Question

A crane lifts a sunk ship of a mass of 20,000 kg out of the ocean. Density of sea water is 1025 kg/m³ and the average density of the ship is 7200 kg/m³. Determine (a) the tension in the crane's cable when the ship is below the surface of the water and (b) the tension when the ship is completely out of the water.

Answer 1

Given,

mass of the ship, m = 20000 kg

Density of the ship, _s = 7200 kg/m³

Density of sea water, _w = 1025 kg/m³

Now,

Volume of the ship, V = mass / density

                                    = m / _s = 20000 / 7200

                                    = 2.78 m³

(a)

Let buoyant force be F _B

Since,

Crane is pulling the ship upwards and buoyant force also pushes the ship upwards

Force acting downwards on the ship is the weight of the ship

Hence,

T + F _B = mg

=> T = mg - F _B = mg - (_w*V*g)

          = 20000*9.8 - (1025*2.78*9.8)

          = 196000 - 27925.1 = 168074.9 N

          = 1.7 * 10⁵ N

or Tension in the cable is 1.7 * 10⁵ N

(b)

When ship is completely out of water,

Tension, T = mg = 20000*9.8

                 = 196000 N

                 = 1.96 * 10⁵ N

Tension in the cable is 1.96 * 10⁵ N