Newton's Second Law / Archimedes Principle

[nxn]

Homework Statement

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.

Homework 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 = $1000 \frac{kg}{m^3}$

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.

 

[cepheid]

Welcome to PF nxn,

Homework Statement

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.

Homework 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 = $1000 \frac{kg}{m^3}$

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.

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.

 

[nxn]

Thanks! This ended up working out just fine.