Percentage of volume of object above water

Question

I'm having difficulty proving that the percentage of the volume of an object above the surface of a liquid is

$$\frac{100(p_f-p_o)}{p_f}$$

given that buoyant force exerted by the liquid on the object is

$$F=p_fg\int_{-h}^0A(y)dy$$

where $A(y)$ is the cross sectional area of the object and the weight of the object is

$$W=p_og\int_{-h}^{L-h}A(y)dy$$

where $L$ is the height of the object and $h$ is the distance the object is submerged into the liquid.

I tried comparing the ratio of two volume of revolution integrals but didn't manage to get anywhere. Any help?

Answer 1

Using Archimedes principle we have $$ F = W $$ or using the equations provided $$ p_fg\int_{-h}^0A(y)dy = p_og\int_{-h}^{L-h}A(y)dy $$ we can remove $g$ from both sides $$ p_f\int_{-h}^0A(y)dy = p_o\int_{-h}^{L-h}A(y)dy $$ The integral on the right hand side is the total Volume of the object. $$ p_f\int_{-h}^0A(y)dy = p_oV_{\text{Total}} $$ whilst the left hand side is $$ p_f\int_{-h}^0A(y)dy = p_f\int_{-h}^{L-h}A(y)dy - p_f\int_{0}^{L-h}A(y)dy = p_fV_{\text{total}} -p_f\int_{0}^{L-h}A(y)dy $$ the integral $$ \int_{0}^{L-h}A(y)dy $$ is the volume above the water, $V_{\text{above water}}$. This means we have $$ p_f\int_{-h}^0A(y)dy = p_fV_{\text{total}}-p_fV_{\text{above water}} = p_oV_{\text{Total}} $$ we can now re-arrange to obtain $$ p_fV_{\text{above water}} = p_fV_{\text{total}}-p_oV_{\text{Total}} = (p_f - p_0)V_{\text{Total}} $$ or $$ \frac{V_{\text{above water}} }{V_{\text{Total}}} = \frac{p_f-p_0}{p_f} $$ or in terms of percentage you can multiply by $100$.

Answer 2

Let $V$ be the volume of the object. Let $V'$ be the volume of water dispaced.

% of the volume above the water $\frac {V-V'}{V}$

$\rho_0$ is the density of the object $\rho_f$ is the density of water.

the weight of the object = the weight of the water displaced

$\rho_0 V = \rho_f V'\\ V' = \frac{\rho_0}{\rho_f} V$

Substituting into what we have above:

$1-\frac{\rho_0}{\rho_f} = \frac{\rho_f-\rho_0}{\rho_f}$