Weight of the object equals the weight of the displaced water

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SUMMARY

The discussion clarifies the principle of buoyancy, specifically that the buoyant force acting on an object submerged in fluid is equal to the weight of the fluid displaced, not the weight of the object itself. In the case of a 70-kg ancient statue with a volume of 3.0 x 104 cm3, the correct calculation for buoyant force involves using the density of water and the volume of the displaced water, leading to a force less than the weight of the statue. The confusion arises from misinterpreting the relationship between the weight of the object and the buoyant force, which is crucial for understanding submerged objects.

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On my book, it says the weight of the object equals the weight of the displaced water. Here is a question: A 70-kg ancient statue lies at the bottom of the sea. Its volume is 3.0*10^4 cm3. How much force is needed to lift it?

Now, Ignore how much force is needed, just focus on the buoyant force. My book says:Fb=m(water)*g=p(water)Vg. Why should we use volume to solve the problem if the weight of the statue equals to the weight of the object? so we can just do:Fb=m(water)g=m(statue)g=70*10=700N. However, this answer is different from the "correct" one. So I am confused at the concept of equal weight of the fluid and the object.

THanks for help.
 
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The weight of the fluid displaced is not equal to the weight of the displaced object. It is equal to the buoyant force on the object. Since the object is not floating, the buoyant force must be less than the weight of the statue. Therefore, the weight of the displaced water is less than the weight of the statue.

So, when finding the buoyant force, you are going to have to work with the volume and density of the displaced water.

HINT: Start by drawing a free body diagram. What forces are involved?
 

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