Question 1.4.1: A cube of side a and relative density S (S < 1) floats in......

Let x be the height of the cube immersed in water. By Archimedes principle,

a³ S w = a² × w × x

→

x=aS                       (A)

Let G and B represent the centre of gravity of the cube and the displaced water; their height from the bottom of the cube is a/2 and x/2, respectively. Therefore, BG = (a – x)/2. Using Eq. 1.4.5 to get BM, we get

∴  \mathrm{BM=\frac{I_{xx}}{V}=\frac{second  moment  of  the  plane  of  floating  about  fore -aft  axis}{Immersrd  volume} }                              (1.4.5)

BM = I/V = (a^4/12)/(a^2 x) = a^2/(12x) and

GM = BM – BG = a²/(12 x) – ½ (a – x) > 0 for stability
or,

\mathrm{\frac{a^2}{12x}\gt \frac{1}{2}(a- x) }

or,

\mathrm{1\gt 6\left\lgroup\frac{x}{a}- \frac{x^2}{a^2} \right\rgroup =6(S-S^2)}

or,

S² – S + 1/6 > 0

or,

(S – 0.789) (S – 0.211) > 0.

We get the factors 0.789 and 0.211 taking equality sign. This condition is satisfied if S > 0.789 or < 0.211; i.e. the specific gravity of the cube should be less than 0.211 or greater than 0.789. If the specific gravity is within these limits then the cube will be unstable. In case the cube has any of these two values then it is in neutral equilibrium.