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The same, I understood my error. Thanks

for me it was the sum of each attraction force multiply (vector) by force from wall (N). For one force of attraction, N works in positive and for another force N works in negative, the sum is zero, so N don't works. For me it was logical to take center of gravity not center of mass. I hope you...

Thanks for links but I can't resolve that. It's not possible to calculate only 2 closed points of the center of gravity for show the trajectory and look of the angle ? Even, I don't know when the disk is on a precise part of circle I know all points. I don't know how to calculate the position of...

No torque, I'm agree with that and the software too, I said that because the software lost energy and must go somewhere, it was an hypothesis. For me, the force from wall is not perpendicular to the trajectory of the center of gravity. Have you seen my images, the last showing center of gravity...

Because Algodoo it's a 2D software ! and when I draw with my paper it's in 2D too. It's not possible to have 2D object in reality ? (I changed my first message because I would like to resolve in 2D not in 3D for show where the software bug).

This would say in 2d (ball = disk) the center of gravity is not in the center of mass but in 3d (ball = sphere) it is ?

Thanks for your help, yes I would like to resolve this system. @mfb: are you sure the center of gravity is always the center of mass, expecially when the attraction is not homogeneous ? http://en.wikipedia.org/wiki/Centers_of_gravity_in_non-uniform_fields I read on Wikipedia the Lagrangian...

I would like to study a pendulum in 2D, not on Earth but only with one fixed mass attract a disk. Considered this fixed mass like a point. No friction on this theoretical study. The disk can move only around a part of circle. There are only 2 forces, F the attraction and N the force from wall...

It's because the pressure of layer just above is lower ? And what's cancel torque from Fb/Fa ? Maybe it's easier with a square section.

1/ This would say the pressure on a dam http://en.wikipedia.org/wiki/Dam is bigger due to these forces ? 2/ If I place a tube full of liquid like the image show: http://imageshack.us/a/img801/3471/3btm.jpg [Broken] Red forces in the curvature are canceled by axis. But like sum of forces...

gravity is perpendicular to the screen, like that it's easier to understand for me. If radius of atom is divided by 2, the angle decrease, the red force too but the number of red forces increase too.

I consider the pressure come from top, like an object is in water (but liquid can be another than water). The red force come from left and right atoms that compress atom. An atom receive pressure from top (in a liquid under gravity like you put an object in a swimming pool). But atom receive...

For compute force from pressure on a surface, do I need take in account the red force ? When an atom press a surface it press atoms at right and at left, this increase the pressure on the curved surface if the volume is closed ? http://imageshack.us/a/img823/3852/66o4.jpg [Broken] Or...

Maybe it's not like I would say but yes, I "see" a force (I know I'm wrong), this force come from the difference between attraction and pressure. Imagine this theoretical case, maybe you can understand my problem: - A universe is only composed of gas, no planet, no sun, etc. just gas with no...

@simon: I will try to find my error in my program. When the torus turn, with good rotational speed (good size, small height for object, etc) it can cancel pressure in the fluid, right ? If necessary it's possible to have 2 object diametrycally opposed, like that no specific pressure from...

Ok, like that I understand, you're a very good teacher Simon ! The best here ! I come back of my first message, with gravitational force. An object composed of half part of water and half part of gas under low pressure. I don't compute gas, I consider like vacuum, it's possible to say the...

In static I can understand but in dynamic no: accelerate the torus, enough for the wall don't give 100 % of red force at big radius. In this case, torus have a net torque in one direction. If ball at small radius is near center, balls have a net torque in the same direction than torus because...

but sum of force for each ball is 0, and the wall apply counter force to the ball not on itself, I don't understand

just for understand the "thing" I don't understand with particles, gravitational attraction is like a spring in fact. With 2 balls it's easier to understand. Why red forces don't apply a torque ?

no gravitational forces here ! just a spring that can be attact 2 balls. the solid can only turn. I done sum of forces and for me red forces don't apply them with same radius, so I see a torque.

I think I understood where is my problem. I change the study: it's a solid torus with 2 solid balls in it like drawing show. Balls attrack each others with a spring. If I done sum of forces, I can see sum of forces on each ball is 0. For the torus too, but red forces don't apply at the same...

yes, it's very important when the study is in a disk. sure, but it's not a problem. I saw in particular liquid helium has no viscosity. I don't understand (english) I prefer compute with forces that I know. That the heart of the problem, I study attraction forces from liquid particles to...

It's not infinite set of problems, just 2 cases, first I would like study internal/external attraction forces like first post say, second I would like to study attraction of particles through object itself. I think these problems are link if I want to understand all pressure/force. Look image in...

Have you seen image in message #32 ? How force from pressure can be perpendicular with a simple asymmetrical object ? For me, particle attrack particle through object.

I said that ? For me water is compressible. You can change water by liquid helium at low temperature. Compressibility of liquid helium is bigger (100 times more than water I think).

You compute the density is different due to the presence of air-filled sphere ? Inside water-filled sphere the density is not the same than at external.

If I changed some parameters in the problem it's because I try to find the best problem (easier to resolve), it's not easy. Ok, in this case it's possible to show me another forces which compensate red forces. Not compute just drawing it. I would like to simulate forces with only gravity...

Even if one atom attrack one atom, the force can be like 1e-60 N, it's very small, but sum of force must be exactly at 0 and if liquid helium has no viscosity this would say the object can accelerate with this force. Yes, why not ? the gravitational attraction is direct (line from atom A to...

Red force is gravitational force: a mass attract a mass, so an atom of helium must attrack an atom of helium, the force is very small. The surface at right = surface at left, I'm agree. But at left: atoms of helium can't attract atoms of helium through walls of solid. It's a liquid helium not a...

Ok, so in this case red forces don't exist ? Atoms of helium don't attract atoms of helium through volume of gas ? Why ? The volume of gas is asymetric and I don't understand why sum of red forces can be at 0. At right there are red forces, not at left (no helium at bottom). Someone in this...

Ok, you accept these forces. So with a shape like that in water, on Earth, red object move alone at left ? Even there is viscosity the lateral force must be 0 (you can replace water by helium).

I would like to calculate sum of torque on object composed of one container of helium II liquid and one container of gas with low pressure. I know you think it's 0, but I would like to be sure (I think it's not 0, but it's only my intuition). It's not a theoretical problem, so I take in account...

exactly is exactly the same value, precision at 1e-3000 if necessary

I would like to know if density is exactly the same everywhere in a disk full of water except at one part where there is gas. All are fixed: disk of water, and container of gas. I don't want the value of density just to know if density is exactly the same everywhere in disk of water. 2 cases...

I don't know, for me it's P1 because additional water don't attract solid when it is in water, I don't want the exact value, I just want to understand what's happen when solid is move down of 1 meter, the weight must be the same P2 (at 1 meter near), but like molecules of water touch walls of...

Sorry, it's not 100001 km, it's 100 km +1 m = 100.001 km :( it is altitude, like that when I put a column of water of 100 km height the solid is just above. The diameter (or square section) of the column is only 1 m, this could change the center of mass of Earth ? If yes, for simplify the...

Ok, I don't knew ! thanks ! For my last message, #9 step3, the weight at step 3 is P1 or P2 ? If it's P2 could you explain how water can do for attract solid in it ? Because, solid attract molecules of water but they are in contact with solid and the force is canceled, no ? I know the force I...

Yes, I speak about gravitationnal attraction. I know water attract solid or anything else but I don't understand the difference of weight when an object is in water or not, (it must be the same, ok) take this case: 1/ On Earth, a solid is placed at 100,001 km of altitude, at this altitude the...

I understood, no additional force for solid due to the presence of water in column. The same for water on water (buoyancy), all forces cancel themselves on the last case. I have 2 anothers questions: 1/ Study 1 a/ Put a solid (height = 1 m, same density than liquid in column) at 11 m of...

it's easier, yes the solid is in water. I think it's possible to think in 2d. not at all. I think it's that I don't understand. Take a molecule at top between wall and solid. What forces on it ? From Earth and from water behind it, no ? For me the column add additional gravity, not ? Edit...

you're right it's easier to calculate: On Earth, put a column of water on the ground. I put at top of the column a solid in water, with the same density of local water, the solid mustn't move. Pressure from Earth (gravity) give a weight to the solid and buoyant force from Earth attraction is...

I'm interesting on the difference between attraction and pressure forces. If you let my study on Earth, supposed that gravity is exactly perpendicular to the image, like that it's easy to understand. 1/ Consider the sphere full of water: what are forces on it ? At right, gas attract near...

I would like study a theoretical problem. A system composed of 2 spheres can move at left of at right in water. One sphere has gas at low pressure and other sphere has water in it. The wall of spheres are very thin and has the same density of water. Sphere of gas don't has force because it has...

Hi, I would like to calculate the pressure inside a liquid planet in rotation. How can I do ? Pressure depend of depth under gravity but it depend of rotational speed too. Is it gravity pressure less centripetal pressure ? Is it \frac{1}{2}ρω^2r^2 - ρgh ?

I had done several tests and I can say the weight decrease a little more than the accuracy (0.001g) of a good balance. But it can be a problem of a gasket or anything else. The rotation of Earth is not in equation ? for 1kg, the rotation decrease pressure of 0.03 N (centripetal forces) at 6400...

A friend has let me a balance with a precision of 0.001g and now all is fine the weight is the same. Even this change a little during bubbles move up, it's not more the accuracy. But now, in a standard bottle, I would like to understand why weight is not greater when bubbles move up, due to...

I measured the pressure at start and after 2 minutes, the difference is 38000 Pa, but how get the differential pressure and take account of speed of sound ? not mass but weight, mass is always the same, but weight can decrease with only gravity like external force, nothing else ? A part of the...

I think there is an error in my study with delay due to the pressure. Pressure is not coming at 833 Pa in each second, like steps, it's not an air cylinder very fast. So: 1/ How could I calculate the good differential pressure with a tablet ? 2/ I can replace tablet by air cylinder very fast...

I measured the additional pressure inside the container (like last drawing show), it's only 38000 Pa with one tablet and a volume big enough for have 0.1 m in height and 36 cm², so the result is 0.034 g. Maybe the lost of weight come from this. This could explain the difference of weight is...

Watch new image please (theoretical system) With height of gas = H = 0.1 m Surface at contact = S = 0.0036 m² Speed of sound = V = 340 m/s (decrease when CO_{2} is added but increase with pressure) Gravity = g = 9.81 m/s² Difference of pressure = P = 100000 Pa Duration of bubbles = T = 120 s...