91Ó°ÊÓ

This problem can be solved by using the Archimedes principle.

Using Archimedes' Principle, the floating force equals to the weight of the liquid displaced. The steel "boat" total weight should be equaled to the water displaced. Assume that the water is very calm, and the level of water can be the same as the heights of the four sides without water leaking into the steel "boat".

The volume of boat is $V_{\text{boat}}=5.0\mathrm{m}×10\mathrm{m}×0.020\mathrm{m}$

$=1.0{\mathrm{m}}^3$

Let the height of the sides be $h$and the thickness of the sides is $t$. The volume of the water displaced is,

$\begin{array}{r}{V}_{\text{displaced}}=\left(5.0\mathrm{m}\right)\left(10\mathrm{m}\right)(h+0.020\mathrm{m})\\ =\left(50{\mathrm{m}}^2\right)(h+0.020\mathrm{m})\end{array}$

Calculate the mass of $m_{\text{water}}$on the bottle as follow

$m_{\text{wuter}}={\mathrm{ÒÏ}}_{\text{water}}{V}_{\text{displaced}}$

Here ${\mathrm{ÒÏ}}_{\text{water}}$is the density of water

Substitute $1000\mathrm{kg}/{\mathrm{m}}^3\text{for}{\mathrm{ÒÏ}}_{\text{water}}\text{and}\left(50{\mathrm{m}}^2\right)(h+0.020\mathrm{m})\text{for}{V}_{\text{dicplaced}}$

$\begin{array}{r}{m}_{\text{water}}=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(\left(50{\mathrm{m}}^2\right)(h+0.020\mathrm{m})\right)\\ =(50000\mathrm{kg}/\mathrm{m})(h+0.020\mathrm{m})\end{array}$

The volume of the boat

$\begin{array}{r}{V}_{\text{total}}=V_{\text{boat}}+V_{\text{obiect}}\\ \end{array}$

$\begin{array}{r}\\ V_{\text{total}}=1.0{\mathrm{m}}^3+\left(2\right(5\mathrm{m}\left)\right(h\left)\right(t\left)\right)+\left(2\right(10\mathrm{m}\left)\right(h\left)\right(t\left)\right)\end{array}$

Mass of the steel

$m_{\text{steel}}={\mathrm{ÒÏ}}_{\text{sseel}}{V}_{\text{total}}$

Here, ${\mathrm{ÒÏ}}_{\text{steel}}$is the density of water.

Calculate the mass of steel as follows:

$m_{\text{steel}}={\mathrm{ÒÏ}}_{\text{steel}}{V}_{\text{total}}$

$\begin{array}{r}=\left(7900\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.0{\mathrm{m}}^3+\left(2\right(5\mathrm{m}\left)\right(h\left)\right(t\left)\right)+\left(2\right(10\mathrm{m}\left)\right(h\left)\right(t\left)\right)\right)\\ \end{array}$

$=7900\mathrm{kg}+237000ht$

From the principle of buoyancy, the mass of displaced water is equal to mass of steel. Thus,

$m_{\text{water}}=m_{\text{steel}}$

$(50000\mathrm{kg}/\mathrm{m})(h+0.020\mathrm{m})=7900\mathrm{kg}+237000ht$

$\begin{array}{r}\\ (50000\mathrm{kg}/\mathrm{m})(h+0.020\mathrm{m})=7900\mathrm{kg}+237000h\left(0.50\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)\end{array}$

$h=0.141\mathrm{m}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right)$

$=14.1\mathrm{cm}$