Floating 2

The center of mass of the block is at an elevation of $a/2$ from the base (red dot on the Figure). The center of the mass of the displaced water is at an elevation of $\rho a/2$ (blue dot). The weight of the block and the weight of the displaced water are equal to each other, $W=\rho g abc$ , where $g$ is the gravitational acceleration.

Imagine that the block is tilted by a small angle $\alpha$ , and the displaced water water does not change its shape. In this situation the weight vector of the block does not line up with the upward force due to the displaced water. There will be a torque of $\tau_1=W \ell$ acting on the block, where $\ell=\alpha \frac{a}{2} (1-\rho)$ is the distance between the two parallel an opposite forces.

This torque is counter-acted by a wedge-shaped piece of water that flows from one side of the block to the other side, so that the water surface remains horizontal. For small $\alpha$ , the height of the water at a distance $x$ from the center line is $h=\alpha x$ . The weight of a slice of water of width $dx$ is $dW=\rho_w g c dx (\alpha x)$ and the torque is $d\tau=x \rho_w g c dx (\alpha x)$ . The total torque due to this water amount is

$\tau_2= 2g \alpha c \int_{0}^{b/2}x^2 dx = \frac{1}{12} g \alpha c b^3$ ,

where the factor 2 stands for the left and the right side from the center line and we assumed that the density of water is $\rho_w=1$ . For the configuration to be stable one needs $\tau_2>\tau_1$ . At the at the critical density the two torques are equal

$\rho a b c \frac {a}{2} (1-\rho)\alpha= \frac{1}{12}\alpha c b^3$ or

$\rho(1-\rho) a^2= \frac{1}{6} b^2$

For any given $b/a$ ratio we can solve this equation for the density to get a non-trivial, tilted, equilibrium:

$\rho= \frac{1\pm\sqrt{1-\frac{4b^2}{6a^2}}}{2}$

As $b/a$ increases at a certain point the solution becomes a complex number. The critical value is

$\frac{b}{a}= \sqrt{\frac{3}{2}}=1.2247$

Answer is derived from the expression,

width > (√6) height, (submerged height).

After correcting for density of 0.5,

Width > (1/2) (√6) full_height.

Answer=(1/2)√6

For derivation of [width > (√6) height], see the objective question with four options in preparation to Entrance examination to Mechanical Engineering Course in a University in India. The link is given below:

https://books.google.co.in/books?id=ZREwDwAAQBAJ&pg=PA39-IA268&lpg=PA39-IA268&dq=rotational+stability+of+floating+of+rectangular+blocks&source=bl&ots=dc-0zUeoq2&sig=4-qqvElQG6rX5fjzA4qJKtEkAPA&hl=en&sa=X&ved=2ahUKEwjYhsnSk_ndAhVEQI8KHTdrBgwQ6AEwGnoECAYQAQ