Direction of pressure in a Fluid

[andyrk]

Have a look at this link guys: http://ocw.mit.edu/courses/mechanic...cs-13-012-fall-2005/readings/2005reading2.pdf.

Could someone explain the term − 1/2 ρ(b∆x∆z) in equation 2.2? I don't understand how does that represent force on the plane along z-axis.

 

[Chestermiller]

That's supposed to be the weight of the fluid inside the wedge (they left out the g, or assumed that it is combined with the rho). They are doing a force balance on the wedge of fluid.

Chet

 

[andyrk]

That's supposed to be the weight of the fluid inside the wedge (they left out the g, or assumed that it is combined with the rho). They are doing a force balance on the wedge of fluid.

Chet

So the 1/2 factor is because ρ(b∆x∆z) is the weight of the whole cuboid and we need the weight of the only one half along the diagonal from upper left point to lower right point? Am I correct? Because if you place another identical triangular volume upside down on the already existing triangular volume, we get a cuboid. Right? So we include the 1/2 into get the desired area.

 

[Chestermiller]

So the 1/2 factor is because ρ(b∆x∆z) is the weight of the whole cuboid and we need the weight of the only one half along the diagonal from upper left point to lower right point? Am I correct? Because if you place another identical triangular volume upside down on the already existing triangular volume, we get a cuboid. Right? So we add the 1/2.

Yes. That's how we arrived at the equation in geometry for the area of a right triangle = (1/2) bh.

 

[andyrk]

So why are we only considering forces perpendicular to each of the faces of the triangular prism? There are many other forces too which are not perpendicular to the faces of the triangular prism. And what about the faces parallel to each other, which exist on the page and are parallel to it?

 

[Chestermiller]

So why are we only considering forces perpendicular to each of the faces of the triangular prism? There are many other forces too which are not perpendicular to the faces of the triangular prism. And what about the faces parallel to each other, which exist on the page and are parallel to it?

Pressure can only exert forces perpendicular to surfaces.

 

[andyrk]

Pressure can only exert forces perpendicular to surfaces.

So what happens to the forces which are not perpendicular. They definitely exist. But then why aren't they shown in the diagram? If you say that they cancel out when we divide them into components, then we are assuming that they are of same magnitude. Otherwise they won't cancel out. But does this assumption have any basis?

 

[SteamKing]

So what happens to the forces which are not perpendicular. They definitely exist.

And exactly which forces are not perpendicular to the wedge?

But then why aren't they shown in the diagram? If you say that they cancel out when we divide them into components, then we are assuming that they are of same magnitude. Otherwise they won't cancel out. But does this assumption have any basis?

Fig. 2.1 can be converted into a free body diagram of the wedge. You can determine if the wedge is in static equilibrium given the weight of the wedge and the hydrostatic pressures as shown.

 

[Chestermiller]

And exactly which forces are not perpendicular to the wedge?

Yes. Where do these alleged forces come from?

Chet

 

[andyrk]

From the fact that pressure is exerted in all directions.

 

[SteamKing]

From the fact that pressure is exerted in all directions.

Pascal's Law states that pressure is exerted equally in all directions. This means that at a given depth of fluid, the static pressure produced by the fluid will be the same in all parts of the fluid's container.

However, just because the pressure exerted is the same in all directions, it does not follow that there are forces acting in all directions which are produced by this pressure.

http://en.wikipedia.org/wiki/Pressure [Specifically, the first section and the section on Liquid Pressure]

http://en.wikipedia.org/wiki/Fluid_statics#Pressure_in_fluids_at_rest

You have alleged that:

So what happens to the forces which are not perpendicular. They definitely exist.

and have been informed by Chet that:

Pressure can only exert forces perpendicular to surfaces.

So, the ball is in your court to show the existence of these oblique forces supposedly caused by pressure.

I still maintain that if you use Fig. 2.1 in your link as a free body diagram and analyze the hydrostatic forces normal to all sides of the wedge, you will see that these oblique forces do not exist.

 

[Chestermiller]

Hi Andyrk,

Unfortunately for you and many other students, they have simplified the concept of pressure to the point where it has caused considerable confusion. Later on, you will learn that pressure is not a scalar, but really part of a second order tensor quantity known as the "stress tensor," which has a certain kind of directional character associated with it. The pressure part of the stress tensor has the directional character that, for any tiny element of surface area within the fluid or at the boundary of the fluid, the pressure force per unit area acts perpendicular to that surface, irrespective of the orientation of the surface element. This is what they mean when they say "the pressure acts equally in all directions."

I wish I could be more precise than this, but to do that, I would have to teach you tensor analysis. For now, it will suffice for you just accept the key feature of pressure that SteamKing and I have conveyed to you (obviously in a better way than your text has done it).

Chet

 

[Puma]

Can you not just say pressure is exerted in all directions - which it is as the molecules move in random directions, but these add up to a perpendicular force on the face of a container? Just do it in a Newtonian ping pong ball analogy way?

 

[Chestermiller]

Can you not just say pressure is exerted in all directions - which it is as the molecules move in random directions, but these add up to a perpendicular force on the face of a container? Just do it in a Newtonian ping pong ball analogy way?

That physical interpretation works great for an ideal gas, but not all fluids are ideal gases. For liquids, it might be less convincing to andyrk.

Chet

 

[andyrk]

Pascal's Law states that pressure is exerted equally in all directions. This means that at a given depth of fluid, the static pressure produced by the fluid will be the same in all parts of the fluid's container.

However, just because the pressure exerted is the same in all directions, it does not follow that there are forces acting in all directions which are produced by this pressure.

http://en.wikipedia.org/wiki/Pressure [Specifically, the first section and the section on Liquid Pressure]

http://en.wikipedia.org/wiki/Fluid_statics#Pressure_in_fluids_at_rest

You have alleged that:
and have been informed by Chet that:
So, the ball is in your court to show the existence of these oblique forces supposedly caused by pressure.

I still maintain that if you use Fig. 2.1 in your link as a free body diagram and analyze the hydrostatic forces normal to all sides of the wedge, you will see that these oblique forces do not exist.

I don't know how to prove it but I feel they should exist because I connect the different directions of pressure to different directions of forces.

 

[A.T.]

I feel they should exist

Even if they exist on the atomic level, they cancel each other on average, because they are uniformly distributed in all directions parallel to the surface.

 

[andyrk]

Even if they exist on the atomic level, they cancel each other on average, because they are uniformly distributed in all directions parallel to the surface.

So why doesn't the perpendicular one also cancel out too? Why does it remain intact despite the fact that all others get completely canceled out?

 

[A.T.]

So why doesn't the perpendicular one also cancel out too?

Because along that axis the colliding molecules are all coming from one direction: from the fluid.

 

[andyrk]

Because along that axis the colliding molecules are all coming from one direction: from the fluid.

So along the opposite direction also the colliding molecules are coming from the fluid. There aren't any other particles in this other than the fluid.

 

[A.T.]

So along the opposite direction also the colliding molecules are coming from the fluid. There aren't any other particles in this other than the fluid.

No idea what you mean. You should draw yourself some diagrams. It's all quite obvious from geometry.

 

[andyrk]

The thing that I don't understand is that why is there is no opposite force to the perpendicular one to cancel it? I couldn't make any sense of the reason you provided. Could you explain it a bit more simply?

 

[A.T.]

Could you explain it a bit more simply?

I don't think so. Hence the advice to draw a diagram, with forces from the fluid on a surface point, uniformly distributed along a hemisphere.

 

[andyrk]

I don't think so. Hence the advice to draw a diagram, with forces from the fluid on a surface point, uniformly distributed along a hemisphere.

Hey, I think I get what you mean. The only area of contact is the wedge. That includes one side of all the 180 degrees where the forces could hit the wedge. The other 180 degrees don't matter because there isn't any area there (and since F = PA, therefore F = 0). But my question is, we assume that pressure is exerted equally in all directions. But this is what we set out to prove in the first place. Then how can we assume it?

 

[Tom_K]

I agree there must be pressure present in all directions within the fluid. Force is only developed when that pressure comes into contact with a surface that resists the pressure and that surface gives the force a direction making force a vector, while pressure is not. The vector always points perpendicular to the surface. Just as a vise with only one jaw cannot develop a force, pressure cannot develop a force unless there is an opposing surface.

 

[A.T.]

The vector always points perpendicular to the surface.

Even if the force from a single collision had a component parallel to the wall, those parallel components from many collisions at the same point would average to zero, because they are uniformly distributed along all directions parallel to the wall.

 

[Chestermiller]

The wall is exerting a force in the opposite direction on the molecules that bounce off of it. This is the opposing force that you are looking for. It is the reaction force of the wall on the fluid. So the Newton's Third Law action-reaction pair are: (1) fluid exerts force per unit area on wall and (2) wall exerts equal and opposite force per unit area on fluid.

Chet

 

[jbriggs444]

There is an argument that can be made from symmetry. For a uniform fluid that is motionless and non-accelerating with respect to a uniform wall there can be no preferred force direction tangent to the wall. All such directions are equivalent. If the laws of physics predict a force in a direction with a non-zero tangential component in one direction then they must also predict a force with the same non-zero tangential component in the opposite direction. But that's silly. So the predicted force direction can only be perpendicular to the wall.

 

[andyrk]

Do the forces parallel to the incline have any effect on the diagonal surface? Do they exert a force on the diagonal surface or not? If they don't then its pretty easy from there because we can decompose all forces acting at all angles from 0 to 180 degrees on the wedge to 2 components, namely perpendicular and parallel. The perpendicular forces keep on adding and that results in Pz (Pressure in z-direction). The same thing happens with Px and Py. The parallel forces don't have any effect so they are not considered. Well, this is the ideal scenario when things happen the way I have put it. But what if they don't? What if parallel forces do actually have an effect on the slant/diagonal? Then how can we be sure that the parallel forces will cancel out? We can't say/assume that each parallel force would have an equal and opposite parallel force to it, so that it cancels out. If we assume it..the whole subsequent proof is based on this assumption, which might or might not be true for all cases.

The answer to this post determines the conclusion of this whole thread.

 

[andyrk]

Anybody there?