Calculating Pressure and Gravity Forces in a Two-Sphere System in Water

[Gh778]

The assumption is that water is incompressible.

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).

 

[K^2]

You could. It would make the problem numerically more complex and won't change the answer. You are still dealing with something that's completely static due to symmetries. So you'll still have pressure gradients that keep everything from moving.

 

[Gh778]

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.

 

[K^2]

We can't discuss an infinite set of problems due to there being infinitely many. You've had a specific definition in your first post. We've covered all possible variations there of. Gravity and pressure balance there.

If you have a different problem, it's a different problem. But you should be working it out the same way. There is only going to be motion if it's going to reduce system's energy. If you start out at the minimum, there will not be any movement. And it's not hard to come up with simple examples that are out of balance. Place a sealed container with air under water and let it go. It comes up to the surface, because that minimizes energy.

If you think you see a situation where you expect motion while energy of the system isn't changing, look for a mistake. That's all there is to it.

 

[Gh778]

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 message #24, what force cancel red force ?

I would like to compute sum of forces on object (that can move or turn) with only gravitationnal attraction forces. I saw N-body simulation but it's for free object. How can I apply this sort of simulation with my examples ? I would like to take in parameters the density of liquid. Someone can help me to do this ?

 

[Simon Bridge]

I said that ? For me water is compressible.

You said it when you specified "water" i your setup - and you seemed to want the simplest model.
http://ga.water.usgs.gov/edu/compressibility.html
... you intended a compressible liquid?

Don't get me wrong - I'm not complaining about the parameters of the problem changing - I was explaining why I, for one, didn't dwell on the initial diagram.

You can change water by liquid helium at low temperature.

Very low temp.

Compressibility of liquid helium is bigger (100 times more than water I think).

Or, indeed any compressible fluid. And you can replace the "air" with any low density fluid.
You understand that things float in air right?

All the arguments prev still hold - the compression of the gas makes them a bit more complicated but the low-density volume acts like an absence of mass with the exact result depending on the geometry of the various elements.

i.e. if you had a large ball of compressible fluid with no external gravitation, then we have to ask why it stays as a ball right? Maybe it is held together under it's own gravitation? Well then the density will vary with radius, most dense at the center.

So the equivalent problem would be to have an enclosed void in the ball right next to an enclosed volume of the same fluid at the same distance from the center. Or, maybe, you have one at a greater radius than the other.

The voided sphere will want to rise of course - head away from the center of mass. It's less dense that the surroundings and the pressure gradient is reinforced by a density gradient.

Or, maybe, you want the classical infinite volume of fluid all at a constant temperature? In that case the density is initially equal everywhere since there is no center - so is the pressure. But if we put an enclosed void someplace, that place automatically becomes a center and then we now have an asymmetry.

If I remember my cosmology properly, the thing with infinite uniform mass distributions is that small asymmetries cause things to move about - the compressible fluid collapses over time - it's no longer a static system.

I don't want to go further without checking with you for what you are talking about.

But K^2 is correct - you need to work the energy description - what will reduce the energy.
Trying to work the forces will just get you confused and it is better than continually coming up with more and more examples ... there are always more, it's the nature of knowledge. Look for ways to group them into classes of problems and solve them like that.

 

[Gh778]

you intended a compressible liquid?

yes, it's very important when the study is in a disk.

Very low temp.

sure, but it's not a problem. I saw in particular liquid helium has no viscosity.

You understand that things float in air right?

I don't understand (english)

But K^2 is correct - you need to work the energy description - what will reduce the energy.

I prefer compute with forces that I know.

you want the classical infinite volume of fluid all at a constant temperature? In that case the density is initially equal everywhere since there is no center

That the heart of the problem, I study attraction forces from liquid particles to liquid particles itself through an object. If the study is on Earth, this don't change the problem if study is perpendicular to Earth's gravity. I prefer to study on Earth because in space I need to study with big dimensions and need a long time to compute.


The best I can do is to compute with a program and understand how forces are in equilibrium state. Like I described a study, someone can help me to compute it ? I would like to compute torque of object around black disk.

 

[Gh778]

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 radius and I see a torque. Could you explain the error I done ?

 

[Simon Bridge]

sure, but [super-cold] not a problem. I saw in particular liquid helium has no viscosity.

Liquid helium introduces all sort of odd effects.
But we can figure for a compressible, negligible viscosity, fluid without specifying the actual material - since we are just making models.

I don't understand (english)

"fluid" does not have to mean "liquid" - gasses are fluids too. Buoyancy rules work for gasses as well.

I prefer compute with forces that I know.

Which just makes it harder for yourself.
In very complicated systems the forces can get quite subtle and they stop being a useful way of analyzing things.

That the heart of the problem, I study attraction forces from liquid particles to liquid particles itself through an object. If the study is on Earth, this don't change the problem if study is perpendicular to Earth's gravity. I prefer to study on Earth because in space I need to study with big dimensions and need a long time to compute.

Earth's gravity will still affect everything - for instance, the gravity will create a vertical pressure gradient in a fluid. If the fluid is compressible, then there will be a density gradient as well.

 

[Simon Bridge]

Looking at your diagrams - it would help if you told me what it is that these studies are supposed to help you with?
What is the problem you are trying to solve?
Because it looks like you are trying to build a perpetual motion machine.

The first one with the liquid helium is meaningless to me - I cannot tell what the parts are supposed to be.
The second one with the solid torus...

Are you still trying to understand gravity?
The gravitational field of a torus is a little difficult.
Your two solid balls inside cavities in the torus are pulled towards each other by the spring - and also by each other's gravity. They also experience the torus' gravity and any external gravity fields.

Why the spring? What is ti supposed to show?
Why a torus? Is the torus supposed to be spinning?
Anyway - in the configuration shown, everything is held static by the spring.

If the red arrows are indicating gravity from the block of solid they point at, then you seem to have forgotten the gravity from the solid material in the other direction. You have to sum all the forces. Gravity acts between centers of mass. All the torus affects the spheres.

The center of mass of the torus is slightly off-center because of the holes the spheres are sitting in but both spheres will try to get closer to that point which will be close to the center of the torus.

 

[Gh778]

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.

 

[Simon Bridge]

If there is no gravity - then the free-body diagrams for each sphere will have the spring force pointing along the line of the spring and two reaction forces from contact with the two solid surfaces.

The balls are being held against a corner by the spring... add in the forces from the walls.

 

[Gh778]

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 ?

 

[Simon Bridge]

I think I can see how you are thinking here ... the tangential components of the spring force on each ball pushes the balls against the walls at different radii - so they apply a different torque. You have to remember that the walls push back which, in that analysis, will produce a net counter-torque the same size.

The balls are being held against a corner by the spring... add in the forces from the walls.

 

[Gh778]

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

 

[Gh778]

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 acceleration are not the same. I hope it's clear

 

[Simon Bridge]

Nope - same thing - you have neglected some of the reaction forces at the walls of the holes the sphere's sit in.
[edit - you changed these to paddles.]

You keep drawing your diagrams as if the spheres push on the walls without the walls pushing back.

In the case where you have started the system turning - you will have an extra centripetal force to include.
Draw the free-body diagrams for each sphere.

[edit] I see you changed the attachment - but it still makes no difference!
You have still made the same mistake.
If you will not follow advise I cannot help you.

 

[Simon Bridge]

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

The ball pushes on the walls, and the walls push on the ball.

Consider:

The brown part is a solid "rotor" - pivoted about the black dot in the center.
The red balls are sitting in square cavities and are held in place by a spring (green).
The assembly is imagined to be mounted horizontally. I included a counterweight so any rotation will be smooth.

The balls are forced into contact with two walls of their respective cavities.
The force of the contact is equal on both walls.
Ball A pushes on the wall at point a at a greater radius than ball B pushes on the wall at point b.
This means that there is a net torque - so the whole setup should rotate clockwise (and accelerate!)

One can imagine putting a similar arrangement of balls in the "counterweight" to make the machine work better ... in fact, have a whole circle of them! Patents and riches beckon...

But that is an inaccurate picture.

Look at ball A - by itself:

It experiences a force from the spring acting into the corner.
It also experiences the two walls it is touching pushing back.
The sum of all these forces is zero - so ball A cannot go anywhere.

Look at ball B - by itself:

It experiences a force from the spring acting into the corner.
It also experiences the two walls it is touching pushing back.
The sum of all these forces is zero - so ball B cannot go anywhere.

If ball B and Ball A are not moving - then the whole contraption has to stay still!

Look at the forces on the walls ... the forces on ball A's cavity, exerted by ball A, add up to the same value as the spring force, and points diagonally. Ball B pushes in exactly the opposite way. So the total force is zero.

Here are the diagrams:

Each object is considered separately - with all the forces included for each one.
Force arrows can be added head-to-tail. I did not draw in the entire brown "rotor".

The mistakes you made were:
(a) neglected the effect of half the forces
(b) added forces that were acting on different bodies

This is why we do free body diagrams.

The only difference that having the contraption spin would make is an extra centripetal force on each ball. This will make a difference to the spring and contact forces but they will all adjust back to zero.

For your next suggestion you must produce free body diagrams if you want the idea to be considered. If you will not do some of the work you are just wasting everyone's time.

If you want to continue to propose more and elaborate devices, you must learn to use conservation of energy more - you may even need to learn to use the principle of least action.

 

[Gh778]

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 pressure is very low. I consider density the same everywhere. I don't compute temperature. Even there is friction the sum of force on object must be to 0. I computed a very small object in other case, I must wait a long time ! I compute only one layer of atoms between walls and object.

I compute force that attact water due to the lack of water in gas-container and I compute force of pressure due to water inside object on gas-container. Maybe it's not good to thinking like that. For me, I imagine an external fluid which don't attract (weight) and don't give pressure. I put an object in it (half water, half gas). Water-container will change the pressure around the air-container, but gas can't do the same. The net force in this case is the force of buoyancy I compute in program.

I'm not sure my program is good enough:

#include <stdio.h>
#include <math.h>


#define L 0.000000001 //Width of column
#define H 0.000001 //Height of column
#define r 0.5e-10 //radius of a molecule m
#define h 0.0000001 //height of solid
#define w 3e-26 //mass of a molecule kg

int main()
{
    long int i,j,k,l; //counter for loop
    double nbL, nbH, nbh, sumB, forceB, sumG, forceG, angle, length, o, c, force, d, rw;
    double w2 = pow((double)w,2.0); //mass*mass

    d = 2.0*r; //diameter of a molecule
    nbL = L/d; //number of molecules in width of column
    nbH = H/d; //number of molecules in height of column
    nbh = h/d; //number of molecules in height of solid
    sumB = 0.0; //sum of buoyancy forces
    sumG = 0.0; //sum of weight forces
    rw = (nbL-2)*nbh*w; //total mass kg

    printf("nbL=%lf\nnbH=%lf\nnbh=%lf\nd=%lf\nsumB=%lf\nw2=%e\nsumG=%lf\ntotal mass=%e", nbL, nbH, nbh, d, sumB, w2, sumG, rw);
    printf("-----------------------------------------------");


/////////////////////////////////////////////////////////////////////////////////
/////////// Loops for buoyancy
/////////////////////////////////////////////////////////////////////////////////
    for(i=0;i<(int)nbL;i++)
    {
        for(j=0;j<(int)nbH;j++)
        {
            for(k=0;k<(int)nbh;k++)
            {
                o = (double)i*d; // length of opposite of triangle
                c = H - (double)j*d - (double)k*d -d; // length of side or triangle
                length = sqrt( pow(c,2.0) + pow(o,2.0) );
                angle = atan(o/c);
                force = w2 / pow(length,2.0) * cos(angle);
                sumB += fabs(force);
                //printf("o=%e\nc=%e\nlength=%e\nangle=%e\nforce=%e\nsumB=%e\n", o, c, length, angle, force, sumB);
                //getchar();

            }
        }printf("i=%i / %i ******** sum = %e\n",i, (int)nbL, sumB);


    }
    forceB = 6.674e-11 * sumB*(L-2*d)/(d);
    printf("buoyancy force = %e\n", forceB);


/////////////////////////////////////////////////////////////////////////////////
/////////// Loops for weight
/////////////////////////////////////////////////////////////////////////////////

    for(i=0;i<(int)nbL;i++)
    {

        for(j=0;j<(int)nbH;j++)
        {
            for(k=0;k<(int)nbh;k++)
            {
                for(l=1;l<(int)nbL-1;l++)
                {
                    o = fabs( (double)i*d+r - (double)l*d+r -d);
                    c = H - fabs ((double)j*d + (double)k*d ) + d;
                    //if(c==0.0) printf("pb 0");
                    length = sqrt( pow(c,2.0) + pow(o,2.0) );
                    angle = atan(o/c);
                    force = w2 / pow(length,2.0) * cos(angle);
                    sumG += fabs(force);
                    //printf("o=%lf\nc=%lf\nlength=%lf\nangle=%e\nforce=%e\nsumG=%e\n", o, c, length, angle, force, sumG);
                    //getchar();

                }
            }
        }
        printf("i=%i / %i\n",i, (int)nbL);
    }
    sumG=6.674e-11 * sumG;


    printf("buoyancy force = %e\nattraction force = %e\n", forceB, sumG);





    return 0;
}

The net force on object is not 0. For me, it's possible to put it in a torus but I don't know how to compute torque.

http://imageshack.us/a/img46/5976/o13h.jpg

 

[Simon Bridge]

If the net force is zero in Nature and non-zero in your computer program - then the computer program is wrong.
I refuse to troubleshoot code.

You seem to be thinking in terms of gravity here - you notice that a void (low density part) in a medium acts like a negative mass - as discussed by K^2 earlier.

Note:
You are distracting yourself by attaching the filled box to the unfilled one - it does nothing.
You appear to have failed to account for gravity acting in both (all) directions and the effect of gravity on the fluid is to create a pressure gradient in the fluid. A compressible fluid will also have a density gradient.

The void has a pseudo anti-gravity effect which tends to repel the surrounding denser medium.
(Technically, the dense medium is more attracted to itself than to the low density region.)

The void is in a container - so it does not change volume. This means that a pressure difference builds up between the void and the far ends of the container - both ends. Think of the water as trying to get away from the void, but it is stopped by the ends of the tube.

The effect is the same as a spirit level.

In a closed tube of water with a void in it, the low density box will just move until it reaches the middle of the tube. Equal mass (and, thus, gravity) and pressure from both directions. This is what happens in a spirit level ... the air bubble is the low-density container.

With very low viscosity and no turbulence, the void may oscillate about the equilibrium point if it starts out displaced like you show.

In a torus, the low density container does not do anything.
You can see what happens if you just loop the diagram - every position is the center.
In terms of forces - the pseudo anti-gravity acts in both directions equally, creating a pressure gradient so the highest pressure is diametrically opposite the void.

 

[Buckleymanor]

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 pressure is very low. I consider density the same everywhere. I don't compute temperature. Even there is friction the sum of force on object must be to 0. I computed a very small object in other case, I must wait a long time ! I compute only one layer of atoms between walls and object.

I compute force that attact water due to the lack of water in gas-container and I compute force of pressure due to water inside object on gas-container. Maybe it's not good to thinking like that. For me, I imagine an external fluid which don't attract (weight) and don't give pressure. I put an object in it (half water, half gas). Water-container will change the pressure around the air-container, but gas can't do the same. The net force in this case is the force of buoyancy I compute in program.

I'm not sure my program is good enough:

#include <stdio.h>
#include <math.h>


#define L 0.000000001 //Width of column
#define H 0.000001 //Height of column
#define r 0.5e-10 //radius of a molecule m
#define h 0.0000001 //height of solid
#define w 3e-26 //mass of a molecule kg

int main()
{
    long int i,j,k,l; //counter for loop
    double nbL, nbH, nbh, sumB, forceB, sumG, forceG, angle, length, o, c, force, d, rw;
    double w2 = pow((double)w,2.0); //mass*mass

    d = 2.0*r; //diameter of a molecule
    nbL = L/d; //number of molecules in width of column
    nbH = H/d; //number of molecules in height of column
    nbh = h/d; //number of molecules in height of solid
    sumB = 0.0; //sum of buoyancy forces
    sumG = 0.0; //sum of weight forces
    rw = (nbL-2)*nbh*w; //total mass kg

    printf("nbL=%lf\nnbH=%lf\nnbh=%lf\nd=%lf\nsumB=%lf\nw2=%e\nsumG=%lf\ntotal mass=%e", nbL, nbH, nbh, d, sumB, w2, sumG, rw);
    printf("-----------------------------------------------");


/////////////////////////////////////////////////////////////////////////////////
/////////// Loops for buoyancy
/////////////////////////////////////////////////////////////////////////////////
    for(i=0;i<(int)nbL;i++)
    {
        for(j=0;j<(int)nbH;j++)
        {
            for(k=0;k<(int)nbh;k++)
            {
                o = (double)i*d; // length of opposite of triangle
                c = H - (double)j*d - (double)k*d -d; // length of side or triangle
                length = sqrt( pow(c,2.0) + pow(o,2.0) );
                angle = atan(o/c);
                force = w2 / pow(length,2.0) * cos(angle);
                sumB += fabs(force);
                //printf("o=%e\nc=%e\nlength=%e\nangle=%e\nforce=%e\nsumB=%e\n", o, c, length, angle, force, sumB);
                //getchar();

            }
        }printf("i=%i / %i ******** sum = %e\n",i, (int)nbL, sumB);


    }
    forceB = 6.674e-11 * sumB*(L-2*d)/(d);
    printf("buoyancy force = %e\n", forceB);


/////////////////////////////////////////////////////////////////////////////////
/////////// Loops for weight
/////////////////////////////////////////////////////////////////////////////////

    for(i=0;i<(int)nbL;i++)
    {

        for(j=0;j<(int)nbH;j++)
        {
            for(k=0;k<(int)nbh;k++)
            {
                for(l=1;l<(int)nbL-1;l++)
                {
                    o = fabs( (double)i*d+r - (double)l*d+r -d);
                    c = H - fabs ((double)j*d + (double)k*d ) + d;
                    //if(c==0.0) printf("pb 0");
                    length = sqrt( pow(c,2.0) + pow(o,2.0) );
                    angle = atan(o/c);
                    force = w2 / pow(length,2.0) * cos(angle);
                    sumG += fabs(force);
                    //printf("o=%lf\nc=%lf\nlength=%lf\nangle=%e\nforce=%e\nsumG=%e\n", o, c, length, angle, force, sumG);
                    //getchar();

                }
            }
        }
        printf("i=%i / %i\n",i, (int)nbL);
    }
    sumG=6.674e-11 * sumG;


    printf("buoyancy force = %e\nattraction force = %e\n", forceB, sumG);





    return 0;
}

The net force on object is not 0. For me, it's possible to put it in a torus but I don't know how to compute torque.

http://imageshack.us/a/img46/5976/o13h.jpg

Or less than 0 for the object to move about and float.
A floating body displaces it's own weight of water so for it to move about within the water it can only be at the maximum the weight of the water it displaces or less than the weight of the water it displaces.
You could argue that the objects moment is different in that it's centre of mass is higher or lower at different points when it's floating around the taurus.If you examine closer though this also cancels out.

 

[Gh778]

@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 liquid.

http://imageshack.us/a/img208/7206/3a0p.jpg
The only pressure around object in it:

- pressure around gas-container = 0 because all is symmetric around it
- pressure around water-container different of 0: force of buoyancy I compute

The object is not attract by gas-container: force of weight I compute, why I find this force bigger ? because it's matrix of particles that attract a matrix of particles. In the buoyancy force it's a matrix that attract only a line of particules. For me, when particles attract particles all forces are canceled by themselves like spring can do. But what's cancel the buoyant force ?

 

[Simon Bridge]

You are still distracting yourself with a water-container attached to the gas-container.
Why do you keep doing that? It does nothing.

Do you think that the filled box will be repelled by the voided one (being attracted to the center of mass) - moving away, pulling the gas-container with it?

When the torus turn, with good rotational speed (good size, small height for object, etc) it can cancel pressure in the fluid, right ?

Wrong - it would add an extra pressure gradient to the fluid.

If necessary it's possible to have 2 object diametrycally opposed, like that no specific pressure from liquid.

There is always pressure from the fluid - that is because the fluid has a temperature above absolute zero.

The only pressure around object in it:

There is pressure everywhere in the fluid.

- pressure around gas-container = 0 because all is symmetric around it

No - the pressure around the gas container is the pressure of the surrounding fluid.
You mean there is no net pressure difference around the container - so zero force?
Also no true - the rotation forces a pressure gradient, with higher pressure at greater distance from the center of rotation.

- pressure around water-container different of 0: force of buoyancy I compute

The water container has the same density as the water - so it has neutral buoyancy.

The object is not attract by gas-container:

What object?
I've been thinking there is a water-filled container next to a gas filled container in a torus filled with water.

force of weight I compute, why I find this force bigger ?

In what way?

because it's matrix of particles that attract a matrix of particles. In the buoyancy force it's a matrix that attract only a line of particles. For me, when particles attract particles all forces are canceled by themselves like spring can do. But what's cancel the buoyant force ?

Remember - the buoyant force comes from a pressure gradient ... in the spinning torus, the pressure is greater along the outside than it is along the inside, so the net force on anything floating in it will point inwards. This is an unbalanced force - which is expected because the system is accelerating.

 

[Gh778]

Why do you keep doing that? It does nothing. Do you think that the filled box will be repelled by the voided one (being attracted to the center of mass) - moving away, pulling the gas-container with it?

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 pressure, just particles everywhere with the same density
- You put in this gas a point with big mass, it's a only a point. At this point is attached an object with vacuum in it
- the point change pressure around vacuum object, this give a force at the vacuum object

http://imageshack.us/a/img32/2773/vv89.jpg


Like all particles attrack by point are canceled by point I don't imagine the force that allow to have sum of force to 0. Like the point is a point, it can't receive difference of pressure because it has no volume. And like vacuum is vacuum it can't receive force from mass, mass can't attract vacuum. I'm sure vacuum receive a force due to the modification of pressure. So the point must receive a force from gas ? But how ?

Imagine the point is strong enough for attract gas just at the end of vacuum object, I can see the point attract gas, gas attract point and what ? how the force on vacuum object can be canceled ? Even the sphere is not a sphere (an ellipsoid or something like that) all forces are canceled themselves.

http://imageshack.us/a/img607/1548/v6yw.jpg

 

[berkeman]

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