7.NA October 2018 Q.9 (B)

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 - (i) The mass of oil used; (ii) The angle which the centre of gravity moves relative to the horizontal

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Given 

△ = 11200 t
Shift in CG = 380 mm

To find 

1. Mass of oil m
2. angle θ

Solution 

From the diagram Gg²  = Ga² + ag²
                  Distance Gg   = Square root of (Ga² + ag²)
                                     = Square root of (16.5² + 6.6²)
                                     = 17.77 m

1. Shift in CG = (mass of oil x Distance) / Final △
                      = (m x Gg)  / (△ - m)
            0.380 = (m x 17.77) / (11200 - m)

(11200 x 0.380) - (0.380 x m) = m x 17.77
                                      4256 = m (0.38 + 17.77)
                                           m = 4256 / 18.15
                         mass of oil m = 234.5 t

2. From diagram tan θ = ag / Ga
                                    = 6.6 / 16.5
                                    = 0.4

                       Angle θ = 21⁰ 80¹