3
$\begingroup$

Pascal's law states that a force applied on a surface of a fluid is transmitted within the fluid in all directions of the fluid with the same intensity on equal surfaces. Similarly, it can be stated that pressure exerted at one point of a fluid mass is transmitted with the same intensity to every other point and in all directions.

Stevin's law states that, if only atmospheric pressure $p_{\text{at}}$ acts on the surface of a fluid of density $\rho$ then at a depth $h$ below the surface we have $$p=p_{\text{at}}+\rho gh$$ Suppose that the atmospheric pressure is increased by an amount $\Delta p$, that is, by $$p_{\text{at}}\to p_{\text{at}}+\Delta p$$ Then at the depth $h$ will be $$p=p_{\text{at}}+\Delta p+\rho gh=(p_{\text{at}}+\rho gh)+\Delta p \tag 1$$ so increasing the pressure at the fluid surface by an amount $\Delta p$ increases the pressure at each point in the fluid by the same amount.

I am looking for a simple demonstration for my high school students (15 years old) Is there another one that is a little better because I did not understand the motivation of (1), for example.

$\endgroup$

5 Answers 5

1
$\begingroup$

This thought-experiment is a little too sophisticated, I think, for 15-year-olds, but I like it, so I thought I'd include it for interest.

(a) At first, assume no pressure acting on the liquid surface.

Take a long narrow cylinder, open at one end and fitted with a movable piston of area $A$ a short distance from its open end. Partially immerse the cylinder at some angle in the liquid (of density $\rho$) so that the height of liquid above the piston is $h$. If the liquid pushes with pressure $p$ on the piston, then if we push the piston outwards through a small distance $\Delta s$, the liquid level rises because a volume $A \Delta s$ of liquid is effectively displaced upwards through a height $h$. If the piston is pushed outwards very slowly so that viscous and accelerating forces are negligible:

Work done pushing piston out through distance $\Delta s$ = Gravitational PE acquired by displaced liquid

So $$pA\Delta s =h (A\Delta s)\rho g$$

From which follows $p=h\rho g$.

Because this holds for any orientation of piston we see that the normal force per unit area on a surface by a liquid is the same whatever the orientation of the surface.

(b) Now assume a pressure $p_0$ to be exerted on the liquid's surface. This is equivalent to adding an extra layer of thickness $h_0$ to the liquid, in which $h_0=p_0/\rho g$. So, from the previous argument, at depth $h$ ...

$$p=(h+h_0)\rho g=h\rho g+p_0.$$

In other words, adding pressure $p_0$ above the liquid adds pressure $p_0$ at all points in the liquid, and all points at the same depth in the liquid are at the same pressure, in accordance with Pascal's principle.

$\endgroup$
0
$\begingroup$

The pressure varies linearly with depth $h$, but the other two pressures are constant without any depth variation transmitted everywhere by Pascal's Law.

School students who understand histograms as changes in height of a graph with respect to a horizontal independent varying quantity may also hopefully follow pressure variation with depth depicted graphically shown alongside a container holding a liquid.

enter image description here

$\endgroup$
0
$\begingroup$

What do you mean by "demonstrated"? Experimentally you will have to perform pressure measurements on columns of fluids. That is the proper "physical proof" of these relations. Mathematically you need a lot of assumptions to "prove" that they are correct. You need an incompressible fluid in hydrostatic equilibrium. These simple relationships do not hold for compressible fluids and fluids in motion. The usual "proof" at the high school level, which is what you have there, is therefor always more of a handwaving argument. There is nothing wrong with that. What we are trying to teach is simple physical intuition and how it can be put into mathematical form. Complicating this will only make it harder for students to understand the topic. What a student needs to learn is that a) the pressure in a weightless fluid is everywhere the same and b) that one can add up the weight of a column by stacking many such weightless cubes and then consider the individual weight in each of them as an additional pressure term on the ones below.

New contributor
FlatterMann is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
0
$\begingroup$

In my opinion, you need a demonstration rather than an equation. This will take a bit of work and expense, but you need to build a device that demonstrates what you are teaching (e.g., Home Depot will have most or all equipment you need).

Ideally the device will be a transparent cylinder, but that's not an absolute requirement. The cylinder should be several centimeters long (e.g., 25-50 cm), sealed at both ends. Install a pressure gauge on the top of the cylinder and 2-4 pressure gauges down the side of the cylinder. Also, install two valves on the cylinder. One valve on the top should be used to attach a compressed air line that allows you to increase the pressure in the top of the cylinder. The other valve, close to the bottom of the cylinder, will allow you to drain water out of the cylinder when the experiment is done.

For the demo, fill the cylinder with enough water to cover all of the pressure gauges that are on the side of the cylinder. Have the students take readings of all the pressure gauges (including the pressure gauge on top of the cylinder) such that they can see that the pressure increases with depth according to the appropriate equation. Then, connect a pressurized air line to the top valve and pump air into the cylinder. Again, have the students take readings of all pressure gauges. When they compare the initial readings to the final readings, they should note that all pressures increased by the same amount, regardless of depth.

$\endgroup$
0
$\begingroup$

If Stevin's law results true, we know that at a depth $h$ the pressure is equal to $$p=p_{\text{at}}+\rho gh$$ and if for hypothesis I would increase the atmospheric pressure by an amount $\Delta p$, i.e. $$p_{\text{at}} \to p_{\text{at}}+\Delta p$$ and for all points at depth $h$ $$p=p_{\text{at}}+\Delta p+\rho gh=(p_{\text{at}}+\rho gh)+\Delta p \tag 1$$ I will have the pressure $p$ changed of a quantity $\Delta p$ $(1)$ i.e. the same pressure when I had increased the atmospheric pressure $p_{\text{at}}+\Delta p$ acting on the surface of the liquid contained in the container. In other words, the pressure applied at one point of a fluid mass is transmitted with the same intensity to every other point and in all directions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.