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Expression for a Floating Cube

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a. The attachment shows a cube floating in a fluid. Derive an expression relating the submerged depth X, the specific weight of the cube, and the specific weight of the fluid.

b. If the cylinder in the second attachment, made of a uniform material, is placed in fresh water at 95 degrees Celsius, how much of its height would be above the surface (ignore kerosene note and 600mm note - refers to another problem?. Assume the specific weight of water is 9.44 kN/m^3.

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Solution Summary

For (a), this solution solves using the equation: volume of the cube Vc = s*s*s = s^3. For (b), the problem is solved in general terms because the specific weight of the cylinder is not given.

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a. Let us assume that the specific weight of the cube and fluid be sc and sf respectively.

Volume of the cube Vc = s*s*s = s^3

The weight of the cube W = sc*Vc
=> W = sc*s^3 ......................(1)

The upthrust of fluid on cube = weight of the fluid displaced W'
W' = sf*V'
where, V' = volume of cube ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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