An open rectangular barge is 40 m long, 22 m wide, and 12 m high, made of 4.0-cm-thick steel plate on its four sides and bottom. What maximum mass of coal (density 1500 kg/m³) can it carry in freshwater without sinking, and is there enough room in the barge to hold that coal?

Answer

At the verge of sinking, the barge can displace $V=40\times22\times12=10560\,\text{m}^3$ of freshwater, so the maximum supported mass is $\rho_w V = 1000\times10560 = 1.056\times10^7\,\text{kg}$. The steel volume is $V_s=10560-(39.92)(21.92)(11.96)=94.45\,\text{m}^3$, so its mass is $m_s\approx7800\times94.45=7.37\times10^5\,\text{kg}$. Therefore the maximum coal mass is $m_c\approx1.056\times10^7-7.37\times10^5=9.82\times10^6\,\text{kg}$, and it easily fits because it would occupy only $m_c/1500\approx6.55\times10^3\,\text{m}^3$ which is less than the barge’s interior volume.

Explanation

What “without sinking” means here

“Without sinking” means the barge can be loaded until the waterline just reaches the top edge. At that point, it cannot displace any more water, so the buoyant force is at its maximum.

Maximum buoyant support from displaced freshwater

Using Archimedes’ principle, the maximum supported total mass equals the mass of displaced freshwater:

Displaced volume at the rim: $$V_{\text{disp}}=LWH=(40)(22)(12)=10560\,\text{m}^3$$

Maximum supported mass: $$m_{\text{total, max}}=\rho_w V_{\text{disp}}=(1000)(10560)=1.056\times10^7\,\text{kg}$$

Mass of the steel barge (4 sides + bottom)

A clean way to get steel volume is: outer box volume minus inner empty volume.

Outer volume: $$V_{\text{outer}}=(40)(22)(12)=10560$$

Inner dimensions (top is open):

Inner length: $40-2t=40-0.08=39.92\,\text{m}$
Inner width: $22-2t=22-0.08=21.92\,\text{m}$
Inner height: $12-t=12-0.04=11.96\,\text{m}$

Inner (hollow) volume: $$V_{\text{inner}}=(39.92)(21.92)(11.96)=10465.55\,\text{m}^3$$

Steel volume: $$V_s=V_{\text{outer}}-V_{\text{inner}}=10560-10465.55=94.45\,\text{m}^3$$

Steel mass (using $\rho_{\text{steel}}\approx7800\,\text{kg/m}^3$): $$m_s=\rho_{\text{steel}}V_s\approx(7800)(94.45)=7.37\times10^5\,\text{kg}$$

Maximum coal mass before the rim goes under

Now subtract the barge’s own mass from the maximum supported mass: $$m_{\text{coal, max}}=m_{\text{total, max}}-m_s$$ $$m_{\text{coal, max}}\approx 1.056\times10^7-7.37\times10^5=9.82\times10^6\,\text{kg}$$

Does that much coal fit inside the barge?

Coal volume required: $$V_{\text{coal}}=\frac{m_{\text{coal, max}}}{\rho_{\text{coal}}}=\frac{9.82\times10^6}{1500}=6.55\times10^3\,\text{m}^3$$

Available interior volume: $$V_{\text{inner}}=10465.55\,\text{m}^3$$

Since $6549 < 10466$, there is enough room. (Equivalently, the coal height would be about $6549/(39.92\cdot21.92)\approx7.49\,\text{m}$, which is less than the interior height $11.96\,\text{m}$.)

Skills You Achive

buoyancy archimedes-principle density 3d-geometry unit-conversion