Finding Angle Between Fluid & Horizontal: Clarified

[brochesspro]

I have been taught that in order to find the angle made by a liquid in a container with the horizontal, we use the fact that the free surface is perpendicular to the net force vector acting on the fluid (including the pseudo force vector for an accelerating container), however, if say the container is moving horizontally with an acceleration, this would lead to the conclusion that the free surface makes a right angle with the floor/horizontal, which is clearly wrong. Where am I going wrong? I think that I make a mistake in the "net force" part of my above statement. Please help me in clarifying my misconception. Thank you.

 

[Baluncore]

Where am I going wrong?

The net force results from the vector sum of the sideways acceleration, and the vertical acceleration due to gravity.

 

[brochesspro]

The net force results from the vector sum of the sideways acceleration, and the vertical acceleration due to gravity.

But what about the normal reaction?

 

[jbriggs444]

But what about the normal reaction?

For purposes of

net force vector acting on the fluid (including the pseudo force vector for an accelerating container)

we are only considering forces acting within the body of the fluid. We are not considering forces from the walls of the container or from neighboring parcels of fluid.

Consider the difficulty that would arise from trying to determine the angle of the surface of a stationary fluid under ordinary gravity. If we were to consider the normal reaction from the container walls, the net force on the fluid (or on any parcel within the fluid) would be zero. What angle does the zero vector make with the horizontal?

 

[A.T.]

I have been taught that in order to find the angle made by a liquid in a container with the horizontal, we use the fact that the free surface is perpendicular to the net force vector acting on the fluid (including the pseudo force vector for an accelerating container),

A better formulation would be:

The free surface is perpendicular to the sum of all inertial/pseudo forces in the rest-frame of the container, which result from its acceleration relative to free fall (thus including gravity).

 

[brochesspro]

Consider the difficulty that would arise from trying to determine the angle of the surface of a stationary fluid under ordinary gravity. If we were to consider the normal reaction from the container walls, the net force on the fluid (or on any parcel within the fluid) would be zero. What angle does the zero vector make with the horizontal?

That was my point, but the problem is that it just seems like a convenience at this point, whether to count normal forces or not.

The free surface is perpendicular to the sum of all inertial/pseudo forces in the rest-frame of the container, which result from its acceleration relative to free fall (thus including gravity).

I do not understand this formulation, what I did understand was (based on solving various problems) that the free surface is perpendicular to the normal force vector (alternatively the vector sum of all the forces on the liquid in the frame of the container save for the normal force.)

 

[jbriggs444]

That was my point, but the problem is that it just seems like a convenience at this point, whether to count normal forces or not.

I do not understand this formulation, what I did understand was (based on solving various problems) that the free surface is perpendicular to the normal force vector (alternatively the vector sum of all the forces on the liquid in the frame of the container save for the normal force.)

It is more useful to consider the body forces right here rather than the normal forces over there. And there. And over there as well. The shape of the container ought not enter in.

That is particularly important when considering curved fluid surfaces. Such as the fluid surface in a spinning bucket.

 

[A.T.]

I do not understand this formulation, what I did understand was (based on solving various problems) that the free surface is perpendicular to the normal force vector (alternatively the vector sum of all the forces on the liquid in the frame of the container save for the normal force.)

This is also correct, but more difficult to apply.

 

[brochesspro]

I do not understand this formulation, what I did understand was (based on solving various problems) that the free surface is perpendicular to the normal force vector (alternatively the vector sum of all the forces on the liquid in the frame of the container save for the normal force.)

Can we prove this though? I just find it difficult to visualise, so any aid is appreciated.

 

[brochesspro]

This is also correct, but more difficult to apply.

I actually feel that this is easier, but just in case, could you explain your formulation?

 

[A.T.]

I actually feel that this is easier, but just in case, could you explain your formulation?

For your method you have to know the full force distribution between the walls and fluid. And then you have to integrate over it.

For mine you just look at the acceleration of the container relative to a free falling reference frame. This is also what an accelerometer attached to it measures. The surface is orthogonal to that acceleration.

 

[Steve4Physics]

FWIW, here’s my favourite way to deal with these sorts of problems.

One way of thinking about what happens inside an accelerating system is to take a leaf out of Einstein’s book and use the ‘equivalence principle’. (This is much easier than it sounds!)

In essence, if you are accelerating in some direction, what you experience is indistinguishable from a gravitational field of the same magnitude in the opposite direction.

For example, if you are accelerating at 5.0m/s² to the left, this is indistinguishable by you from being subject to a gravitational field of 5N/kg (≡5m/s²) to the right.

If you are also experiencing the earth’s gravitational field (9.8N/kg downwards) then overall, it seems as if you are not accelerating but are experiencing a net gravitational field of $\sqrt{9.8^2 +5.0^2} = 11N/kg$ at an angle to the vertical of $tan^{-1} \frac{5.0}{9.8} = 27^o$. The surface of any fluid accelerating with you will be perpendicular to this net field.

And you can also claim that you solved the problem using General Relativity!

Of course, this is necessarily equivalent to other possible approaches (e.g. see @A.T.'s).

 

[Lnewqban]

... however, if say the container is moving horizontally with an acceleration, this would lead to the conclusion that the free surface makes a right angle with the floor/horizontal, which is clearly wrong....

That would be correct in the total absence of the acceleration of gravity.
For a rectangular container being accelerated by an external force in a direction parallel to its bottom surface, you will see exactly what you see for liquid inside a static container on Earth, only that all would be rotated 90°, including a free pressure and a hydrostatic pressure gradient within the liquid.

If our contained liquid is under the influence of two different accelerations (gravity and lateral horizontal), having orthogonal directions, the angle of the surface, as well as the planes of similar hydrostatic pressures, should be within those two extremes (perfectly horizontal or perfectly vertical).
Regarding the magnitude of each acceleration, the formed angle should be equal to the tangent of the ratio of both accelerations.

When the magnitude of the lateral horizontal acceleration of the container reaches 1 g (a = g), that surface angle should be 45°.

 

[brochesspro]

Hey, I am sorry, but I have been swamped lately, so I have not been able to reply to any of your guys' posts. Thank you for all the suggestions. They were fantastic and were all incredibly helpful. I think I have understood why we do not consider the normal force while calculating the angle of the free surface. We look at everything with respect to the container.

Taking the simplest case, the forces on the liquid will be the force of gravity, the normal reaction and the pseudo force vector. We know that the free surface will be perpendicular to the normal reaction vector. We also know that the fluid will be at rest with respect to the container. So, we find the resultant of the force of gravity and the pseudo force and so, the free surface will be perpendicular to the resultant. Am I correct?

 

[Lnewqban]

Consider as well that planes of same static pressure will be always parallel to the surface of the fluid, regardless of the angle that it adopts respect to the horizon.
Each differential volume of liquid tends naturally to take a position of balance respect to the surrounding ones, if there is no continuos movement or flow.

Please, see:
https://www.me.psu.edu/cimbala/Learning/Fluid/Rigid_body/rigid_body.htm