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I simply do not understand how point P2 can be at the same pressure as point P1.

As I understand it, the pressure at those points is due to the column of water directly above the point. Well, in a tilted tube, the column of water directly above P2 is not very large at all.

I know that in other shapes of tubes that this difference is made up for with the normal force reaction of the tube itself pushing down on the water, but I don't see how that applies in this case.

EDIT: I think I mixed a couple things up, since the hydrostatic paradox is about the pressure on the bottom surface not matching up with the weight of the tube, and that discrepancy is resolved by considering the pressure against other surfaces of the tube and the consequent normal force reaction.

My problem, I suppose, is that pressure is always described as being due to the weight of the column of water above the point, but there are many situations in which that makes no sense. Can anybody speak to whether or not the "weight of the column of water" idea is nonsensical or not?

  • $\begingroup$ if h and h are the same the pressure will be the same, if h and h are different the pressure is also different $\endgroup$ Commented Sep 27, 2019 at 6:22
  • $\begingroup$ are they supposed to be the same tube? If so the scale is off, tilting the same tube will lower pressure at the bottom $\endgroup$ Commented Sep 27, 2019 at 6:28
  • $\begingroup$ No, they are different tubes. I knew I should have made that clear. $\endgroup$
    – Pixel S
    Commented Sep 27, 2019 at 6:46
  • $\begingroup$ you are right, that expression only applies on simple configurations $\endgroup$
    – user65081
    Commented Sep 27, 2019 at 14:29
  • $\begingroup$ Yeah, math dudes are bad at explaining things, this is humanitarian field. They spoil it in a funnel experiment calling it "height above". The correct word related to pressure maybe "depth". $\endgroup$ Commented Nov 24, 2021 at 11:32

3 Answers 3


What you have neglected to consider is the effect of the walls of the container which exert forces on the water.

Look at the sequence below where on the left there is just water and then as you progress to the right the walls of the container are added nd the hole closed which does not change the pressure at point $P$ and finally the water outside the container is removed.

enter image description here

The pressure at point $P$ is till the same.

The water at point $P$ can only interact with the water which immediately surrounds it and it has no "knowledge" of the rest of the water (and walls of the container).


Since they represent different tubes, h and h are probably the same height, otherwise they would be labeled differently. Water pressure increases with depth from the surface, not volume. So with water of the same purity and temperature, the pressure at the same depth from the surface will be equal in both tubes.


Consider a small cube (of size $\Delta x$) of fluid at equilibrium. If you draw a free body diagram you see that there are 7 forces acting on the box of fluid: one pressure force acting on each face and the weight. Since it is in equilibrium the forces on the horizontal faces are all equal and therefore the pressure does not change horizontally. However, to maintain equilibrium the pressure force on the bottom must be equal to the sum of the pressure force on the top and the weight of the box, and therefore pressure does change vertically.

Next, let’s determine how the pressure changes vertically. The pressure at the bottom of the cube is $P(x)$ and the pressure at the top of the cube is $P(x+\Delta x)$. The weight of the cube is $\rho g \Delta x^3$, and the force on the bottom is $P(x) \Delta x^2$ so $$P(x+\Delta x) \Delta x^2 - P(x) \Delta x^2 = -\rho g \Delta x^3$$ dividing both sides by $\Delta x^3$ and taking the limit as $\Delta x \to 0$ we get $$\frac{dP}{dx}=-\rho g$$ which we can solve to get $$P(x)=-\rho g x + P(0)$$

So, we can set the surface of the water to $x=0$ and then $-\rho g x$ is the weight of a column of water above $x$. Most explanations stop there, however it is important to not neglect the other term $P(x)=-\rho g x$ only if $P(0)=0$. In other words, the weight of the water does not give the pressure, it gives the change in pressure! This is unfortunately neglected in many explanations leading to the confusion you express.

The column of water above P2 is indeed quite short, but the pressure at the point where the column reaches the cylinder is not zero. So the weight of that column of water does not give you the pressure but only the change in pressure from the top of the column to P2. Since the pressure does not change horizontally you can see that the usual formula applies regardless of the shape of the container provided $x$ is placed so that $P(0)=0$.

So the “weight of the column of water" idea is not nonsensical, but it equals the change in pressure from the top to the bottom, not the pressure at the bottom.


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