Question

What is the minimum area (in square meters) of the top surface of an ice slab $\mathbf{0}\mathbf{.441}\mathbf{m}\mathbf{.}$thick floating on fresh water that will hold up to $\mathbf{938}\mathbf{kg}$automobile? Take the densities of ice and fresh water to be${\mathit{\rho }}_{\mathbf{i}}\mathbf{=}\mathbf{917}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ and $\mathbf{998}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$, respectively.

Short answer

The minimum surface area of the top surface of the ice slab is$26.3{\mathrm{m}}^2$

Step-by-step solution

Listing the given quantities

• The thickness of an ice slab is,$d=0.441\mathrm{m}.$

• The mass of an automobile held on an ice slab is,$M=938\mathrm{kg}$

• The density of ice is,${\rho }_i=917\mathrm{kg}/{\mathrm{m}}^3$

• The density of water is,${\rho }_{\mathrm{w}}=998\mathrm{kg}/{\mathrm{m}}^3$

Understanding the Archimedes Principle

Archimedes Principle states that when a body is fully or partially submerged in a fluid, a buoyant force ${\mathfrak{F}}_{\mathbf{b}}$from the surrounding fluid acts on the body. The force is directed upward and has a magnitude given by,

${\mathit{F}}_{\mathbf{b}}\mathbf{=}{\mathit{m}}_{\mathbf{f}}\mathit{g}$

Where ${\mathit{m}}_{\mathbf{F}}$ is the mass of the fluid that has been displaced by the body.

Using Archimedes' principle, we can find the volume of the ice slab floating on the water. From that, we can find the mass of the slab.

Formulae:

From Archimedes’ principle, when an object floats in a liquid, then

$\begin{array}{r}{F}_i=F_i\\ m=\rho V\end{array}$

Calculation for minimum surface area 

According to Archimedes’ principle, the ice slab of volume V with an automobile will float on water if

$\begin{array}{r}{F}_g=F_i\\ (M+m)g={\rho }_{w}Vg\\ Mg+{\rho }_{i}Vg={\rho }_{w}Vg\\ M+{\rho }_{i}V={\rho }_{iN}V\end{array}$

Substitute the values in the above equation.

$\begin{array}{r}938\mathrm{kg}+\left(917\mathrm{kg}/{\mathrm{m}}^3\right)\mathrm{V}=\left(998\mathrm{kg}/{\mathrm{m}}^3\right)\mathrm{V}\\ 81\mathrm{V}=938\\ V=11.58{\mathrm{m}}^3\end{array}$

But,

$V=Ad$

So,

$\begin{array}{r}A=\frac{11.58{\mathrm{m}}^3}{0.441\mathrm{m}}\\ =26.26{\mathrm{m}}^2\\ \approx 26.3{\mathrm{m}}^2\end{array}$

Therefore, the minimum surface area of the top surface of the ice slab is$26.3{\mathrm{m}}^2$