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Consider an arbitrarily shaped body immersed partly in one fluid and partly In another. Can an expression be derived for two different forces applied by the the two fluids on the immersed volumes by using Archimedes principle separately for the displaced fluids?

If so how will an upward force be applied by the upper fluid layer

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    $\begingroup$ HINT: What if fluid 1 was air? $\endgroup$
    – Sam
    Commented Jan 16, 2020 at 16:04
  • $\begingroup$ The question mentioned in the link is no way related to the one @Aditya Ahuja has asked. The OP wants to know how a liquid layer above the object exerts an upward force . The link mentioned is a problem based on this fact and doesn't proves it. The OP wants the proof, justification of the fact. $\endgroup$
    – Guji2203
    Commented Jan 1, 2021 at 7:13
  • $\begingroup$ Buoyant force when object is between two liquids physics.stackexchange.com/questions/467610/…. This can be related. $\endgroup$
    – Guji2203
    Commented Jan 1, 2021 at 8:04

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The total buoyant force on the object comes out of integrating all of the forces per unit area acting on the surface of the object—in this case, due only to pressure.

Because all the forces per unit area add up—the physics of buoyancy is linear—you can think of this physical system as two separate blocks immersed in each fluid, bridged by an infinitesimally thin horizontal layer of fluid at the interface. Using Archimedes' principle on each of these blocks, correctly determining the pressure distribution of the liquid, and adding each of these resultant forces up, you can determine the buoyant force on the single block.

Notice that in that calculation, there is indeed an upwards-pointing contribution to the force on the top block coming from the "fictional" bottom face. This contribution isn't "real", since all of the actual upwards-pointing forces per unit area come from the true bottom of the block, but it generates an equivalent mathematical solution when combined with the forces on the "fictional" bottom block due to the linearity of buoyancy.

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  • $\begingroup$ My physics professor explained this to me by assuming the rigid body to be a uniform cube and calculating the expression for buoyant force by calculating the pressure on the upper and lower faces. The result came out to be the sum of the expressions for buoyant force on the parts of the cube immersed in the liquids. He then said that this result could be generalised for any body. But what about a non-symmetric one? If I integrate over the surface above and below the Fluid interface will I get the same result ? $\endgroup$ Commented Jan 17, 2020 at 6:50
  • $\begingroup$ When you assume a fictional bottom face above the interface you assume that the upper Fluid layer is in contact with it right ? $\endgroup$ Commented Jan 17, 2020 at 6:54
  • $\begingroup$ The principle of splitting up the immersed object works even when the object is non-symmetric; you can think of the asymmetric object as being made up of tiny cubes, for example. And yes, you assume the fluid is making contact with both “fictional” faces. $\endgroup$ Commented Jan 17, 2020 at 10:36

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