# Why don't we take air pressure into account in calculations?

• Dishsoap
In summary, when calculating normal forces between two objects, we disregard the weight of the air on top of the block.
Dishsoap
I once was explained the answer to this question - something about both the air and the object being incompressible. However, for the life of me I cannot seem to reproduce the answer of why we ignore air pressure when doing basic calculations.

If it matters, the particular system I'm interested in is that of a submerged particle in a fluid. Does the frictional coefficient of the particle to the surface depend on the amount of water above it? My intuition tells me two things:

a) that a particle that is even slightly more positively buoyant (with respect to the fluid) will float to the top regardless of how much fluid is on top of it, and
b) the frictional coefficient is not dependent on the amount of fluid above it (pushing a penny at the bottom of a pool and at the bottom of a beaker requires an identical amount of work).

PeroK and FactChecker
I don't understand your question. A penny at the bottom of the pool is not a penny at the surface. You are using "water on top of it" and "at the surface" in the same sentence.

Are you thinking of something like a floating log? A very light log will have less wetted surface than a log just slightly less dense than water.

Dishsoap said:
Summary: When calculating normal forces of, e.g., a block sliding down an incline (with friction), we disregard the weight of the air on top of the block. Why?
Because air pressure acts on all sides of the block.
a) that a particle that is even slightly more positively buoyant (with respect to the fluid) will float to the top regardless of how much fluid is on top of it,
That doesn't appear to be the same issue. But yes, an object with a positive net force in a fluid (buoyancy minus weight) will float.
b) the frictional coefficient is not dependent on the amount of fluid above it (pushing a penny at the bottom of a pool and at the bottom of a beaker requires an identical amount of work).
Friction between what and what?

For a block sitting on a table or incline, it is true that there is air pressure pushing down on the top of the block toward the contact surface. But there is also air in the interstices of the interface region between the block and the table pushing back in the opposite direction. The table and the block are not perfectly flat. They only make actual contact at the tips of the surface asperities. So the air in the interstices balances the pressure on the top of the block, and the actual contact force between the block and the table is just the weight of the block. This is the same force as if the system of block and table were sitting in vacuum.

FactChecker
...we disregard the weight of the air on top of the block...

You regard the weight of the air on top of the block - for something like a suction cup.

russ_watters and FactChecker
anorlunda said:
I don't understand your question. A penny at the bottom of the pool is not a penny at the surface. You are using "water on top of it" and "at the surface" in the same sentence.

Are you thinking of something like a floating log? A very light log will have less wetted surface than a log just slightly less dense than water.

To be clear, I'm not talking about the air-water interface, I'm talking about the solid surface on which the object sits (the bottom of the pool, in this case).

russ_watters said:
Friction between what and what?

In the thought experiment, the friction between the "block" and whatever it's moving on. In my experiment, I have an SiO2 device submerged in water on top of an SiO2 substrate. I'm curious about the friction between the device and the substrate.

Chestermiller said:
For a block sitting on a table or incline, it is true that there is air pressure pushing down on the top of the block toward the contact surface. But there is also air in the interstices of the interface region between the block and the table pushing back in the opposite direction. The table and the block are not perfectly flat. They only make actual contact at the tips of the surface asperities. So the air in the interstices balances the pressure on the top of the block, and the actual contact force between the block and the table is just the weight of the block. This is the same force as if the system of block and table were sitting in vacuum.

If our block is atomically flat, does this change things?

Dishsoap said:
If our block is atomically flat, does this change things?
If you can expel that air, or in your case the water, from the interface between the block and the surface it is resting on, say by using a suction cup or perfectly flat surfaces, that does change things.

Chestermiller said:
If you can expel that air, or in your case the water, from the interface between the block and the surface it is resting on, say by using a suction cup or perfectly flat surfaces, that does change things.
In this case, is the normal force (or frictional force) dependent on the amount of water in the dish?

Dishsoap said:
In this case, is the normal force (or frictional force) dependent on the amount of water in the dish?
No. Similarly, divers standing on the bottom of a lake do not feel a "downward push" of the water above them. This is the case whether the lake is 30 feet deep or 100 feet deep.

What they feel is a uniform pressure from all sides.

Dishsoap said:
In this case, is the normal force (or frictional force) dependent on the amount of water in the dish?
If you expel the air from under the block then, as in the case of a suction cup, the normal force and frictional force depends on the pressure at depth.

Dishsoap
Chestermiller said:
If you expel the air water from under the block...

Except for the suction cup effect described above, in general the effect of the atmosphere, by Archimedes' principle, is to add a slight amount of buoyancy to the object in these problems, like a block on an incline. In general, the buoyancy can usually be neglected, but it is this buoyancy that makes a helium balloon or hot air balloon rise in the air. ## \\## Edit: And to see some calculations where this buoyancy is taken into account, see https://www.physicsforums.com/threads/lifting-a-man-with-100m-3-of-helium.943829/#post-5972396 and the "links" in post 3 of this "link".

Last edited:
Suppose there is a complete lack of water/air between the object and the ground and none can enter. Then there would be no additional force from above, but there would be a lack of force from below and the net force would increase downward.

DaveC426913 said:
As ChesterMiller points out:

If you did this same experiment in a vacuum jar, the block would weigh exactly the same.(Would you expect otherwise?)

The buoyant force due to air on a block is actually pretty easy to measure. It does make a small difference.

JT Smith said:
The buoyant force due to air on a block is actually pretty easy to measure. It does make a small difference.
Yeah. I rethought that.
Deleted as apocryphal.

## 1. Why is air pressure not typically considered in calculations?

The primary reason air pressure is not typically considered in calculations is because it is constantly changing. Air pressure is affected by factors such as elevation, temperature, and weather, making it difficult to accurately measure and account for in calculations. Additionally, for most everyday calculations, the effects of air pressure are negligible and can be safely ignored.

## 2. How does air pressure affect calculations?

Air pressure affects calculations by exerting a force on objects and substances. This force, known as atmospheric pressure, can impact the behavior of gases, liquids, and solids. However, for most calculations, the effects of air pressure are minimal and can be ignored without significantly impacting the accuracy of the results.

## 3. Are there any situations where air pressure should be taken into account in calculations?

Yes, there are certain situations where air pressure should be considered in calculations. For example, in aviation and meteorology, air pressure is a crucial factor in determining weather patterns and flight paths. In these fields, precise measurements of air pressure are necessary for accurate calculations and predictions.

## 4. How is air pressure measured in scientific calculations?

Air pressure is typically measured using a device called a barometer. A barometer works by measuring the weight of the atmosphere above it, which is then converted into a numerical value, usually in units of pressure such as pounds per square inch (psi) or pascals (Pa). Other devices, such as aneroid barometers and electronic pressure sensors, can also be used to measure air pressure.

## 5. Can air pressure be manipulated in calculations?

In most cases, air pressure cannot be manipulated in calculations. As mentioned before, air pressure is constantly changing and can be affected by a variety of factors. However, in controlled laboratory environments, scientists can manipulate air pressure to study its effects on certain substances or phenomena. This involves using specialized equipment such as vacuum chambers or pressure chambers to create specific air pressure conditions.

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