In seawater, a life preserver with a volume of $\mathbf{0}\mathbf{.}\mathbf{0400}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$will support a 75.0-kg person (average density $\mathbf{9}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$), with 20% of the person’s volume above the water surface when the life preserver is fully submerged. What is the density of the material composing the life preserver?

The given data can be listed below,

• The volume of life preserver is,${\mathrm{V}}_{\mathrm{LP}}=0.0400{\mathrm{m}}^3$ .
• The mass of a person is, $\mathrm{m}=75\mathrm{kg}$.
• The average density of person is,${\mathrm{p}}_{\mathrm{p}}=980\mathrm{kg}/{\mathrm{m}}^3$
• The volume of a person above the water surface is 20% .

Archimedes’ principle assumes that fluids have a pressure gradient; the pressure gradient is a gradual rise in pressure from top to bottom. The pressure gradient causes buoyancy, resulting in a net up-thrust force as more pressure is exerted below than above.

This force is opposite in direction to its weight. It is a matter of density. Floating objects are those whose density exceeds water density.

The buoyant force is equal to the weight of the seawater displaced by the life preserver and 80% of the person’s volume.

Balance the forces on the person-life preserver system gives,

$\mathrm{B}={\mathrm{W}}_{\mathrm{P}}+{\mathrm{W}}_{\mathrm{lp}}$

Here, B is the buoyant force,${\mathrm{W}}_{\mathrm{p}}$ is the weight of the person and${\mathrm{W}}_{\mathrm{LP}}$ is the weight of the life preserver.

By mass-density relation,

${\mathrm{\rho }}_{\mathrm{w}}({\mathrm{V}}_{\mathrm{LP}}+0.80{\mathrm{V}}_{\mathrm{P}})={\mathrm{\rho }}_{\mathrm{P}}{\mathrm{V}}_{\mathrm{P}}+{\mathrm{\rho }}_{\mathrm{LP}}{\mathrm{V}}_{\mathrm{LP}}$

Here,${\mathrm{\rho }}_{\mathrm{W}}$ is the density of sea water,$\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)$,${\mathrm{V}}_{\mathrm{LP}}$ is the volume of life preserver, ${\mathrm{V}}_{\mathrm{P}}$ is the volume of person,${\rho }_P$ is the density of person, and ${\rho }_{LP}$ is the density of life preserver.

By conversion of person’s volume above expression can be written as,

$$\begin{gathered} {\rho }_W=\left(V_{LP}+0.800m_p+{\rho }_P\right)=m_p+{\rho }_{LP}{V}_{LP} \\ {\rho }_{LP}{V}_{LP}={\rho }_W\left(V_{LP}+0.800m_p+{\rho }_P\right)-m_p \\ {\rho }_{LP}=\frac{{\rho }_W\left(V_{LP}+0.800m_p+{\rho }_P\right)-m_p}{V_{LP}} \end{gathered}$$

Substitute all the values in the above expression,

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{LP}}=\frac{{\mathrm{\rho }}_{\mathrm{w}}\left({\mathrm{V}}_{\mathrm{LP}}+0.800{\mathrm{m}}_{\mathrm{p}}/{\mathrm{\rho }}_{\mathrm{p}}\right)}{{\mathrm{V}}_{\mathrm{LP}}} \\ {\mathrm{\rho }}_{\mathrm{LP}}=\frac{\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)\left[0.0400{\mathrm{m}}^3+{\displaystyle \frac{(0.800)(75\mathrm{kg})}{980\mathrm{kg}/{\mathrm{m}}^3}}\right]-75\mathrm{kg}}{0.400{\mathrm{m}}^3} \\ =\frac{\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)\left(0.400{\mathrm{m}}^3+0.061{\mathrm{m}}^3\right)-75\mathrm{kg}}{0.400{\mathrm{m}}^3} \\ =\frac{104.26\mathrm{kg}-75\mathrm{kg}}{0.0400{\mathrm{m}}^3} \\ =732\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Thus, the density of life preserver is $732\hspace{0.33em}\mathrm{kg}/{\mathrm{m}}^3.$.