Solution Naval Numerical 20

Solution Naval Numerical 20

Numerical 20: A ship of 11200 tonne displacement has a double bottom tank containing oil. Whose centre of gravity is 16.5m forward and 6.6m below the centre of gravity of the ship? When the oil is used the ship’s centre of gravity moves 380mm calculate – (a) The mass of oil used; (b) The angle which the centre of gravity moves relative to the horizontal.

Solution:

Gg = $\sqrt{16.5^2 +6.6^2}$ =17.77

This is the distance from the centre of gravity of the tank to the original centre of gravity of the ship.

(a) Let m= mass of oil used

then shift in centre of gravity = $\frac{m x Gg}{\Delta _f}$

0.380 = $\frac{mx17.77}{11200-m}$

m = $\frac{11200 x 0.38}{17.77 + 0.38}$ =234.5 t

(b) $tan\theta = \frac{6.6}{16.5}$ =0.4

Angle of shift $\theta = 21^o48'$

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