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Pressure and attraction

  1. May 14, 2012 #1
    I would like to know where is the force that prevent the system to turn around. I use balls which are attrack with springs, the force of the spring is like 1/d or 1/d² like gravity for example. I see the force FR in the system, like the circle can only put a force in the center C I don't see the force which cancel the force FR, have you an idea ?
     

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  3. May 17, 2012 #2
    Nobody can help me ? I don't need all the calculations only what force cancel FR ?
     
  4. May 17, 2012 #3

    Ken G

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    I don't think anyone understands why you imagine there being any forces in the direction FR. Are the balls pressing against the circular wall because the springs are compressed? If so, then the balls would tend to be pushed to the right along FR, but the base that they are anchored to would tend to be pushed downward. That's the opposite direction of rotation, so there's no net angular momentum there.
     
  5. May 17, 2012 #4
    It's the contrary, the springs want to attrack. At bottom another force exist ?
     
  6. May 17, 2012 #5

    russ_watters

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    What is preventing them from contracting? As usual, your drawing is very inclear.
     
  7. May 18, 2012 #6
    Sorry for the poor drawing :( I don't draw all balls. Imagine with a lot of balls, the pressure is like the pressure in water. We can take the pressure of springs like 1/d or 1/d² for have the same pressure in water. Here, the difference it's we can turn the system which attrack (it's not possible with water, we can't turn Earth on the system). The sides retain balls. The circular side can only put a radial force (this side is fixed).

    If I add the radius:

    At bottom, we have 2F with a mean radius of R/2 so the momentum is F*R

    At right, we have 0->2F this give a momentum of 2*F*R/3

    Attraction give: integrate(2*F/R*sqrt(R²-x²)*x dx) from 0 to R this give 2*F*R/3

    The sum is 0.333*F*R

    old part of message:
    If I think like water, at top pressure = 0, at bottom pressure = P. FR at right. 2FR at bottom, the pressure depend only of the high. The total weight is 2FR-0.215FR (the 1/4 of the surface of the circle compared of a square surface), this is this force which attrack "water" or balls. At top we have 0.215FR but this force is combined with left force and don't interfere in the system because this is a radial force. So at right we have FR, at bottom we have 0.215FR. The total moment is not equal to 0. Sure I made mistake but where ?
     
    Last edited: May 18, 2012
  8. May 19, 2012 #7

    Ken G

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    I can't follow the mistake because I can't follow the argument. It doesn't help that you are doing things like calling F*R a "momentum", when that actually has units of energy not momentum. Could you possibly imagine a much simpler system that exhibits the surprising property you are wondering about, and just describe the system and ask what such a system would do? It's easier for us to calculate the forces in some well-described system that to find errors in a calculation that basically just doesn't make any sense to us.

    Maybe it would help if I said that if you set up a complicated system of springs, it is not required that the forces add up to zero at every point in the system. All that is required is that all the forces, at all the points, add up to zero in total, but not at every point. Also, the system cannot start rotating as a whole, but some parts can start rotating one way as other parts rotate the other way. Also, these statements apply only to an isolated system-- if the system is fixed to something external that is taken to be unmoving, then you can get both net forces and net rotation in that system.
     
  9. May 19, 2012 #8
    Ok, I will try to explain a more simple system.

    See the drawing. I try to find the torque on the blue part. The blue color is water, I have put the forces from pressure. And now, imagine you put the source that attrack water in the system (like Earth for example, but smaller ! this is for that I have put springs for simulate Earth attraction). I have show the attraction forces too on the drawing. I have choose a small radius for see where is the center of rotation but imagine a big radius, the blue part is like a triangle. When I calculate the torque on the blue system I don't find 0.


    [itex]\frac{2}{d}*\int_0^d (R+x)*\frac{F*x}{d} {d}x-\frac{1}{d}\int_0^d(R+x)*F {d}x[/itex]

    with R the big radius, d the lenght of a side of triangle, F the basic force.

    Even I increase R, the result is always 1/6. I don't find the momentum equal to zero. And without calculations, I can understand because 2F is apply when the radius is bigger this must give a difference. Sure the blue area is not a perfect triangle but with R enough I find always 1/6.
     

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  10. May 19, 2012 #9

    Ken G

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    The way you have set up the attraction forces, there is no reason to expect the torque to be zero. Why do you think it should be?
     
  11. May 19, 2012 #10

    russ_watters

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    Sounds like you think pressure can be different in different directions for the same point. It can't. As well, when two objects are in contact with each other, neglecting friction, the force between them can only be perpendicular to the contact plane. In other words, your forces here will always be directly away from the center of the circle. No torque.

    It appears that these spring-ball devices are contained in a wedge-shaped container inside the circle. This wedge shaped container has forces pressing against the side walls, but no torque.
     
  12. May 19, 2012 #11

    Ken G

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    The pressure forces cannot produce torque, but the mysterious "attraction" forces here can. These forces are shown along the bottom of the figure, and of course must come with action/reaction forces on the sphere. Those reaction forces on the sphere seem like they would torque the sphere, so we should expect there to be a compensating torque on the blue "water", just as you find, though you do seem to be doing something strange with the pressure, which should not be 2F in one direction and F in the other.
     
  13. May 20, 2012 #12
    the system to rotate it's the blue part, this system has forces from pressure and forces from attraction (spring or gravity). My drawing is false ?

    why ?

    this is not a mysterious, it's springs that attrack small balls, these forces are not like that ?

    where i'm wrong in my drawing ?
     

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  14. May 20, 2012 #13

    Ken G

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    There's just no need for the pressure forces, the torque comes from the springs.
    Pressure can't produce net torque, or even net force, on an enclosed liquid. That's because pressure will produce an equal force everywhere on the enclosing boundary, and that creates a net force that adds up to zero.
    What's mysterious is why you want to combine spring forces on balls with pressure. The pressure is not doing anything important in your scenario, the torque is coming from the springs and is equal and opposite to the torque on the sphere (at the other end of the springs).
    In my opinion, you are wrong in thinking that the springs should not produce torque on the balls. From your diagram, I would expect that they would.
     
  15. May 23, 2012 #14
    I would like to show something easier to understand. In this case F1=-F2, all forces of attraction cancel themselves. But the force from FP exist always. Sure if I use gravity, FP it's the weight but with springs I never change the potential energy and FP exist always ? Blue color is water or balls.

    Why there is a relation between attraction force and pressure when balls are not round but like an ellipse ? The radius is not the same in x or y axis and a difference of momentum exist.
     

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