According to Archimedes Principle,
We know that buoyant force on object ${F_{buoy}}$ is given by,
$F_{buoy} = V_{immersed} * {\rho}_{fluid} * g $
where, $V_{immersed}$ is the volume of object that is immersed in the fluid.
Now, let us consider an object that completely floats and is in static equilibrium
on the fluid surface (more precisely a liquid)
then
$V_{immersed} = 0$
$ \implies F_{buoy} = 0$.
But , we know from given situation that it is in static equilibrium,
Then, is it apt to say that,
$ F_{buoy} = F_{normal}$. where $ F_{normal} =$ normal force on the object in equlibrium
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$\begingroup$ I may not understand completely, but what makes you think such a situation is possible? $\endgroup$– Kristoffer SjööCommented Apr 19, 2021 at 11:52
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$\begingroup$ @KristofferSjöö I think you did not understand the problem fully . Try imagining a balloon floating on water, and assume not a slightest part of its volume is immersed. $\endgroup$– HarshCommented Apr 19, 2021 at 11:57
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2$\begingroup$ It is not possible for no part of the object to be immersed when it is floating -- even a baloon.. $\endgroup$– mike stoneCommented Apr 19, 2021 at 12:36
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4 Answers
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I think it's true that this is not a realistic description of how fluids behave. But it's an interesting limiting situation nonetheless. The key to it is that Archimedes' principle completely ignores surface tension. Surface tension is always there, but when buoyancy is the most important feature, it can be safely ignored.
Not so in the case you're picturing, in which we would have a fluid with enormous surface tension. So much so that it would completely stop the body from affecting its internal molecular cohesion, and override any other hydrostatic effect. In fact, your fluid would be a solid, and Archimedes' principle no longer holds. The normal force that holds the body in place is not buoyancy; it's surface tension.
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It may appear that a body floats completely in water but there must be at least a very small part of the body submerged in water that provides buoyant force.This is mostly observed in case of baddies with large volumes and low densities.
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Whenever you operate in a transition zone, you will find other things messing up your expectations. In the set up you suggest, there is more going on than just buoyancy, which starts to matter, the closer you get to zero buoyancy. You could even have an object actually pull a liquid up, instead of pushing it down. Your objective then gets lost somewhere in the middle between that and ordinary buoyancy, without any particularly interesting results.
Therefore the equation you are trying to put into practice, doesn't have a reality equivalent. Reality doesn't allow you to ignore anything. On your test bench, you will never ever find a truly static equilibrium the way they exist in theory.
But yes, on paper it looks great.
If there is any particular practical problem you need to solve, please get more specific.
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If the object (with non-zero mass) is just touching the fluid's surface, then the fluid doesn't coerce any buoyancy force on the object, meaning that the object's non-zero weight must be compensated by some other force, and that can be (e.g. for a helium balloon) the buoyancy in air.
If there is no other source of upward force besides the fluid buoyancy, then the object will get immersed into the fluid, at least to a tiny degree, just enough to compensate its weight.
P.S. With real-world fluids, there are effects like surface tension that can exert upward as well as downward forces on objects touching the surface, but I deliberately ignored that effect and stayed within the Archimedes model.
