Why not crushed by air?

pivoxa15

1. Homework Statement
It has been estimated that the mass of nitrogen alone in the column of atomsphere above each square metre of the Earth's surface amounts to about 8 tonnes. Why aren't we crushed by this?

2. Homework Equations
none

3. The Attempt at a Solution
Air in the atomsphere is not just nitrogen and they are dense as the question suggested. Are we are not crushed by them essentially because of Newton's third law?

We are surrounded by dense air in ground level. What kind of molcules are they? They surround us and collide with our bodies and make us not as cold if without them.

The different densities of air all exert forces to the next layer and due to the third law there is an upward force from the succeeding layers. These layers does the same and you get somewhat of a cancellation of forces on each layer. Hence equilibrium is reached for each layer as there is no net force and we are not crushed by the upper level air. So a cushioning effect is passed down through the layers.

The image tries to illustrate this.

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We can't be crushed by earth's atmosphere because air spreads out equally. The only way you could be crushed by air is if you were in a tank and someone kept pumping in air.

chroot

Staff Emeritus
Gold Member
The air pressure at the surface of the earth is 14.7 psi. This force does not act in a single direction (downwards) -- it acts in all directions. In other words, the air pushes on your left side with the same force that it pushes on your right side. The force is balanced on both sides, so you don't go sliding sideways. The same is true of the forces in any two directions, including up and down.

This is easily understood when you realize that pressure is caused by molecules of air moving in all directions, randomly, imparting forces upon things they strike from any direction.

Water is essentially imcompressible, meaning that it does not change volume much with changing pressure. Thus, we stay the same size even when the air pressure changes due to weather or when we walk up hills.

- Warren

pivoxa15

The air pressure at the surface of the earth is 14.7 psi. This force does not act in a single direction (downwards) -- it acts in all directions. In other words, the air pushes on your left side with the same force that it pushes on your right side. The force is balanced on both sides, so you don't go sliding sideways. The same is true of the forces in any two directions, including up and down.

This is easily understood when you realize that pressure is caused by molecules of air moving in all directions, randomly, imparting forces upon things they strike from any direction.

Water is essentially imcompressible, meaning that it does not change volume much with changing pressure. Thus, we stay the same size even when the air pressure changes due to weather or when we walk up hills.

- Warren
But the net force of the air in the atomsphere is towards the earth because of gravity. I was trying to illustrate the net effect with the arrows. Is my diagram correct? If so it seems that we are not crushed by air above us because the air the surrounds us cusions the above air. Correct? And the air that surrounds us dosen't crush us because of the reason you suggested.

chroot

Staff Emeritus
Gold Member
The net force due to the air pressure is zero, because the air pushes on you from every direction. I cannot understand your 'cushioning' concept, but I can tell you immediately that it's not useful.

- Warren

DaveC426913

Gold Member
Have you ever tried to suck all the air out of a plastic coke bottle? Yup, it gets crushed. Our bodies would too if we weren't filled with air and liquid.

pivoxa15

The net force due to the air pressure is zero, because the air pushes on you from every direction. I cannot understand your 'cushioning' concept, but I can tell you immediately that it's not useful.

- Warren
Imagine most of the nitrogen is on the top few layers on my diagram. It's dense as specified in the question - although maybe not because it is spread out over a large volume. Gravity acts it towards the earth. So why doesn't it all come down and crush us all at once? The reason is as I suggested in my previous post.

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chroot

Staff Emeritus
Gold Member
No. No offense, but your notion is nonsense. We understand gas very well via a theory known as statistical mechanics, which is derived immediately from Newton's laws of motion. The reason the gas doesn't all come down at once is:

Thermal energy

The individual gas particles have thermal energy, and move about randomly. They rise and fall in ballistic trajectories, just like baseballs (except there are many, many collisions).

- Warren

pivoxa15

No. No offense, but your notion is nonsense. We understand gas very well via a theory known as statistical mechanics, which is derived immediately from Newton's laws of motion. The reason the gas doesn't all come down at once is:

Thermal energy

The individual gas particles have thermal energy, and move about randomly. They rise and fall in ballistic trajectories, just like baseballs (except there are many, many collisions).

- Warren
So you think the gravitational force on the air molecules are neglible compared to their random forces from thermal energy. Although gravity is still a factor because air dosen't just escape into space. However hydrogen does because it's even lighter. So there is still a net force toward the surface of the earth from the atomsphere. I am talking in a very macroscopic way.

Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.

As for the original question, the answers in the book suggested: "While the atmosphere exerts a pressure on us, we exert an equal pressure back." The pressure on us must be very small because the effect of gravity is tiny. The overriding force are random forces due to thermal motion as you say. That is why they don't crush us because it's also good that they don't escape into space as would be predicted without gravity.

DaveC426913

Gold Member
Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.
Helium actually, but that's just bifurcating bunnies...

Doc Al

Mentor
So you think the gravitational force on the air molecules are neglible compared to their random forces from thermal energy. Although gravity is still a factor because air dosen't just escape into space. However hydrogen does because it's even lighter. So there is still a net force toward the surface of the earth from the atomsphere. I am talking in a very macroscopic way.
Of course there is a force on the earth's surface due to the weight of the atmosphere. But there's no net force on you due to atmospheric pressure, since the air is pressing on you from all sides. (Actually, there is a small upward buoyant force due to the change in pressure with height--the upward buoyant force on you will equal the weight of the air you displace.)

Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.
The reason some balloons rise is because they are light enough (they are filled with a gas less dense than air) that the buoyant force from the surrounding air is greater than the weight of the balloon plus its contents.

As for the original question, the answers in the book suggested: "While the atmosphere exerts a pressure on us, we exert an equal pressure back."
That's not a very good answer, since that statement would be true whether we were crushed or not! (Imagine an elephant stepping on a peanut. The peanut exerts the same force on the elephant that the elephant exerts on the peanut--but the peanut is surely crushed.)
The pressure on us must be very small because the effect of gravity is tiny. The overriding force are random forces due to thermal motion as you say. That is why they don't crush us because it's also good that they don't escape into space as would be predicted without gravity.
The pressure on you is not small--it's about 14.7 lbs per square inch (at sea level). But we are perfectly capable of handling it. (For one thing, we wisely keep our lungs full of air at the same pressure as the outside air.)

pivoxa15

The pressure on you is not small--it's about 14.7 lbs per square inch (at sea level). But we are perfectly capable of handling it. (For one thing, we wisely keep our lungs full of air at the same pressure as the outside air.)
But the net pressure is virtually 0 as net force is 0.

But imagine a person standing on the ground. He is more exposed to air on top of him than from below him so there is a net force downwards from the atomsphere? And if you factor in gravity than that is another tiny bit more net force directed towards your feet.

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Doc Al

Mentor
But the net pressure is virtually 0 as net force is 0.
Sorry, but this is incorrect. The fact that the net force might be zero tells you nothing about the pressure. Recall my example of the elephant crushing the peanut. The net force on the peanut is zero, but the pressure is very high. Pressure is a force per unit area.

Further, air pressure is not zero!

But imagine a person standing on the ground. He is more exposed to air on top of him than from below him so there is a net force downwards from the atomsphere?
That would make some sense if your feet made an airtight seal with some surface, but that's unlikely to be the case. Rather than a person standing, imagine a suction cup stuck on a smooth surface. It sticks to the surface because air pressure pushes it. When you break the seal, air (at atmospheric pressure) gets inside and neutralizes the net force from the air.

Note that the orientation of the suction cup and surface is irrelevant: the surface can be horizontal, vertical, sideways, upside down--it doesn't matter. Air pressure acts in all directions.
And if you factor in gravity than that is another tiny bit more net force directed towards your feet.
I don't know what you mean by "factor in gravity"; gravity is the source of air pressure.

pivoxa15

Sorry, but this is incorrect. The fact that the net force might be zero tells you nothing about the pressure. Recall my example of the elephant crushing the peanut. The net force on the peanut is zero, but the pressure is very high. Pressure is a force per unit area.

Further, air pressure is not zero!
Force is a vector but area is a scalar. So if we assume net force is 0 than how can there be nonzero pressure=force/area?

Or are you saying even though if you add all the forces up, it equals 0, the area these forces act on may not be equal. i.e. There are two forces 8N and -8N. The 8N force act on 2m^2 of the surface only whereas the -8N force act on 3m^2 of the surface only. In this way net pressure is nonzero.

But the mouse could also get completely crushed even if total force and pressure on it is 0. i.e. there could be opposite force and pressure (hence net 0) on top of it and from the bottom. The top exerted by the elephant and the bottom by the ground.

That would make some sense if your feet made an airtight seal with some surface, but that's unlikely to be the case. Rather than a person standing, imagine a suction cup stuck on a smooth surface. It sticks to the surface because air pressure pushes it. When you break the seal, air (at atmospheric pressure) gets inside and neutralizes the net force from the air.

Note that the orientation of the suction cup and surface is irrelevant: the surface can be horizontal, vertical, sideways, upside down--it doesn't matter. Air pressure acts in all directions.
Those things are amazing. There was a time when I didn't know how these thigns worked and thought that new physics was needed to explain them. Then I realised it was only the air around us that was causing it to suck.

I don't know what you mean by "factor in gravity"; gravity is the source of air pressure.
But a weak source? I was trying to say that on the surface of the earth, the net force of air is towards the earth is not 0 but not much higher either.

Gokul43201

Staff Emeritus
Gold Member
Force is a vector but area is a scalar. So if we assume net force is 0 than how can there be nonzero pressure=force/area?
Area is a vector quantity. Pressure is a scalar.

Note: If the edges of the soles of your shoes make airtight seals with the floor below them, then there will be a net downward force (because there's no air below your feet) of about 1000-2000lbs.

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pivoxa15

Area is a vector quantity. Pressure is a scalar.

Note: If the edges of the soles of your shoes make airtight seals with the floor below them, then there will be a net downward force (because there's no air below your feet) of about 1000-2000lbs.
Force is a vector quantity and if area is a vector quantitiy than Pressure=Force/Area meaning a vector divide another vector. How does that make sense

Gokul43201

Staff Emeritus
Gold Member
Force is a vector quantity and if area is a vector quantitiy than Pressure=Force/Area meaning a vector divide another vector. How does that make sense
You can not divide vectors. I believe the way you relate the quantities would be:

$$P = \frac{\vec{F} \cdot \vec{A}}{|A|^2}$$

Also, if these quantites change over the scale of the object of interest, you will have to take elemental areas and then integrate over the entire object. Perhaps a better way of writing this would be:

$$\vec{dF}=P~\vec{dA}$$

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Doc Al

Mentor
Force is a vector but area is a scalar. So if we assume net force is 0 than how can there be nonzero pressure=force/area?
You are confusing net force on something with the individual forces that might act on something. Just because the net force is zero, does not mean that no force acts.

Imagine two dogs having a tug of war over a bone. The net force on the bone may be zero, but you sure wouldn't say there are no forces acting on the bone. Same thing with air pressure: the air is most definitely pushing on every square inch of you with a significant force. But since the air surrounds you, the net force from the air is just about zero.

Or are you saying even though if you add all the forces up, it equals 0, the area these forces act on may not be equal. i.e. There are two forces 8N and -8N. The 8N force act on 2m^2 of the surface only whereas the -8N force act on 3m^2 of the surface only. In this way net pressure is nonzero.
Again, don't confuse net force with "net pressure".

But the mouse could also get completely crushed even if total force and pressure on it is 0. i.e. there could be opposite force and pressure (hence net 0) on top of it and from the bottom. The top exerted by the elephant and the bottom by the ground.
Again, if the pressure on the mouse is zero (meaning NO force acts to squash it) it will not be crushed. But even if the pressure is great and the mouse gets crushed, the net force may well be zero.
But a weak source? I was trying to say that on the surface of the earth, the net force of air is towards the earth is not 0 but not much higher either.
I wouldn't call 14.7 pounds per square inch a weak pressure. And that pressure is caused by gravity--it's the weight of the atmosphere that creates the air pressure near the Earth's surface.

pivoxa15

You are confusing net force on something with the individual forces that might act on something. Just because the net force is zero, does not mean that no force acts.

Imagine two dogs having a tug of war over a bone. The net force on the bone may be zero, but you sure wouldn't say there are no forces acting on the bone. Same thing with air pressure: the air is most definitely pushing on every square inch of you with a significant force. But since the air surrounds you, the net force from the air is just about zero.

Again, don't confuse net force with "net pressure".

Again, if the pressure on the mouse is zero (meaning NO force acts to squash it) it will not be crushed. But even if the pressure is great and the mouse gets crushed, the net force may well be zero.

Could you rigorously illustrate an example where net force is 0 but net pressure is non zero. How about my one?

"Or are you saying even though if you add all the forces up, it equals 0, the area these forces act on may not be equal. i.e. There are two forces 8N and -8N. The 8N force act on 2m^2 of the surface only whereas the -8N force act on 3m^2 of the surface only. In this way net pressure is nonzero."

Is it correct?

If you take Gokul43201's definition of pressure

$$P = \frac{\vec{F} \cdot \vec{A}}{|A|^2}$$

than you could have the case where net pressure is 0 but make each area vector antiparallel to each force vector hence making pressure negative and also it now cannot be 0.

I wouldn't call 14.7 pounds per square inch a weak pressure. And that pressure is caused by gravity--it's the weight of the atmosphere that creates the air pressure near the Earth's surface.
I thought most of that 14.7 was due to random thermal motion of the air molecules? The fact that they don't all escape into space and denser near the surface of the earth is due to gravity.

Doc Al

Mentor
Could you rigorously illustrate an example where net force is 0 but net pressure is non zero. How about my one?

"Or are you saying even though if you add all the forces up, it equals 0, the area these forces act on may not be equal. i.e. There are two forces 8N and -8N. The 8N force act on 2m^2 of the surface only whereas the -8N force act on 3m^2 of the surface only. In this way net pressure is nonzero."

Is it correct?
This thing you call "net pressure" is just a term you made up. What are really referring to is better called net force. The pressure is nonzero, but the net force that it delivers to an object can certainly be zero.

If you take Gokul43201's definition of pressure

$$P = \frac{\vec{F} \cdot \vec{A}}{|A|^2}$$

than you could have the case where net pressure is 0 but make each area vector antiparallel to each force vector hence making pressure negative and also it now cannot be 0.
You misunderstand the use of that definition. That defines the pressure delivered by a force on an area; the force in that expression is not the net force on the object, it is the force delivered by whatever is creating the pressure. To find the net force on the object you need to add up the forces acting on each element of the object's surface.

I thought most of that 14.7 was due to random thermal motion of the air molecules? The fact that they don't all escape into space and denser near the surface of the earth is due to gravity.
No. The 14.7 pounds per square inch is due to the weight of the air. (The random thermal motion of the air molecules prevents the atmosphere from settling into a shallow puddle of liquid air.)

DaveC426913

Gold Member
Could you rigorously illustrate an example where net force is 0 but net pressure is non zero. How about my one?
Last time I went diving. I was at neutral bouyancy, meaning the net forces on me were zero, yet I was under a fair amount of pressure.

I thought most of that 14.7 was due to random thermal motion of the air molecules? The fact that they don't all escape into space and denser near the surface of the earth is due to gravity.
If you separated out a column of air with one inch cross section (say, by erecting a 100 mile tall hollow straw), it would weight 14.7 pounds. It would weigh 14.7 pounds even if you froze it to its component liquids, stopping the thermal motion.

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pivoxa15

It sounds as if you people insist that pressure shouldn't have a direction and should just be taken as a magnitude value hence when net force is 0, net pressure will remain positive. (despite that pressure = force/area and force is a vector).

However, If you take Gokoul's definition of pressure than it definitely is a magnitude value but I've never seen it in textbooks yet.

I think I am starting to understand that air weighs 14.7pounds/inch^2 soley due to gravity. So it seems we are crushed by air but not to a serious extent as stated in the question because air unlike liquid or gas has translational motion so will not likely all fall to the ground due to thermodynamics.

What do you think about the explanation in the OP and the diagram? I admit it's fairly crude as it can apply to liquid or solid in the 'cloud' in the diagram. For example the explanation I gave could also explain why a tall layer of brickwall just dosen't all crush down to the ground. Unless if a layer or two couldn't stand the forces.

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Hootenanny

Staff Emeritus
Gold Member
It sounds as if you people insist that pressure shouldn't have a direction and should just be taken as a magnitude value hence when net force is 0, net pressure will remain positive. (despite that pressure = force/area and force is a vector).

However, If you take Gokul's definition of pressure than it definitely is a magnitude value but I've never seen it in textbooks yet.
That's because pressure is a scalar, as many people have already told you. Per Gokul's definition;

$$P = \frac{\vec{F}\cdot\vec{A}}{|A|^2}$$

Note that while both force and area are vector quantities (since the pressure exerted depends on the direction of the force compared to the surface of the object), the numerator of the ratio is a dot / scalar product which returns a scalar quantity. When you find the dot product of two vectors the result is a scalar, roughly speaking in the above case, $\vec{F}\cdot\vec{A}$ is the length of the projection of $\vec{F}$ on $\vec{A}$.

Please either show me a reference which states that pressure is a vector quantity or accept what the very knowledgeable people here are telling you. absit iniuria verbis

pivoxa15

Please either show me a reference which states that pressure is a vector quantity or accept what the very knowledgeable people here are telling you. absit iniuria verbis
I have not seen anywhere pressure referred to as a vector. But I have always seen pressure defined as F/A and I naturally, I was led to believe that Pressure was a vector because Force is normally a vector. Note I have never seen F with a magnitude sign nor a vector tidle in P=F/A. So the latter might suggest it is never a vector in this situation.

Hootenanny

Staff Emeritus
If force is not written with a vector tidle, nor bolded or underlined one must assume that it is the magnitude of the force vector $\left(F = \left\|\vec{F}\right\|\right)$. Note however, that force is always a vector, but it is possible just to consider its magnitude in calculations.