physics, buoyance

A wood cube 0.20m on each side has a density of 730kg/m3 and floats levelly in water.
Part A
What is the distance from the top of the wood to the water surface?

Part B
What mass has to be placed on top of the wood so that its top is just at the water level?

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  1. A. Hb=(Do/Dw)*Ho=(730/1000)*0.2=0.146 m
    Below the the water.
    Ha = 0.2 - 0.146 = 0.054 m Above water.

    B.Mc=V*D = 0.2^3m^3 * 730kg/m^3=5.84 kg.
    = Mass of cube.

    (Mc+Mo)/Vc = Dw(Density of water).
    (5.84+Mo)/0.008 = 1000kg/m^3
    5.84 + Mo = 8
    Mo = 2.16 kg = Mass of object to be added.

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