Pascal’s principle - the very basics Watch

Leah.J
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Can someone help me understand Pascal’s principle please ?
the definition says that if a pressure is applied on a liquid it is transmitted undiminished through the liquid but I don’t understand what that means ?
does the internal pressure of the fluid itself increase ? Does Pascal’s principle just say that the pressure of every particle in the fluid will increase by the same amount ?

also, in this picture
Name:  4B914092-AEE2-47CA-942B-28A5390C8FFC.jpeg
Views: 14
Size:  128.9 KB
the article says that the equation only works if the height of the pistons is the same, why ? To me it makes perfect sense that in the diagram I drewName:  FF6EB1ED-121E-490F-A906-2A8DCA803355.jpeg
Views: 13
Size:  131.5 KB although the pistons aren’t at the same height, F1/A1=F2/A2
why not ? Help please, I know it’s simple and I don’t know why I’m struggling
Last edited by Leah.J; 1 month ago
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ghostwalker
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(Original post by Leah.J)
Can someone help me understand Pascal’s principle please ?
the definition says that if a pressure is applied on a liquid it is transmitted undiminished through the liquid but I don’t understand what that means ?
does the internal pressure of the fluid itself increase ? Does Pascal’s principle just say that the pressure of every particle in the fluid will increase by the same amount ?

also, in this picture
Name:  4B914092-AEE2-47CA-942B-28A5390C8FFC.jpeg
Views: 14
Size:  128.9 KB
the article says that the equation only works if the height of the pistons is the same, why ? To me it makes perfect sense that in the diagram I drewName:  FF6EB1ED-121E-490F-A906-2A8DCA803355.jpeg
Views: 13
Size:  131.5 KB although the pistons aren’t at the same height, F1/A1=F2/A2
why not ? Help please, I know it’s simple and I don’t know why I’m struggling
Assuming it's the document I found online, if you go back a couple of paragraphs it says, "Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth."

If the pistons were at different depths then you'd need to take into account the difference in pressure due to the density of the fluid.

E.g A ~10m column of water has a pressure of 1 atmosphere at the bottom (above atmospheric pressure). Enclose the system and put a piston at the top and a piston at the bottom. Then the pressure at the lower piston will always be 1 atmosphere greater than the pressure at the upper piston.
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Leah.J
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(Original post by ghostwalker)
Assuming it's the document I found online, if you go back a couple of paragraphs it says, "Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth."

If the pistons were at different depths then you'd need to take into account the difference in pressure due to the density of the fluid.

E.g A ~10m column of water has a pressure of 1 atmosphere at the bottom (above atmospheric pressure). Enclose the system and put a piston at the top and a piston at the bottom. Then the pressure at the lower piston will always be 1 atmosphere greater than the pressure at the upper piston.
Oh, so they're just saying that if 1 of the pistons is underwater and the other isn't, or of they're both under water (or whatever the fluid is) but at different heights, the equation won't work ?
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ghostwalker
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(Original post by Leah.J)
Oh, so they're just saying that if 1 of the pistons is underwater and the other isn't, or of they're both under water (or whatever the fluid is) but at different heights, the equation won't work ?
That's correct, the basic equation won't work.

The important thing is that if they are at differing heights, you have to account for the extra pressure on the lower piston due to the density of fluid between those two heights.
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0le
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Imagine you have a cup of tea and you look closely at a very small piece of fluid from the middle of the cup of tea.

If you draw a force diagram on the small piece of fluid, called a fluid element, on each surface you will find that there is a force of pressure acting perpendicular to each surface. However, there is an additional force, the weight of the fluid element (the red arrow in the figure). This ultimately means the pressure in the vertical direction varies with height (and increases in the direction of increasing gravitational attraction) to account for the weight of the fluid.
Last edited by 0le; 1 month ago
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Leah.J
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(Original post by ghostwalker)
That's correct, the basic equation won't work.

The important thing is that if they are at differing heights, you have to account for the extra pressure on the lower piston due to the density of fluid between those two heights.
I have 1 more question. Does the pressure of a liquid column only depend on pgh ? so No matter how wide of a glass tube I use, I'm still gonna need 760 mmHg to exert 1 atm ?
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(Original post by ghostwalker)
Assuming it's the document I found online, if you go back a couple of paragraphs it says, "Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth."

If the pistons were at different depths then you'd need to take into account the difference in pressure due to the density of the fluid.

E.g A ~10m column of water has a pressure of 1 atmosphere at the bottom (above atmospheric pressure). Enclose the system and put a piston at the top and a piston at the bottom. Then the pressure at the lower piston will always be 1 atmosphere greater than the pressure at the upper piston.
this right here, 100%
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Leah.J
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(Original post by 0le)
Imagine you have a cup of tea and you look closely at a very small piece of fluid from the middle of the cup of tea.

If you draw a force diagram on the small piece of fluid, called a fluid element, on each surface you will find that there is a force of pressure acting perpendicular to each surface. However, there is an additional force, the weight of the fluid element (the red arrow in the figure). This ultimately means the pressure in the vertical direction varies with height (and increases in the direction of increasing gravitational attraction) to account for the weight of the fluid.
oh, yeah I understand why the pressure varies with depth but I thought they meant that the pistols had to be at the same height outside the fluid
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ghostwalker
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(Original post by Leah.J)
I have 1 more question. Does the pressure of a liquid column only depend on pgh ? so No matter how wide of a glass tube I use, I'm still gonna need 760 mmHg to exert 1 atm ?
Yes, that's correct.
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Leah.J
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(Original post by ghostwalker)
Yes, that's correct.
(Original post by 0le)
Imagine you have a cup of tea and you look closely at a very small piece of fluid from the middle of the cup of tea.

If you draw a force diagram on the small piece of fluid, called a fluid element, on each surface you will find that there is a force of pressure acting perpendicular to each surface. However, there is an additional force, the weight of the fluid element (the red arrow in the figure). This ultimately means the pressure in the vertical direction varies with height (and increases in the direction of increasing gravitational attraction) to account for the weight of the fluid.
following on that
Name:  Screenshot (136).png
Views: 19
Size:  135.3 KB I don't understand why P1=P2 here . I know P2 should equal the atmospheric pressure but shouldn't P1 = atm pressure + pg(y2-y1) ?
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0le
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(Original post by Leah.J)
following on that
Name:  Screenshot (136).png
Views: 19
Size:  135.3 KB I don't understand why P1=P2 here . I know P2 should equal the atmospheric pressure but shouldn't P1 = atm pressure + pg(y2-y1) ?
This is a problem related to flowing fluids, a topic referred to as fluid dynamics. Previously we were dealing with static fluids, a topic referred to as hydrostatics. In essence, yes, the pressure still varies with height, but in flowing fluids it can now also vary due to changes in the velocity field.

Also note that Bernoulli's principle is simply an application of conservation of energy, with some assumptions. If I remember correctly, one assumption was to neglect heat losses. In that equation, we have (static pressure + dynamic pressure + GPE) at point one is equivalent to (static pressure + dynamic pressure + GPE) at point 2. The author has stated assumptions with have reduced the equation. The static pressures in this case are the same. The dynamic pressure is given by:

\frac{1}{2}\rho v^2

It appears from that equation in the figure that the fluid moves because gravitational potential energy is in effect, converted to a moving energy. As for WHY the pressures are the same, think of it first like doing a force diagram, where we gather all the forces first. In the previous example, the local thermodynamic pressure acting on a fluid element was the same, but there was an additional weight of the fluid element to be considered. Similarly, in this case, we first collect all the energy terms and then simplify the balance equation - everything on the left equals everything on the right because in an isolated system, energy is conserved, i.e. it cannot be created or destroyed, it just remains the same, albeit in different forms. In essence, Bernoulli's equation then states that at point 2, the pressure energy + motion energy (=0) + GPE is equivalent to the pressure energy + motion energy + GPE at point 2. Now that we have constructed the full equation, then we simplify.
Last edited by 0le; 1 month ago
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Leah.J
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(Original post by 0le)
This is a problem related to flowing fluids, a topic referred to as fluid dynamics. Previously we were dealing with static fluids, a topic referred to as hydrostatics. In essence, yes, the pressure still varies with height, but in flowing fluids it can now also vary due to changes in the velocity field.

Also note that Bernoulli's principle is simply an application of conservation of energy, with some assumptions. If I remember correctly, one assumption was to neglect heat losses. In that equation, we have (static pressure + dynamic pressure + GPE) at point one is equivalent to (static pressure + dynamic pressure + GPE) at point 2. The author has stated assumptions with have reduced the equation. The static pressures in this case are the same. The dynamic pressure is given by:

\frac{1}{2}\rho v^2

It appears from that equation in the figure that the fluid moves because gravitational potential energy is in effect, converted to a moving energy. As for WHY the pressures are the same, think of it first like doing a force diagram, where we gather all the forces first. In the previous example, the local thermodynamic pressure acting on a fluid element was the same, but there was an additional weight of the fluid element to be considered. Similarly, in this case, we first collect all the energy terms and then simplify the balance equation - everything on the left equals everything on the right because in an isolated system, energy is conserved, i.e. it cannot be created or destroyed, it just remains the same, albeit in different forms. In essence, Bernoulli's equation then states that at point 2, the pressure energy + motion energy (=0) + GPE is equivalent to the pressure energy + motion energy + GPE at point 2. Now that we have constructed the full equation, then we simplify.
Ohhh, this makes sense, thanks. But, why are the static pressures the same if the heights are different, ( static just means at equilibrium right ?)
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0le
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(Original post by Leah.J)
Ohhh, this makes sense, thanks. But, why are the static pressures the same if the heights are different, ( static just means at equilibrium right ?)
In the example I gave, it is fluid element, which is infinitesmally small. In essence, its just a "point" in space, and so the pressure acting on that point is equal in all directions. The point itself has an infinitesmally small weight associated with it as well. The height is negligibly small and only becomes important when you integrate (add up) a number of these points together.

The way to picture it is that at a point in space in a static fluid, the pressure in all directions is the same. At a finite distance away from that point, but in the same fluid, the pressure acting on that second point will also be the same in all directions but its magnitude will differ compared to that of the first point.
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