Question 2.14: A rectangular barge of width b and a submerged depth of H ha...

Let B, G and M be the centre of buoyancy, centre of gravity and metacentre of the burge (Fig. 2.37) respectively.

Now, O B=H / 2

and, OG = H (as given in the problem)

Hence B G=O G-O B=H-\frac{H}{2}=\frac{H}{2}

Again B M=\frac{I}{V}=\frac{L b^{3}}{12 \times L \times b \times H}=\frac{b^{2}}{12 H}

where L is the length of the barge in a direction perpendicular to the plane of the Fig. 2.37.

Therefore, M G=B M-B G=\frac{b^{2}}{12 H}-\frac{H}{2}=\frac{H}{2}\left\{\frac{1}{6}\left(\frac{b}{H}\right)^{2}-1\right\}

For stable equilibrium of the burge, MG ≥ 0

Hence, \frac{H}{2}\left\{\frac{1}{6}\left(\frac{b}{H}\right)^{2}-1\right\} \geq 0

which gives b / H \geq \sqrt{6}