"A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid"

"A pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere"

I'm having a hard time getting the intuition behind Pascal's Law. I've already read many kinds of explanations, involving energy, work and displacement and read almost all the posts on this subject on Physics.SE but I can't convince myself.

The main problems for me are to understand the force amplification as a consequence of Pascal's Law and the reason why the pressure is transmitted undiminished. I made a drawing to illustrate what's going on in my mind (where the balls represent atoms):

What my intuition tells me

What Pascal's Law says

Assumptions in this post:

So, in a more explicit way, Pascal's Law says WHAT happen, but I want to know WHY it happen:

  • In terms of forces, at molecular level, how pressure is transmitted undiminished to all points in the fluid, causing the amplification of the force when area increases?

$$\bbox[5px,border:1px solid black] {F_{top}=100\,N\xrightarrow{\,\uparrow{A}\,} F_{bottom}=1000\,N}$$

  • In terms of energy, how does the energy associated with the force applied at the top spreads and turns into the force applied at the bottom? (Note: I know that energy, given my assumptions, is conserved | Note 2: just to remember - there are no displacements involved)

$$\bbox[5px,border:1px solid black] {100\,N=F_{top}\rightarrow E_{top}=E_{bottom}\rightarrow F_{bottom}=1000\,N}$$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Sep 24, 2022 at 22:55

6 Answers 6


$\def\vg{\vec g} \def \vF{\vec F} \def \vR{\vec R}$ Warning: This answer does not take into account the long series of comments preceding it. Therefore some repetition will be unavoidable. My aim is to give - I hope - an orderly treatment of the matter.

External forces and stress

Your figures are mistaken for two reasons. First, you draw many small arrows directed downward, as if pressure were a vector. It isn't. Pressure has no direction.

Second, your first figure applies to a solid, not to a fluid (liquid or gas). For solids Pascal's law doesn't hold, nor you may simply speak of pressure: the useful concept is stress tensor, a much more complicated thing.

Let's see however how reasoning would work in the solid case. You have a block of solid matter, in form of a truncated cone. Suppose it is resting on a table. Then resultant force on it must vanish. Which are the forces acting on the block? The first is its weight, vertical downward, resultant $m\vg$. The second is table's "reaction" $\vR$, actually a distributed force, applied all over the contact surface. Since net force is vanishing, we know that $\vR=-m\vg$. No forces exist on the oblique lateral face.

In a solid atoms cannot move freely. They may only effect small displacements from their equilibrium positions: thermal vibrations and average net displacements if external forces are acting as in our case. That the latter exist means that no solid is absolutely rigid. Such displacements are necessary to produce forces between atoms, which globally counteract external forces.

Let me explain better this delicate point. In static conditions not only the resultant of external forces must vanish, but the same must happen for whichever portion of the body. If you mentally isolate that portion, you will see it to be subjected to two kinds of forces:

  • force distributed all over the portion's volume (typically it is weight)

  • surface force, acting between atoms residing at the two sides of the portion's boundary.

It is the second kind of forces which at a macroscopic level are represented by the stress tensor. They are generally not normal to the boundary's surface, nor have the same intensity and direction from one point to another of the body.

A solid block

Suppose now you apply a force $\vF$, vertically downward, to the upper base of block. It is easy to understand that at equilibrium $\vR$ will change, becoming $$\vR = -m \vg - \vF.$$ Note that this is true independently of where exactly the force is applied. You may also split it in several sub-forces applied in different points. Or you may think of a force continuously distributed over the entire base. In the latter case it is customary to speak of an applied "pressure" and this the reason why pressure is often thought of (wrongly) as a vector.

What can be said about $\vR$? I mean, $\vR$ is the resultant of the forces the table applies to block's lower base. But how are these forces distributed? The question cannot be answered with the given data. In mechanics jargon, this is a "statically undetermined problem".

Caution: something more could be said by keeping into account the moments of applied forces. However my last statement remains true.

Of course since we do not know the exact distribution of forces in the lower base, the same happens for internal stresses in the block, both when $\vF=0$ and when $\vF\ne0$. This shows the intrinsic complexity of our problem when a solid body is concerned.

Fluid and pressure

And now let's come back (finally! you will say) to the case of a fluid. There are two main differences between this case and that of a solid. First, a fluid needs to be contained in a vessel (a closed one for a gas). Second, what defines a fluid for our problem is that surface forces can only be orthogonal to the surface.

The first difference entails a complication: we have to take into account forces applied by the vessel's lateral wall to the fluid. And if these are not given, it could seem that the problem becomes insoluble.

The second difference instead amounts to a great simplification, because internal stress can be wholly described by a single scalar: pressure. Let's see why is it so, from a microscopic point of view. A solid can transmit a shear force, i.e. a force parallel to the surface where it is applied. The reason is that atoms are only allowed small displacements around their equilibrium positions, but there is no constraint as to the displacement's direction. Therefore atoms near boundary between two portions of the body (see above) may well displace parallel to the surface and in opposite directions on opposite sides of it.

On the contrary atoms (or molecules) in a fluid are more or less free to move around, not being constrained in the vicinity of some point. As a consequence the fluid's bulk cannot resist to a shear stress: it immediately gives in and cancels the stress.

Then in a fluid only normal stresses are allowed. But there is more: it can be shown that when this situation prevails, then in a given point the intensity of (normal) force is always the same, whichever direction you choose for the boundary between two portions of the body. Shortly, we say that stresses are isotropic. Thus we have arrived at the pressure concept.

Pressure is not a force, has no direction (is a scalar). It does not act at the surface of a body, but is present in every internal point. In fact, it is well known that we may have pressure in an unbounded gas. The omnipresent instance is our atmosphere, but think of stars too: they are gigantic gas masses with no bounds, held together only thanks their own gravity.

The case of gases

It is well to open a short parenthesis to mark an important difference between liquids and gases. In liquids atoms are very near to each other, and what we see as macroscopic forces may be correctly interpreted as the resultant of microscopical forces between them.

Not so for gases, where distances are much higher, so that macroscopic forces are better seen as the effect of a myriad of collisions in which atoms exchange momentum one with another or with the vessel's walls.

No rôle for viscosity

Another clarification is in order. I spoke of shearing stresses and of their absence in fluids. Someone could think this is not so for a viscous fluid. After all, viscosity is just defined as the ability of real fluids to transmit shear stresses!

The answer is that we strictly bounded ourselves to static situations. Viscosity only acts when a fluid is moving; it is a force arising because parts of one fluid flow one wrt to another. The classic example is a river, whose water runs faster at centre, whereas grows slower nearing the bank, where it's still.

Therefore in static problems there is no need to restrict to non-viscous fluids. Viscosity has no effect.

To begin with, let's neglect gravity

To understand Pascal's law it's well to neglect gravity, at least initially. This could appear disconcerting, as there is a common misconception that pressure is due to gravity. Many people believe that atmospheric pressure is due to the weight of air above us (which in a sense is true) and conclude (erroneously) that pressure acts "from above". I'm afraid that not all introductory physics books are free from such sin.

Curiously enough, those people forget that they are continuously using objects which are counterexamples to that idea. I'm alluding to tires: of cars, bicycles, and so on. All these are inflated with a pump pressing air within. No rôle is played by atmosphere's weight.

A more exotic example is given by ISS, where a pressure is maintained to keep astronauts' breathing comfortably. Yet there is almost no air outside!

It is easy to show that in such situations pressure is the same in the whole volume of your block, irrespective of the force you can apply to movable parts of the vessel. "Wait a moment!" - I feel like I'm hearing - "Force? Which force? Why should be a force?

It's better to begin with gases, easier to understand, I believe. If the vessel is rigid no action is required by the experimenter. Everything stands still, nothing happens. But we know (I said it before) that the gas molecules continually hit the walls and rebound, giving them some momentum. More exactly, a definite amount of momentum per unit time and surface area. Momentum per unit time equals force. Force per unit area equals pressure. Therefore this momentum exchange is a measure of the gas pressure.

You may not notice that the wall is subjected to that force since usually, if vessel is sufficiently rigid, it automatically develops internal forces that counterbalance those due to gas and keep walls to move or deform. But sometimes things go differently: a balloon inflated at too high a pressure may blow. A welding in a metal tank may leak...

In other cases the vessel is built with a movable part to make experiments (the famous cylinder-with-piston of thermodynamics). This case is obvious: the piston is steady only if force due to gas pressure (force = pressure x area) is contrasted by an equal and opposite force applied from outside.

For a liquid too things go in an analogous way. Instead of collisions exchanging momentum we have forces between neighbouring molecules. Those near a wall interact with piston's molecules and directly apply forces to them. Result is the same: to keep piston steady an opposite external force is needed. The bigger the piston's area, the stronger the force.

If there are two pistons, the same argument applies to both, and we easily conclude that the force required to keep a piston still is proportional to its area. This is what was improperly called "force amplification".

Increasing pressure

We can also see things the other way around. Gas or liquid pressure is determined by the forces applied to piston(s). If you increase external force(s) the fluid will momentarily give in. If liquid, molecules will slightly get nearer one to another; this will increase the repulsive forces between, until a new equilibrium is reached. For a gas a volume reduction will result in a greater number of molecules per unit volume, thus augmenting the number of collisions per unit time against piston, i.e. augmenting pressure. Again, compression will halt when equilibrium is attained.

Gravity comes into play

We may not always neglect gravity. Not for atmosphere only: scuba divers know very well that underwater pressure increases by one atmosphere every ten meters of depth. This contradicts what I said before about pressure being the same at all points in a (still) fluid. The increment of pressure is too easily attributed to weight of the water column above. Sometimes this works, other times doesn't: see hydrostatic paradox in the internet.

Actually what can be experimentally verified is the following law (Stevin): in a fluid at equilibrium in a uniform gravitational field $\vg$ the pressure difference between two any points is $$p_1 - p_2 = \rho\,g\,(z_1 - z_2)$$ if $z$-axis is oriented like $\vg$. Just a simple example: if fluid is water ($\rho=10^3\,\mathrm{kg/m^3}$) and $z_1-z_2=10\,\mathrm m\,$ then $p_1-p_2=98\,\mathrm{kPa}$ which is about 1 atm.

Of course Stevin's law is a consequence of fluid equilibrium under internal and external forces already discussed, with gravity added. I can't dwell on the proof, however.

What about energy?

I cant't close this extra long post without answering the above question. It would be off topic if I'd used really static arguments, but this is not so, since in several places I spoke of "displacements". And when something is displaced with a force applied to it, work is involved and therefore energy.

Here again a distinction must be made between gases and liquids, because of their very different compressibilities. Under ordinary pressures liquids may be assumed incompressible without significant error. This is far from true for gases.

Of course, even an incompressible liquid can accomplish important displacements. Incompressibility only means that overall volume does not change. If the vessel has two pistons of different surface areas it's easy to see that proportionality of force to area, derived above, together with invariable volume, entail that works of external forces on pistons are equal, save for sign. So total work done is zero and liquid's energy does not change.

We could have reasoned in the converse: since energy must be conserved, total work by external forces must be zero, then force is proportional to area. But this argument has a flaw: work is done not only by external forces. Internal forces too can do work. So we have to prove that work of internal forces vanishes. This is not too easy, and requires an inquiry at microscopic level. It is better to assume it as a characteristic property of an incompressible fluid: no work is required to displace it in any way (until kinetic energy is negligible).

Note: I hope reader did notice that my energy argument rested on a hypothesis: equality of pressures on both pistons. But we have seen that this is not true in presence of gravity, if pistons are located at different heights. Let me set aside this complication for now.

As to gases, constant volume cannot be assumed. Nothing forbids to compress or dilate a gas. Furthermore, for gases temperature too becomes important in this respect. But you are not expecting from me a treatise on mechanics and thermodynamics of fluids ... did you?

When gas volume changes, work is done on it by external forces. The relevant formula is well known: $W=-p\,\Delta V$. As I wrote it, this formula is not generally true: it holds if $p$ stays constant during displacement. Otherwise we should write an integral: $$W = -\int_A^B \!\!p\,dV.$$ And this too requires that during transformation $p$, even if not constant, is well defined and the same all over gas volume. Also remember that if volume varies and work is done, to keep $p$ constant energy must be drawn from or given to gas as heat flowing through walls.

Nothing more about gases. A short comment on what happens if pistons are placed at different heights in a vessel containing a heavy fluid. In this case pressures are different, and proportionality between force and area is not respected. Then if pistons are moved, although liquid's volume stays constant, work is done. A positive work if lower piston is moved into liquid, higher piston in the opposite direction.

Question: positive work means liquid gained energy. Where is it to be found? Easy answer: liquid was generally lifted. More precisely, its c.o.m. was lifted. Then liquid's potential energy in the gravity field has increased. For a simple geometry, e.g. a parallelepiped vessel, it is easy to prove that work equates increment of P.E. For a general shape this holds still true, but the proof is more involved.

To conclude. We may say that energy plays no relevant rôle in relation to Pascal's law. There is no "pressure energy" in a liquid, although you will easily find such expression about Bernoulli's theorem (which is out of my actual aim). Just to unravel the mystery: what is improperly called pressure energy is enthalpy density.

  • $\begingroup$ First of all, I am extremely thankful that you have written such an incredible answer. Since about an hour that you posted, I read it over and over again. I even translated it word by word to my native language to ensure that I focused only on physics. I now understand that the pressure is a scalar associated with the intensity of the isotropic stress of each point of the fluid. Even so, I still have doubts, and I apologize for even after this very well done and detailed answer I still have not been able to fully understand Pascal's Law. I feel dumb for not being able to... $\endgroup$ Nov 11, 2018 at 23:46
  • $\begingroup$ ..understand something that seems so obvious to everyone. Now, at least, I know precisely what my brain refuses to accept: My brain thinks that Pascal's law is a violation of Newton's 3rd law. I'll use an example similar to what I used in BowlOfRed's answer comments. Suppose that initially, the internal pressure of the vessel of my drawing is due only to a liquid. Let's say that it's $90\,Pa$. So the force at the bottom, by the contribution of all surfaces will be $90\,Pa\cdot 10\,m^2=900\,N$. $\endgroup$ Nov 11, 2018 at 23:49
  • $\begingroup$ Assuming that the piston has $1\,m^2$, if I push the piston down with, for example, $300\,N$, we have an additional pressure of $300\,Pa$. According to Pascal's Law, the force at the bottom will be $330\,Pa\cdot 10\,m^2=3300\,N$. I now know that this additional force of $2400\,N$ is not a "force amplification", but the reaction produced by the inclined walls. The problem is: if I am injecting $300\,N$ of pressure into the system, how the reaction forces of the walls can be more than $300\,N$ ? $\endgroup$ Nov 11, 2018 at 23:52
  • 1
    $\begingroup$ @ViniciusACP 300N isn't a pressure; it's a force and forces aren't conserved. We can use levers (or a set of pulleys or a hydraulic system) to increase or decrease the magnitude of a force. Your question is identical to asking how pushing one side of a lever with 100N makes the other side push up with 200N. It's a hydraulic lever. $\endgroup$
    – BowlOfRed
    Nov 13, 2018 at 18:16
  • $\begingroup$ @BowlOfRed I wrote "pressure" by mistake. I meant force $\endgroup$ Nov 13, 2018 at 20:31

Fundamentally, you seem think of force as conserved stuff. I put 100N of force in, so I should get 100N of force out. The answer is to simply stop thinking of force that way. It's not stuff. It isn't conserved.

Suppose you have a long lever

lever - By CR, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=1532333

The 5 kg mass lifts the 100 kg mass. But that's impossible! Where did all that extra force come from! Tell me at the molecular level! And none of this garbage about "displacement"!

Yeah, you could try to make up some overly-complicated story that fits those requirements. But a much better explanation is that if the 5kg mass falls 1cm, the 100 kg mass rises only 1/20 cm, so the total gravitational energy is unchanged. Energy behaves like stuff and an "amount in equals amount out" sort of rule applies. Force doesn't work that way.

For an analogy, imagine someone sells apples for \$1 in the market. In their house, they have some chairs that cost \$20 each. Would you demand that they explain how anyone could conceivably buy something that costs \$20 when they only get paid \$1 for what they sell? Would you demand they explain it at the level of how each penny moves? Of course not, because it's not a mystery and not hard to understand. What must be balanced is not the money per unit good, but the total money they take in. (money they make per apple)*(number of apples sold) > (money to buy a chair)*(number of chairs bought).

Force is like money per apple or money per chair. When we multiply force by displacement, we get energy, and that's conserved. If you see someone creating a large force somewhere, whether with an ordinary lever or with hydraulics, it doesn't make any more sense to demand an explanation of how they could possibly create such a large force than it does to demand an explanation of how someone could ever generate a large sum of money. They can generate a large force from a small force as long as they pay for it with a large displacement of the small force, the same as they can pay for an expensive chair with a large volume of apples sold.

But your post demands that no one is allowed to talk about displacement. Sorry, but that's wrongheaded. Displacement is a useful and important concept here. Your post further demands to know about the force at the molecular level, while stipulating that the fluid is incompressible. This just doesn't make sense. The pressure at the molecular level is explained by the repulsion of molecules from each other, which depends on how far apart they are. You can't have "incompressible" and "explain from the molecular level" at the same time.

The reason you were getting nowhere in your reading of previous answers isn't that those answers were bad. It's that you were making strong assumptions going into them - assumptions about what sort of thing force is - and those assumptions were wrong. It will be much more productive to revise your idea of force than to demand that everyone else's explanation kowtow to it.

  • $\begingroup$ I know the answers given were good, so much so that I attribute my lack of understanding to my "dumbness" and not to the quality of the answers. You're right, maybe I should try to revise my concept of "force". $\endgroup$ Nov 13, 2018 at 20:41
  • $\begingroup$ About the displacement assumption, when I made the assumption of no displacement, I was referring to a displacement like the displacement made in the hydraulic press (where the piston moves). In my drawings, there are no displacements involved at MACROscopic level. $\endgroup$ Nov 13, 2018 at 20:43
  • 1
    $\begingroup$ Even when no displacement occurs, displacement is a useful concept to analyze force. See en.wikipedia.org/wiki/Virtual_displacement $\endgroup$ Nov 13, 2018 at 21:01

Your graphic is not showing the contribution from the vessel itself. It might be clearer to see that the angled walls can be approximated by a lot of small stairsteps. Every little step sideways is accompanied by the vessel wall supplying pressure downward.

Or you can imagine a section of a wall supplying pressure inward, and decompose that pressure into sideways and downward components. Either way is the same. The walls will be found above 9m^2 of the bottom, and are pushing down on the fluid with 900N of force.

But that doesn't explain why the force is amplified.

Let me see if I can add some things so that it doesn't seem so strange.

First of all, we are only considering this to be a static system. Because of this, you shouldn't think of the piston as a special source of force here. It's not. Every wall is pushing with the same pressure. The force on the piston isn't causing the entire force on bottom surface, instead all of the surfaces are pushing at the same time.

In fact, maybe you could imagine a vessel with 10 pistons on top. 9 are locked in place, and you push down on one. All the other pistons will have the same pressure (and if the same size, the same force). The only difference is that for the other pistons, the force is coming from the strength of the lock and the attachment to the vessel, not from your hand pushing down.

In this case and in your case, there's 1000N pushing up on the fluid from below and 1000N pushing down on the fluid from above. So there's no amplification of force.

We can change the question to: "why doesn't it matter how small the piston is"?

In this sense it does. In order to get the fluid inside to a pressure of 100Pa, you had to compress it and do work. (In the case of water or a substance that has very low compressibility, the work is very small, but is not zero). The smaller the piston, the further you have to push it to get the same work and the same compression. If the piston were tiny, you would need less force, but more distance and the amount of work done would be the same.

  • $\begingroup$ But that doesn't explain why the force is amplified. For the angled walls to push down the fluid with $900\,N$ of force, by Newton's Third Law, the wall must be pushed with $900\,N$ of force by the fluid. $\endgroup$ Nov 7, 2018 at 14:44
  • $\begingroup$ Thanks for the edit, but I'm still confused. Before my hand pushes the piston, the internal pressure of the vessel is due only to the fluid. Let's say that it's $90\,Pa$. So, the force at the bottom, by the contribution of all surfaces, will be $90\,Pa\cdot 10\,m^2=900\,N$. $\endgroup$ Nov 8, 2018 at 12:01
  • $\begingroup$ After pushing it down with, for example, $300\,N$, according to what I understood from you answer, the force at the bottom will be $900\,N+300\,N=1200\,N$. But, assuming that the piston has $1\,m^2$,we have an additional pressure of $\frac{300\,N}{1\,m^2}=300\,Pa$. By Pascal's Law, the force at the bottom will be $330\,Pa\cdot 10\,m^2=3300\,N$ $\endgroup$ Nov 8, 2018 at 12:02
  • $\begingroup$ Based on what I said, what is my misconception? I'm sorry. I'm trying hard to understand, but I'm still confused. Please, help me. $\endgroup$ Nov 9, 2018 at 14:52

In terms of forces, at molecular level, how pressure is transmitted undiminished to all points in the fluid, causing the amplification of the force when area increases?

Remember your stipulation that there are no displacements. Then, for there to be no displacement within the fluid each drop of fluid must experience a balanced force. That means the force on the left must be equal and opposite to the force on the right, and so forth*. Since the cross sectional area is the same on the left and on the right, that implies that the pressure is also the same. Therefore the pressure is the same throughout the fluid. Since the pressure is undiminished this leads directly to the usual hydraulic force multiplication.

*neglecting the weight of the fluid

In terms of energy, how does the energy associated with the force applied at the top spreads and turns into the force applied at the bottom? (Note: I know that energy, given my assumptions, is conserved | Note 2: just to remember - there are no displacements involved)

With no displacements there is no work and the energy is not relevant. You need to relax the no displacements stipulation to have a meaningful answer about energy.

  • $\begingroup$ "Without displacements there is no work". I made a comment about that in the other answers. I don't know if it's right. I'll post here too. $\endgroup$ Nov 9, 2018 at 3:24
  • $\begingroup$ To connect FORCE with ENERGY it's mandatory that we have a displacement involved. So, even when no apparent displacement is involved, there are "molecular level displacements" involved. This does not violate the assumption of incompressibility, because incompressibility does not mean that the molecules of the fluid are static, as the balls in my drawings, but that the volume of the fluid doesn't change. $\endgroup$ Nov 9, 2018 at 3:28
  • $\begingroup$ So when I apply a force in an incompressible fluid, there is a displacement involved: the displacement between the molecules of the fluid, a "change of seats" between the molecules, so there is energy involved. This energy is transferred to the bottom and the sides of the container by exchange of momentum between molecules. Then, assuming that the sides of the container of my drawing have $10\,m^2$, they will also experience a force of $1000\,N$, just like the bottom. $\endgroup$ Nov 9, 2018 at 3:29
  • $\begingroup$ Is this reasoning valid? $\endgroup$ Nov 9, 2018 at 3:30
  • 1
    $\begingroup$ Again, even allowing internal displacements there is no change in energy. The force is multiplied, but since no displacements are involved there is no energy transfer associated with the forces on either surface and no net energy transfer within the fluid either $\endgroup$
    – Dale
    Nov 9, 2018 at 4:04

Pressure is omnidirecitonal because it is caused by the tranmission of energy between large numbers of tiny particles and the boundary under pressure.

When momentum is imparted from one tiny particle to the other the direction changes, based on the specifics of the interaction (like billiard balls on a pool table)

Each interaction is governed by conservation of momentum $m_1v_1+m_2v_2=m^{'}_1v^{'}_1+m^{'}_2v^{'}_2$

This, occuring randomly over large numbers of interactions, distributes the momentum (and therefore the kinetic energy) evenly in all directions.

Hence Pascal's Principle, pressure is the equal throughout the medium.

  • $\begingroup$ Fancy physics indeed! $\endgroup$
    – Elio Fabri
    Nov 7, 2018 at 10:33
  • $\begingroup$ @ElioFabri Why fancy? $\endgroup$ Nov 7, 2018 at 14:39
  • 1
    $\begingroup$ @Vinicius ACP I'm writing an answer which could also answer to "why fancy". But it's getting rather long and I'm afraid I'll not be able to finish it this evening (here it's 18 pm). $\endgroup$
    – Elio Fabri
    Nov 7, 2018 at 16:59
  • $\begingroup$ @ElioFabri Thank you so much in advance! Just take your time, I'll wait your answer. $\endgroup$ Nov 7, 2018 at 17:21
  • $\begingroup$ @JMLCarter And I thank you very much for the answer, patience and help too. Based on what we've discussed, could you tell me if the reasoning I did in the last two comments is correct? (I reproduced them below) $\endgroup$ Nov 8, 2018 at 6:37

@Vinicius ACP wrote:

if I am injecting 300N of pressure into the system, how the reaction forces of the walls can be more than 300N?

First of all, you didn't inject a pressure, you applied a force. Even units are different. You applied that force on a portion of vessel's walls. So doing, you forced liquid's molecules to get closer one to another, in the whole volume of liquid. This is unavoidable, as I already explained, since molecules can reach a new equilibrium position only when each feels (in the average) a net zero force from its neighbours.

Incidentally, this doesn't happen instantaneously: a compression wave propagates within liquid and when rest is attained liquid's volume is slightly reduced. If you could measure liquid's density in various points, you would see that it's increased, although very little, of the same amount everywhere. Just to give you a concrete figure: if liquid is water, by increasing its pressure by $900\,\mathrm{Pa}$ you cause a relative density increment of about $4\cdot10^{-7}$.

Now molecules all around near vessel's walls are pushed by other molecules, but on the wall's side there are no other liquid molecules to contrast this push. In order to get equilibrium, push must be acted by the wall's molecules. And just to obey Newton's third law, those liquid's molecules must press the wall with an opposite force: $p\,dS$ on every surface element $dS$ of wall.

Bottom included, of course.

  • $\begingroup$ Elio, I wrote "pressure" by mistake. I had not realized it until now. I meant force. $\endgroup$ Nov 13, 2018 at 16:48
  • $\begingroup$ I think now everything is making sense to me!!! As the force increases, the atoms/molecules are approximated more and more. This approximation results in a very little decrease in density (and hence in volume) and causes an increase in repulsive forces between them. And this increase is not equal to the force that I applied, its greater than. Since there is no room for the atoms/molecules to move away, all this repulsion force is suffered by the walls of the container. $\endgroup$ Nov 14, 2018 at 18:21
  • $\begingroup$ And all this interaction between molecules/atoms and between molecules/atoms and the container can be described - as you explained in the other answer - by a single scalar: pressure, because stresses are isotropic in any point of the fluid. And one of the reasons why the stresses are isotropic in fluids is that the repulsive forces depend on the radial distance between atoms/molecules. $\endgroup$ Nov 14, 2018 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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