How does liquid starts moving when there is Pascal's law

[Logic hunter]

Consider a conventional U-tube with both the vertical tubes having the same uniform cross section area A and the horizontal tube of length L, connecting those tubes containing an ideal liquid. Now the free surfaces in both the vertical tubes will be at the same height and will have pressure equal to P₀. Now if put a piston on anyone of the free surface and apply a force on it such that the pressure due to it on the surface of liquid just below it is F/A, then since at t=0 the liquid will still have 0 instaneous velocity so i can apply pascal's law according to which the pressure of each of the points will increase by F/A - P₀ (P₀ being the atompheric pressure.) So the pressure of the free surface in the other tube will also become F/A and on making FBD of any continuous liquid section the net force acting on it due to pressure will be same as the net force that was acting on it before the piston was used since pressure increased by same magnitude for all of the points i.e. net force would be zero so it will not have acceleration in any direction and won't start moving. So does this imply that pascals law would not hold true at this instant even if the liquid is at instantaneous rest. So what will happen? Also can i apply Bernoulli theorem once the liquid starts moving to analyse its properties ?

 

[scottdave]

If you are saying the other side of the tube (free end) is pushing up with F, has the downward force from atmospheric pressure changed? Will the net force still be nonzero?

 

[jambaugh]

I think you are neglecting the fact that the fluid undergoing an acceleration will have a pressure gradient. (unless an external force can be applied to the whole fluid as with gravity or MHD forces). It is no different than the question of how an extensive rigid block accelerates when a force is applied to one side.

I would add also that this is a limiting case for an idealized incompressible fluid which is pragmatically impossible. It implies and Pascal's law can be restated as the premise that the speed of sound in the medium is infinite.

[EDIT] Let me add that you get the same pressure gradient in a gravitational field... i.e. the deeper you go the more pressure in the fluid.

 

[Chestermiller]

Consider a conventional U-tube with both the vertical tubes having the same uniform cross section area A and the horizontal tube of length L, connecting those tubes containing an ideal liquid. Now the free surfaces in both the vertical tubes will be at the same height and will have pressure equal to P₀. Now if put a piston on anyone of the free surface and apply a force on it such that the pressure due to it on the surface of liquid just below it is F/A, then since at t=0 the liquid will still have 0 instaneous velocity so i can apply pascal's law according to which the pressure of each of the points will increase by F/A - P₀ (P₀ being the atompheric pressure.)

This is an incorrect statement of Pascal's law. Pascal's law does not say that the pressure changes by the same amount at all locations in a fluid. It says that, at a given location in a fluid, the pressure is the same in all directions. These are two very different things.

 

[Logic hunter]

This is an incorrect statement of Pascal's law. Pascal's law does not say that the pressure changes by the same amount at all locations in a fluid. It says that, at a given location in a fluid, the pressure is the same in all directions. These are two very different things.

refer to the second paragraph, in bold text.
https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/Pascals_principle.html

 

[vela]

So the pressure of the free surface in the other tube will also become F/A.

This would violate Newton's third law. The fluid can't push on the atmosphere with any more force than the atmosphere pushes on the fluid.

The problem you're running into is, as @jambaugh noted, that the approximation that the fluid is incompressible is not valid in this situation, so Pascal's principle doesn't quite hold. It takes time for the increase in pressure in one part of the fluid to be transmitted to other parts of the fluid—that is, the speed of sound in the fluid is finite. So force imbalances can exist within the fluid, and the fluid will move in response.

 

[Chestermiller]

This would violate Newton's third law. The fluid can't push on the atmosphere with any more force than the atmosphere pushes on the fluid.

The problem you're running into is, as @jambaugh noted, that the approximation that the fluid is incompressible is not valid in this situation, so Pascal's principle doesn't quite hold. It takes time for the increase in pressure in one part of the fluid to be transmitted to other parts of the fluid—that is, the speed of sound in the fluid is finite. So force imbalances can exist within the fluid, and the fluid will move in response.

This problem can certainly be done under the assumption that the fluid is incompressible.

If you push on a rigid (incompressible) block with a force, you can certainly determine that the acceleration of the block is given by a = F/m.

 

[Chestermiller]

refer to the second paragraph, in bold text.
https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/Pascals_principle.html

With all due respect to NASA, this is an incorrect statement of Pascal's principle. I stand by what I said.

 

[Chestermiller]

In your problem, the pressure on the top of the fluid the one of the U tube is $F/A$. The pressure on the top of the fluid on the other side of the U tube is $P_0$. So the driving force for acceleration is $F-P_0A$. The tangential acceleration of the fluid is just the force $F-P_0A$ divided by the mass of fluid.

 

[vela]

With all due respect to NASA, this is an incorrect statement of Pascal's principle. I stand by what I said.

NASA's statement agrees with the intro physics textbooks I've used.

 

[Chestermiller]

I'm going to assume that F/A is the gauge pressure (rather than the absolute pressure), so the absolute pressure at the top left is $P_0+F/A$ and the absolute pressure on the top right is $p_0$. In the vertical section to the left, a differential force balance on the fluid between elevation z and elevation $z + \Delta z$ is
$$[P(z+\Delta z)-P(z)]A+\rho gA\Delta z=\rho Aa\Delta z$$ where, a is the downward acceleration. If we divide this by $\rho A\Delta z$, we obtain:$$\frac{1}{\rho}\frac{dP}{dz}+g=a$$Similarly, for the vertical section on the right, we have $$[P(z)-P(z+\Delta z)]A-\rho g A\Delta z=\rho Aa \Delta z$$, where a is the upward acceleration on the right. So this reduces to $$-\frac{1}{\rho}\frac{dP}{dz}-g=a$$For the horizontal section of length L, we have $$-\frac{1}{\rho}\frac{dP}{dx}=a$$where a is the acceleration from left to right. If we integrate these equations along the path from the top of the left column to the top of the right column, we have
$$\int_H^0{\left[\frac{1}{\rho}\frac{dP}{dz}+g-a\right]dz}+\int_0^L{\left[\frac{1}{\rho}\frac{dP}{dx}+a\right]dx}+\int_0^H{\left[\frac{1}{\rho}\frac{dP}{dz}+g+a\right]dz}=0$$This integrates to $$-\frac{1}{\rho A}F-gH+aH+La+gH+aH=0$$or equivalently $$F=\rho g A (2H+L)a=ma$$where m is the total mass of liquid in the U tube.

 

[Chestermiller]

NASA's statement agrees with the intro physics textbooks I've used.

Well the Material Science and Continuum Mechanics textbooks I've used state Pascal's Law as: At a given location in a fluid, the pressure is equal in all directions (i.e., it is isotropic). So I stand by what I said. In the case of the problem at hand, as soon as the force is applied, the pressure will vary linearly with position along each section of the channel. it will never be the same everywhere. The linear variation in the three sections, however, will differ, depending on the direction of gravity and the direction of the acceleration. See my previous post.

The thought that the pressure ever increases at the interface with the atmosphere is silly. The pressure there is always atmospheric.

 

[vela]

so i can apply pascal's law…. So the pressure of the free surface in the other tube will also become F/A

Pascal's Law applies to enclosed fluids (the NASA page refers to a "confined" fluid), so that's another issue with your reasoning.

 

[Chestermiller]

Pascal's Law applies to enclosed fluids (the NASA page refers to a "confined" fluid), so that's another issue with your reasoning.

Like I said. I stand by what I said. Moreover, in the problem at hand, the fluid is not confined. So your version of Pascal’s law wouldn’t even apply to it.

 

[Chestermiller]

http://www.cns.gatech.edu/~predrag/courses/PHYS-4421-13/Lautrup/pressure.pdf

Check out p 26: This conclusion is part of broader theorem, called Pascal’s law, which says that the pressure in a fluid at rest is the same in all directions.

Actually, even in a fluid not at rest, the pressure (which represents the isotropic part of the stress tensor) is the same in all directions.

 

[vela]

Like I said. I stand by what I said. Moreover, in the problem at hand, the fluid is not confined. So your version of Pascal’s law wouldn’t even apply to it.

Which was my point. I was replying to the OP, not you, in that post.

Anyway, the name Pascal's Law apparently applies to both concepts.

https://www.britannica.com/science/Pascals-principle