A cube of wood having an edge dimension of 19.1 cm and a density of 653 kg/m3 fl

Question

A cube of wood having an edge dimension of 19.1 cm and a density of 653 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? Please provide numerical answer in cm (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface? Please provide numerical answer in kg

Explanation / Answer

a)given side of cube = 19.1 cm = 0.191 m = a Density = 653 kg/m^3 = x Volume of cube = a^3 = (0.191^3) m^3 Mass of cube = Density*volume = (0.191^3)*653 = Let the distance from the horizontal top surface of the cube to the water level = y m =>the distance from the horizontal bottom surface of the cube to the water level = 0.191-y m It floats => Buoyancy = Weight =>Density of water*g*Volume of water displaced = Mass*g =>1000*g*0.191*0.191*(0.191-y) = (0.191^3)*653*g =>1000*0.191*0.191*(0.191-y) = (0.191^3)*653 =>6.968 - 36.481*y = 4.550 =>36.481*y = 6.968 - 4.550 =>y = 0.06628 m = 6.628 cm =>the distance from the horizontal top surface of the cube to the water level = 6.628 cm b)Let the mass of lead to be placed on the cube = z. =>Buoyancy = Weight =>Density of water*g*Volume of water displaced = (Mass*g)_Cube + (Mass*g)_Lead =>1000*g*0.191*0.191*(0.191) = (0.191^3)*653*g + z*g =>1000*0.191*0.191*(0.191) = (0.191^3)*653 + z =>z = 2.418 Kg =>the mass of lead to be placed on the cube = z = 2.418 Kg

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