Question

(II) A crane lifts a sunken ship's 18000 kg steel hull out of the water. Determine (a) the tension in the crane’s cable when the hull is fully submerged in water and (b) the tension when the hull is completely out of the water.

Short answer

(a) The tension in the crane’s cable is 1.53 x 10⁵ N.

(b) The tension when the hull is completely in water is 1.76 x 10⁵ N.

Step-by-step solution

Concept of Buoyancy

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

Given the data

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

Calculation the tension for a fully submerged hull

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

The density of steel isρ_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.

Calculation the tension when the hull is out of the water

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.