(Solved): Here how I can get CG Because the metacentric height is negative, the block is not stable ...

Here how I can get CG

Because the metacentric height is negative, the block is not stable about the longitudinal axis. Thus a slight disturbance will make it tip to the orientation shown below. 3. Equilibrium (vertical direction-see above figure) $\begin{array}{c}\text{-weight }+\text{ buoyant force }=0\\ -(318\mathrm{N})+\left(9810\mathrm{N}/{\mathrm{m}}^3\right)\left(F_D\right)=0\\ d{F}_D=0.0324{\mathrm{m}}^3\end{array}$ 4. Find the dimension $w$. * 0 (Displaced volume $)=($ Block volume $)-($ Volume above the waterline). $\begin{array}{rl}{\forall }_D=0.0324{\mathrm{m}}^3& =\left(0.3^2\right)(0.6){\mathrm{m}}^3-\frac{w^2}{4}(0.6\mathrm{m})\\ w& =0.379\mathrm{m}\end{array}$ 5. Moment of inertia at the waterline $I_{00}=\frac{b{h}^3}{12}=\frac{(0.6\mathrm{m})(0.379\mathrm{m}{)}^3}{12}=0.00273{\mathrm{m}}^4$ 6. Metacentric height $GM=\frac{I_{00}}{V}-CG=\frac{0.00273{\mathrm{m}}^4}{0.0324{\mathrm{m}}^3}-0.0573\mathrm{m}=0.027\mathrm{m}$ Since the metacentric height is positive in this position. EVAMPLE 3.13 STABLITY OF A FLOATING BLOCK A block of wood $30\mathrm{cm}$ square in cross section and $60\mathrm{cm}$ long weighs $318\mathrm{N}$. Will the block float with sides vertical as shown?