How does Archimedes' Principle work?

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SUMMARY

Archimedes' Principle asserts that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by that object. This principle can be visualized using a container of water, where the upward force exerted by the fluid equals the weight of the fluid that the object displaces. The discussion emphasizes that the upward force remains constant regardless of whether the displaced fluid is removed, as long as the object is submerged. Additionally, the pressure difference between the top and bottom of the submerged object contributes to the net upward force.

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  • Knowledge of buoyancy and density
  • Basic grasp of forces and equilibrium
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AudioFlux
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the Archimedes' Principle states that the weight of the fluid displaced is equal to the buoyant force. why/how does that happen?
 
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The fluid is trying to fall below you, and has to push you up and out of the way to do so.
 
Imagine a container of water and picture a volume of the water at some depth in the container. It might help to imagine this volume of water in a plastic bag (ignore the weight of the bag...it is only to help your imagination).
This volume of water will not be sinking or rising so the effect of the rest of the water in the container is to provide a force upwards equal to the weight of the water in the plastic bag.
Now replace the plastic bag of water with an identical volume plastic bag of steel. The water in the container continues to provide an upward force equal to the weight of water that was originally in place. Upward force on steel equals upwards force on weight of water that steel has replaced.
No equations, no complicated maths... hope it helps you cope with any equations and maths that you may meet.
 
jetwaterluffy said:
The fluid is trying to fall below you, and has to push you up and out of the way to do so.

but if that happens, then the diagram on the right has buoyancy=0, because the liquid displaced has been removed.
 

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technician said:
Imagine a container of water and picture a volume of the water at some depth in the container. It might help to imagine this volume of water in a plastic bag (ignore the weight of the bag...it is only to help your imagination).
This volume of water will not be sinking or rising so the effect of the rest of the water in the container is to provide a force upwards equal to the weight of the water in the plastic bag.
Now replace the plastic bag of water with an identical volume plastic bag of steel. The water in the container continues to provide an upward force equal to the weight of water that was originally in place. Upward force on steel equals upwards force on weight of water that steel has replaced.
No equations, no complicated maths... hope it helps you cope with any equations and maths that you may meet.

that's cool, but i have already understood how buoyancy is the same for equal displacement of the volume of a fluid. i wanted to know what the displacement of the volume of a fluid has to do with buoyancy and how the mechanism works.
 
The only extra that I can add involves talking about pressure due to a depth of liquid.
If the top of your container is h1 below the surface then the pressure on the top of the container is P = h1ρg and the force on the top of the container is P x A = h1ρgA
Similarly the force on the bottom is h2ρg if the bottom is at a depth of h2.
The upwards force is therefore greater than the downwards for by (h2-h1)ρgA
but this is the weight of the liquid that would occupy the same volume as the oject.
So the upthrust = weight of displaced liquid...(it does not matter whether the displaced liquid came out of the container or not)
I hope this adds to the explanation for you
 
technician said:
The only extra that I can add involves talking about pressure due to a depth of liquid.
If the top of your container is h1 below the surface then the pressure on the top of the container is P = h1ρg and the force on the top of the container is P x A = h1ρgA
Similarly the force on the bottom is h2ρg if the bottom is at a depth of h2.
The upwards force is therefore greater than the downwards for by (h2-h1)ρgA
but this is the weight of the liquid that would occupy the same volume as the oject.
So the upthrust = weight of displaced liquid...(it does not matter whether the displaced liquid came out of the container or not)
I hope this adds to the explanation for you

thanks you, that answered my question :)
 
AudioFlux said:
but if that happens, then the diagram on the right has buoyancy=0, because the liquid displaced has been removed.

In that diagram the ball still has water above it. This water is attracted towards the ground.
 
jetwaterluffy said:
In that diagram the ball still has water above it. This water is attracted towards the ground.

oh. right
 

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