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Fluid Mechanics metal cube

  1. Dec 12, 2011 #1
    A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube :

    Ans: Will remain the same.

    But how? Won't the force exerted in the second case,i.e when the vessel is filled with water be less because of the buoyant force?


    Please help!
     
  2. jcsd
  3. Dec 12, 2011 #2

    Doc Al

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    Staff: Mentor

    Presumably: There is no water layer beneath the cube, so it remains in direct contact with the surface of the vessel. And the water level is just up to the top of the cube, but does not cover it.
     
  4. Dec 13, 2011 #3
    I have another doubt.
    The force which the cube will exert on the bottom will be equal to : pressure*area , i.e., (height*density*gravitational acceleration) * area of the bottom.
    Won't this force vary in the two cases? where in case 1, air is the fluid(density almost negligible) and in case 2 water is the fluid (much more dense)
     
  5. Dec 13, 2011 #4

    Doc Al

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    You're going to have to describe the situation in more detail. (See the presumptions of my last post.) Assuming that the water does not cover the cube, why would the presence of the water increase the pressure that the cube exerts on the bottom?
     
  6. Dec 13, 2011 #5
    This is the downward force due to gravity: F=mg. It stays constant.
    The upward force due to buoyancy is obtained from the pressure difference between the top and the bottom of your object. The density on the bottom of your solid is the density of the 'floor'. The density on the top is the density of air. These stay the same, regardless of any density changes at the sides of the cube.

    In practice however, it will be difficult to prevent the occurrence of a micro layer of water between the solid and the floor. As soon as the solid is detached from the floor, you will get buoyancy.
     
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