Newton's Second Law / Archimedes Principle

1. Nov 15, 2013

nxn

1. The problem statement, all variables and given/known data
A block of wood floats in fresh water with 0.722 of its volume V submerged and in oil with 0.895 V submerged. Find the density of (a) the wood and (b) the oil.

2. Relevant equations

mass = density x volume

f = ma

conservation of mass

$F_{weight} - F_{buoyancy} = 0$ when an object floats on the surface of a fluid.

density of water = $1000 \frac{kg}{m^3}$

3. The attempt at a solution

$ρ_{wood}(V)g - ρ_{water}(.722V)g = m\cdot0$ when the block floats

so

$ρ_{wood}(V)g = ρ_{water}(.722V)g$

and

$ρ_{wood} = ρ_{water}\frac{(.722V)g}{(V)g}$

we know density of water = $1000 \frac{kg}{m^3}$ so

$ρ_{wood} = 1000(.722) = 722 \frac{kg}{m^3}$

Can you confirm that this looks correct? I've been waiting for a chance to practice with LaTeX and decided to jump in with this problem. I appreciate your time.

2. Nov 15, 2013

cepheid

Staff Emeritus
Welcome to PF nxn,

Yes, this looks correct. If the wood floats, it must be less dense than water, and it makes perfect sense that it would be less dense than water by a factor of 0.722, the fraction of the wood volume that had to be displaced.

3. Nov 15, 2013

nxn

Thanks! This ended up working out just fine.