Question

A cube of wood having an edge dimension of 20.1 cm and a density of 645 kg/m3 floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level?

Answer 1

7.1355 cm

Step-by-step explanation:

Principle of flotation

If an object of density
`\rho_2` is floating on a liquid of density
`\rho_1`, the weight of the displaced liquid equals the weight of the object. It's also known as the Archimedes' principle.

The object will float if its density is less than the density of the liquid and will sink if it's greater. In the special case it floats, part of its volume (and therefore, mass) will keep floating over the liquid's surface (let's call it
`V_f`), and the rest will be sunk (
`V_s`). The total volume of the object will be:

`V_2=V_f+V_s`

Applying the principle of flotation

`m_2=m_1`

`Since\ m=\rho V`

`\rho_2 V_2=\rho_1 V_1`

But
`V_1` equals the sunk volume of the object

`\rho_2 V_2=\rho_1 V_s`

This relation gives us

`\displaystyle V_s=(\rho_2 V_2)/(\rho_1)`

This is the proportion of the object sunk in liquid .

The wooden cube has an edge dimension of L=20.1 cm and a density
`\rho_2=645\ kg/m^3`. The density of water is
`\rho_2=1000\ kg/m^3`. So the underwater volume of the cube is

`\displaystyle V_s=(645 V_2)/(1000)=0.645V_2`

`V_2 = L^3`

Assuming the cube floats with one side parallel to the face of the water

`V_s=L^2h`

Where h is the length of the side of the cube submerged in the water. Replacing those formulas we get

`L^2h=0.645L^3`

Simplifying

`h=0.645L=12.9645\ cm`

Finally, the distance from the horizontal top surface of the cube to the water level is

`L-h=20.1-12.9645=7.1355\ cm`