Question

A wood cube 0.26 m on each side has a density of 730 kg/m3 and floats levelly in water.

What is the distance from the top of the wood to the water surface?​ 7.0 x 10^-2m

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

Answer 1

Density of the wood cube, Pwd = 730 kg/m^3

Density of water, Pwt = 1000 kg/m^3

So, length of the vertical face of the cube under water = 0.26*(730 / 1000) = 0.1898 m

So, distance from the top of the wood to the water surface = 0.26 - 0.1898 = 0.0702 = 0.07 m

Now, the volume of the cube is (0.26 m)^3, which is 0.017576 m^3

So, the mass of the cube = 0.017576 * 750 kg/m^3 = 13.18 kg

Weight of the same volume of water = 0.017576 * 1000 = 17.58 kg.

Hence (17.58 - 13.18) kg = 4.40 kg has to be placed on the top of the cube