Calculating the Density of Wood with Archimedes' Question

[jimithing]

A wooden cylinder 30.0 cm high floats vertically in a tub of water (density 1.00 g/cm^3). The top of the cylinder is 14.1 cm above the surface of the liquid. What is the density of the wood?

Does the radius of the cylinder need to be known in order to relate height and volume? Obviously the 15.9 cm that is under water comes into play, but where exactly?

 

[Leong]

The up thrust force acts on the wooden cyclinder is equal to the weight of the wooden cyclinder when the cyclinder floats on the water.

The up thrust force is equal to the weight of the water displaced when the cyclinder floats on the water.

My answer : $$5.3 X 10^{-3} kg/m^3$$

 

[wisky40]

for this example, the radius of the cylinder is not important.

$$m\vec{g}=\vec{B}$$ where B is buoyancy force.

=> $$V_c=\pi{r^2}H$$ and $$V_l=\pi{r^2}{(H-h)}$$

=> $$\rho_c{V_c}g=\rho_w{V_w}g$$

=> $$\rho_c=\rho_w{\frac{\pi{r^2}{(H-h)}}{\pi{r^2}{H}$$

finally $$\rho_c={(1-\frac{h}{H}0}g/cm^3$$

 

[wisky40]

well when you divide V_c by V_w the factor (pi)r^2 is removed.

 

[jimithing]

for this example, the radius of the cylinder is not important.

$$m\vec{g}=\vec{B}$$ where B is buoyancy force.

=> $$V_c=\pi{r^2}H$$ and $$V_l=\pi{r^2}{(H-h)}$$

=> $$\rho_c{V_c}g=\rho_w{V_w}g$$

=> $$\rho_c=\rho_w{\frac{\pi{r^2}{(H-h)}}{\pi{r^2}{H}$$

finally $$\rho_c={(1-\frac{h}{H}0}g/cm^3$$

H being the total height of the cylinder and h being the height of the cylinder under water, correct?

 

[wisky40]

yes sir, that's correct, but be careful the answer is 530 kg/m^3 or .53 g/cm^3, and I think that is logic because the density of wood is < the density of water, and also see the proportion of h vs H ~.5 which agrees with this result.