Question

(III) A scuba tank, when fully submerged, displaces $15.7\mathrm{L}$ of seawater. The tank itself has a mass of $14.0\mathrm{k}\mathrm{g}$ and, when “full,” contains $3.00\mathrm{k}\mathrm{g}$ of air. Assuming only its weight and the buoyant force act on the tank, determine the net force (magnitude and direction) on the fully submerged tank at the beginning of a dive (when it is full of air) and at the end of a dive (when it no longer contains any air).

Short answer

The net force on the fully submerged tank at the beginning of a dive is $9\mathrm{N}\left(\mathrm{d}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\right)$ and the net force on the fully submerged tank at the end of a dive is $21\mathrm{N}\left(\mathrm{u}\mathrm{p}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}\right)$.

Step-by-step solution

Concept of net force

Net force is the reduced forces applied on a system where major forces take over the minor forces. The direction of major forces shows the direction of acceleration of the system as well.

Given data

The volume of the tank is $V_{\mathrm{t}}=15.7\mathrm{L}$.

The mass of the tank is $m_{\mathrm{t}}=14\mathrm{k}\mathrm{g}$.

The mass of the air is $m_{\mathrm{a}}=3\mathrm{k}\mathrm{g}$.

Calculation of net force acting on the tank in the beginning

The buoyant force is calculated as:

$F_{Bouyant}=\rho Vg$

Here,$\rho$is density of fluid, $V$ is volume occupied and $\mathrm{g}$ is acceleration due to gravity.

The density of a fluid is calculated as:

$\rho =\frac{m}{V}$

Here,$m$is the mass of the fluid.

Using the law of equilibrium of forces, calculate the net forces.

$\begin{array}{rl}{F}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}& =F_{\mathrm{B}}-m_{\mathrm{T}}g\\ & ={\rho }_{\mathrm{w}}{V}_{\mathrm{t}}\mathrm{g}-\left(m_{\mathrm{t}}+m_{\mathrm{a}}\right)g\end{array}$

Here, ${\rho }_{\mathrm{w}}$is the density of water and $m_{\mathrm{T}}$ is the total mass.

Substitute the known values in the above equation to find the net force.

$\begin{array}{rl}{F}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}& =\left({\rho }_{\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}{V}_{\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{k}}-\left(m_{\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{k}}+m_{\mathrm{a}\mathrm{i}\mathrm{r}}\right)\right)g\\ & =\left(\left(1025\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)\left(15.7\mathrm{L}\times \frac{1{\mathrm{m}}^3}{1000\mathrm{L}}\right)-\left(14\mathrm{k}\mathrm{g}+3\mathrm{k}\mathrm{g}\right)\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ & =-8.89\mathrm{N}\\ & \approx -9\mathrm{N}\end{array}$

Hence, the net force is $9\mathrm{N}$ downwards.

Calculation of net force acting on the tank in the end

Using law of equilibrium of forces,

$\begin{array}{rl}{F}_{\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{t}\mathrm{y}}& =F_{\mathrm{B}}-m_{\mathrm{T}}g\\ & ={\rho }_{\mathrm{w}}{V}_{\mathrm{t}}g-\left(m_{\mathrm{t}}\right)g\end{array}$

Substitute the known valuesin the above equation to find the net force.

$\begin{array}{rl}{F}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}& =\left({\rho }_{\mathrm{w}}{V}_{\mathrm{t}}-\left(m_{\mathrm{t}}\right)\right)g\\ & =\left(\left(1025\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)\left(15.7\mathrm{L}\times \frac{1{\mathrm{m}}^3}{1000\mathrm{L}}\right)-14\mathrm{k}\mathrm{g}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ & =20.51\mathrm{N}\\ & \approx 21\mathrm{N}\end{array}$

Hence, the net force is $21\mathrm{N}$ upwards.