I saw in this answer that the atmospheric pressure is conveniently canceled mathematically when taking the atmospheric pressure at the level of the top of the object.

However, when I was inventing cases to practice buoyancy it occurred to me to derive the same thing taking into account the buoyant force of the air on the object, assuming that the density of the air is constant as the change of height is very very low.

Assuming that the point at the top of the object has pressure $ P_{top} $ and in the waterline the pressure is atmospheric. I can express the buoyant force of the air on the object as.

$$ P_{atm} = P_{top} + \rho_{air} g h \\ A(P_{atm} - P_{top}) = A\rho_{air} g h \\ F_{atm} - F_{top} = A\rho_{air} g h \\\ F_{buoyant-air} = A\rho_{air} g h $$

Using some random numbers, as assuming a cross-area of $ 1 m^2 $ and that the height of the part of the object not submerged in water is also $ 1 m $ the force exerted by the air at the top is $ ~11.76 N $

If the other half of the object, of also $ 1 m $ is underwater, the buoyant force of the water is around $ ~9800 N $ so one could say that the one by the air is negligible in comparison.

Otherwise, the net force of the fluids on the object would be:

$$ F_{air-top} = A(P_{Atm} - \rho_{air}gh) \\ F_{water-bottom} = A(P_{Atm} + \rho_{water}gh) \\ F_{buoyant-air-water} = F_{water-bottom} - F_{air-top} = A\rho_{water}gh + A\rho_{air}gh $$

I'm wondering if this actually the case? Are we implicitly neglecting the buoyant force of the air and focusing on only the water buoyancy when calculating the buoyancy of a half-submerged object?

  • $\begingroup$ Buoyant force of air is not neglected . The formula which says that upward force due to water is equal to weight of displaced water includes the effect of atmospheric pressure. You need to look at the derivation of how the buoyancy formula is calculated to understand that the force exerted by air pressure on the top portion of the body which is outside of water is taken into account as well. $\endgroup$ Commented Apr 21, 2021 at 15:28
  • $\begingroup$ @silverrahul Could you point out the flaws in my derivations then? The usual derivations of buoyancy assume that the pressure at the top of the object is the same as the pressure at the waterline. $\endgroup$
    – Jon
    Commented Apr 21, 2021 at 21:30
  • $\begingroup$ i did not understand your derivation. A simple diagram could be helpful. or explain those 4 equations better $\endgroup$ Commented Apr 21, 2021 at 22:19
  • $\begingroup$ I looked at your question again, and i think you have got a point $\endgroup$ Commented Apr 22, 2021 at 5:24

2 Answers 2


Clearly yes. Suppose the half submerged object was a boat containing a cylinder of compressed helium and a large uninflated balloon tethered to the deck. If the balloon is inflated, the buoyancy offered by the air becomes more significant and the boat will sit higher in the water, eventually leaving the water if the balloon is sufficiently large. The weight of the boat and balloon has not changed, but the uplift from buoyancy due to the air has increased,

  • $\begingroup$ If you inflate the balloon the weight will change, but I assume your point is that the average density will decrease and at some point the buoyancy of the air will matter? $\endgroup$
    – Jon
    Commented Apr 21, 2021 at 22:25
  • $\begingroup$ The density will decrease, but the mass will not. $\endgroup$ Commented Apr 23, 2021 at 5:32
  • $\begingroup$ How come? Doesn't an inflated balloon weight more than an non inflated balloon? $\endgroup$
    – Jon
    Commented Apr 23, 2021 at 7:36
  • 1
    $\begingroup$ The cylinder of gas was already on the boat! $\endgroup$ Commented Apr 23, 2021 at 8:26
  • $\begingroup$ Ah! Fair enough! $\endgroup$
    – Jon
    Commented Apr 23, 2021 at 8:52

Since the air pressure drops slowly with increasing altitude, the pressure on the top of a partially submerged object is usually assumed to be constant (and the same as at water level).


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