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Dynamics and displacement problem

  1. Jun 17, 2006 #1
    hello i have trouble understanding this problem... hope u guys can help. heres the question:

    'when a beaker of water rests of a balance, the weight indicated in X. a solid object of weight Y in air displaces water of weight Z, when it is floating. what will be the final reading of the balance?'

    according to the answer the reading should be X+Y+Z

    I understand that X should be included (duh) but im confused about Y and Z. Why is the weight of the water displaced included? The water that is displaced comes from the water that is already in the beaker so shouldnt its weight be already included in the X? Also i understand that for an object to float upthrust must be equals to the weight of the object, if they are equal shouldnt the two forces cancel each other out and hence, there is no Y?
  2. jcsd
  3. Jun 17, 2006 #2


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    Assume you have 20lbs of beaker and water, and you add 1 lb of ice (air weight) to the beaker.

    Water is about 800 times as dense as air (dry air at sea level); ice is about 736 times as dense, so the buoyancy of the air on either is mostly insignificant in this case (less than 1%).

    So the bottom line is you have very close to 1lb mass of ice (within 1%), added to the 20lbs of beaker and water. Gravity is pulling downwards on 21lbs of mass, so the scale should read 21lbs, the total of the beaker, water and ice (X+Y).

    Assuming the ice is 8% less dense than water (this depends on the temperature of both), then the ice displaces about .92 lbs of water, but the total weight isn't going to be 21.92 lbs.
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