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Is it possible for something that can't float in a rectangular container to float in a triangular container?

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  • $\begingroup$ You may need to assume that the containers have the same volume $\endgroup$
    – Henry
    Nov 15, 2023 at 2:41
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    $\begingroup$ This problem was solved about 2200 years ago by a Syracusian called Archimedes. You might even have heard the cry "Eureka!" $\endgroup$ Nov 15, 2023 at 10:45
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/316715/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 15, 2023 at 11:36

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No--not if the containers when submerged both displace the same amount of fluid, and if the empty containers have the same weight. If one sinks the other will sink.

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  • $\begingroup$ I think the question uses the word "something" for the object that is about to float or sink, the word "container" for the object containing the fluid (and thus containing the "something" as well). Your "container" is the OP's "something". IMO you answered a different question. In the actual question the variable is the geometry of the "container", not the geometry of "something". $\endgroup$ Nov 15, 2023 at 10:37
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Let's an imaginary vertical cylinder, with a very small radius, crossing the bottom and top surface of an object. The net upward force, disregarding internal stresses inside the object is: $$\frac{P_{bottom}\pi r^2 cos(\theta_1)}{cos(\theta_1)} - \frac{P_{top}\pi r^2 cos(\theta_2)}{cos(\theta_2)} - \rho_o gh \pi r^2 = \pi r^2 (\Delta P - \rho_o gh)$$ On the other hand, the pressure gradient of the liquid is: $\Delta P = \rho_l gh$. As it is valid for any very thin cylinder with volume $\pi r^2 h$ we can say for the whole object, with any form: $$F_{up} = (\rho_l - \rho_o)gV$$

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In a word, "No." Buoyancy depends only on the relative densities of the object and the fluid in which it is immersed.


I confess, I don't know how to prove it. But I'm pretty sure that the proof does not involve any mention of a container. It only mentions the geometry of the object itself (ultimately, in order to prove that the geometry does not matter) and, on the fact that the pressure of the fluid at any given point is directly proportional to the depth at that point.

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    $\begingroup$ Consider the centre of gravity of the combination of the fluid and the floating object. The system will be in its least energy state when the CoG is lowest. $\endgroup$ Nov 15, 2023 at 10:43
  • $\begingroup$ @MartinKealey, are you hinting at a proof in which there is no need to mention either the shape of the object, or the forces acting on it? $\endgroup$ Nov 15, 2023 at 13:01
  • $\begingroup$ That kinda depends on what you take as your axioms. If we're taking "uniform density regardless of shape" as an axiom, then the shape is clearly irrelevant: the water displacement is equal to the volume of the submerged part of the object, and the mass and total volume of the object is fixed. You do kinda need to take the forces into account, but you can simplify those down to pressure in the vertical direction (as a function of depth), since forces in any horizontal direction must cancel out. There are several ways to perform an integration that gets you gravity per volume of displaced water. $\endgroup$ Nov 15, 2023 at 18:57
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The generally accepted answer is "no", simply by application of Archimedes buoyancy principle, but there are limits.

Firstly, if you make your container long enough and thin enough, and place it end-on in the water, the bottom could be at a depth where the pressure exceeds the structural integrity of the container, resulting in it being crushed, and sinking (a bit).

(This is a bit easier to analyse if you simply split your container into two spheres connected by a incompressible rigid rod of negligible volume.)

But if we assume that the bottom of your container is infinitely strong and incompressible...

Secondly, the rate of increase in water pressure is not quite uniform, as gravity decreases the further below the earth's surface you get. This results in slightly less buoyancy.

Thirdly, the composition of the fluid may not be uniform. The bottom is likely to be more dense, resulting in greater buoyancy. (Earth's oceans are generally colder and more saline in the depths than at the surface, except under the ice sheets.)

Fourthly, if you touch the bottom of the ocean then all bets are off.

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