91影视

In this problem, the concept of buoyancy will be utilized in which the weight of a substance decreases due to the upward buoyant force.

The actual mass of the hull is m_ac = 18000kg.

The density of water is 蟻_w = 1.0 x 10³ kg/m³.

The density of steel is蟻_s = 7.8 x 10³ kg/m³.

The tension when the hull is fully submerged is,

\begin{aligned}T=mg-{F_{\rm{B}}}\\T=mg-V{\rho_{\rm{w}}}g\\T=mg-\frac{m}{{{\rho_{\rm{s}}}}}{\rho_{\rm{w}}}g\\T=mg\left({1-\frac{{{\rho_{\rm{w}}}}}{{{\rho _{\rm{s}}}}}}\right)\end{aligned}

Here, F_B is the buoyant force, V is the submerged volume of the hull, m is the mass of the hull, and g is the acceleration due to gravity.

Now substituting and the values give,

\begin{aligned}T=\left({18000\;{\rm{kg}}}\right)\times\left({9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}\right)\left({1-\frac{{1.0\times{{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}}{{7.8\times{{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}}}\right)\\\approx 1.53 \times{10^5}\;{\rm{N}}\end{aligned}

Hence, the tension is 1.53 x 10⁵ N.

The tension when the hull is completely out of water is,

T=mg

Substitute the values in the above relation.

\begin{aligned}T=\left({18000\;{\rm{kg}}}\right)\times\left({9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}\right)\\\approx1.76\times{10^5}\;{\rm{N}}\end{aligned}

Hence, the tension is 1.76 x 10⁵ N.