Force acting on one side of a fluid

[palaphys]

Homework Statement

As shown in the figure below, there is a beaker of radius R . Water is filled in the beaker up to a height h . The density of water is ρ , the surface tension of water is T and the atmospheric pressure is P_0 . Consider a vertical section ABCD of the water column through a diameter of the beaker. What is the magnitude of the force on the water on one side of this section by the water on the other side of this section ?:

Relevant Equations

P= pgh

My attempt:
Using the equation to find excess pressure across two surfaces, we get
$P_{inside}= P_0 + S( 1/r_1 + 1/r_2 )$ where r1 and r2 are the curvatures of the cylinder and the surface perpendicular to it( so infinite curvature)

hence, $P_{inside}= P_0 + S/R= P_{AB}$

however, this varies with depth, and it is better to say that the expression for P_inside represents the pressure at the surface AB (see figure)

Hence, the pressure at CD should be,
$P_{CD} = P_0 + \rho gH + S/R$

now as is force is acting on the projected area of $A= 2RH$, we take the average pressure on the area to find the net force:

$P_{avg} = (P_{CD} + P_{AB} )/2 = P_0 + S/R +\rho g H$

And multiplying by the area to get the net force, I get

$F_{net} = 2P_ 0 RH + 2SR + \rho g RH^2$

note: I have assumed surface tension of water as S.
however, my answer closely matches option B (which is the right answer),but differs by 4SH.
Why is there this difference? I have not included surface tension as a force as it is already taken into account by the excess pressure.

 

[Chestermiller]

For the fluid in one half of the beaker, the surface tension acts within the upper surface in the direction normal to AB. The magnitude of the surface tension force is 2RT, and is pointing in the opposite direction of the pressure force. The radius of curvature of the upper surface is infinite, so the pressure above the upper surface is equal to the pressure just below the upper surface.

 

[palaphys]

The radius of curvature of the upper surface is infinite, so the pressure above the upper surface is equal to the pressure just below the upper surface.

https://en.wikipedia.org/wiki/Laplace_pressure
but using the equation (see above link), I get $P_ {just below upper surface} = P_0 + S/R$ ..

 

[Chestermiller]

https://en.wikipedia.org/wiki/Laplace_pressure
but using the equation (see above link), I get $P_ {just below upper surface} = P_0 + S/R$ ..

Like I said, the radius of curvature of the upper surface is infinite.

 

[palaphys]

Like I said, the radius of curvature of the upper surface is infinite.

still , there is another circular surface (of the cylinder, with radius R)

 

[Lnewqban]

$P_{avg} = (P_{CD} + P_{AB} )/2 = P_0 + S/R +\rho g H$

It seems to me that your error is excluding the value of the atmospheric pressure from the total pressure acting at the bottom.
Perhaps the last term should be $P_0\rho g H$

 

[Chestermiller]

still , there is another circular surface (of the cylinder, with radius R)

In your equation, the radii are the local principal radii of curvature of the surface, not of the container

 

[Chestermiller]

It seems to me that your error is excluding the value of the atmospheric pressure from the total pressure acting at the bottom.
Perhaps the last term should be $P_0\rho g H$

This is not correct.

 

[palaphys]

In your equation, the radii are the local principal radii of curvature of the surface, not of the container

ahh okay. So that means I have to include an additional surface tension force.. correct?

 

[Chestermiller]

ahh okay. So that means I have to include an additional surface tension force.. correct?

The surface tension has no effect on the pressure distribution within the beaker. The force from hydrostatic pressure is independent of the surface tension force.

The force ascribable to the hydrostatic pressure distribution is $$(2Rh)\left(P_0+\frac{\rho gh}{2}\right)$$The force ascribable to the surface tension occurs on the upper surface, and is equal to $$(2R)S$$ (in the opposite direction of the pressure force). So the total force exerted by the fluid on the left half of the beaker on the fluid in the fluid in the right half of the beaker is $$(2Rh)\left(P_0+\frac{\rho gh}{2}\right)-(2R)S$$

 

[Steve4Physics]

Is there a mistake/omission in the original question?

There seems to be an unstated assumption that the liquid-beaker contact-angle is 90$^{\circ}$, i.e. there is no curved meniscus where the liquid’s surface meets the beaker. Without this assumption there would be additional tension forces on sides BC, CD and DA.

Edit - typo'.

 

[palaphys]

The surface tension has no effect on the pressure distribution within the beaker. The force from hydrostatic pressure is independent of the surface tension force.

The force ascribable to the hydrostatic pressure distribution is $$(2Rh)\left(P_0+\frac{\rho gh}{2}\right)$$The force ascribable to the surface tension occurs on the upper surface, and is equal to $$(2R)S$$ (in the opposite direction of the pressure force). So the total force exerted by the fluid on the left half of the beaker on the fluid in the fluid in the right half of the beaker is $$(2Rh)\left(P_0+\frac{\rho gh}{2}\right)-(2R)S$$

yes, this was what I meant. I wanted to say that when we take the force on one of the cylindrical segments, then we include surface tension apart from pressure.
However, you say:
The surface tension has no effect on the pressure distribution within the beaker. The force from hydrostatic pressure is independent of the surface tension force.

Take the case of a bubble though. If we go through the derivation for excess pressure inside a bubble, it is actually due to the surface tension force. Then how can it be said that the surface tension force has no effect on pressure?

 

[palaphys]

Is there a mistake/omission in the original question?

There seems to be an unstated assumption that the liquid-beaker contact-angle is 90$^{\circ}$, i.e. there is no curved meniscus where the liquid’s surface meets the beaker. Without this assumption there would be additional tension forces on sides BC, CD and DA.

Edit - typo'.

I don't think so, this is a JEE advanced question picked right from the official paper so must be right, I think what you are saying is already depicted in the diagram..

 

[Steve4Physics]

If you rely on the diagram - which shows no curved meniscus - then yes. But note that 'official' examination questions do occasionally have flaws - it happens! Just my opinion though.

For information, it may be worth adding that $S( 1/r_1 + 1/r_2 )$ (from Post #1) is inappropriate here. It is called the Laplace pressure and is the pressure-difference across the boundary-surface between two different fluids.

It is not helpful in this problem as ABCD is not a boundary-surface: it has the same fluid on both sides and, anyway it is flat – so $r_1=r_2=\infty$.

 

[Steve4Physics]

I don't think so, this is a JEE advanced question picked right from the official paper so must be right, I think what you are saying is already depicted in the diagram..

If you rely on the diagram - which shows no curved meniscus - then yes. But note that 'official' examination questions do occasionally have flaws - it happens! Just my opinion though.

For information, it may be worth adding that $S( 1/r_1 + 1/r_2 )$ (from Post #1) is inappropriate here. It is called the Laplace pressure and is the pressure-difference across the boundary-surface between two different fluids.

It is not helpful in this problem as ABCD is not a boundary-surface: it has the same fluid on both sides and, anyway it is flat – so $r_1=r_2=\infty$.

Edit - typo's fixed.

 

[palaphys]

If you rely on the diagram - which shows no curved meniscus - then yes. But note that 'official' examination questions do occasionally have flaws - it happens! Just my opinion though.

For information, it may be worth adding that $S( 1/r_1 + 1/r_2 )$ (from Post #1) is inappropriate here. It is called the Laplace pressure and is the pressure-difference across the boundary-surface between two different fluids.

It is not helpful in this problem as ABCD is not a boundary-surface: it has the same fluid on both sides and, anyway it is flat – so $r_1=r_2=\infty$.

Edit - typo's fixed.

got it thanks.

 

[haruspex]

The question asks for the net force each half exerts on the other.

Surface tension results from cohesive forces between the molecules of the liquid. So it acts throughout the volume, but is usually only evident at a surface.
But of course, when the molecules get too close together these are balanced by repulsive forces, so other than the pressure, as calculated the usual way, there is no net force between the two halves.

This applies at the surface too. Yes, there is extra horizontal attraction because of the absence of water molecules above, but having pulled the molecules in the surface a bit closer together it is balanced by an increase in the repulsive forces.
In short, there should not be a surface tension component in the answer.

 

[Steve4Physics]

This applies at the surface too. Yes, there is extra horizontal attraction because of the absence of water molecules above, but having pulled the molecules in the surface a bit closer together it is balanced by an increase in the repulsive forces.
In short, there should not be a surface tension component in the answer.

Is there something wrong with the following argument?

1. We start with a beaker of liquid (Fig.1). ‘L’ is the liquid to the left of AD and ‘R’ is the liquid to the right of AD:

2. Using magic, we instantaneously remove R (Fig.2):

3. The surface tension will cause the corner of L becomes curved (Fig. 3):

(Admittedly this situation will only last for a short time until bulk flow of the liquid becomes significant but we could be using a viscous liquid in an environment with a low gravity/low atmospheric pressure.)

4. The removal of R allowed the liquid at the corner of L to move left, to form the curve. This implies that before removal, R was exerting a force (acting right) on the liquid surface at the corner of L. The magnitude of this force is the surface tension term (2RT).

(I won't be able to reply to any comments for some hours as it's past [edit - whoops, I mean 'passed'] my bedtime.)

 

[haruspex]

Is there something wrong with the following argument?

Seems a bit of, ahem, a stretch to me.
One possible mistake is in thinking of surface tension as an actual unresolved tension in the surface layer, as would apply in the surface of a solid if that surface were cooled. In a liquid, any such tension would be resolved by the movement of the molecules. (The density of the surface layer must be a bit more than in the body.)
As an analogy, consider two magnets, static in space, resting in contact at opposite poles. There is no net force on either, so neither can be exerting a net force on the other.
Now apply equal and opposite forces F to the far ends of each to represent the pressure in the water. Since there is still no net force on either, they must each exert a repulsive force F on the other.

 

[Steve4Physics]

Seems a bit of, ahem, a stretch to me.

Haha!

One possible mistake is in thinking of surface tension as an actual unresolved tension in the surface layer, as would apply in the surface of a solid if that surface were cooled. ...

Will think about it some more.

 

[Chestermiller]

The curvature of the free surface is negligible, except in close proximity to the cylindrical boundary of the beaker. This pressure deviation dies out very rapidly with distance from the beaker wall, and is important only in establishing the small meniscus.

As for the effect of surface tension in the free surface contributing to the answer, in the first figure of response #18, the fluid on the right exerts a tensile surface tension force on the fluid on the left (within the overall free surface). This tensile force is (2R)S.

 

[haruspex]

As for the effect of surface tension in the free surface contributing to the answer, in the first figure of response #18, the fluid on the right exerts a tensile surface tension force on the fluid on the left (within the overall free surface). This tensile force is (2R)S.

I do not dispute that. My issue is that the mean proximity of water molecules is a balance between polar attraction and (I assume) electron shell repulsion. Surface tension arises because molecules at the surface have some "spare" polar attraction. I reasoned that this should draw those molecules a bit closer together than in the fluid body, but still only to the point of establishing a new equilibrium with the repulsive forces. Thus, the 2RS would be exactly balanced, leaving no net attraction between the two halves of the surface.
But I am now unsure of this…

I arrived at this view by thinking about the whole contact area between the two halves. Polar attraction acts there too, but has no contribution to the net forces exerted because it is balanced by electron shell repulsion. It seemed to me the same should happen at the surface.

 

[Chestermiller]

I do not dispute that. My issue is that the mean proximity of water molecules is a balance between polar attraction and (I assume) electron shell repulsion. Surface tension arises because molecules at the surface have some "spare" polar attraction. I reasoned that this should draw those molecules a bit closer together than in the fluid body, but still only to the point of establishing a new equilibrium with the repulsive forces. Thus, the 2RS would be exactly balanced, leaving no net attraction between the two halves of the surface.
But I am now unsure of this…

I arrived at this view by thinking about the whole contact area between the two halves. Polar attraction acts there too, but has no contribution to the net forces exerted because it is balanced by electron shell repulsion. It seemed to me the same should happen at the surface.

I’m unaware of how the molecular interpretation of surface tension comes about. I’m mostly a continuum mechanics guy. From a continuum mechanics perspective, a fluid with a free surface behaves as if a membrane is stretched over the free surface with an internal force per unit length that is transversely isotropic and of magnitude equal to the surface tension.