Why the fluid pressure does not affected by the object submerged?

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Discussion Overview

The discussion centers around the concept of fluid pressure in relation to submerged objects, specifically questioning why the pressure below a submerged object is not calculated using the object's density. Participants explore the principles of buoyancy, equilibrium, and the definitions of pressure in fluids.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why the pressure below a submerged object is not calculated using the object's density, suggesting it may relate to equilibrium or other fluid properties.
  • Another participant asserts that the water pressure on an object is defined by the weight of the water above it, independent of the object's density.
  • A different viewpoint emphasizes that the derivation of the pressure-depth relation relies on the weight of the water column and questions the applicability of this relation when the object is not in equilibrium.
  • One participant introduces an intuitive explanation of buoyancy by comparing the object to an equivalent volume of water, suggesting that the forces acting on both should be similar.
  • Another participant reiterates that the pressure difference between the top and bottom of the submerged object leads to buoyancy, noting that atmospheric pressure does not affect this difference in the first approximation.
  • One participant discusses the indirect effect of an object on water pressure, mentioning that submerging a floating object increases the water height slightly.
  • A later reply expresses gratitude for the insights shared in the discussion, indicating that the conversation has been helpful.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the pressure exerted by water and the density of submerged objects. The discussion remains unresolved with multiple competing perspectives on the principles of buoyancy and fluid pressure.

Contextual Notes

Some participants highlight limitations in the assumptions made during the derivation of pressure relations, particularly regarding equilibrium conditions and the role of atmospheric pressure.

tsw99
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As we know P=P_{0}+\rho gh
If now I put an object of height h into the water, just completely submerged,
why the pressure below the object not P=P_{0}+\rho_{obj}gh instead?
Is it because it may not be in equilibrium or is it due to some other properties of fluid?
Thanks!
 
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Why do you think it is not?
 
The water pressure acting on an object is, by definition, the force (per unit area) the water is applying to the object, which is just the weight of the water above the object, not the other way around. The density of the object has nothing to do with the weight of the the water.
 
But if we see the derivation of the first equation, we use the relation of the weight of the column of water (not other objects) equals to the net force arises from the water above and below.
If the density of the object is greater than that of water, it will be sinking obviously, so the system is not in equilibrium. Why can we still use the relation P=P_{0}+\rho gh which comes from the equilibrium condition of water?
 
In my opinion, here's the most intuitive way to think about "buoyancy." First, take a bit of water with the same shape as the object being put in, and put it in instead of the object. We'll assume the system will be in equilibrium. The weight of this bit of water will be pulling down on it, so an equal force must be pushing it up for the system to be in equilibrium. We call this force buoyancy. Now replace the water with the object again. The same force should be being applied. This should help you understand what's going on.

Anyways, no fluid will ever be in equilibrium; these formulae are more of an approximation anyways.
 
tsw99 said:
But if we see the derivation of the first equation, we use the relation of the weight of the column of water (not other objects) equals to the net force arises from the water above and below.
If the density of the object is greater than that of water, it will be sinking obviously, so the system is not in equilibrium. Why can we still use the relation P=P_{0}+\rho gh which comes from the equilibrium condition of water?

The pressure at the bottom of the (barely) submerged object is
P_{bottom}=P_{0}+\rho gh
The pressure on the top surgace of the object is
P_{top}=P_{0}
The difference is \rho gh
The buoyant force is due to the difference so it does not matter if you include atmospheric pressure or not, at least in the first approximation. And the difference is the same at any depth, again in the first approximation.
So including the atmospheric pressure does not change the buoyancy of the object.
 
Whovian said:
In my opinion, here's the most intuitive way to think about "buoyancy." First, take a bit of water with the same shape as the object being put in, and put it in instead of the object. We'll assume the system will be in equilibrium. The weight of this bit of water will be pulling down on it, so an equal force must be pushing it up for the system to be in equilibrium. We call this force buoyancy. Now replace the water with the object again. The same force should be being applied. This should help you understand what's going on.

Anyways, no fluid will ever be in equilibrium; these formulae are more of an approximation anyways.
Yes, I understand what you say, but as I said when we derive the Pressure-depth relation, we consider the weight of the water to find out the pressure. But if we replace the water by an object with same mass, my question is, why we can say that the pressure exerted by the water is still the same?
 
Any increase in pressure at some specific point in the container will correspond to the increase in the height of the surface of the water due to the object being placed onto or in the water, which is an indirect effect of the object being placed in the water. If a floating object is forced to be submerged, it only increases the height of the water by a small amount.
 
Thanks a lot guys..
this helped me a lot.. I Googled this and didn't expect finding great answer like this one.

And I registered specifically to thank you guys
 

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