# Can anyone explain Pascal's principles?

• B

## Main Question or Discussion Point

According to pascal's principle a change in the pressure applied to enclosed fluid is transmitted and diminished to every point of fluid and to was of the containing vessel and my question is how this happens.

Related Classical Physics News on Phys.org
Hi.
Fluid flows from high pressure side to low pressure side. As the result of this flow pressure goes down in high pressure side and it goes up in low pressure side. In final equilibrium the both side have the same pressure.

Hi.
Fluid flows from high pressure side to low pressure side. As the result of this flow pressure goes down in high pressure side and it goes up in low pressure side. In final equilibrium the both side have the same pressure.
If one side of hydraulic lever is narrow and other is broad why does not pressure decreases as it transmits to broader side from narrower side.

Mark44
Mentor
If one side of hydraulic lever is narrow and other is broad why does not pressure decreases as it transmits to broader side from narrower side.
Because the fluid is incompressible. If you have a thick steel bar that is 2 m. long, and you push on one end, you're exerting a force on one end. Is that force dissipated through the bar so that there is no force on the other end?

Because the fluid is incompressible. If you have a thick steel bar that is 2 m. long, and you push on one end, you're exerting a force on one end. Is that force dissipated through the bar so that there is no force on the other end?
Sir but isn't it like that , as I applied a force and that force transmits and as area( the area of liquid through which it was transmited) increases the pressure should decrease

Mark44
Mentor
Sir but isn't it like that , as I applied a force and that force transmits and as area( the area of liquid through which it was transmited) increases the pressure should decrease
But that's not how it works, which is what Pascal's Principle is saying. Suppose you have a full tube of toothpaste, and take the cap off. If you squeeze the bottom (apply a force to increase the pressure), does the pressure merely dissipate, or does something else happen?

The toothpaste, like the fluid in a hydraulic jack, and like the metal in a steel bar, is incompressible. If you apply a force at one end, that force is carried through to the other end.

... and that force transmits and as area( the area of liquid through which it was transmited)
That makes no sense. The units of pressure are $\frac{\text{Force}}{\text{Area}}$, such as $\frac{\text{Nt.}}{m^2}$ or $\frac{\text{lb}}{in^2}$. The pressure has nothing to do with the volume through which the force is transmitted.

But that's not how it works, which is what Pascal's Principle is saying. Suppose you have a full tube of toothpaste, and take the cap off. If you squeeze the bottom (apply a force to increase the pressure), does the pressure merely dissipate, or does something else happen?

The toothpaste, like the fluid in a hydraulic jack, and like the metal in a steel bar, is incompressible. If you apply a force at one end, that force is carried through to the other end.

That makes no sense. The units of pressure are $\frac{\text{Force}}{\text{Area}}$, such as $\frac{\text{Nt.}}{m^2}$ or $\frac{\text{lb}}{in^2}$. The pressure has nothing to do with the volume through which the force is transmitted.
Sorry sir as I am again questioning but anyhow in nutshell I want to know how the pressure remains same when force is applied to liquid and why it is independent of height and area of container.

Mark44
Mentor
Sorry sir as I am again questioning but anyhow in nutshell I want to know how the pressure remains same when force is applied to liquid and why it is independent of height and area of container.

The pressure remains the same because the fluid in an enclosed container is incompressible, similar to the molecules in a steel bar. If the cross-sectional area of the bar is $1 cm^2$ and you apply a force of 10 Nt. at one end, the pressure is $\frac {10 Nt.}{cm^2}$. The force at the other end of the bar is exactly the same, and so is the pressure.

Now consider that you have a tube with a cross-section area of 1 cm^2, with pistons at each end, and filled with oil. If you apply the same 10 Nt. at one end, the pressure is is $\frac {10 Nt.}{cm^2}$. There will be a force of 10 Nt. at the other piston, and the same pressure.
and why it is independent of height and area of container.
Pressure is **not** independent of height if you're talking about the pressure at a certain depth. The pressure at a depth of 2 m. in a swimming pool is exactly the same as the pressure at the same depth in the ocean. The pressure is due to the weight of the column of air above the pool or ocean, plus the weight of the water above that depth.

When you say "area of the container" you aren't being clear. Are you talking about the cross-sectional area of a horizontal tube, or are you confusing this with the volume? If you are talking about a vertical container, its area is irrelevant. What is relevant in this case is the depth of the fluid.

The pressure remains the same because the fluid in an enclosed container is incompressible, similar to the molecules in a steel bar. If the cross-sectional area of the bar is $1 cm^2$ and you apply a force of 10 Nt. at one end, the pressure is $\frac {10 Nt.}{cm^2}$. The force at the other end of the bar is exactly the same, and so is the pressure.

Now consider that you have a tube with a cross-section area of 1 cm^2, with pistons at each end, and filled with oil. If you apply the same 10 Nt. at one end, the pressure is is $\frac {10 Nt.}{cm^2}$. There will be a force of 10 Nt. at the other piston, and the same pressure.
Pressure is **not** independent of height if you're talking about the pressure at a certain depth. The pressure at a depth of 2 m. in a swimming pool is exactly the same as the pressure at the same depth in the ocean. The pressure is due to the weight of the column of air above the pool or ocean, plus the weight of the water above that depth.

When you say "area of the container" you aren't being clear. Are you talking about the cross-sectional area of a horizontal tube, or are you confusing this with the volume? If you are talking about a vertical container, its area is irrelevant. What is relevant in this case is the depth of the fluid.
Sir please tell me where I am wrong,

anwsering to question #4:-
The force will not be dissipated but but some work done by force will get converted to some other type of energies but if we ignore the loss in other type of energies the force will remain same,I agree with you.

as when we applied a force on one end of toothpaste tube on less area a pressure will be generated but as the force transmits and remains same and the area increases so as area increases the pressure should decrease as I can think.

Thanks for giving me a lot of your precious time.

Mark44
Mentor
as when we applied a force on one end of toothpaste tube on less area a pressure will be generated but as the force transmits and remains same and the area increases so as area increases the pressure should decrease as I can think.
Again, I don't think you have a clear idea of area vs. volume. If you squeeze on one end of the toothpaste tube, you increase the pressure on the toothpaste, and it comes out the open end.

A hydraulic jack works in a similar way, but with a small piston at one end and a much larger piston at the other end, the end that does the lifting.

When you work the handle of the jack, you are exerting a force on the hydraulic fluid, raising the pressure on the fluid. The pressure on the larger piston is the same, but over a larger cross-section area, so the force is much larger. So you can lift a heavy car with a small applied force. The tradeoff is that you have to pump on the smaller piston numerous times to get the large piston to move a small distance. That's because the volume of fluid that you move by pumping the small piston can move the large piston only a short distance.

A.T.
Sir but isn't it like that , as I applied a force and that force transmits and as area( the area of liquid through which it was transmited) increases the pressure should decrease
If the applied force changes very quickly there will be a propagation delay. But for most applications you can treat the fluid as in-compressible and the propagation of pressure changes as instantaneous.

If the applied force changes very quickly there will be a propagation delay. But for most applications you can treat the fluid as in-compressible and the propagation of pressure changes as instantaneous.
I can't understand that when we apply some force on one side of hydraulic lever which has less area and that force exerts a pressure which will be more because of less area and then the force will be propagated but when this force reaches another side which has larger area so that force act on large area so pressure should become less but we do not see that,please tell me where I am wrong.

A.T.
....force will be propagated but when this force reaches another side ....
Pressure is propagating, not force.

Mark44
Mentor
I can't understand that when we apply some force on one side of hydraulic lever which has less area and that force exerts a pressure which will be more because of less area and then the force will be propagated but when this force reaches another side which has larger area so that force act on large area so pressure should become less but we do not see that,please tell me where I am wrong.
Where you are wrong is in thinking that pressure and force are the same thing.
Here is an image from the wikipedia article on jacks, https://en.wikipedia.org/wiki/Jack_(device), in the section on hydraulic jacks.
The operator exerts a force $F_{in}$ on the piston on the left, raising the pressure on the hydraulic fluid to $P \frac{lb}{in^2}$. The pressure on the piston on the right, under the car, is the same, but due to the pressure acting against a much larger area, $F_{car}$ will be much larger.

Let's assume that the cross-section area of the small piston is 1 sq. in., and the large piston's cross-section area is 40 sq. in. If I apply a force of 50 lb, the pressure is 50 lb/in^2. The pressure on the large piston is also 50 lb/in^2, but this pressure is acting on each square inch of the 40 sq. inches, so the total force on the car is 50 x 40 = 2000 lb.

Of course, when the small piston moves down by 1 inch, the large piston moves up by only 1/40 inch, so I have to press down on the handle a long way to raise the car by an appreciable amount.

To　Mr.　Hemant

Say there are two same lever A and B. Put 1kg weights on A and B. The two balance because of symmetry.
Say there are one more same lever C next to B with thin wall between. Put 1kg weight on C also. The three balance because of symmetry.

Then say we remove the thin wall carefully to make B+C joint lever of double area.
The joint lever with total weights 2kg,
(A) goes down because same force with lever A needs only 1kg weight.
(B)stays still because same pressure with lever A needs 2kg weight.

(B) seems reasonable to me. What about you?

To　Mr.　Hemant

Say there are two same lever A and B. Put 1kg weights on A and B. The two balance because of symmetry.
Say there are one more same lever C next to B with thin wall between. Put 1kg weight on C also. The three balance because of symmetry.

Then say we remove the thin wall carefully to make B+C joint lever of double area.
The joint lever with total weights 2kg,
(A) goes down because same force with lever A needs only 1kg weight.
(B)stays still because same pressure with lever A needs 2kg weight.

(B) seems reasonable to me. What about you?
Thanks thanks a lot

Where you are wrong is in thinking that pressure and force are the same thing.
Here is an image from the wikipedia article on jacks, https://en.wikipedia.org/wiki/Jack_(device), in the section on hydraulic jacks.
The operator exerts a force $F_{in}$ on the piston on the left, raising the pressure on the hydraulic fluid to $P \frac{lb}{in^2}$. The pressure on the piston on the right, under the car, is the same, but due to the pressure acting against a much larger area, $F_{car}$ will be much larger.

Let's assume that the cross-section area of the small piston is 1 sq. in., and the large piston's cross-section area is 40 sq. in. If I apply a force of 50 lb, the pressure is 50 lb/in^2. The pressure on the large piston is also 50 lb/in^2, but this pressure is acting on each square inch of the 40 sq. inches, so the total force on the car is 50 x 40 = 2000 lb.

Of course, when the small piston moves down by 1 inch, the large piston moves up by only 1/40 inch, so I have to press down on the handle a long way to raise the car by an appreciable amount.

View attachment 250559
Thanks sir you helped me to figure out my mistake