Efficiency to move a curve shape


by Gh778
Tags: curve, efficiency, shape
Gh778
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#1
Feb22-13, 07:37 AM
P: 365
Fig 1 : the blue volume has a lot of very small spherical balls in it. Balls are under pressure from external system (weight or other), the potential energy is always the same because the blue volume is constant. When I move the blue volume from position 1 to position 2, I can understand that energy needed for move surface S2 is equal to the energy needed to move surface S1. The force on S1 is smaller but the distance to move is bigger than S2.

Fig 2 : now, in this case, the shape to move is curve. Like balls are in the blue volume there are forces like black arrows showing. In this drawing I drawn absolute forces from pressure and relative forces from the curve. The S1 and S2 surfaces have forces "up". So it seems the energy needed for move from position 1 to position 2 can recover energy, in this case where I lost energy ?
Attached Thumbnails
e1.png   e2.png  
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DaleSpam
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#2
Feb22-13, 08:00 AM
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Quote Quote by Gh778 View Post
So it seems the energy needed for move from position 1 to position 2 can recover energy, in this case where I lost energy ?
From the pressure and volume inside the blue balls. If you prevent their volume from changing then there is no net force and no energy to be recovered.

We do not discuss perpetual motion on this forum. Please don't wander that way in the discussion here.
Gh778
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#3
Feb22-13, 08:07 AM
P: 365
Don't worry, I'm interresting about efficiency not perpetual motion.
If I prevent the blue volume to change and move from pos1 to pos2, in this case, my black forces are wrong, could you explain at least this point ?

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#4
Feb22-13, 08:34 AM
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Efficiency to move a curve shape


Quote Quote by Gh778 View Post
If I prevent the blue volume to change and move from pos1 to pos2, in this case, my black forces are wrong, could you explain at least this point ?
You should draw a pair of free-body diagrams, one for a small element of the outer surface and the other for a small element of the inner surface. Next, solve for the motion as a function of the external pressure. Then use the condition that the volume is constant to fix the relationship between the external pressures on the outer surface and the inner surface. Then integrate that across the surface to get the net force.
Gh778
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#5
Feb23-13, 05:59 AM
P: 365
It's a little too difficult for me. I'm thinking an easier shape for understand where is my error. Imagine the black shape in water (new drawing), like that F1+F2+F3 = 0 (vectors), I'm ok with that. But with small balls (not molecules of water), I "see" for each ball in contact with the curve shape fb1+fb2 force, for me this force decrease F3. If fb1+fb2 exist for each ball in contact with curve part, it must be another force that cancel these but I don't find them.
Attached Thumbnails
e7.png  
DaleSpam
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#6
Feb23-13, 07:37 AM
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Draw a free-body diagram of one of the balls. Label all of the forces (don't forget the second row of balls if there is one and don't forget the weight of the ball). Now determine which forces are known and which are unknown (for simplicity, I would consider the force from the black object to be known).

That is the easy part. Once you have done that, think of all of the equations that you have about those forces. If you have as many knowns as unknowns then solve for all unknowns in terms of knowns. If you have more unknowns than knowns then solve for the most unknowns you can in terms of the remaining unknowns and the knowns.

Once you have done that you will either have exact values for fb1 and fb2 or you will have something like fb1 as a function of fb2.
Gh778
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#7
Feb23-13, 09:17 AM
P: 365
Can I consider balls small enough for use the law of pressure in liquid ?

F1 = (h2-h1)*g*σ
F2 = h1*g*σ
F3x = (h2-h1)*g*σ

with σ the density of balls, I consider only F3x because F3y can be cancel by a mechanical system and in this case F2 don't help/prevent the movement (this for reduce the problem).

We have only:

F1 = (h2-h1)*g*σ
F3x = (h2-h1)*g*σ

in this case how fb1+fb2 come in these equations ?
Attached Thumbnails
e8.png  
DaleSpam
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#8
Feb23-13, 11:35 AM
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Quote Quote by Gh778 View Post
Can I consider balls small enough for use the law of pressure in liquid ?
If you want to do that then you are not analyzing balls, you are analyzing liquid. That is certainly fine by me, but it didn't seem to be what you were trying to do.

Quote Quote by Gh778 View Post
in this case how fb1+fb2 come in these equations ?
They don't. You did a free-body diagram on the black object, not the ball. If you want to analyze fb1 and fb2 then you need to do a free-body diagram on the ball since the ball is the object that fb1 and fb2 act on.
Gh778
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#9
Feb23-13, 01:37 PM
P: 365
For simplify the study, I imagine the black last shape but gravity is not down, it is through the screen and I have only one layer of balls. This is the weight of ball that give pressure (ball can be deformable).

I hope the FBD is good ? For me, each force has the same value and depend of weight of ball and the deformation.
Attached Thumbnails
fbd.png  
DaleSpam
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#10
Feb23-13, 08:29 PM
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The diagram looks good, but you can immediately see that the forces are not all equal.
Gh778
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#11
Feb24-13, 01:44 AM
P: 365
"not equal", the sum is not zero ?
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#12
Feb24-13, 07:46 AM
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No, they are not equal precisely because their sum IS zero and their directions are not symmetric.

Suppose that you have three forces which sum to zero and they are separated by 120 deg each. In that case, they must all be equal magnitude. However, suppose that instead of being separated by 120 deg the first two are separated by 180 deg and then the third is 90 deg from the others. Then the first two must be equal magnitude and the third must be 0.
Gh778
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#13
Feb24-13, 08:16 AM
P: 365
ok, so it's Fsb or Fb3 that must be lower/greater, but why, can you explain ?
DaleSpam
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#14
Feb24-13, 02:55 PM
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Quote Quote by Gh778 View Post
ok, so it's Fsb or Fb3 that must be lower/greater, but why, can you explain ?
From what you have drawn Fsb and Fb3 look like they are colinear, so let's use x to represent the direction of Fb3. Fb1 and Fb2 each have a component of their force along x, specifically in the negative x direction, and Fsb is entirely in the negative x direction. Therefore, in order for their sum to be 0 we must have |Fb3|>|Fsb|.


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