Question 1.4.3: A barge in the shape of a rectangular block which is 8 m wid......

Centre of buoyancy will lie below the free surface at a distance = 1.8/2 = 0.9 m.
Therefore, BG = 0.9 + 0.3 = 1.2 m. Further, BM = \mathrm{\frac{12.8 \times 8^3/12}{(8\times 12.8 \times 1.8)} } = 2.963 m and GM = 1.763 m. We observe that BM > BG and hence the barge is in stable equilibrium.
The righting moment for 10° heel is given by

F_b × GM \sin 10,  where  F_b = 9.81 × 8 × 12.8 × 1.8 = 1808   kN

→

righting moment = 553.56 kNm