Archimedes Principle: B = δVg + T

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When a metal object is submerged in water, it experiences two upward forces: buoyancy (B = δVg) and the tension in the string (T). The downward force acting on the object is its weight, which is the gravitational force. The buoyant force arises from the displaced water, counteracting the weight of the object. Clarification is needed regarding the concept of "taking out an imaginary area," as it may confuse the understanding of forces acting on submerged objects. Understanding buoyancy is crucial for grasping the balance of forces in fluid mechanics.
Kork
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Hi :)

When I have a container with water and put a piece of metal down in the water while it's hanging on the string, I know that there will be two forces pulling it upwords:

B = δVg
and the string force T

But the force pulling it down, is that the gravitation? Or is it a force that equals to the weight of the piece of metal?
 
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Kork said:
But the force pulling it down, is that the gravitation? Or is it a force that equals to the weight of the piece of metal?
Yes and yes. Gravity pulls the metal downward. (That gravitational force is the weight of the object.)
 
I don't think I quite understand the buoyancy part.

If I tak a cointainer with water and take out an imaginaray area then the forces on the area would be the buoyancy and the gravitationa pulling down? Right?
 
Thank you very much!
 
Kork said:
I don't think I quite understand the buoyancy part.

If I tak a cointainer with water and take out an imaginaray area then the forces on the area would be the buoyancy and the gravitationa pulling down? Right?
Not sure what you mean by 'take out an imaginary area'. The forces on the submerged object will include the buoyant force and the weight of the object. (Maybe you can restate the question.)
 
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