Consider this point $A$, just at the surface of any liquid exposed to atmosphere. The hydrostatic pressure exerted by this liquid of height $h$ is $h\rho g$. My confusion is tha that this pressure should be transmitted in all directions, so shouldn't Point A experience this pressure as well?

The net pressure at A should be : "$h\rho g$ $-$ $P_a$". But, this obviously sounds absurd, the pressure at $A$ is considered to be just atmospheric pressure.

Where could I have gone wrong (excuse me for the not so good diagram)?

enter image description here

  • 1
    $\begingroup$ Pascal's law is stated as $\Delta p = \rho g \Delta h$. I don't see how you would conclude from that that $A$ should experience any pressure from the fluid at all. $\endgroup$ – ACuriousMind Nov 6 '14 at 10:49
  • $\begingroup$ Let's not involve pascals' law then. There is no change of pressure, just this hydrostatic pressure. Can we now continue the discussion? $\endgroup$ – Swami Nov 6 '14 at 10:51
  • $\begingroup$ Pascal's law means that the pressure force on any elementary surface is independent of the direction of the surface. It does not means that the pressure can not vary from point to point. $\endgroup$ – Whelp Nov 6 '14 at 12:10

The net pressure on the liquid is just the atmospheric pressure. Pressure in a fluid acts in every direction, but as the point is on the surface, $\text{P}_{water}=h\rho g=0$ as $h=0$. So only atmospheric pressure will be acting on point A.

The height of the liquid column doesn't affect the pressure on top. Pressure in a liquid is affected by the weight of the liquid above it.


Consider this diagram, :

enter image description here

  • The pressure on a cross section inside the fluid that height is given by the weight of fluid above that height, divided by horizontal cross sectional area of the fluid column. We get the pressure by $$P=\frac{F}{A}=\frac{mg}{A}$$ $$=\frac{\rho Vg}{A}=\frac{\rho (A\times h)g}{A}$$ $$\boxed{P=h\rho g}$$

  • When we calculate the pressure at the surface of the fluid, we take the weight of fluid above the cross sectional area. But as there is no fluid above the surface, $\text{Weight}=0,~~ \therefore \text{P}=0$

  • $\begingroup$ as ACuriousMind has already stated in a comment above. $\endgroup$ – user49111 Nov 6 '14 at 10:55
  • $\begingroup$ That's what I am asking, why does it not? As indicated by arrows, pressure is being exerted by the liquid on point 'A', why has it vanished, mathematically or physically? Since pressure does not have any direction, why can't it act upwards? $\endgroup$ – Swami Nov 6 '14 at 11:00
  • $\begingroup$ It's too different, we can't expect buoyancy for a point on the surface. $\endgroup$ – Swami Nov 6 '14 at 11:06
  • $\begingroup$ No, my friend, it's still the same explanation. Why do we not expect the liquid below the surface to exert a pressure in upward direction? $\endgroup$ – Swami Nov 6 '14 at 11:22
  • $\begingroup$ because weight is exerted downwards. $\endgroup$ – user49111 Nov 6 '14 at 11:25

when you say that pressure is exerted everywhere that is correct. at the surface the force direction that you have shown will exert pressure but that will be nullified by atmospheric pressure that is why probably you are getting a negative sign in your expression. on molecular level Surface pressure can also refer to the change of surface tension as a function of the area of water surface available to each molecule in a solution. The measurement of surface pressure is key in determining the monolayer properties of a liquid.

  • $\begingroup$ if that is nullified, it would mean net pressure at point A should be zero. $\endgroup$ – Swami Nov 6 '14 at 11:24
  • $\begingroup$ Correct that is Zero just like net force is zero however forces are still acting.... So the same way you can say it for pressure also.... $\endgroup$ – Einstein'sLover Nov 6 '14 at 11:28
  • $\begingroup$ But the pressure at the surface, at point A, as the textbooks say, is atmospheric. $\endgroup$ – Swami Nov 6 '14 at 11:28
  • $\begingroup$ seems you are presenting half truths..ha.. surface will have two sides the upper has atmospheric pressure and lower has also some pressure due to fluid. Had this pressure not been there your fluid would be compresses. Another proof for this forces always act in pair this is an universal law so if there is a force up something has to balance that and it is this force that is coming from the fluid. $\endgroup$ – Einstein'sLover Nov 6 '14 at 11:36
  • $\begingroup$ Your argument is logical, but any surface that is exposed to atmosphere is said to have an atmospheric pressure only as in the case of Torricelli's barometer. $\endgroup$ – Swami Nov 6 '14 at 11:38

In Pascal's pressure transmission law, pressure which is transmitted appears due to external work-done on the continuous liquid and liquid being in-compressible this work done the liquid on the other external system. In this way pressure is transmitted through the continuous liquid. But in your system pressure appears due to the weight of the liquid not by external work done. That's why here seems the violation of Pascal's law.

  • $\begingroup$ Can we openly say that pressure that is exerted by the weight of this liquid column is exerted in all directions? (as pressure has no fixed direction) $\endgroup$ – Swami Nov 6 '14 at 11:29
  • $\begingroup$ It's not only exerted by weight but also by any external agency. I mean for transmission of pressure through out the liquid we must have to done work on the liquid by any external agency. But in this case the pressure appears due to weight of the liquid itself.-Swami $\endgroup$ – Rajesh Sardar Nov 7 '14 at 6:35

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