Physics-Wood block in water

Question

A 10cm×10cm×10cm wood block with a density of 680 kg/m3 floats in water.

1. What is the distance from the top of the block to the water if the water is fresh?
2. If it's seawater? Suppose that ρ=1030 kg/m3

Answer 1

The block is in equilibrium when the weight of the block is equal to the displacement force (=equal to the weight of displaced water, according to Archimede's principle).

If h is the height of the submerged portion of the block, then equilibrium requires:

M_b g = (M_w)g (weight of block = weight of displaced water)

V_block ρ_block = (V_submerged ρ_water) = (0.10m^2) h ρ_water

0.1^3 ρ_block = 0.1^2 h ρ_water

0.1 ρ_block = h ρ_water ---> h = 0.1 ρ_block/ρ_water

So, if you have standard water or sea water, you'd have different densities for them, and so the submerged fraction of the block h would also change