Question

A block of wood floats in fresh water with 0.600 of is volume V submerged and in oil with 0.912 V submerged. Find the density of a.) the wood and b.) the oil

Answer 1

By Archimede's Principle, the buoyant force acted upon the block is equal to the weight of liquid displaced. So, since the block displaces 0.600V of fluid in water:
ma = (weight of block) - (buoyant force) = p(wood)Vg - 0.600p(water)Vg.

Since the block floats, a = 0, and:
0 = p(wood)Vg - 0.600 p(water)Vg ==> p(wood) = 0.600 p(water).

Applying Archimede's Principle to the block in oil:
ma = (weight of block) - (buoyant force) = p(wood)Vg - 0.912p(oil)Vg.

a = 0, so:
0 = p(wood)Vg - 0.912p(water)Vg ==> p(wood) = 0.912p(oil).

By comparing the values of p(wood):
0.600p(water) = 0.912p(oil)
==> p(oil) = (0.600/0.912)p(water)
= (0.600/0.912)(1000 kg/m^3)
= 658 kg/m^3.

Then, since p(wood) = 0.0.600p(water):
p(wood) = (0.600)(1000 kg/m^3) = 600kg/m^3.

I hope this helps!