Question

An iceberg floats with approximately 1/7 of its volume in the air and
the 6/7 of its volume in water. If the wind speed is U and the water is
stationary, estimate the speed at which the wind forces the iceberg
through the water.

Answer 1

Given that,

The volume of iceberg in air = 1/7

The volume of iceberg in water = 6/7

We know that,

The drag force is

$F_{d}=\dfrac{1}{2}\rho v^2AC_{d}$

Where, $C_{d}$ = drag cofficient

$\rho$ = density

A = area

v = velocity

Let the speed at which the wind forces the iceberg through the water $U_{b}$.

The drag force on the iceberg in air is equal to the drag force on the iceberg in the water

We need to calculate the speed at which the wind forces the iceberg through the water

Using the formula of drag force

$F_{a}=F_{w}$

$\dfrac{1}{2}\rho_{air} (U-U_{b})^2 A_{air}C_{d}=\dfrac{1}{2}\rho_{w} U_{b}^2A_{w}C_{d}$

$\dfrac{(U-U_{b})^2}{U_{b}^2}=\dfrac{\rho_{w}A_{w}}{\rho_{air}A_{air}}$

$\dfrac{U-U_{b}}{U_{b}}=\sqrt{\dfrac{\rho_{w}A_{w}}{\rho_{air}A_{air}}}$

$\dfrac{U}{U_{b}}-1=\sqrt{\dfrac{\rho_{w}A_{w}}{\rho_{air}A_{air}}}$

$\dfrac{U}{U_{b}}=\sqrt{\dfrac{\rho_{w}A_{w}}{\rho_{air}A_{air}}}+1$

$U_{b}=\dfrac{U}{\sqrt{\dfrac{\rho_{w}A_{w}}{\rho_{air}A_{air}}}+1}$....(I)

We need to calculate the value of the ratio of the area

Using formula of area

$\dfrac{A_{air}}{A_{w}}=(\dfrac{V_{air}}{V_{w}})^2$

Put the value of volume

$\dfrac{A_{air}}{A_{w}}=(\dfrac{\dfrac{V}{7}}{\dfrac{6V}{7}})^{\frac{2}{3}}$

$\dfrac{A_{air}}{A_{w}}=(\dfrac{1}{6})^{\frac{2}{3}}$

$\dfrac{A_{air}}{A_{w}}=0.3028$

$\dfrac{A_{w}}{air}=3.3019$

Now, put the value in the equation (I)

$U_{b}=\dfrac{U}{\sqrt{\dfrac{1000\times3.3019}{1.2}}+1}$

$U_{b}=0.0187U=\approx0.02U$

Hence,  The speed at which the wind forces the iceberg through the water is 0.02U.