Problem source

A tall container made of light material of negligible thickness has the form of a prism, with a square base of area $a^2$. It contains a volume $ka^3$ of fluid of uniform density. The container is held so that it stands on a rough plane, which is inclined at angle  $\theta$ to the horizontal, with two of the edges of the base of the container horizontal. 
In the case $k > \frac12 \tan\theta$, show that the centre of mass of the fluid is at a distance $x$ from the lower side of the container and at a distance $y$ from the base of the container, where
\[
\frac x a = \frac12 - \frac {\tan\theta}{12k}\;,
\ \ \ \ \ \
\frac y a = \frac k 2 + \frac{\tan^2\theta}{24k}\;.
\]
Determine the corresponding coordinates in the case $k < \frac12 \tan\theta$.
The container is now released.
Given that  $k < \frac12$, show that the container will topple if $\theta >45^\circ$.

Solution source

\begin{center}
\begin{tikzpicture}[scale=0.8]
    \coordinate (O) at (0, 0);
    \coordinate (X) at (6, 2.5);
    \coordinate (Y) at (5, 0);

    \coordinate (A) at ($(O)!0.2!(X)$);
    \coordinate (B) at ($(O)!0.6!(X)$);
    \coordinate (C) at ($(A) + ({-2.5*.6}, {6*.6})$);
    \coordinate (D) at ($(C) + (B)-(A)$);
    \coordinate (E) at ($(A)!0.8!(C)$);
    \coordinate (F) at ($(E) + ({6 * .47}, 0)$);

    \fill[blue, opacity = 0.1] (A) -- (E) -- (F) -- (B) -- cycle;
    
    \draw (O) -- (X);
    \draw (A) -- (C);
    \draw (B) -- (D);
    \draw (E) -- (F);

    \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = Y--O--X};

    \draw[dashed] ($(E) + ({6 * .47}, 0)$) -- ($(E) + ({6 * .47}, 0) + (A) - (B)$);
    
\end{tikzpicture}
\end{center}

The fluid can be divided into a cuboid parallel to the slope and a right-angled triangle. If the height of the water on the longer side is $\ell a$, then we have $ka^3 = (\ell a - a\tan \theta)a^2 + \frac12 a^3\tan \theta \Rightarrow \ell = k + \frac12 \tan \theta$

This is acceptable when $k > \frac12 \tan \theta$.

The centre of mass of the cuboid will be $(\frac{a}{2},  \frac12 (k - \frac12 \tan \theta))$ and of the triangle will be $(\frac13 a, \frac13 \tan \theta + (k - \frac12 \tan \theta) )$

Therefore we have:

\begin{align*}
    && \text{COM} && \text{mass} \\
    \text{cuboid} && (\frac{a}{2},  \frac{a}2 (k - \frac12 \tan \theta)) && a^3(k - \frac12 \tan \theta) \\
    \text{triangle} && (\frac13 a, \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) ) && a^3\frac12 \tan \theta \\
    \text{whole system} && (x, y) && a^3k
\end{align*}

Therefore

\begin{align*}
   && a^3k x &= \frac{a}{2} \cdot a^3(k - \frac12 \tan \theta)  + \frac13 a \cdot a^3\frac12 \tan \theta \\
    &&&= a^4 \frac{k}{2} - \frac{1}{12}a^4 \tan \theta \\
    \Rightarrow && \frac{x}{a} &= \frac12 - \frac{\tan \theta}{12 k} \\
    \\
    && a^3k y &= \frac{a}2 (k - \frac12 \tan \theta) \cdot a^3(k - \frac12 \tan \theta) + \\
    &&& \qquad\qquad \cdots + \l \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) \r \cdot a^3\frac12 \tan \theta  \\
    &&&= \frac{a^4k^2}{2} -\frac{a^4k \tan \theta}{2} + \frac{a^4 \tan^2 \theta}{8} - \frac{a^4 \tan^2 \theta}{12} + \frac{a^4k \tan \theta}{2} \\
    \Rightarrow && \frac{y}{a} &= \frac{k}2 + \frac{\tan^2 \theta}{24k}
\end{align*}

\begin{center}
\begin{tikzpicture}[scale=0.8]
    \coordinate (O) at (0, 0);
    \coordinate (X) at (3, 4);
    \coordinate (Y) at (5, 0);

    \coordinate (A) at ($(O)!0.3!(X)$);
    \coordinate (B) at ($(O)!0.7!(X)$);
    \coordinate (C) at ($(A) + ({-4*.6}, {3*.6})$);
    \coordinate (D) at ($(C) + (B)-(A)$);
    \coordinate (E) at ($(A)!0.8!(C)$);
    \coordinate (F) at ($(E) + ({6 * .5}, 0)$);

    \fill[blue, opacity = 0.1] (A) -- (E) -- (F) -- cycle;
    
    \draw (O) -- (X);
    \draw (A) -- (C);
    \draw (B) -- (D);
    \draw (E) -- (F);

    \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = Y--O--X};
\end{tikzpicture}
\end{center}

If the water only fills up a prism, it's sides must be $b$ and $b\tan \theta $, therefore the volume is $\frac12 ab^2 \tan \theta = ka^3 \Rightarrow b = a\sqrt{\frac{2k}{\tan \theta}}$

The centre of mass will be at $\l \frac13 a\sqrt{\frac{2k}{\tan \theta}}, a\sqrt{2k \tan \theta}\r$

The container will topple if the centre of mass is outside the base, ie if the centre of mass $(x,y)$ lies above the line $y = \tan (90^\circ- x) = \frac{1}{\tan \theta} x$. If $\theta > 45^\circ$ then $\tan \theta > 1$ and so we are in the $\frac12 \tan \theta > \frac12 > k$ and so we are in the second case.

\begin{align*}
    \frac{y}{x} &= \frac{\frac 13 a\sqrt{2k \tan \theta}}{\frac13 a\sqrt{\frac{2k}{\tan \theta}}} \\
    &= \tan \theta 
\end{align*}

$\tan \theta > \frac{1}{\tan \theta} \Leftrightarrow \tan \theta > 1 \Leftrightarrow \theta > 45^\circ$.