How is pressure on an object ?

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  • #1
Gh778
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I would like to know how is the pressure on an object (black on figure). The pressure is not a classical pressure like gas but is given by macroscopic balls. Balls are like magnets, North pole outside and South pole inside with iron for give roads to magnetic fields (see figure). All balls repuls themselves with a force like f(1/d²). All balls are free to move. I put black object and balls in a recipient and reduce volume for have pressure. But after all volumes are constants. Black object is fixed. Imagine balls like a homogenous North pole outside, it's more complex than I drawn, sure. I drawn big balls but imagine smaller in reality. Balls on figure are not exactly like that in reality, pressure is higher at external, the pressure inside "circle" is not homogous. Fixed black object don't have magnetic field, imagine it like wood for example. My question is : is the pressure in each point the same on the black fixed object ? (point where there is a contact).
 

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  • #2
Drakkith
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Are you asking if the pressure on the object is equal at all points on its surface?
I cannot follow most of your example with the magnets and all, but that seems to be your basic questions. Is that correct?
 
  • #3
Gh778
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Yes, I would like to know the pressure in each point where there is a contact ball/object and especially if this pressure is equal in fact. I don't need the exact value of pressure. The figure is very simple, a black fixed object in the center and a lot of balls (all balls are free to move) that repuls themselves. The difference with a fluid is important, here a ball receive forces from all others balls, the presence of the black object change the forces on each balls like the shape of recipient too. I know it must be impossible to calculate, difficult to simulate but I think it's a problem that be studied before but I don't find something in Internet. It's possible to imagine balls like a point for simplify the problem.
 
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  • #4
Drakkith
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I don't understand the significance of the balls. Why not just use a normal fluid or something instead of making the example more complicated?
 
  • #5
Gh778
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Oh, fluid is not the same: object don't influence the pressure with a fluid. Here the presence of the black object change the pressure, it's not the same case. Imagine a ball at contact, what forces on this ball (except black object) ? the ball want to move in one direction due to the back balls BUT there are balls in front this ball that repuls in other direction.
 
  • #6
Drakkith
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I'm not saying you're wrong, but I honestly don't see the difference between your example and an example with a fluid instead of balls. Why would the object somehow alter the pressure of the balls? After all, a fluid is made up of many many particles, which are analogous to small balls.
 
  • #7
Gh778
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in a fluid, a particule at contact with black object receive forces from only close particles not particles at distance, because atoms are neutral, the pressure is the same in each point, the pressure is given close to close. Here a ball receive forces from all other forces, close and far balls. Do you understand ? Figure show pressure in circle without black object.
 

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  • #8
Drakkith
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First, are you aware that magnetic fields do not work the way you have assumed them to in your example? It is not possible to have a single pole facing outwards in your balls. The field lines MUST be closed. You will have a sphere with many different poles of both types on its surface. Thus they will not actually repel each other.
 
  • #9
Gh778
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Magnetic lines are closed, see the detail of a ball, there are iron parts (grey color) for close magnetic lines. North pole is outside and South pole is inside. Sure, if I want homogenous external field, it's more complex than I drawn. I choose magnetic technology but it's an example. This is not my real problem in fact, I would like to know if the pressure is the same in each point of contact with repuls balls.
 
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  • #10
russ_watters
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Unless gravity is a factor in the setup, the pressure is the same everywhere. As described, it works exactly like fluid pressure.
 
  • #11
Gh778
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No gravity here. If pressure is the same everywhere, density is the same everywhere. So, take one ball at outer circle, the pressure is not give by the sum of all forces receive from all others balls ? And at third radius (at object), the pressure is not the sum of all forces from all others balls ?
 
  • #12
russ_watters
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Pressure and density in a static fluid with no gravity are the same everywhere, yes.

And yes, the force exerted by a single particle on the wall is equal to the sum of the forces on it from adjacent particles.
 
  • #13
Gh778
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Pressure and density in a static fluid with no gravity are the same everywhere, yes.
In a standard fluid I'm ok :)

And yes, the force exerted by a single particle on the wall is equal to the sum of the forces on it from adjacent particles.
why adjacent balls, why not all balls in the circle ?
 
  • #14
russ_watters
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why adjacent balls, why not all balls in the circle ?
In order to exert a force on each other, they have to be "touching".
 
  • #15
Gh778
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but they use a repulsive force with a law of 1/d², for me balls don't touch balls only circle and black object, I'm wrong ?
 
  • #16
russ_watters
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I put "touching" in quotes because the macroscopic and microscopic perceptions do not match. Regardless, the magnetic force drops off so fast that other balls further away can't exert enough force to matter.
 
  • #17
Gh778
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ok, but in this case at center pressure is 0 ?
 
  • #18
russ_watters
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No, pressure is the same everywhere.
 
  • #19
Gh778
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Is it possible to replace balls with electrostatic points ? it's a theorical problem in this case. Maybe like this I can estimate the sum of forces. You said, it's only close points that give pressure. But in reality, it's all points. If d is the mean distance between 2 points. Sum of forces depend of 1/d², first layer give 100%, second layer 25%, third layer 11%, etc. At outer circle the sum add a lot of far layers, this give more pressure than close to the center. Could you explain how you have same pressure everywhere ?
 
  • #20
russ_watters
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Is it possible to replace balls with electrostatic points ?
You mean like electrons? Sure.
You said, it's only close points that give pressure. But in reality, it's all points. If d is the mean distance between 2 points. Sum of forces depend of 1/d², first layer give 100%, second layer 25%, third layer 11%...
Sure*, but that's a significant change in the scenario since the objects before had a size that was much larger than the distance between the objects.

*Unless you are suggesting that the force is different at different places (ie, getting stronger in the center or at the edge). It still doesn't.
At outer circle the sum add a lot of far layers, this give more pressure than close to the center. Could you explain how you have same pressure everywhere?
Your assessment is wrong. You are measuring your distance with respect to the center of the vessel instead of with respect to each particle. Each particle has the same number of particles at a distance d, 2d, 3d, 4d, etc away from it, except those at the edge in which case the edge provides the force in the other direction.

What is the point of all of this anyway? It might help if we understood the big picture of what you are asking.
 
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  • #21
Gh778
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yes, like electron but without temperature, gravity, or other, it's just for study

Each particle has the same number of particles at a distance d, 2d, 3d, 4d,
it's that I don't understand, for me outer particule has only all particule in front of, no particule behind. A the center, a particule has 50% af all particle behind (push it) and 50% (push it in other direction) in front of. It's for that I think pressure is not the same everywhere because the ratio of molecule behind/front is not the same.
 
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  • #22
russ_watters
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it's that I don't understand, for me outer particule has only all particule in front of, no particule behind. A the center, a particule has 50% af all particle behind (push it) and 50% (push it in other direction) in front of.
A particle to the left can only push to the right. A particle to the right can only push to the left. If instead of another particle there is a wall to the right, the wall pushes with the same force the particle would have.
It's for that I think pressure is not the same everywhere because the ratio of molecule behind/front is not the same.
You've apparently been questioning the accepted understanding of pressure for several years now. This is really basic, common and important stuff: it isn't wrong. You need to accept that the scientific mainstream knows what it is doing.
 
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  • #23
Gh778
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In specific case of Coulomb interaction ([itex]F=\gamma\frac{q_1q_2}{r^2}[/itex]) we can try to obtain exact expression for it.
At the marginal case when size of a ball tends to zero we can introduce volumetric charge density [itex]\rho[/itex], balls form a layer of charged material.
Let [itex]R[/itex] be a radius of the outer spheric shell and [itex]h[/itex] - net thickness of charged layer. Let [itex]p(x)[/itex] be a pressure at radius [itex]x[/itex]. Obviously, [itex]p(x)=0[/itex] if [itex]0 \leqslant x <R-h[/itex].
Now let us consider the case when [itex]R-h \leqslant x \leqslant R[/itex].
Inside of radius [itex]x[/itex] we have charge [itex]q(x)=4\pi \rho\int_{R-h}^{x}t^2 dt=\frac{4\pi \rho}{3}(x^3-{(R-h)}^3)[/itex].
So, spherical layer of thickness [itex]dt[/itex] at radius [itex]t[/itex] contributes to the pressure in outer layers
[itex]dp=\rho \gamma \frac{q(t)}{t^2}dt=\frac{4}{3} \pi \rho^2 \gamma \frac {t^3-{(R-h)}^3}{t^2}dt[/itex]
Let [itex]\frac{4}{3} \pi \rho^2 \gamma=k[/itex]
Finally,
[itex]p(x)=\int_{R-h}^x k \frac {t^3-(R-h)^3}{t^2}dt=k (\frac{t^2}{2}+\frac{{(R-h)}^3}{t})\vert_{R-h}^x=k (\frac{x^2-{(R-h)}^2}{2}+{(R-h)}^3(\frac{1}{x}-\frac{1}{R-h}))=k(\frac{x^2}{2}+\frac{{(R-h)}^3}{x}-\frac{3}{2}{(R-h)}^2).[/itex]

so, the pressure will increase from inner layers to outer and reaches its maximum at the outer bound ?

'sender
 
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  • #24
russ_watters
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Without scrutinizing the math, what you are still missing about the logic is that EVERY particle has other particles surrounding it, not just the particle at the center.
 
  • #25
Gh778
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You don't watch the math formulas ? In this case Gauss's flux theorem is applicable and it is easy to show that all balls will be located at the surface of the outer spheric shell.
 
  • #26
russ_watters
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You don't watch the math formulas ?
They look wrong/nonsensical but in any case if you haven't properly conceptualized what happens, you have no hope of correctly applying math to it.
In this case Gauss's flux theorem is applicable and it is easy to show that all balls will be located at the surface of the outer spheric shell.
Huh? I thought the shell is packed with balls. You seem to be randomly changing the scenario as you go.
 
  • #27
Gh778
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I give up, bye
 

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