How is pressure on an object ?

[m441]

I'm french teacher and I would like to understand how works pressure with repuls forces. I don't receive reply in France or In USA so I try in Russia, maybe here people are more open

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).





Я учитель французского языка, и я хотел бы понять, как работает с давлением сил repuls. Я не получаю ответа во Франции или в США, поэтому я стараюсь в России, может быть, здесь люди более открыты
Я хотел бы знать, как это давление на объекте (черный на рисунке). Давление не является классическим давление, как газ, но задается макроскопических шаров. Шары, как магниты, Северный полюс снаружи и внутри Южной полюсе с железными дорогами, простите к магнитным полям (см. рисунок). Все шары repuls себя с силой, как F (1 / D ²). Все шары могут свободно перемещаться. Я ставлю черный предмет и мячи в получателем и уменьшить громкость есть давление. Но ведь объемы постоянные. Черный объект является фиксированным. Представьте себе, как шары однородными за пределами Северного полюса, это сложнее, чем меня тянет, конечно. Меня тянет большие шары но представьте себе, на самом деле меньше. Шары на рисунке не совсем так на самом деле, давление выше на внешних, давление внутри «круга» не homogous. Исправлена ​​черный объект не магнитное поле, представить это, как дерево, например. Мой вопрос: не является ли давление в каждой точке же на черной фиксированного объекта? (точка, где есть контакт).

[Munin]

"Google Translate" is a bad translation service. Using it makes text incomprehensible. Please:
1. if you are fluent in English, then write in English. This forum is very OK with English, and many people will answer you.
2. if you are NOT fluent in English, then please write in BOTH English and French. Your original French text would help to understand vague places in the English text. Some people understand French here. Certaines personnes ici comprennent le français.

m441 в сообщении #760161 писал(а):

My question is : is the pressure in each point the same on the black fixed object ? (point where there is a contact).

Approximately, yes.

But, I want to warn you: real magnets cannot be made the way they have only North pole outside, and all the South pole is inside. That is because North pole and South pole are connected with continuous magnetic lines, and since they go out of the ball (from the North pole), they have to go in somewhere (that makes a South pole).

I take your question as an abstract question about some particles (balls) that repulse with a uniform central force by some law, say, $F\propto 1/d^2$. This would be a good wording for a physical question. You don't have to be specific and describe some way to make such balls. In theoretical physics, you can postulate any law and pose a question about it, without saying how you would get such a law (or how you got it).

[m441]

ok, thanks Munin :) I don't knew if this forum would be ok with english. Sure, google translation is very bad...

ok too for replace magnet and say it's f(1/d²), just a theorical study replace ball by point particle ;) At France, when I said that all people say: It's not possible ! even I say it's a theorical study ...

With drawings I thought at first the pressure is not the same, but I'm not sure, I'm looking on internet, I asked at collegues and what I can say :

1/ It's not possible to calculate because it's too complex
2/ I contacted Comsol sofware, and for now it's not possible to simulate with the last software, nobody has done this type of simulation. A collegue said too it's not possible to simulate now. And without magnet only with simple electrostatic point particule !
3/ I don't find research study where I can have the result or even a study in this field, I'm not sure this study has done
4/ It's not possible to test easily, need a lot of magnets balls and if I want a near homogenous field I need a lot of small magnets, irons, very complex study, this is very difficult to build one ball so 1000 or more...

So, why I think the pressure is different ? because the black object change forces in each particule. For have the pressure in each point, I take a particule (after all the system is stable, sure), I calculate without black objets all forces from all others particules, and like the geometry is different why the sum of forces will be equal ? Ok, there is lateral force that move particule a little for have only a perpendiculary force from black object but why this force must be the same in each point ?

Another problem: the shape of the recipient, here I drawn a circle (or sphere in 3d), but imagine something asymetric, in this case it will be very strange to have the same pressure in each point.

It's not easy but I think it's very interesting problem, I would like to know if the pressure is equal or not in each point. Maybe someone know a very old study ?

Tell me what do you think about what I say :)

[m441]

Je viens juste de lire anglais et français ! cool !! j'espère que le forum accepte les caractères français :)

Alors voilà:

D'accord pour remplacer les balles magnétiques par des points, c'est plus simple. En France, impossible de faire une étude théorique, il faut toujours que la chose existe...

Je pensais que la pression n'était pas la même mais rien de certain, je me suis bien renseigné, et là eh bien rien !! car:

1/ Il n'est pas possible de calculer car c'est trop complexe
2/ Il n'est pas possible de simuler car même un logiciel comme Comsol ne le fait pas. Un collègue m'a dit la même chose...cela même si fait l'étude à partir de particules
3/ J'ai cherché au niveau recherches scientifiques et je n'ai rien trouvé, même pas quelque chose dans le domaine similaire
4/ Ce n'est pas facile de tester facilement, en effet, il faut arriver à concevoir une balle magnétique quasi parfaite, ce qui impose d'avoir une multitude de différentes parties aimantées et en fer, bon déjà en faire un ce n'est pas donné mais alors en faire 1000...

Pourquoi je pense que la pression n'est pas identique ? Parce que, admettons que j'attende que le système soit stable, pour avoir la pression d'un point, je fait la somme de tous les forces dues à l'ensemble des autres points et comme le solide crée un asymétrie (tous les solides excéptés le sphère), je ne vois pas comment la somme des forces serait la même. Il y aura une force latérale, cela fera bouger un peu le point car le solide ne peut fournir qu'une force perpendiculaire à sa surface. Mais une fois le système stable, pourquoi cette somme des forces serait identique ?

Autre problème, celui du récipient qui contient l'ensemble, dans mon exemple, j'ai pris un cercle (ou une sphère en 3d) mais si je prends une force asymétrique (tétraèdre par ex) je ne vois pas comment les forces de pressions peuvent être les mêmes.

Voilà si vous pouvez me dire ce que vous en pensez. Il doit bien y avoir quelqu'un qui a fait ce genre de recherche même si c'est un peu viellot, style 16° siècle...

Bonne soirée à tous de France

ps. il est peut être possible de faire un programme spécifique ? Vous pourriez m'aider à faire cela ? je suis prof en informatique !!

[Munin]

m441 в сообщении #760215 писал(а):

ok too for replace magnet and say it's f(1/d²), just a theorical study replace ball by point particle ;) At France, when I said that all people say: It's not possible ! even I say it's a theorical study ...

That is strange. Such kind of problems is very usual for theoretical physics. You could also say this is theoretical model, abstract model, or even toy model.

By the way, on this forum we use LaTeX to write formulas. Using non-latin sybmols is unwanted. For example, you could just write in your text:

...say it's $f(1/d^2)$, just a...

and it would give

...say it's $f(1/d^2)$, just a...

For more examples and notations, you can look in «Как набирать формулы?» and «FAQ по тегу [ math ]».

m441 в сообщении #760215 писал(а):

With drawings I thought at first the pressure is not the same, but I'm not sure, I'm looking on internet, I asked at collegues and what I can say :

1/ It's not possible to calculate because it's too complex
2/ I contacted Comsol sofware, and for now it's not possible to simulate with the last software, nobody has done this type of simulation. A collegue said too it's not possible to simulate now. And without magnet only with simple electrostatic point particule !
3/ I don't find research study where I can have the result or even a study in this field, I'm not sure this study has done
4/ It's not possible to test easily, need a lot of magnets balls and if I want a near homogenous field I need a lot of small magnets, irons, very complex study, this is very difficult to build one ball so 1000 or more...

This all is correct, to some degree. Yes, the question is hard to give an exact answer. Yes, it is hard even for numerical software, and it is hard to research experimentally. Yes, it is unlikely to find the research of exactly this problem setting. But.

This class of problems is very well-known in physics. It is called the many body problem or $N$-body problem (where $N$ is considered large). It is usually impossible to give an exact solution, but sometimes it is possible to find an approximate answer, which is good enough for qualitative questions, and for quantitative ones at some precision.

For the systems of very many similar bodies, there is a powerful method of consideration, called statistical physics. It gives approximate answers, which becomes more accurate as the number of bodies becomes larger. And it is relatively easy to obtain these answers.

Also, it is useful to try to get the answer in the analytical form (as a formula). Playing with formulas is sometimes much more powerful than using numerical methos and software.

m441 в сообщении #760215 писал(а):

So, why I think the pressure is different ? because the black object change forces in each particule. For have the pressure in each point, I take a particule, I calculate without black objets all forces from all others particules, and like the geometry is different why the sum of forces will be equal ? Ok, there is lateral force that move particule a little for have only a perpendiculary force from black object but why this force must be the same in each point ?

Another problem: the shape of the recipient, here I drawn a circle (or sphere in 3d), but imagine something asymetric, in this case it will be very strange to have the same pressure in each point.

I think that easiest way to deal with the problem is to consider the statistical limit. That is, we imagine that the number of balls increases while the balls themselves become smaller (their repulsive forces also diminish in scale, that is, for example, instead of $f\propto d^{-\alpha}$ they become like $f\propto (kd)^{-\alpha},$ where $k>1$ is the scale factor). Size of balls and their numbers are changed so as to keep the constant density. Then we can go to the limit $N\to\infty.$ In this limit, the mass of balls starts to behave like some continuous media, that is, substance.

Now, would it become like a liquid or like a solid state substance? Solids resist to the sheer deformation. But you have drawn balls rather sparce, and it seems they won't resist if one tries to shift them by layers. So I think, these balls would be like a liquid, either viscous or not (that does not matter now). And then I recall the Pascal's law, that says that the pressure onto the object, immersed into the liquid, would be the same onto every point of the object, whatever size and shape it has.

And one last note:

m441 в сообщении #760215 писал(а):

3/ I don't find research study where I can have the result or even a study in this field, I'm not sure this study has done

m441 в сообщении #760238 писал(а):

3/ J'ai cherché au niveau recherches scientifiques et je n'ai rien trouvé, même pas quelque chose dans le domaine similaire

Actually, the search for scientific researches is not easy. It needs some practical skills, and if you don't have them, you can easily fail. Please consult your colleagues and teacher about how to perform this search more thoroughly and effectively. Here are some tips:
- search for the scientific papers in scientific journals and proceedings; Google Scholar, arXiv
- know the right keywords to use in the search, which describe the domain and concepts;
- make your search muli-stage. Look for review articles, which give you some basic knowledge about the domain, read them, make more clear idea of what you want, and then search further, based on that idea and new keywords (and maybe particular journals, authors, paper titles).

[Sender]

Well, if $f(\frac{1}{d^2})$ is just $\frac{\gamma}{d^2}$ where $\gamma$ is some constant then we can consider all these balls to be charged with electric charge of the same value, so the force of repulsion is just Coulomb force. 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. More generally, the same arrangement takes place when the field of each ball has non-negative divergence in the considered area.

[m441]

Many thanks for your help :)

Цитата:

I think that easiest way to deal with the problem is to consider the statistical limit. That is, we imagine that the number of balls increases while the balls themselves become smaller (their repulsive forces also diminish in scale, that is, for example, instead of $f\propto d^{-\alpha}$ they become like $f\propto (kd)^{-\alpha},$ where $k>1$ is the scale factor). Size of balls and their numbers are changed so as to keep the constant density. Then we can go to the limit $N\to\infty.$ In this limit, the mass of balls starts to behave like some continuous media, that is, substance.

I try to explain how I understand what's happen, like that maybe you can explain what's wrong in my thoughts:

That point it's very difficult to understand for me (not an english problem). I can't understand how the density can be the same ! Surface of recipient and surface of black object are not charged, this is only neutral material. So, if I imagine the density equal everywhere, if I take one ball at external radius the repulsive force on it depend of all balls behind. But near center, the sum of forces is near 0 (I don't consider the force from surfaces, circle or black obect) With a black object, the sum of forces on surfaces of the black object change due to the position of the calculation (example with the rectangle). I imagine this system like amtosphere on Earth but with "gravity" in the contrary direction.

Цитата:

that all balls will be located at the surface of the outer spheric shell

It's possible to consider a smaller number than infinite with a "very high" repulsive force, I don't understand how all balls can be only at external circle (or sphere in 3d) ?

thanks for links

[Oleg Zubelevich]

this possibly resembles something like " magneto fluid", "magneto-hydrodynamics"

such liquids have interior moments so their stress tensor need not to be necessarily symmetric

[Sender]

m441 в сообщении #760354 писал(а):

I don't understand how all balls can be only at external circle (or sphere in 3d)

Please take a look at the proof of Earnshaw's theorem.

[m441]

ok, thanks, so what, all balls move all the time ? I don't understand with this theory why balls would be at external radius only. And if all the systeme is unstable, if the central object if free to turn, this would say you can recover energy (if there is not infinity of balls), it's not possible. But maybe I don't understand all your message.

[Sender]

If we consider balls to be electrically charged, the system can be stable only when they are affected by forces of non-electric kind. In our case it can be only the reaction of the outer spheric surface.

[m441]

ok, the system can be stable but all balls are at external radius, can you explain please ?

[Sender]

If a ball is affected only by electrostatic field induced by other balls, it can's stay in stable equillibrium according to Earnshaw's theorem.

[m441]

yeah, but there is an external circle and and object in the center.

[Sender]

Do you mean that these external circle and object in the center can somehow affect charged balls distantly?