91Ó°ÊÓ

The given data can be listed below,

• The length of the barge is, $I=40\mathrm{m}$.
• The width of the barge is, $b=22\mathrm{m}$.
• The height of the barge is, $h=12\mathrm{m}$.
• The thickness of the barge is, $t=4\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.04\mathrm{m}$.
• The density of coal is, ${\mathrm{ÒÏ}}_{\mathrm{coal}}=1500\mathrm{kg}/{\mathrm{m}}^3$.

Archimedes’s principle describes that the buoyancy force equals the water’s weight or mass displaced. The integral of the pressure is equal to the weight of the water.

In other words, the volume of the object that is displaced by the fluid is the volume required for the buoyant force to counteract the force of gravity.

The buoyant force acting on the system can be given by,

$B={\mathrm{ÒÏ}}_w{V}_{barge}g$

Here, ${\mathrm{ÒÏ}}_w$is the density of water whose value is $\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)$and $V_{barge}$is the volume of the barge.

The net force acting on the system can begivenby,

$$\begin{gathered} B-\left(m_{barge}+m_{coal}\right)g=0 \\ B=\left(m_{barge}+m_{coal}\right)g \end{gathered}$$

Here, B is the buoyant force, $m_{barge}$is the mass of barge, $m_{coal}$ is the mass of coal, g is the acceleration due to gravity.

Substitute the value of buoyant force in the above equation.

$$\begin{gathered} {\mathrm{ÒÏ}}_w{V}_{barge}g=\left(m_{barge}+m_{coal}\right)g \\ {m}_{coal}={\mathrm{ÒÏ}}_w{V}_{barge}-m_{barge} \end{gathered}$$

The volume of the barge can be calculated as,

$V_{barge}=I×b×h$

Here, l is the length of the barge, b is the width of the barge, and h is the height of the height.

Substitute all the values above equation.

$$\begin{gathered} V_{barge}=40\mathrm{m}×22\mathrm{m}×12\mathrm{m} \\ =1.056×{10}^4{\mathrm{m}}^3 \end{gathered}$$

The volume of the steel is $0.040\mathrm{m}$times the total area of the five pieces of steel that the barge made.

The volume of steel is calculated as,

$$\begin{gathered} V_s=\left(0.040\mathrm{m}\right)\left[2\left(22\right)\left(12\right)+2\left(40\right)\left(12\right)+\left(40\right)\left(22\right)\right]{\mathrm{m}}^2 \\ =94.7{\mathrm{m}}^3 \end{gathered}$$

Therefore, the mass of the barge can be given by,

$m_{barge}={\mathrm{ÒÏ}}_s{V}_s$

Here, ${\mathrm{ÒÏ}}_s$is the density of steel whose value is $\left(7800\mathrm{kg}/{\mathrm{m}}^3\right)$and $V_s$is the volume of steel.

Substitute all the values above equation.

$$\begin{gathered} m_{barge}=\left(7800\mathrm{kg}/{\mathrm{m}}^3\right)\left(94.7{\mathrm{m}}^3\right) \\ =7.39×{10}^5\mathrm{kg} \end{gathered}$$

So, the mass of coal the barge can contain can be given as,

$m_{coal}={\mathrm{ÒÏ}}_w{V}_{barge}-m_{barge}$

Substitute all the values above equation.

$$\begin{gathered} m_{coal}=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.056×{10}^4{\mathrm{m}}^3\right)-7.39×{10}^5\mathrm{kg} \\ =9.8×{10}^6\mathrm{kg} \end{gathered}$$

Thus, the mass of coal the barge carries without sinking is $9.8×{10}^6\mathrm{kg}$.

The volume of this mass of coal can be expressed as,

$V_{coal}=\frac{m_{coal}}{{\mathrm{ÒÏ}}_{coal}}$

Substitute all the values above equation.

$$\begin{gathered} V_{coal}=\frac{9.8×{10}^6\mathrm{kg}}{1500\mathrm{kg}/{\mathrm{m}}^3} \\ =6533.33{\mathrm{m}}^3 \end{gathered}$$

Thus, the volume of coal is $6533.33{\mathrm{m}}^3$.

This volume is less than the volume of the barge. So, it will fit into the barge.