, 2012
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