The application of Newton's laws to fluid equilibrium

  • #1
Florian Geyer
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TL;DR Summary
Why is fluid equilibrium based on Newton's first and third laws?
Hello respected members,

I have been studying Sears and Zemansky's university physics, and I have encountered the following statement.
We begin our study with fluid statics, the study of fluids at rest in equilibrium situations. Like other equilibrium situations, it is based on Newton’s first and third laws. We will explore the key concepts of density, pressure, and buoyancy.
I could not understand it well, so my question is: Why is fluid equilibrium based on Newton's first and third laws?

Here is my try:
- Newton's first law: like the case of rigid body equilibrium the net external force exerted on the fluid must be zero, so that the fluid remains in its state of liner motion.
- Newton's third law: I am really lost here nor I could understand why this law is important in this regard.
- Also, may I ask what is the equivalent of the condition of zero torque like the case of rigid body equilibrium? you may say that there is such thing as torque in the case of fluids since there is no "lever arm" but how about the case of vortexes the net force -as I understand- is zero, but the fluid still in motion, or shall we consider a fluid with vortexes as a fluid in a state of equilibrium?
 
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  • #2
Florian Geyer said:
TL;DR Summary: Why is fluid equilibrium based on Newton's first and third laws?

Hello respected members,

I have been studying Sears and Zemansky's university physics, and I have encountered the following statement.

I could not understand it well, so my question is: Why is fluid equilibrium based on Newton's first and third laws?

Here is my try:
- Newton's first law: like the case of rigid body equilibrium the net external force exerted on the fluid must be zero, so that the fluid remains in its state of liner motion.
- Newton's third law: I am really lost here nor I could understand why this law is important in this regard.
- Also, may I ask what is the equivalent of the condition of zero torque like the case of rigid body equilibrium? you may say that there is such thing as torque in the case of fluids since there is no "lever arm" but how about the case of vortexes the net force -as I understand- is zero, but the fluid still in motion, or shall we consider a fluid with vortexes as a fluid in a state of equilibrium?
You are overthinking this.
There are no accelerations, so the second law is trivially satisfied. There are no reasons to ”ignore” the others laws.
I am not quite sure what you are meaning, but a vortex involves a moving fluid which is not in STATIC equilibrium.
 
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  • #3
Florian Geyer said:
- Newton's third law: I am really lost here nor I could understand why this law is important in this regard.
When a liquid is contained in a reservoir, pairs of forces (action-reaction) exist between both.

pressure-vessels.jpg


maxresdefault.jpg
 
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  • #4
Florian Geyer said:
I could not understand it well, so my question is: Why is fluid equilibrium based on Newton's first and third laws?
Have you tried reading the chapter beyond that introductory statement?
 
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  • #5
A.T. said:
Have you tried reading the chapter beyond that introductory statement?
Yes, I am studying it right now.
I think I do not understand you appropriately, do you mean:
- Introductory statements in each chapter are arbitrary and does not have any importance, thus we do not need to read them.
- Or I shall read the entire chapter before I ask any question about any part in it?

May you explain please?
 
  • #6
I have reached to the following conclusions:
1- Newton's first law: it means the net force on the fluid is zero so there is no movement, which was the obvious part.
2- Newton's third law explains how the fluid come into its current shape by the force (from the fluid) and the counter-force by the walls of the container, or the force due to atmospheric pressure in case of pen liquid... I came to this understanding with the help of Mr. Lnewqban in his previous respond.
3- With the regard of the condition equivalent to the condition of zero torque in the case of rigid body, I reached to the following conclusion:
we cannot apply the condition of zero torque, because it suggests that the whole body should be affected as a whole, but each part of the body (volume elements) should not be affected separately. which is certainly not the case of fluids...
Thus we deal with each part of the fluid and apply Newton's laws on each volume element of the fluid instead of dealing with it as a whole... I think this can solve the problem of internal motion in the fluid which -as one of the respected members Mr. (Frabjous) has suggested cannot be considered as a fluid equilibrium state.

Now is my previous understanding correct?
 
  • #7
Florian Geyer said:
- Also, may I ask what is the equivalent of the condition of zero torque like the case of rigid body equilibrium? you may say that there is such thing as torque in the case of fluids since there is no "lever arm" but how about the case of vortexes the net force -as I understand- is zero, but the fluid still in motion, or shall we consider a fluid with vortexes as a fluid in a state of equilibrium?

Note that zero-torque is just an application of Newton's laws (specifically for rigid bodies), it by itself is not a new law. There is no equivalence in Fluid mechanics (i.e. in the Navier-Stokes equations, which is just Newton's second law applied to a fluid).

Since a vortex requires energy to make and even to keep on existing, the net force on the fluid that generates a vortex is not zero (also, it is of course not static).
 
  • #8
A vortex is rotating, which means that it is accelerating radially, which means it is not in static equilibrium.
 
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  • #9
Chestermiller said:
A vortex is rotating, which means that it is accelerating radially, which means it is not in static equilibrium.
Aha... I have forgotten the idea that acceleration can also mean a change in direction. Yes, yes. However, I think there is some argumentation can be added here so that I can reach to the final conclusion.
How about if we take the following scenario: a pool of water in which there are two opposite forces acting on each side of the pool towards the center -I think this will be more obvious if I uploaded a sketch- in this case the resultant force is zero on the fluid, but still the fluid is not in a state of equilibrium as I think.Now, I think the first two conditions of equilibrium in fluids (Newton's first and third law) are not enough, because of the previous scenario... I have two ideas but do not which one is more accurate.
1- We shall add the condition that the pressure is the same in all parts of the fluid.
2- We shall apply Newton's laws on each little segment of the fluid.

Which one is more accurate and why?
 

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  • #10
Arjan82 said:
Note that zero-torque is just an application of Newton's laws (specifically for rigid bodies), it by itself is not a new law. There is no equivalence in Fluid mechanics (i.e. in the Navier-Stokes equations, which is just Newton's second law applied to a fluid).

Since a vortex requires energy to make and even to keep on existing, the net force on the fluid that generates a vortex is not zero (also, it is of course not static).
- First, I have not said that zero-torque is a new law... it is just a condition for equilibrium in the case of rigid bodies.
- Second, I do not know what is the relation between energy and equilibrium, please may you explain to me further?
 
  • #11
Florian Geyer said:
- First, I have not said that zero-torque is a new law... it is just a condition for equilibrium in the case of rigid bodies.
I didn't say you did. I just wanted you to note that since you were asking for an equivalence. Why would you ask for an equivalence if there is nothing to justify that?

Florian Geyer said:
- Second, I do not know what is the relation between energy and equilibrium, please may you explain to me further?
You write that you've understood the net force on a vortex is zero. Zero force means zero energy added to the flow. This cannot hold since it requires a force to keep a vortex in existence.
 
  • #12
Florian Geyer said:
Now, I think the first two conditions of equilibrium in fluids (Newton's first and third law) are not enough, because of the previous scenario... I have two ideas but do not which one is more accurate.
Newton’s laws are not conditions for equilibrium. They also hold when things are not in equilibrium.
 
  • #13
Arjan82 said:
I didn't say you did. I just wanted you to note that since you were asking for an equivalence. Why would you ask for an equivalence if there is nothing to justify that?
Well, but I still struggle to find a full list of conditions for equilibrium in fluids, which can satisfy any scenario I think about. Thus I am trying to find them... and I thought comparing the two situation is a good way to do this (or at least to help in doing this).
Arjan82 said:
You write that you've understood the net force on a vortex is zero. Zero force means zero energy added to the flow. This cannot hold since it requires a force to keep a vortex in existance.
Aha, well I did not include the concept of energy since you can imagine some cases where the energy exerted on the system is not zero, and still the system is in a state of equilibrium.
How about if you exerted a force on each surface of a book? in this case you exert energy but the system as a whole is in equilibrium. May you explain if I made a mistake in this regard?
 
  • #14
Frabjous said:
Newton’s laws are not conditions for equilibrium. They also hold when things are not in equilibrium.
mmm, as far as I understand they hold in both cases... or more generally they explain both static and dynamic states... However, how about if I asked what are the conditions of equilibrium in fluids?
 
  • #15
Florian Geyer said:
mmm, as far as I understand they hold in both cases... or more generally they explain both static and dynamic states... However, how about if I asked what are the conditions of equilibrium in fluids?
In terms of freshman physics, it is when the fluid is at rest.
 
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  • #16
Frabjous said:
In terms of freshman physics, it is when there is no motion.
Aha, now I understand. I will try to read about this from the internet. I have some good understanding of vector analysis, and plasma phys. with some PDEs... Or maybe I will delay this until another time.
 
  • #17
Arjan82 said:
Note that zero-torque is just an application of Newton's laws (specifically for rigid bodies), it by itself is not a new law. There is no equivalence in Fluid mechanics (i.e. in the Navier-Stokes equations, which is just Newton's second law applied to a fluid).
But the Navier-Stokes equation is Newton's equation of motion for a fluid element:
$$\rho a_j=\rho \mathrm{D}_t v_j = \rho (\partial_t v_j +v_k \partial_k v_j)=-\partial_j \vec{p} + \partial_k [\eta (\partial_k v_j + \partial_j v_k -2/3 \partial_j \partial_k v_k)] + \partial_j (\zeta \partial_k v_k).$$
The forces per unit volume on the right-hand side are due to the pressure and the shear- and bulk viscosity. The latter are dissipative corrections to ideal fluid dynamics.
 
  • #18
Florian Geyer said:
Aha, well I did not include the concept of energy since you can imagine some cases where the energy exerted on the system is not zero, and still the system is in a state of equilibrium.

You are right, I was imprecise in my wording. A fluid at rest can have normal forces applied to it (e.g. to counteract gravity). This means no energy has to be exerted. But in case of a vortex you will always have shear forces. A fluid cannot hold a shear force without continuous deformation, i.e. movement. So a vortex (and many other flows) will always need energy to maintain. This is the picture I had in mind.
 
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