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If an object is dissolved in a container of two liquid layers so that volume $V_1$ is in liquid 1 and volume $V_2$ is in liquid 2. $ρ_1>ρ_2$ be given where it refers to respective densities.

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I want to know why the buoyant force liquid layer 1 gets included though it's above the object. I have tried myself that the liquid layer 1 would exert a pressure on top of the object and also a pressure on layer 2 which transmits to the bottom of object by Pascal's law and the net is $ρ_1V_1g$. I have proven it for some regular objects but can't do it for irregular any rigid object. I know that layer 1 doesn't directly acts buoyant force it's a pure mathematical coincidence . I want to know the theoretical proof of this coincidence for any irregular, rigid body(i have tried integrating but couldn't proceed). Also if anyone can help in extending the concept to n layers would also helpful.

Gert
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  • https://physics.stackexchange.com/questions/467610/buoyant-force-when-object-is-between-two-liquids –  Jan 01 '21 at 18:02
  • https://physics.stackexchange.com/questions/467610/buoyant-force-when-object-is-between-two-liquids –  Jan 01 '21 at 18:03
  • https://physics.stackexchange.com/questions/525363/archimedes-principle-in-case-of-two-fluids. K. L https://physics.stackexchange.com/questions/433529/buoyancy-force-on-object-between-three-liquids –  Jan 01 '21 at 18:06
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    All of the above mentioned links are related but don't answer my doubt . I want the quantitative treatment and proof of the mathematical coincidence. The others treat it qualitatively. –  Jan 01 '21 at 18:08
  • Any help would be appreciated. Thank you –  Jan 01 '21 at 18:13
  • Hint: buoyant force equals area multiplied by pressure. Also, pressure only acts perpendicular to surfaces. For your cube, calculate the force pushing up on the bottom of the cube and the force pushing down on the top of the cube. Then calculate the sum of the weight of each fluid that is displaced. – David White Jan 01 '21 at 23:32
  • I can calculate for the regular objects like cube. I am unable to prove for any generalized irregular rigid body. Thanks –  Jan 02 '21 at 05:14
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    A plus b, note that physics uses math to draw a conclusion, but physics is NOT math. If you can't accept the intuitive fact that Archimedes principle applies to submerged bodies of any shape, you are going to have a difficult time. I suggest you run experiments on submerged bodies of various "odd" shapes to convince yourself that the buoyant force is actually equal to the weight of the fluid displaced and is independent of the shape of the body. – David White Jan 02 '21 at 05:29
  • @David White you are right thanks –  Jan 02 '21 at 05:36
  • I am getting difficulty here that if we consider a mixture of n fluids say 47 one over the other with $v_1,v_2,...v_{47} $ volumes respectively in each fluid the Archimedes principle of fluid displacement won't directly apply here because there's no direct displacement of fluid in the middle layers(like 34th,26th...) it just would surround the object. Can you help me in how to extend Archimedes principle to this. Any physical intuition is there? –  Jan 02 '21 at 18:08

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