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Calculate the buoyant force exerted by the water on the sphere

  1. Jun 13, 2005 #1
    A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 760 N.

    Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
    **for this I set up Bouyant = density *Volume*gravity = 6860 N

    What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
    **here I used Buoyant = mg+T and solved for m, so answer for m= 622 kg

    The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
    **For the sphere to be at rest, mg = B. There is no tension anymore, since the cord was broken. I express B as a function of the volume of the sphere that is still submerged:

    mg = B --> rho*V*g = 0.7 how do I express it as a percentage? I tried 70%
     
  2. jcsd
  3. Jun 13, 2005 #2

    OlderDan

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    The first two look good. The mg on the left side of your last equation involves the mass of the sphere you already calculated. The buoyant force will be rho*V*g, but rho is the density of water and V is the volume of water displaced. This does not equal .700. Use the mg = B part to find B and use B = rho*V*g to find V. The ratio of that V (volume of displaced water) to the volume of the sphere is the fraction submerged.
     
  4. Jun 13, 2005 #3
    622 kg * 9.8 = 6095.6 N

    6095.6N / (1000 * 9.8) = 0.622

    0.622 / 0.700 = 88.9% is this correct?
     
  5. Jun 13, 2005 #4

    OlderDan

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    Looks good to me.
     
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