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Homework Help: Weight of the object equals the weight of the displaced water

  1. May 11, 2008 #1

    MIA6

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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.
     
    Last edited: May 11, 2008
    MIA6, May 11, 2008
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  3. May 11, 2008 #2

    G01

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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 bouyant 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 bouyant 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?
     
    G01, May 11, 2008
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