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Why is atmospheric pressure it not taken into account when finding the force exerted by incompressible fluids onto a surface within the fluid? If you don't understand my question, then consider the example below:

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  • $\begingroup$ it looks like there's air all around the swimming pool, so that atmospheric pressure acts on the other side (the dry side) of the floor of the swimming pool $\endgroup$
    – basics
    Nov 24 at 21:45

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I find this text a bit confusing. The pressure field: $$P(z)=P_0+\rho gz$$ assuming that $\rho$ is constant. If $z=0$ is the surface of the water, then $P_0$ is the atmospheric pressure (the pressure field is continuous).

The text you're quoting says "due to the water". The author probably sees the above formula as a sum: $$P(z)=\underbrace{P_0}_{\text{atm}}+\underbrace{\rho gz}_{\text{water}}$$ In other words, $\rho gz$ is the additional pressure due to the water.

Assuming that the bottom of the pool is exposed to atmospheric pressure under it (for example because there's a tunnel letting air in from the surface), there yes, as you said, $P_0$ is present on both sides of the bottom and vanishes from the equation.

This reasoning also assumes that atmospheric pressure is constant. Since air has a very low density, this is a good approximation as long as the height difference remains reasonably small (a few dozen meters, perhaps more).

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  • $\begingroup$ Thank you for your answer! $\endgroup$ Nov 25 at 7:54
  • $\begingroup$ That additional pressure (difference from atmospheric) is sometimes called gauge pressure (e.g., that seen on the gauge of your bicycle pump). I hadn't thought about it that way, but yes it makes some sense that gauge pressure is what you'd be worried about when filling a pool. $\endgroup$
    – Ben H
    Nov 25 at 12:23

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