# Buoyancy and the Archimedes Principle

tomtomtom1
Homework Statement:
Buoyancy and Archimedes Principle
Relevant Equations:
Buoyancy and Archimedes Principle
Hello all

I was hoping someone could help me with some intuition regarding Buoyancy and Archimedes Principle.

I thought i understood Buoyancy in terms of pressure but when i came across Archimedes Principle I realized I did not fully understand what Buoyancy is.

I have an object that is a cube 3 x 3 x 3 meters with a density of 900kg/m^3.

This object is submerged in water such that the top of the object is 15m below the surface and the bottom of the object is 18m below the surface.

I have calculated the pressure at the top & bottom of the object and since pressure increases with depth i knew the pressure at the bottom would be greater than the top.

So i thought that Buoyancy is the difference in pressure between the top and bottom of the submerged object and since the pressure was greater at the bottom then the fluid would want to push the object up to the surface.

I drew a diagram of my thinking:-

https://physicshelpforum.com/attachments/or-jpg.2935/

Based on the above my buoyancy force is the difference in pressure and in this case 2940.

However i have read about Archimedes Principle which states that the buoyancy force a submerged object experiences is equal to the weight of the fluid it displaces.

With this in mind I added the weight of the object due to gravity and included the buoyancy force the object would experience into my diagram to get:-

https://physicshelpforum.com/attachments/ap-jpg.2937/

Factoring in Archimedes Principle i know that this object will float because the buoyancy force is greater than the weight of the object - the bit i am struggling to understand is how does the pressure at the top and bottom of the object get factored in this?

I hope this makes sense.

Can anyone shed any light?

Thank you.

Mentor
Factoring in Archimedes Principle i know that this object will float because the buoyancy force is greater than the weight of the object - the bit i am struggling to understand is how does the pressure at the top and bottom of the object get factored in this?
The buoyant force is the net force on the submerged object due to the pressure of the surrounding fluid. Since pressure increases with depth, there's a net upward force from the fluid. Archimedes' Principle is the realization that this buoyant force happens to equal the weight of the displaced fluid. (You can also find the buoyant force by calculating the force on each surface and adding them to find the net force.)

• tomtomtom1
Homework Helper
Gold Member
2022 Award
With this in mind I added the weight of the object due to gravity and included the buoyancy force the object would experience into my diagram
The links to your diagrams are broken, so I'm not sure what you did there.
You say you "added" the weight (and buoyancy force according to Archimedes?) to your earlier analysis using pressures. The pressure analysis and Archimedes' principle are two ways of analysing the same problem. You cannot combine them in one analysis. Maybe you did not mean that.

But there is anyway a subtlety with Archimedes' principle. It assumes that the fluid can reach all surfaces of the body that are below the surface of the fluid. That is not the case for, e.g., a rubber suction pad stuck to the bottom of the vessel. In that situation you have to subtract the missing pressure x area.

• tomtomtom1
Cutter Ketch
I will redundantly state what others have already said, but hopefully also add a little intuition regarding why.

The difference in the force of pressure on the bottom and the top and the weight of the displaced water are two sides of the same coin. You don’t need to consider both. They are the same thing.

To understand intuitively why this is true ask yourself why the pressure increases with depth. The increase of pressure with depth is because of the weight of the water above, so, naturally, the pressure differential is related to the weight of the displaced water.

• tomtomtom1
tomtomtom1
I will redundantly state what others have already said, but hopefully also add a little intuition regarding why.

The difference in the force of pressure on the bottom and the top and the weight of the displaced water are two sides of the same coin. You don’t need to consider both. They are the same thing.

To understand intuitively why this is true ask yourself why the pressure increases with depth. The increase of pressure with depth is because of the weight of the water above, so, naturally, the pressure differential is related to the weight of the displaced water.

Thanks

tomtomtom1
The buoyant force is the net force on the submerged object due to the pressure of the surrounding fluid. Since pressure increases with depth, there's a net upward force from the fluid. Archimedes' Principle is the realization that this buoyant force happens to equal the weight of the displaced fluid. (You can also find the buoyant force by calculating the force on each surface and adding them to find the net force.)

Got it thanks - good feeling when the penny drops :).

Thank you.

tomtomtom1
The links to your diagrams are broken, so I'm not sure what you did there.
You say you "added" the weight (and buoyancy force according to Archimedes?) to your earlier analysis using pressures. The pressure analysis and Archimedes' principle are two ways of analysing the same problem. You cannot combine them in one analysis. Maybe you did not mean that.

But there is anyway a subtlety with Archimedes' principle. It assumes that the fluid can reach all surfaces of the body that are below the surface of the fluid. That is not the case for, e.g., a rubber suction pad stuck to the bottom of the vessel. In that situation you have to subtract the missing pressure x area.

Haruspex as always thank you.