Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum Sep 4th 2008, 09:16 PM #1 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 P=pgh For P representing the pressure and p representing the density of medium. I ask my physics teacher a question: For gravity which is only pointing downwards. For planes that is parallel to the gravity, would the equation fails to work as normal force on plane( as perpendicular to gravity) ? I know the answer is no and the teacher also said no. He explained that we should regard the area dA as a point and pressure exists in all direction. After the explanation, I still don't understand why gravity(g) can be used to calculate pressure on one plane which is parallel to gravity. Could anyone help?   Sep 5th 2008, 08:37 AM   #2

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 Originally Posted by werehk For P representing the pressure and p representing the density of medium. I ask my physics teacher a question: For gravity which is only pointing downwards. For planes that is parallel to the gravity, would the equation fails to work as normal force on plane( as perpendicular to gravity) ? I know the answer is no and the teacher also said no. He explained that we should regard the area dA as a point and pressure exists in all direction. After the explanation, I still don't understand why gravity(g) can be used to calculate pressure on one plane which is parallel to gravity. Could anyone help?
I think this will answer your question, but I'm not sure why your professor would bring up pressure so I'm guessing I'm missing something.

The derivation of the form for the gravitational potential energy (GPE) is given by the amount of work done in moving a mass around in a gravitational field. One result of this is that there is always work done, and hence a change in the GPE, when an object moves vertically. The other point is that because work is a scalar quantity then so is the GPE, so the direction of the field does not matter in our calculation of the GPE.

I hope this clears things up, if not let me know and I'll take another shot at it.

-Dan
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See the forum rules here.   Sep 6th 2008, 07:18 PM #3 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 I guess it is really talking about pressure. As pressure = Force/ Area pressure = =ma/A= Vp/A= Ahpg/A = pgh where A is uniform cross-section, h the depth below the liquid surface, p the density of liquid   Sep 7th 2008, 11:23 AM   #4

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 Originally Posted by werehk I guess it is really talking about pressure. As pressure = Force/ Area pressure = =ma/A= Vp/A= Ahpg/A = pgh where A is uniform cross-section, h the depth below the liquid surface, p the density of liquid
Ahhhh, sorry. I didn't look at the title. Yes the direction of gravity has disappeared here. If it helps an area can be defined to have a direction, in which case
$\displaystyle \vec{F} = P \vec{A}$
so the force is in the direction of the area element. Otherwise you are going to have to content yourself with the concept that pressure is simply a scalar measurement of a property of a fluid. (But you can easily come up with the concept of a pressure gradient, which is similar in concept to a force, which is a vector.)

-Dan
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See the forum rules here.   Sep 8th 2008, 06:27 AM   #5
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 Originally Posted by topsquark Ahhhh, sorry. I didn't look at the title. Yes the direction of gravity has disappeared here. If it helps an area can be defined to have a direction, in which case $\displaystyle \vec{F} = P \vec{A}$ so the force is in the direction of the area element. Otherwise you are going to have to content yourself with the concept that pressure is simply a scalar measurement of a property of a fluid. (But you can easily come up with the concept of a pressure gradient, which is similar in concept to a force, which is a vector.) -Dan
But why the direction of gravity would disappear?

Is pressure a scalar because of some methods of differentiation?   Sep 8th 2008, 07:48 AM   #6

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 Originally Posted by werehk But why the direction of gravity would disappear? Is pressure a scalar because of some methods of differentiation?
It's just the nature of the beast. Pressure is a useful concept because it is a scalar. We don't need to worry about a direction in order to see what's going on with it. In this sense it is much like a potential in electricity. It is sometimes easier to deal with the potential rather than the electric field directly simply because the potential is a scalar quantity. In fluid dynamics I would suspect there is a similar simplification in using pressure rather than force. Of course, all of this depends on the applications that you are applying them to.

As to the mathematical reasoning if we introduced a vector pressure then we'd have to have a relationship between $\displaystyle \vec{P}$, $\displaystyle \vec{F}$, and $\displaystyle \vec{A}$. The only way to relate three vectors with different units is by the cross product so we'd need $\displaystyle \vec{F} = \vec{P} \times \vec{A}$. We can certainly defined a vector pressure in this manner, but notice that the pressure is now required to be pointed in one specific direction depending on the direction of area, and this direction is perpendicular to the applied force. It would appear to have none of the properties familiar to the pressure concept that we know, not even the magnitude would be the same. Perhaps this concept has some use and is worth investigating. I leave that to you.

-Dan
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