Floating 1

Classical Mechanics Level 3

A cuboid block of dimensions $a \times a \times c$ $($ with $a\ll c$ ) and uniform density $\rho<0.5$ floats on water. Depending on the density $\rho$ , there are three different equilibrium possibilities, illustrated in the figure.

A . The bottom side is horizontal.

B . The bottom side makes an angle of $\alpha<45^{\circ}$ to the horizontal. Note that there are two possibilities, one with a leftward tilt and another with a rightward tilt, and these two configurations are equally stable.

C . The block floats so that two bottom sides make $45^{\circ}$ to the horizontal.

For low densities, A correctly describes the stable equilibrium. Configurations B and C follow as the density increases and approaches $\rho=0.5$ .

What is the maximum density for configuration A to be stable?

Notes : The density $\rho$ is measured relative to the water. For $\rho>0.5,$ the stable configurations follow in reverse order, with configuration A being stable for $\rho$ approaching 1, except for the fact that the block is much more submerged.

The answer is 0.2113.

2 solutions

Laszlo Mihaly
Sep 27, 2018

Let us look at a slightly more general problem, where the sides of the block are $a$ (vertical) and $b$ (horizontal).

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 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

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

In the current problem $a=b$ and we get

$\rho= \frac{1\pm\sqrt{1/3}}{2} = 0.2113$

when we take the smaller root.

Note that the solution is symmetric around $\rho=0.5$ .

Vinod Kumar
Oct 9, 2018

Using the following internet link, the answer for parallel bottom, the value of relative density < (3-√3)/6. There is symmetry around 0.5 with another value (3+√3)/6.

Answer=(3-√3)/6=0.211

http://datagenetics.com/blog/june22016/index.html

As we discussed in the context of an other problem related to floating bodies, the link you provided gives the number, but does not derive it. I think a large part of the fun in the problems posted on Brilliant is finding a way to derive the solution from first principles.

Having said that, I agree that this problem is solved completely and fully in many textbooks and scientific papers. The problem of stability of floating bodies is so important (to ships and boats) that there is a HUGE amount of literature about it. It is interesting that in the USA College level introductory physics courses this topic is barely mentioned.

Laszlo Mihaly - 2 years, 8 months ago

Floating 1

The answer is 0.2113.

2 solutions

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