Confused about a very basic concept: Pressure.

In summary: So, in summary, pressure is defined as force per area and is applicable to a wide range of scenarios, from a plate with a force acting on it to the pressure within a star. It is important to note that even if the net force is zero, there can still be pressure, as seen in the example of an elephant stepping on a toe. The definition of pressure in this scenario is not the total force, but rather the net force per area.
  • #1
Tomer
202
0
Hello everybody,

this is slightly embarrassing (for a physics student), but I realized in the last couple of days that I am somewhat confused with the concept of "pressure".

Pressure is defined in "school-level" as "force per area". So if I have a plate of 2m^2 on which a downwards force of 3N is exerted, the pressure on the plate is 3/2 N/m^2. Right?

First question: What happens if, in the scenario above, I exert the same force in the opposite direction (3N upwards) as well? Is my pressure 0 N/m^2 or 3 N/m^2 (double than before)?

The motivation for this question is really somewhat more complex: One says that the pressure on the surface of a star vanishes (P=0) and maximal in its center.

Now if I assume I have a massive ball, and I put aside all gas/radiation/whatever pressures and concentrate only on the gravitational pressure, i.e., the force/area exerted by gravitation-
Second question:
Why should the pressure in the center be maximal? All the forces cancel one another in the "exact center", so that there's practically 0 Force - why not therefore 0 pressure?
Furthermore, why is the pressure on the outer layer 0? The force there (on a certain gas element) is GMm/r^2, thus non-vanishing - why shouldn't there be any pressure? What if I'm simply holding a plate floating above the star - doesn't it have gravitation pressure on it?

So I've been working with pressure for years now, using it in thermodynamic equations as this quantity that tells me of the force transferred by gas molecules to normal surfaces, ideal gas equation and so one, but I realize, I don't really "dig" the definition of pressure.

Can someone enlighten me?

Thanks a lot.
 
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  • #2
Tomer said:
Pressure is defined in "school-level" as "force per area". So if I have a plate of 2m^2 on which a downwards force of 3N is exerted, the pressure on the plate is 3/2 N/m^2. Right?
Makes sense to me. Assuming that the load is evenly distributed over the plate.

First question: What happens if, in the scenario above, I exert the same force in the opposite direction (3N upwards) as well? Is my pressure 0 N/m^2 or 3 N/m^2 (double than before)?
I think what you're describing is this: A plate with a force of 3 N inward on both sides, giving a net force of zero. Correct?

Realize that this is really the same scenario as above, unless you meant for the plate to be floating in air or something. The plate was resting on a table, say, and you pushed down with a force of 3 N. Of course the table is pushing up with an equal force. The pressure is still 3/2 N/m^2.

The motivation for this question is really somewhat more complex: One says that the pressure on the surface of a star vanishes (P=0) and maximal in its center.

Now if I assume I have a massive ball, and I put aside all gas/radiation/whatever pressures and concentrate only on the gravitational pressure, i.e., the force/area exerted by gravitation-
Second question:
Why should the pressure in the center be maximal? All the forces cancel one another in the "exact center", so that there's practically 0 Force - why not therefore 0 pressure?
Furthermore, why is the pressure on the outer layer 0? The force there (on a certain gas element) is GMm/r^2, thus non-vanishing - why shouldn't there be any pressure? What if I'm simply holding a plate floating above the star - doesn't it have gravitation pressure on it?
A couple of points:

(1) Just because the net force is zero, doesn't mean there's no pressure. Imagine an elephant stepping on your toe, squishing it against the ground. The net force is zero--your toe isn't accelerating, is it? Yet I'm sure you'll agree that the pressure is far greater than zero.

(2) Imagine a tall stack of thin books. Thousands of them stacked up. The pressure on the bottom book is due to the weight of all the books piled on top of it. As you go higher in the stack, the pressure is reduced. At the highest point, the pressure goes to zero. (Imagine a very thin book.) That's why the pressure increases as you get closer to the center of the star--you have the weight of the entire star to support.
 
  • #3
Thanks so much for the fast reply!

Doc Al said:
Makes sense to me. Assuming that the load is evenly distributed over the plate.


I think what you're describing is this: A plate with a force of 3 N inward on both sides, giving a net force of zero. Correct?

Realize that this is really the same scenario as above, unless you meant for the plate to be floating in air or something. The plate was resting on a table, say, and you pushed down with a force of 3 N. Of course the table is pushing up with an equal force. The pressure is still 3/2 N/m^2.

Well, in the first scenario I actually did mean a "floating plate" in this sense. I did not want to write it, but I wanted to imagine the situation of a force acting on only one side of the plate, as opposed to the second scenario.

What happens if the plate just "floats"? Is there in this case no pressure? Why do we say that pressure is force/area then?
And if there is pressure (3/2 N/m^2) - how come the existence of a force from the opposite direction doesn't affect (cancel?) it?

A couple of points:

(1) Just because the net force is zero, doesn't mean there's no pressure. Imagine an elephant stepping on your toe, squishing it against the ground. The net force is zero--your toe isn't accelerating, is it? Yet I'm sure you'll agree that the pressure is far greater than zero.
I agree of course with your example, but what does it mean? What is the definition of pressure in this scenario? Obviously, not the total force (0!) per area of my toe? The net force is zero, and yet my toe being squashed. How can this be described if we're dealing an ideal 2D plate?

(2) Imagine a tall stack of thin books. Thousands of them stacked up. The pressure on the bottom book is due to the weight of all the books piled on top of it. As you go higher in the stack, the pressure is reduced. At the highest point, the pressure goes to zero. (Imagine a very thin book.) That's why the pressure increases as you get closer to the center of the star--you have the weight of the entire star to support.


Again, I understand the analogy. But here we have the Earth pulling all the books downwards. Going back to the star, how do the facts that:
1. The force outside of the star is maximal, i.e. GMm/r^2, and
2. The force in the center is 0,
agree with the fact that the pressure outside is 0, and the force in the center is maximal?
Again - if I hold a plate in my hands (forgetting about the "normal force" for a second) - does the gravitational field of Earth not exert a pressure on this plate?

I hope you'll (or others) have the patience to get me out of this disturbing thought-loop, I realize what I'm saying is contra-intuitive, and yet I don't understand how these fact add up from a theoretical/definition point of view.

Thanks!
 
  • #4
Tomer said:
Well, in the first scenario I actually did mean a "floating plate" in this sense. I did not want to write it, but I wanted to imagine the situation of a force acting on only one side of the plate, as opposed to the second scenario.
Well, if you only push one side of the plate, it will go flying. But of course there would still be pressure on it.

What happens if the plate just "floats"? Is there in this case no pressure?
Sure there is pressure. You're pushing on the surface.

Why do we say that pressure is force/area then?
And if there is pressure (3/2 N/m^2) - how come the existence of a force from the opposite direction doesn't affect (cancel?) it?
Pressure is force on a surface. Pushing a different surface doesn't directly affect the pressure on first surface.


I agree of course with your example, but what does it mean? What is the definition of pressure in this scenario? Obviously, not the total force (0!) per area of my toe? The net force is zero, and yet my toe being squashed. How can this be described if we're dealing an ideal 2D plate?
You need to describe the force exerted on the surface. Not the net force on the object (toe or plate) as a whole.



Again, I understand the analogy. But here we have the Earth pulling all the books downwards. Going back to the star, how do the facts that:
1. The force outside of the star is maximal, i.e. GMm/r^2, and
2. The force in the center is 0,
agree with the fact that the pressure outside is 0, and the force in the center is maximal?
You are describing the gravitational field strength, not the pressure. The gravitational force on a given mass will be greatest on the surface and zero at the center. But that's different than the total pressure pushing down on any layer.

Again - if I hold a plate in my hands (forgetting about the "normal force" for a second) - does the gravitational field of Earth not exert a pressure on this plate?
Well, depends what you mean. At what point in the plate? Imagine the plate is an inch thick. At the bottom surface, you must exert an average upward pressure to support the entire weight of the plate. But what about in the middle of the plate? That "surface" only has to exert an upward force to support half the weight of the plate. See how the pressure varies with depth? What pressure would you expect on the top surface of the plate?

Note: You can't forget about normal force. You are holding the plate. If you just let it go, then there won't be any pressure as it would be in free fall. (Ignoring complications such as air resistance.)
 
  • #5
Thanks!
Doc Al said:
Well, if you only push one side of the plate, it will go flying. But of course there would still be pressure on it.

Sure there is pressure. You're pushing on the surface.

Ok. Let's remember this statement.

Pressure is force on a surface. Pushing a different surface doesn't directly affect the pressure on first surface.
You need to describe the force exerted on the surface. Not the net force on the object (toe or plate) as a whole.

So the two sides of an ideal 2D plates are considered to be 2 different surfaces? If this is the case, it definitely helps me understand why the two forces from both ends don't add up/cancel the pressures.

You are describing the gravitational field strength, not the pressure. The gravitational force on a given mass will be greatest on the surface and zero at the center. But that's different than the total pressure pushing down on any layer.
Why? Why is the pressure a layer "feels" due to gravitation not the gravitational force divided by the area?

Well, depends what you mean. At what point in the plate? Imagine the plate is an inch thick. At the bottom surface, you must exert an average upward pressure to support the entire weight of the plate. But what about in the middle of the plate? That "surface" only has to exert an upward force to support half the weight of the plate. See how the pressure varies with depth? What pressure would you expect on the top surface of the plate?
I understand what you mean. and yet, what you're describing is how the different layers of the plate react to the weight of the layers above them, which are attracted to the star due to its gravitational field. But what I think about is pressure the gravitational force itself exerts on a certain layer, without considering the other layers pressing on it. In other words, I'm imagining a flat 2D plate. Wouldn't the gravitational force exert a pressure on it, just because of the fact there's a force acting on it?

Note: You can't forget about normal force. You are holding the plate. If you just let it go, then there won't be any pressure as it would be in free fall. (Ignoring complications such as air resistance.)
How does this agree with your first claim? You wrote that if the plate would just be flying and accelerating due to a one-sided force (me pushing it, gravitation or whatever) - there would be pressure. Now you wrote that there's no pressure in free fall. :cry:

Thank you, waiting for your response :)
 
  • #6
Tomer said:
You wrote that if the plate would just be flying and accelerating due to a one-sided force (me pushing it, gravitation or whatever) - there would be pressure. Now you wrote that there's no pressure in free fall. :cry:
[..]
An example is a rocket with the pressure of the rocket engine accelerating the rocket. That is a contact force that slightly compresses the rocket. However with gravity, and if we regard gravity as a force, this force is acting on all atoms the same (neglecting inhomogeneous fields). As a result there is no compression in free fall. And surely you know that intuitively:
- free fall = weightlessness -> you feel no force
- if you are in a car that takes off very fast you feel pressed against the seat.
 
  • #7
For a star, the condition P = 0 at the surface is a mathematical statement that the star is surrounded by the vacuum of space. P is a maximum at the center due to the fact that the entire mass of the star is attracted gravitationally to the center of the star.
 
  • #8
Right. Pressure is always pressure DIFFERENCE. Saying there is only pressure on one side is equivalent to saying the other side is exposed to a vacuum.
 
  • #9
Tomer said:
Thanks!


Ok. Let's remember this statement.
How can I forget?


So the two sides of an ideal 2D plates are considered to be 2 different surfaces? If this is the case, it definitely helps me understand why the two forces from both ends don't add up/cancel the pressures.
Again, pressure is defined on a surface. The net force on the object may be zero, but the pressure on each side is not.


Why? Why is the pressure a layer "feels" due to gravitation not the gravitational force divided by the area?
If I bury you in a pile of dirt, the pressure you feel is the contact force of stuff above you pushing down on you, not your weight.


I understand what you mean. and yet, what you're describing is how the different layers of the plate react to the weight of the layers above them, which are attracted to the star due to its gravitational field. But what I think about is pressure the gravitational force itself exerts on a certain layer, without considering the other layers pressing on it. In other words, I'm imagining a flat 2D plate. Wouldn't the gravitational force exert a pressure on it, just because of the fact there's a force acting on it?
See my answer directly above.


How does this agree with your first claim? You wrote that if the plate would just be flying and accelerating due to a one-sided force (me pushing it, gravitation or whatever) - there would be pressure. Now you wrote that there's no pressure in free fall. :cry:
Pushing a plate is not the same as free fall. In free fall, the only force acting is gravity.

(Note that pressure is a contact force per unit area, not a gravitational force. It's often due to there being gravity, of course, but is not the same as gravity.)
 
  • #10
russ_watters said:
Right. Pressure is always pressure DIFFERENCE. Saying there is only pressure on one side is equivalent to saying the other side is exposed to a vacuum.
Good point. For simplicity, in my examples I ignored things like atmospheric pressure acting on the plate.
 
  • #11
Doc Al said:
...

If I bury you in a pile of dirt, the pressure you feel is the contact force of stuff above you pushing down on you, not your weight.

I get stepped on by elephants, buried under dirt... :)
Intuitively, of course I understand. From a definition point-of-view, I have a plate, a force F is exerted on it due to gravitation, and thus the pressure should be F/A, only it's not. Perhaps the reason for the vanishing pressure is the fact that both sides of the 2D plate feel exactly the same "gravitational" pressure, and as our friend russ_watters suggested, the pressure difference in this case is 0?

(Note that pressure is a contact force per unit area, not a gravitational force. It's often due to there being gravity, of course, but is not the same as gravity.)

Ok, that's something I didn't know. I didn't know pressure is defined only through contact force. (whatever contact is? Is it not merely an electromagnetic force?).

Thanks everyone for your answers!
 
  • #12
Tomer said:
I get stepped on by elephants, buried under dirt... :)
Intuitively, of course I understand. From a definition point-of-view, I have a plate, a force F is exerted on it due to gravitation, and thus the pressure should be F/A, only it's not.
If the plate is laying on a table, then the average pressure on the bottom surface due to its own weight would equal F/A. There's nothing pushing down on the top surface, so why would there be a pressure there? (As before, I am ignoring atmospheric pressure.)

Perhaps the reason for the vanishing pressure is the fact that both sides of the 2D plate feel exactly the same "gravitational" pressure, and as our friend russ_watters suggested, the pressure difference in this case is 0?
No. In the example I discuss above, there is pressure on the bottom of the plate but not the top. If you have another scenario in mind, describe it.
 
  • #13
Doc Al said:
If the plate is laying on a table, then the average pressure on the bottom surface due to its own weight would equal F/A. There's nothing pushing down on the top surface, so why would there be a pressure there? (As before, I am ignoring atmospheric pressure.)No. In the example I discuss above, there is pressure on the bottom of the plate but not the top. If you have another scenario in mind, describe it.

The scenario is the same one as described before: a 2D plate is free-falling due to a gravitational force. A force F is thus exerted on
the plate. And yet there's no pressure.

You then mentioned that pressure is only defined for "contact" forces (something I'm not seeing written in Wiki), which then of course means that the gravitational force directly cannot exert any pressure. But I was never aware of this distinction.
 
  • #14
Tomer said:
The scenario is the same one as described before: a 2D plate is free-falling due to a gravitational force. A force F is thus exerted on
the plate. And yet there's no pressure.
That's right, because the plate is in free fall and thus accelerating. Time for another example with books.

Put two books, one atop the other, on a table. Each book weighs W. So, the contact force between the bottom book and the table is 2W, and between the two books is W.

Take those same two books and drop them. Now the books are in free fall. What is the force between them? (When they were on the table, they exerted a force equal to W on each other. What about when they are falling?)

Yet another example. You are on an elevator, standing on a scale that reads the force you exert on the elevator floor. When the elevator is stationary (or moving with constant speed) the scale reads a force equal to your weight. When the elevator accelerates upward, the scale reads higher than your weight; when accelerating downward, it reads lower. And if all the cables (and brakes) were removed, so that you and the elevator were in free fall, what would the scale read then?
 
  • #15
:)
I understand of course where you're going at and I agree that the pressure would be intuitively zero in these free-falling scenarios. What I still don't understand is, how does this agree with the definition " Pressure is the amount of force acting per unit area" (Wiki. Not that Wiki must be the best source of knowledge, but in such basic cases it's usually accurate enough). When the 2D plate is free-falling, the gravitational force is exerted on it. If I stick to the naive definition, the pressure would be the force divided by the area of the plate.

If the resolution to this discrepancy is "but gravitation isn't exerting any contact force": ok. But I would love to know what the definition for a contact force is. Does radiation exert contact force? Is the electromagnetic force a "contact force"? What is the difference?

If the resolution is another one, I'd really love it if you tell me what it is. I think we had enough examples with elephants and books :) My intuition tells me I'm wrong, the definitions don't.
 
  • #16
The gravitational force is not being applied to the surface, so it can't be a pressure.
 
  • #17
What is it applied to, if the plate is being accelerated by it? :)
 
  • #18
I think that the problem here is a bad definition. Clearly when something is freefalling there is a force on it or it would not accelerate. There is no pressure despite this. It seems like in every example here the force is a normal force.

So I think the definition should be Normal Force/Area, not Force/Area
 
  • #19
Tomer said:
What is it applied to, if the plate is being accelerated by it? :)
Every individual particle inside the plate.
 
  • #20
Tomer said:
:)
I understand of course where you're going at and I agree that the pressure would be intuitively zero in these free-falling scenarios. What I still don't understand is, how does this agree with the definition " Pressure is the amount of force acting per unit area" (Wiki. Not that Wiki must be the best source of knowledge, but in such basic cases it's usually accurate enough). When the 2D plate is free-falling, the gravitational force is exerted on it. If I stick to the naive definition, the pressure would be the force divided by the area of the plate.
Implicit in that definition is that the force must be applied to a surface.

Before worrying about the accelerating plate, let's be sure you understand how pressure varies in a non-accelerating plate.

Imagine you have a (thick) plate resting on a table. Total weight W, area A. Describe to me the pressure at various points throughout the plate. How would you calculate that pressure?

(Too bad you didn't like the books and elephants. If you understood those, you'd understand this.)
 
  • #21
I think he understands completely. The problem is with the F/A definition.
 
  • #22
I think Dmobb's right, and I wish we'd stick to the definition. Exactly like Dmobb said, the free-falling 2D plate has, as far as I know, a force F applied on it, and yet there's no pressure here.
ress_watters, you say the force is not applied on the "surface" in this case but on every individual particle of the plate. Why is a normal force, which is an electromagnetic force pushing the individual particles of the plate away, different?
Doc Al: Since I'm obviously confused regarding the gravitational force itself not exerting any pressure, I am also confused regarding your thick plate/books/elephants example. What you want to hear is: the bottom layer feels the weight of practically the whole plate, so the bottom layer feels W/A, while the upper layer has nothing pushing it from above, so the pressure is 0. But my question would be: Why is the gravitational force, which is exerted on the upper layer (as well as the other ones) does not exert any pressure on it?
I find it much simpler to discuss the 2D plate. If I understand that, I'll cope with thick plates and elephants.
 
  • #23
Tomer said:
I find it much simpler to discuss the 2D plate. If I understand that, I'll cope with thick plates and elephants.
I actually think you'd be better off going in the other direction. You need to understand what pressure means within a thick plate (or pile of books). Then I think you'll see how gravitational force relates to pressure.

But my question would be: Why is the gravitational force, which is exerted on the upper layer (as well as the other ones) does not exert any pressure on it?
Again, you must think in terms of forces between surfaces. Gravity can result in forces between surfaces, but it is not one itself. (Gravity is the force of the Earth pulling down on it.)

Divide that upper layer into 100 sections. (Like a pile of books!)
 
  • #24
Ok, so what you're saying is that the pressure on a surface is the force applied by another surface on it, or something of the sort, which excludes the gravitational force from being able to exert pressure (unless it indirectly causes a normal force).
I can accept that, I guess. It's just that I've never heard of this distinction before. Everywhere you read, you see that pressure is the force applied on a surface divided by its area. Since the gravitational force exerts a force F on the surface, I didn't see why this shouldn't apply here. But the gravitation isn't arriving from a "surface", whatever this exactly means, and thus cannot exert pressure.
Is this what you're saying?
 
  • #25
Tomer said:
Everywhere you read, you see that pressure is the force applied on a surface divided by its area.
Right.

Since the gravitational force exerts a force F on the surface, I didn't see why this shouldn't apply here.
But gravity exerts a force on mass, not on a surface. A surface is 2D and has no mass.
 
  • #26
Tomer said:
Ok, so what you're saying is that the pressure on a surface is the force applied by another surface on it, or something of the sort, which excludes the gravitational force from being able to exert pressure (unless it indirectly causes a normal force).
I can accept that, I guess. It's just that I've never heard of this distinction before. Everywhere you read, you see that pressure is the force applied on a surface divided by its area. Since the gravitational force exerts a force F on the surface, I didn't see why this shouldn't apply here. But the gravitation isn't arriving from a "surface", whatever this exactly means, and thus cannot exert pressure.
Is this what you're saying?

I think your problem is that you have not drawn a free body diagram of the system. In the static case (no acceleration) of a plate, there are 3 forces acting: the gravitational force (weight of plate), the pressure force on the top of the plate (pushing downward) and the pressure force on the bottom of the plate (pushing upward). So,

(pbottom) A - (ptop) A - mg = 0
 
  • #27
Doc Al -- Why can a point mass (0D) have mass and a surface (2D) not? Why should the plate be accelerated by gravity if it has no mass? I don't understand what you mean. We often describe mass distributions of 2D objects in physics.

Chestermiller: funny, I must have written such equations for static bodies a thousand times, and it always was:
N(normal force upwards) - mg (gravitation downwards) = 0.
In your example, there's a normal force exerting pressure upwards, I take it (P_top * A), but what is exerting the pressure downwards? Gravity? But you already accounted for it. What do I learn from your equation.
 
Last edited:
  • #28
Chestermiller said:
I think your problem is that you have not drawn a free body diagram of the system. In the static case (no acceleration) of a plate, there are 3 forces acting: the gravitational force (weight of plate), the pressure force on the top of the plate (pushing downward) and the pressure force on the bottom of the plate (pushing upward). So,

(pbottom) A - (ptop) A - mg = 0
Exactly! I've been trying to get Tomer to do that basic analysis all along, but no luck.

Tomer said:
Chestermiller: funny, I must have written such equations for static bodies a thousand times, and it always was:
N(normal force upwards) - mg (gravitation downwards) = 0.
In your example, there's a normal force exerting pressure upwards, I take it (P_top * A), but what is exerting the pressure downwards? Gravity? But you already accounted for it. What do I learn from your equation.
No, (P_top * A) is the force acting down on the top of the plate, due to whatever is pushing down on it (if anything). (P_bottom * A) would be the force pushing up on the bottom of the plate by whatever is pushing up on it. The weight of the plate is a different force entirely.

If you want to analyze the pressure on a plate, you must define your surface. A plate is not a surface (unless you want to just ignore its mass). It has two surfaces, each with a different pressure.
 
  • #29
Hmm. Ok. So a plate has two surfaces. The gravitational force cannot be said to apply on one (or both) of the surfaces, since a "surface" has no mass (though the plate is made out of two surfaces which together have a mass!). The key to my mistake is probably the fact that gravitation is not applied to a surface, but I find it very tricky to understand why. I don't see how this equation enlightens my misunderstanding.
For simplicity, I imagine a 2D plate being pushed by nothing above it, just laying on the ground (no air pressure and so on). Every school pupil would say there's a normal force: N, and a gravitational force: mg. Since there's no acceleration, N = mg.
The normal force is illustrated with an arrow pointing upwards, with its starting point on the plate. The gravitational force is illustrated with an arrow pointing downwards, starting at the plate as well. And yet there's a big difference between them, if I understand you guys: The gravitational force is applied to the whole plate. The normal force is applied on the bottom surface. The gravitational force does not exert any pressure, because it is not applied to a specific surface but to the whole plate. The normal force exerts a pressure, because it is applied to the bottom surface.
Is this description correct?
 
  • #30
Tomer said:
Hmm. Ok. So a plate has two surfaces. The gravitational force cannot be said to apply on one (or both) of the surfaces, since a "surface" has no mass (though the plate is made out of two surfaces which together have a mass!).
In this context, you really should think of a surface as a mathematical construct. That plate has two obvious surfaces of interest, but you can draw an imaginary surface anywhere and ask what force does one side exert on the other. That's where pressure will come in.

The key to my mistake is probably the fact that gravitation is not applied to a surface, but I find it very tricky to understand why. I don't see how this equation enlightens my misunderstanding.
For simplicity, I imagine a 2D plate being pushed by nothing above it, just laying on the ground (no air pressure and so on). Every school pupil would say there's a normal force: N, and a gravitational force: mg. Since there's no acceleration, N = mg.
If you keep imagining a 2D plate, I think you'll stall out. The plate has mass and thickness, although we might not care about the thickness. But a surface has zero thickness. Pick your surface of interest and figure out the forces exerted across that surface. That's where this analysis comes in handy.

The normal force is illustrated with an arrow pointing upwards, with its starting point on the plate. The gravitational force is illustrated with an arrow pointing downwards, starting at the plate as well. And yet there's a big difference between them, if I understand you guys: The gravitational force is applied to the whole plate. The normal force is applied on the bottom surface. The gravitational force does not exert any pressure, because it is not applied to a specific surface but to the whole plate. The normal force exerts a pressure, because it is applied to the bottom surface.
The normal force is the force that the plate and ground exert on each other. That's what's creating any pressure between them. Take away the normal force and there's no more pressure on the bottom of the plate.
 
  • #31
Ok. I think I understand the difference, although I still find it very confusing. But I need to let it sink.

Thanks a lot for all the different explanations!
 

1. What is pressure?

Pressure is defined as the amount of force applied per unit area. It is a measure of how much force is distributed over a given area.

2. What are the units of pressure?

The SI unit for pressure is Pascal (Pa), which is equivalent to one Newton per square meter (N/m^2). Other commonly used units include atmospheres (atm), pounds per square inch (psi), and millimeters of mercury (mmHg).

3. How is pressure measured?

Pressure can be measured using instruments such as barometers, manometers, and pressure gauges. These devices typically use a liquid or gas to measure the amount of force being applied.

4. What factors affect pressure?

Pressure is affected by factors such as the amount of force applied, the surface area over which the force is distributed, and the properties of the material being compressed. Temperature and altitude can also affect pressure.

5. What are some real-life examples of pressure?

Some common examples of pressure include the pressure exerted by a person's body weight on the ground, the pressure inside a car tire, and the pressure in a soda can or bottle. Atmospheric pressure, which is the pressure exerted by the Earth's atmosphere, is also a real-life example of pressure.

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