Question 2.13: A cube of side a floats with one of its axes vertical in a l...
A cube of side a floats with one of its axes vertical in a liquid of specific gravity S_{L}. If the specific gravity of the cube material is S_{c}, find the values of S_{L} / S_{c} for the metacentric height to be zero.
Let the cube float with h as the submerged depth as shown in Fig. 2.36. For equilibrium of the cube,
Weight = Buoyant force
a^{3} S_{c} \times 10^{3} \times 9.81=h a^{2} \times S_{L} \times 10^{3} \times 9.81or, h=a\left(S_{c} / S_{L}\right)=a / x
where S_{L} / S_{c}=x
The distance between the centre of buoyancy B and centre of gravity G becomes
B G=\frac{a}{2}-\frac{h}{2}=\frac{a}{2}\left(1-\frac{1}{x}\right)Let M be the metacentre, then
B M=\frac{I}{\forall}=\frac{a\left(\frac{a^{3}}{12}\right)}{a^{2} h}=\frac{a^{4}}{12 a^{2}\left(\frac{a}{x}\right)}=\frac{a x}{12}The metacentric height M G=B M-B G=\frac{a x}{12}-\frac{a}{2}\left(1-\frac{1}{x}\right)
According to the given condition
M G=\frac{a x}{12}-\frac{a}{2}\left(1-\frac{1}{x}\right)=0or, x^{2}-6 x+6=0
which gives x=\frac{6 \pm \sqrt{12}}{2}=4.732,1.268
Hence S_{L} / S_{c}=4.732 \quad \text { or } \quad 1.268