A cube-shaped container with edge a is filled with water to…

A cube-shaped container with edge $a$ is filled with water to half its height and stands with one face on a horizontal floor. How much work is needed to tip it over about one of its bottom edges? (Treat the container's own mass as negligible; its effect is added at the end.)

Explanation of Method

Work to tip over a half-filled cubic container.

Step 1: The water stays level

As the container turns, the water surface stays horizontal and the water flows toward the lower side, i.e. toward the edge it turns about. The container tips over by itself once the centre of mass of the water is vertically above that edge. By symmetry, this happens when the cube stands on the edge with a face diagonal vertical (45°).

Step 2: Centre of mass of the water

Upright: the water is a box of height $a/2$; its centre of mass is at $a/4$.
On the edge: the water fills half the cross-section, the triangle below the horizontal diagonal. A triangle's centre of mass is $2/3$ of the way from the tip to the base, so it lies $\tfrac23 \cdot \tfrac{a}{\sqrt2} = \tfrac{\sqrt2}{3}a$ above the edge.

\[ \Delta h_w = \frac{\sqrt2}{3}a - \frac{a}{4} = \frac{4\sqrt2 - 3}{12}\,a \approx 0.221\,a. \]

Step 3: Work

With $m_w = \rho a^3/2$: $A = m_w g\,\Delta h_w = \frac{\rho a^3}{2}\,g \cdot \frac{4\sqrt2 - 3}{12}\,a$.

\[A = \frac{4\sqrt2 - 3}{24}\,\rho g a^4 \approx 0.111\,\rho g a^4\]

If the empty container has mass $M$, its own centre of mass rises by $\tfrac{\sqrt2 - 1}{2}a$ and the tipping point is still at 45°, so add $Mg\,\tfrac{\sqrt2 - 1}{2}a$.

Check

Example $a = 0.5\ \mathrm{m}$: water 62.5 kg, $A \approx 0.111 \cdot 1000 \cdot 9.81 \cdot 0.0625 \approx 68\ \mathrm{J}$

$\text{upright}$ $\text{on the edge}$ $\tfrac{a}{4}$ $\Delta h$ $\text{water CM: } \tfrac{a}{4} \to \tfrac{\sqrt2}{3}a$

The water stays level as the cube turns. When the cube stands on its edge with a diagonal vertical, the water fills the lower triangle, whose centre of mass (red) is $\sqrt2 a/3$ above the edge, directly over it.

Result

\[\boxed{\,A = \frac{4\sqrt2 - 3}{24}\,\rho g a^4 \approx 0.111\,\rho g a^4\,}\]

Remark

About the original solution. It moved the water as if it were a frozen block (final height $\sqrt2 a/4$), but the water flows and stays level, ending up as the triangle whose centre of mass is at $\sqrt2 a/3$. It also set the container's mass equal to $\rho a^3$, which the problem does not give. Its result $\tfrac58(\sqrt2 - 1)\rho g a^4$ is therefore not correct; for a light container the work is $\tfrac{4\sqrt2 - 3}{24}\rho g a^4$.