How Does Archimedes' Paradox Explain Buoyancy in Immiscible Liquids?

  • Context: Undergrad 
  • Thread starter Thread starter aim1732
  • Start date Start date
  • Tags Tags
    Archimedes Paradox
Click For Summary

Discussion Overview

The discussion revolves around Archimedes' principle and its application to buoyancy in a scenario involving a wooden block balanced at the junction of two immiscible liquids. Participants explore the forces acting on the block and the implications of removing one of the liquids.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes a hypothetical situation where removing the upper liquid would increase pressure in the lower liquid, suggesting that the upper liquid indirectly contributes to buoyancy.
  • Another participant emphasizes the importance of considering vertical forces on the block, stating that the buoyant force remains equal to the weight of the fluid displaced, regardless of the upper liquid's presence.
  • A third participant argues that Archimedes' principle is not a paradox but rather a result of pressure differences, detailing the forces acting on the upper and lower surfaces of the block.
  • There is a mention of the relationship between pressures at different heights in the liquids, indicating that the net upward force is derived from the pressure difference across the block.

Areas of Agreement / Disagreement

Participants express differing views on the nature of buoyancy and the role of the upper liquid, with some asserting that it contributes indirectly while others argue that it does not affect the buoyant force. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Some participants suggest a need for a deeper understanding of fluid mechanics, indicating that foundational concepts may be necessary to fully grasp the discussion. There are also references to specific assumptions about the geometry of the block and the pressures involved.

aim1732
Messages
428
Reaction score
2
We are preparing a physcs quiz and are planning to use the following problem

If we have a wooden block balanced at the junction of two immiscible liquids then we have the buoyancy of the two liquids balance the weight of the block.The lower liquid will exert your normal upthrust but the upper liquid also exerts an upthrust even though it is above the block.How does it happen?

We are thinking of a hypothetical situation in which we remove the upper liquid.If we now pour the other liquid then it tends to increase the hitting of the molecules of the lower liquid hence increasing the pressure.That is to say the upper liquid indirectly exerts that pressure using the lower liquid.

I am looking for an alternative approach or verification of the above explanation.
 
Physics news on Phys.org
I am looking for an alternative approach or verification of the above explanation.

If I've got the question right,
make your block have vertical sides so the geometry is simple, i.e. there's vertical force applied only to the bottom and top surfaces.

What's force up on bottom?
What's force down on top?

P X A both places , of course.
Pbottom = ρ1gh1 +ρ2gh2

Dont forget that the liquid above the block also exerts force down on the top of block, so raising level above block does not increase buoyant force. It remains equal to weight of fluid displaced.
 
Last edited:
first thing this is not a paradox at all... u shud go thru d basics of fluid mechanics,,,, Archimedes principle is nothing but net upward force due to pressure difference.,,,
in ur problem the net force on the curved surface is zero (balanced),,, net force on the upper part is downward which is equal to p1xarea; where p1 is the pressure at upper surface,,,,
while net force on the lower surface is upward which is equal to p2xarea ; where p2 is the pressure at the lower surface,,,

the relation between p1 and p2 is :
p2 = p1+ρ1gh1+ρ2gh2

so eventually net upward force(thrust force) comes out is
Fthrust = Fupward - Fdownward
Fthrust = (ρ1gh1+ρ2gh2) x area
where
h1 = height of cylinder in liquid A (upper liquid)
h2 = height of cylinder in liquid B
take obvious meaning of other symbols
 
vikrambij said:
f... u shud go thru d basics ...

[rant]
Sorry, Vik, I didn't read past that. I find the real newspeak to be double-plus ungood...
[/rant]
 
gmax137 said:
[rant]
Sorry, Vik, I didn't read past that. I find the real newspeak to be double-plus ungood...
[/rant]

didn't mean dat,,only if u catch that thing,,,,
 

Similar threads

How does buoyant force relate to weight and pressure?
  • · Replies 5 ·
Replies
5
Views
6K