Stability of a rectangular box floating on a surface of a liquid

This problem, as many of the previous discussed here, is coming from the Savchenko et al book. It took me around one month to wrap my head around this one. I have to admit that I improved (or refreshed) my knowledge of Archimedes' force thanks to this one. Specifically, this problem highlights the fact that Archimedes' force is applied to the center of mass of the water replaced by a floating object. Therefore, even if the object's density is uniform, it can happen that the torques of the gravity force and the Archimedes' force are not compensated even though the modules of the forces are equal. This fact makes more interesting the question about stability of a floating object in a liquid.

Problem statement

Given a rectangular body of length $a$, width $a$, and height $b$. Its density is $\rho$ and it floats in a liquid of density $\rho_0$. Determine the ratio $\frac{a}{b}$ for which the object's equillibrium state is stable. It is assumed that the side $b$ of the box is vertical in the equillibrium position.

Discussion

It is obvious that the object's equllibrium is stable with respect to vertical displacements, because if we push it down, the Archimedes' force increases and pushes it back. Similarly, if we pull the object up, the Archimedes' force becomes smaller and the gravity pushes it back to its initial equillibrium position where the gravity force is compensated by the Archimedes' force. Therefore, further down we consider only stability of the object with respect to rotations.

We won't consider the case with the box completely submerged into the liquid as in this case the center of mass of the box coincides with the center of mass of the liquid pushed out by the box, due to the uniform density of the box, and therefore the net torque (exerted by the gravity and Archimedes' forces) will always be zero. I guess this case is not stable with respect to rotations, as the box won't be restored to its original vertical position. Zero gravity, like in the outer space, all orientations are equivalent.

Here we can speculate a bit. Intuitively we would expect that the larger base (i.e. greater $a$) with respect to the height $b$ would give us a more stable equillibrium. Hence we can expect the result to be in a form $\frac{a}{b} > r$ or $\frac{a}{b} \geq r$.

Let's denote the height of submerged part of the box as $h$ and the pivot point as $O$. The pivot point is the intersection of the vertical axis of symmetry of the box and the horizontal plane at the level of the surface of the liquid (the updated level after the box is submerged), see the figure below. We consider rotations at a small angle $\varphi$ around this pivot point $O$.

The latex source for the plot can be found here.

There are two ways to find out if the vertical upright orientation of the box is stable: using the change in the height of the center of mass of the system liquid and box, i.e., if the above rotations lower the center of mass of the system then its vertical orientation is not stable, otherwise it is stbale, and using the net torque exerted by the gravity and Archimedes' forces, i.e., in the stable case the torque would act in the direction opposite to the direction of rotation,otherwise it is not stable. We'll use both methods and compare the results.

But first let's prepare a few useful expressions. If we denote as $A$ the center of mass of the liquid pushed out by the box (this is also the point where the Archimedes' force is applied to the box). Let's determine its coordinates $\left(x_{A\varphi}, y_{A\varphi}\right)$ when the box is rotated, in the system of coordinates with the center at the point $O$ and the $y$-axis pointing downward and the $x$-axis pointing to the right. It is easy to see that the mass ($m_A=\rho_0 a^2 h$) of the liquid pushed out by the box does not change during the rotation as well as the module of the Archimede's force $F_A=m_A g$. By definition, expressions for the coordinates of the center of mass of the liquid pushed out by the box can be written as:

$$\begin{cases} m_A x_{A\varphi} = m_A x_{A'\varphi} + m_L x_L + m_R x_R \\ m_A y_{A\varphi} = m_A y_{A'\varphi} + m_L y_L + m_R y_R \end{cases}$$

where

$$\begin{align} m_L = -m_R\\ x_L = -x_R\\ y_L = -y_R\\ \end{align}$$

and the $A'\varphi$ is the center of mass of the fictious liquid volume when it is rotated from the upright position as a solid body, $m_L$ and $m_R$ are the masses of the prisms (triangles in 2D, see the above figure) to be removed from and added to the fictious liquid body respectively to get the real shape of the liquid pushed out by the box during rotation. $\left(x_R, y_R\right)$ and $\left(x_L, y_L\right)$ are the centers of mass of the prisms of masses $m_R$ and $m_L$ respectively (red and blue triangles in the figure, the blue one has negative mass and the red one has positive mass). Since $m_L$ is the mass to be removed, it is negative in the equations to avoid confusion with the signs. We use these triangles (prismes in 3D) to facilitate the calculation of the position of the center of mass of the liquid pushed out by the box $A\varphi$. It is easy to compute coordinates of the center of mass of the fictious rectangular volume $B_1B_2RL$ and then to find coordinates of $A\varphi$ (of the complex shape of the liquid pushed out by the rotated box) we compute the center of mass of a system consisting of $B_1B_2RL$, $m_R$ and the negative mass $m_L$. The center of mass of such a system is $A\varphi$. If we substitude the relations between $m_L$ and $m_R$ into the equations for $x_{A\varphi}$, $y_{A\varphi}$ we get the following expressions:

$$\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \end{cases}$$

It is easy to see that

$$\begin{align} \frac{m_R}{m_A} & = \frac{\rho_0 \frac{1}{2}\cdot \frac{1}{2} a \cdot \frac{1}{2} a \tan\varphi\cdot a}{\rho_0 a^2 h} = \ \frac{a}{8h}\tan\varphi \approx \frac{a}{8h}\varphi \\ x_{A'\varphi} & = -\frac{h}{2}\sin\phi \approx -\frac{h}{2}\phi\\ y_{A'\varphi} & = \frac{h}{2}\cos\phi \end{align}$$

We keep exact values for the $y$ coordinates here as the changes in $y$ due to the rotation are on the order of $\sim\varphi^2$ and the variations of $x$ coordinates are on the order of $\sim\varphi$.

From the geometry and the properties of the centre of mass of a triangle we can determine $x_R$ and $y_R$: $$\begin{cases} x_R & = \frac{a}{2\cos\varphi}\cdot\frac{1}{2} + \frac{1}{3} \frac{a}{2\cos\varphi} \cos2\varphi = \frac{a}{4\cos\varphi}\left(1 + \frac{\cos2\varphi}{3}\right)\\ y_R & = \frac{1}{3}\left(\frac{a}{2\cdot2\cos\varphi}\right)\sin2\varphi = \frac{1}{6}a\sin\varphi \end{cases}$$

For small angles $\varphi$ (which is the case in the problem at hand), we get the following expressions for $x_R$ and $y_R$:

$$\begin{cases} x_R & \approx \frac{a}{3} \\ y_R & \approx \frac{a}{6}\varphi \end{cases}$$

Using the above relation we can express $\left(x_{A\varphi}, y_{A\varphi}\right)$ as functions of $h$, $a$, and $\varphi$:

$$\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \approx -\frac{h}{2}\phi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{3} = \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right)\\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \approx \frac{h}{2}\cos\varphi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{6}\varphi = \ \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 \end{cases}$$

We did not replace $\cos\varphi$ with one for the $y$ coordinate since the accuracy of the terms is $\varphi ^ 2$.

Let's note the coordinates of the center of mass of the pushed out liquid for $\varphi=0$:

$$\begin{cases} x_{A0} = 0\\ y_{A0} = \frac{h}{2} \end{cases}$$

It is easy to derive how the coordinates of the center of mass of the box ($C$) change during rotation, as it is always in the middle of the box:

$$\begin{cases} x_{C\varphi} = \left( \frac{b}{2} - h \right)\sin\varphi\\ y_{C\varphi} = \left( h - \frac{b}{2} \right)\cos\varphi \end{cases}$$

Finally, let's express $h$ using equillibrium between the Archimedes' ($F_A$) and the gravity ($F_C = m_C g$) forces:

$$\begin{align} F_C-F_A=0 \Rightarrow \rho a^2 b - \rho_0 a^2 h = 0 \Rightarrow h=\frac{\rho}{\rho_0}b \end{align}$$

Method 1: analysis of torques

Without the loss of generality we can consider clockwise rotation as shown in the plot. Torques acting in the counter-clockwise direction are positive and torques acting in the clockwise direction are negative. With this convention, for stability it is required that the net torque ($\tau$) should act in the direction opposite to the rotation (i.e. anti-clockwise). Therefore the condition for stability is $\tau > 0$.

Let's consider two cases and impose the stability condition:

a) the center of mass is below the liquid surface $h\geq b/2$

If $x_{A\varphi} \geq 0$, then $\tau = F_C |x_{C\varphi}| + F_A |x_{A\varphi}|$, which is always positive. Both $\tau_C$ and $\tau_A$ act to stabilize the box against rotation.

If $x_{A\varphi} < 0$, then stability condition is:

$$\begin{align} \tau = F_C |x_{C\varphi}| - F_A |x_{A\varphi}| > 0 \end{align}$$

Since $F_C = F_A$ (as well as $m_A = m_C$), for stability we need the following:

$$\begin{align} |x_{C\varphi}| - |x_{A\varphi}| > 0 &\Rightarrow - \left( \frac{b}{2} - h \right)\varphi - (-1)\cdot\frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -\frac{b}{2} + h + \frac{1}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -2bh + 4h^2 + \frac{a^2}{3} - 2 h^2 > 0 \\ & \Rightarrow -6bh + a^2 + 6 h^2 > 0 \\ & \Rightarrow -6b^2\frac{\rho}{\rho_0} + a^2 + 6 \left(\frac{\rho}{\rho_0} b\right)^2 > 0 \\ & \Rightarrow \left(\frac{a}{b}\right)^2 > 6\frac{\rho}{\rho_0} - 6 \left(\frac{\rho}{\rho_0} \right)^2 \\ & \Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}$$

b) the center of mass is above the liquid surface $h < b/2$

In this case the gravity force acts to destabilize the box with its torque acting in the direction of rotation. So the stability in this case is possible when $|x_{A\varphi}| > |x_{C\varphi}|$ and $x_{A\varphi} \cdot x_{C\varphi} > 0$ (i.e. A and C are from the same side from the pivot point $O$) which is equivalent to:

$$\begin{align} \tau = -F_C |x_{C\varphi}| + F_A |x_{A\varphi}| > 0 &\Rightarrow \ - (-1) \cdot \left(h - \frac{b}{2}\right)\varphi + \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ &\Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}$$

Let's check if the conditions we imposed do actually make sense.

$$\begin{align} h < b/2 \Rightarrow \rho/\rho_0 < 1/2 \Rightarrow a/b > \sqrt{3/2} \Rightarrow \frac{a^2}{3} - \frac{b^2}{2} > 0 \Rightarrow x_{A\varphi} > 0 \end{align}$$

remembering that we consider only $\varphi > 0$.

Method 2: minimum of the potential energy of the system

I like this method more as it requires less logic branching. Let's denote as $y$ the $y$-coordinate of the center of mass of the system liquid and box. Since the positive direction of the axis is downwards, the change $\Delta y < 0$ during the rotation of the box would mean that the center of mass of the system is lifted and therefore the system's potential energy would be increasing and the system itself will tend to minimize its potential energy, therefore $\Delta y < 0$ is our stability condition. By the definition of the center of mass we have:

$$\begin{align} (M - m_A + m_C) y = m_C y_{C\varphi} - m_A y_{A\varphi} + M y_0 \end{align}$$

where $M$ is the mass the liquid would have when the immersed part of the box is replaced by the liquid. $y_0$ is the $y$-coordinate of the center of mass of $M$.

Let's examine the change in the $y$-coordinate of the center of mass of the system during rotation:

$$\begin{align} \Delta y = \frac{1}{M - m_A + m_C}\left ( m_C \Delta y_{C} - m_A \Delta y_{A} \right) \Rightarrow \Delta y \propto m_C \Delta y_{C} - m_A \Delta y_{A} \end{align}$$

Noting that $m_C = m_A$, we get:

$$\begin{align} \Delta y \propto \Delta y_{C} - \Delta y_{A} \end{align}$$

Therefore the condition for the system stability is equivalent to:

$$\begin{align*} \Delta y < 0 \Rightarrow \Delta y_{C} - \Delta y_{A} < 0 \end{align*}$$

From the relations prepared above we can express $\Delta y_{C}$ and $\Delta y_{A}$ as follows:

$$\begin{align*} \Delta y_{A} & = y_{A\varphi} - y_{A0} = \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 - \frac{h}{2} = \frac{a^2}{24h}\varphi^2 - \frac{h}{4} \varphi^2 \\ \Delta y_{C} & = y_{C\varphi} - y_{C0} = \left(h - \frac{b}{2}\right)\left(\cos\varphi - 1 \right) \approx -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} \end{align*}$$

Now plugging the above expressions for $\Delta y_{C}$ and $\Delta y_{A}$ into the stability inequality, we get:

$$\begin{align*} & -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} - \frac{a^2}{24h}\varphi^2 + \frac{h}{4} \varphi^2 < 0 \Rightarrow \\ & \Rightarrow \frac{b}{4}\varphi^2 - \frac{h}{4} \varphi^2 - \frac{a^2}{24h}\varphi^2 < 0 \Rightarrow \\ & \Rightarrow 6 (b - h) h - a^2 < 0 \Rightarrow 6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right) b^2 - a^2 < 0 \Rightarrow \\ & \Rightarrow \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{align*}$$

In conclusion, we are happy that the minimum potential energy and the torque methods give the same result.

$$\begin{equation*} \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{equation*}$$