I have seen this problem in one textbook (Physics for Scientists and Engineers, Raymond A. Serway, 2018) and in several places on The Internet, where we have a dam and we need to calculate the total force that the water exerts on the dam.

The problem starts setting $ y = 0 $ at the bottom of the dam and proceeds to define the pressure at depth $ h $ as $ P = \rho g(H-y) $

Then goes on with, $ dF = P dA $, etc, etc. Integrates and gets $ \frac{1}{2} \rho g w H^2 $.

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What it's puzzling me is that the premise is that $ P = \rho g(H-y) $. From my point of view, if we set $ y = H $, the top of the dam, they are implying that the pressure at the top is 0 when it should be the atmospheric pressure, 1 atm.

Are they wrong assuming 0 pressure at the top of the dam or am I wrong assuming 1 atm?

  • 2
    $\begingroup$ They are probably just consider it negligible, or implicitly cancelling it out (since you'll have the same 1atm pressure the other side as well (neglecting height difference)). Either seems like good-enough approximations for a rough calculation like this. $\endgroup$ Commented Apr 16, 2021 at 9:31
  • $\begingroup$ Thanks for your comment. I get that, but then what we are really getting, as a result, is the net force on the dam, water force on one side, and air force on the other one. Not the force exerted by the water. $\endgroup$
    – Jon
    Commented Apr 16, 2021 at 9:49

1 Answer 1


Short answer:

  • Whatever the atmospheric pressure is at the top of the dam, it acts on both sides and thus does not produce a net force.
  • Then, you have a pressure form the water on one side and from the air below the top of the dam on the other. Since the density of air is about 1000 times lower than that of water, you can completely neglect this.
  • In general, atmospheric pressure (about 1 bar) corresponds to a water column of 10 m. So for a larger dam, that pressure (that results form the entire atmosphere above you) is again negligible.

Update following Jon's comment:

If you're interested in the "total force that the water exerts on the dam", then you would presumably include atmospheric pressure, sicne the air pushes on the water and the water transfers this pressure to the dam. The fact that the air also pushes on the outside of the dam is then irrelevant, as is the variation of atmospheric pressure with height, i.e. what counts is the atmospheric pressure at the surface of the lake. (Then again, it's not clear whether the "total force that the water exerts on the dam" is very useful, or whether the question was just worded sloppily.)

  • 1
    $\begingroup$ Another possible answer I have found in electron6.phys.utk.edu/PhysicsProblems/Mechanics/… is that the result really is the net force in the dam. And the air atmosphere cancels mathematically on both sides of the dam. $\endgroup$
    – Jon
    Commented Apr 16, 2021 at 19:05
  • $\begingroup$ @Jon in that page,m they state "...force on the side of the dam facing air is $ *P_0 150$...", i.e. they assume constant air pressure. In other words, they neglected the pressure increase width depth. In reality, pressure is lower on mountain tops and higher at sea level, but for the dam, that is totally negligible. $\endgroup$
    – Toffomat
    Commented Apr 16, 2021 at 19:36
  • $\begingroup$ Indeed. I got that. What I tried to say in my comment is that there's another explanation that makes sense mathematically rather than just neglecting atmospheric pressure. Although in that case, it answers to "net force on the dam" rather than "force of the water on the dam". Although as you can say, both are kinda the same. $\endgroup$
    – Jon
    Commented Apr 16, 2021 at 20:42
  • $\begingroup$ I realized that I didn't make myself very clear in the question. The problem asks for "total force that the water exerts on the dam", no "the net force on the dam". In that case, the atmospheric pressure should be considered as the water produces more force because it has more pressure that if it wasn't air on top of it. $\endgroup$
    – Jon
    Commented Apr 21, 2021 at 11:31

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