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Why is air pressure force not considered while weighing?

  1. Jun 17, 2015 #1
    we know that air pressure on our earth is 1 atm.
    Also 1 atm = 10^5 P
    Also we know Pressure(P) equation = P = F/A So, F = P*A
    So if small area(A) in which we are standing is also taken then pressure force is that area(A) times 10^5(Atmospheric pressure in Pascal) i.e 10^5*A
    So why don't we subtract this force from our weight(W) as W - F
    Why do we only consider gravity?
    Why we don't consider air pressure force?

    if I have any misconceptions then tell me..........
    Thank You
  2. jcsd
  3. Jun 17, 2015 #2
    Air pressure isn't solely pushing down on the human body. Rather, it's pushing all around us, (almost, I believe) balancing out, and so there's no increase in weight from it.
  4. Jun 17, 2015 #3
    What would the scale read if you were not standing on it?
  5. Jun 17, 2015 #4


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    That only applies to flat surface elements. Otherwise force is the surface integral of pressure * surface normal, which for will yield a small buoyant force on the human body.
  6. Jun 17, 2015 #5
    As others have pointed out, it's not just the top pushing down on you, but also the bottom pushing you up, and in fact with more force from below than from above (that's how things swim).
    If you view a human as a cylinder (spherical cow in a vacuum, lol) of 2m height and say 1m^2 surface area at the ends, you get:

    Force on top: 1m^2 * 100kPa = 100kN

    2 meters lower the pressure has increased by 24 Pascal (1.2kPa per 100m per Wikipedia)
    Force at the bottom: 1m^2 * 100.024kPa = 100.024kN

    So, the net force (bottom minus top) is 24N. That's worth a mass of 3kg or so. But because we're not cylinders, it's actually significantly less than that.

    It's interesting to note though that, if you prevent air from pushing from the bottom (e.g. by having an airtight seal at the bottom), then you actually notice the full weight of the air column when trying to lift it. The moment you break the air seal (thus allowing the air to push from the bottom again) it becomes easy to lift it.
    Last edited: Jun 17, 2015
  7. Jun 17, 2015 #6


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    The pressure of air is related to its density: maybe, you should take into account the corrections of bouyancy

    Let's take an example: if you weigh a 1kg of Helium (contained in a ballon) in the surface, will the result be the same if you repeat the measure if the pressure of air if different?

    As it's been said, it depends on what you are weighing :)

  8. Jun 17, 2015 #7


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    Staff: Mentor

    Consider a 1 meter cube, sitting on and vacuum-sealed to a scale.

    What is the difference between the air pressure force the scale feels with or without the object sitting on it?
  9. Jun 17, 2015 #8
    The density of water is 1000 kg/m3 while that of air at the surface of the earth is about 1.3 kg/m3. So the buoyant force amounts to only about 0.13% of the body weight, and thus this "red herring" can safely be neglected. Now, with this realization, what is the answer to Russ' question?

  10. Jun 18, 2015 #9
    The problem of the difference between mass and weight due to buoyancy is encountered with vertical seismometers. The weight is less than the mass by the volume x density of air that is displaced. Unless the vertical seismometer is sealed in an air tight container it will bob up and down with the weather.
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