30.NA January 2019 Q.10(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

If you are noticing some error in problems kindly comment below.Thanks

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