Why is air pressure force not considered while weighing?

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Discussion Overview

The discussion revolves around the question of why air pressure force is not considered when weighing objects, particularly in the context of how atmospheric pressure interacts with weight and buoyancy. Participants explore the implications of air pressure on scales and the forces acting on objects in a fluid medium, including theoretical and practical considerations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that air pressure acts equally from all directions, suggesting that it does not contribute to an increase in weight when standing on a scale.
  • One participant emphasizes that the pressure force on a body is not just from above but also from below, leading to a net buoyant force that could affect weight measurements.
  • Another participant discusses the concept of buoyancy, noting that the buoyant force from air is minimal compared to the weight of a human body, thus can often be neglected.
  • A participant raises a hypothetical scenario involving a sealed object in a vacuum, questioning how the absence of air pressure would affect weight readings.
  • One participant mentions the relationship between air density and buoyancy, suggesting that variations in air pressure could influence weight measurements under certain conditions.
  • Another participant provides a mathematical example involving the pressure difference between the top and bottom of a human body, calculating a small net force due to buoyancy.
  • Concerns are raised about how the weight of objects may vary based on the surrounding air pressure, particularly in specific contexts like weighing helium balloons.

Areas of Agreement / Disagreement

Participants express differing views on the significance of air pressure in weight measurements, with some arguing it can be neglected while others highlight its potential effects. The discussion remains unresolved regarding the extent to which air pressure should be considered in weighing scenarios.

Contextual Notes

Participants note that the buoyant force due to air pressure is significantly less than that of water, suggesting that its impact on weight may be minimal but not negligible in all contexts. The discussion also touches on the complexities of measuring weight in varying atmospheric conditions.

tanmay
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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
 
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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.
 
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tanmay said:
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
What would the scale read if you were not standing on it?
 
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tanmay said:
force is that area(A) times 10^5(Atmospheric pressure in Pascal)
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.
 
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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.
 
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Hello

The pressure of air is related to its density: maybe, you should take into account the corrections of bouyancy
http://www.nist.gov/calibrations/upload/ac53-6.pdf

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 :)

Greetings!
 
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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?
 
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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?

Chet
 
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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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