# Normal Force and Atmospheric Pressure

## Main Question or Discussion Point

1. Suppose a cube of mass m is lying at rest on a horizontal surface. The area of one face of the cube is A. For the equlibrium of cube no net force acts on it. If we consider the vertical direction then the contact force N should be equal to the force due to gravity plus the force due to atmospheric pressure acting on the top face. There is no air between bottom face and the horizontal surface so atmospheric pressure is not acting on the bottom face. The atmospheric pressure is substantial too, so we can't ignore it. Still we say that N = mg instead of N = mg + PA, where P is the atmospheric pressure. Why ?

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If there were actually no air between the bottom of the object and the surface, then the air pressure would be significant. In fact, that is how suction cups work (there isn't "no air" under a suction cup, but the pressure is lowered when you pull upwards on the suction cup and increase the volume of the region between the cup and the table).

This should convince you that most real objects do not seal tightly enough against the table to exclude air.

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CWatters
Homework Helper
Gold Member
Perhaps worth a look at "Wringability" of gauge blocks. (edit...but it's not just air pressure that keeps them together).

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1 person
D H
Staff Emeritus
If we consider the vertical direction then the contact force N should be equal to the force due to gravity plus the force due to atmospheric pressure acting on the top face. There is no air between bottom face and the horizontal surface so atmospheric pressure is not acting on the bottom face. The atmospheric pressure is substantial too, so we can't ignore it. Still we say that N = mg instead of N = mg + PA, where P is the atmospheric pressure. Why ?
You have the sign wrong. Atmospheric pressure buoys objects upwards rather than pushing them down. There is air underneath your block unless you take extreme measures to seal the bottom of your cube.

So why don't we use this buoyancy-corrected weight? Simple: It's a tiny effect. You ignored a couple of other tiny effects when you said "N=mg". Gravitation acceleration is not constant everywhere. You should be using N=mg(x), where g(x) is the gravitational acceleration at the location of interest. Even that is not correct because the Earth is rotating at one revolution per sidereal day. The net force on your cube has to be just the amount needed to make the cube rotate at this rate at about the Earth's rotation axis.

1 person
You have the sign wrong. Atmospheric pressure buoys objects upwards rather than pushing them down. There is air underneath your block unless you take extreme measures to seal the bottom of your cube.

So why don't we use this buoyancy-corrected weight? Simple: It's a tiny effect. You ignored a couple of other tiny effects when you said "N=mg". Gravitation acceleration is not constant everywhere. You should be using N=mg(x), where g(x) is the gravitational acceleration at the location of interest. Even that is not correct because the Earth is rotating at one revolution per sidereal day. The net force on your cube has to be just the amount needed to make the cube rotate at this rate at about the Earth's rotation axis.
What I wasn't able to convince myself was that there was air between the block and the surface, which tantamounts to air only pushing it down. Of course the g we take is the effective g taking into consideration of earth's rotational effects as well as variation with height/depths etc.

256bits
Gold Member
Perhaps worth a look at "Wringability" of gauge blocks. (edit...but it's not just air pressure that keeps them together).
I looked up wringability just to see what they say.
and I got more sites on Writing Ability than wringability.
I encountered the blocks only once many eons ago.