Object in equilibrium between 2 fluids

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    Equilibrium Fluids
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Homework Help Overview

The discussion revolves around the concept of buoyancy and equilibrium of an object situated at the interface of two fluids in an open tank. Participants are exploring why both fluids exert an upward buoyant force on the object despite the upper fluid not being in direct contact with the lower part of the object.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the pressure differences acting on the object and how these relate to buoyant forces. Questions are raised about the role of pressure at the lower end of the object and the implications of fluid densities and heights on buoyancy.

Discussion Status

Some participants have suggested examining the pressure at different points on the object to clarify the reasoning behind the buoyant forces. There is ongoing exploration of how pressure differences contribute to the forces acting on the object, with no explicit consensus reached yet.

Contextual Notes

Participants are discussing the assumption that the weight of the block does not factor into the pressure calculations, which may influence their understanding of buoyancy in this scenario.

Epiclightning
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Homework Statement


if an object is in equilibrium at the interface of 2 fluids in an open tank, why do both provide upthrust?

Homework Equations


P=P' +dgh

The Attempt at a Solution


The upper fluid is not in contact with the lower part of the object, so why would the fluid provide an upward buoyant force? There is no area at which the fluid could exert a force.
Yet almost all the questions I have seen that deal with such a principle equate mg with the buoyant force due to the lower and upper liquids.
 
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Find the pressure at the lower end of the block and you will understand where you are going wrong.
 
CrazyNinja said:
Find the pressure at the lower end of the block and you will understand where you are going wrong.

That would be P0 + h1d1g + h2d2g + (h1+h2)(dobjectg), right?

h1, h2 = heights of object in each liquid
d1, d2 = their densities
but how would this simplification help further?
 
Epiclightning said:
That would be P0 + h1d1g + h2d2g + (h1+h2)(dobjectg), right?

h1, h2 = heights of object in each liquid
d1, d2 = their densities
but how would this simplification help further?
The pressure on the top of the block is P0.

The pressure on the bottom of the block is P0 + h1d1g + h2d2g

The difference in pressure between the bottom of the block and the top of the block is h1d1g + h2d2g.
So the force is A(h1d1g + h2d2g)=V1d1g+V2d2g. But this is just equal to the sum of the weights of the displaced fluids.
 
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So we don't include the weight of the block when we write the pressure?
 
Epiclightning said:
So we don't include the weight of the block when we write the pressure?

No you don't.
 

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