Problem 8: Thickness of Residual Water Layer
Analysis:
When the water stops spreading and spilling, the system is in equilibrium at the edge. We analyze the forces acting on a unit length of the water boundary.
1. Hydrostatic Force (Pushing Outward): The water pressure increases with depth $z$ as $P = \rho g z$. The total average force pushing outwards on the edge of height $h$ is the integral of pressure over the area. $$ F_{hydro} = \int_{0}^{h} (\rho g z) dz = \left[ \frac{\rho g z^2}{2} \right]_0^h = \frac{1}{2}\rho g h^2 $$
2. Surface Tension Force (Holding Inward): Surface tension acts along the surface of the liquid.
- At the top surface, the force is $\sigma$ pulling horizontally.
- At the contact point with the glass, the force has a horizontal component $\sigma \cos \theta$ (acting in opposition to the spread).
3. Solving for Thickness ($h$): $$ h^2 = \frac{2\sigma(1 – \cos \theta)}{\rho g} $$ $$ h = \sqrt{\frac{2\sigma(1 – \cos \theta)}{\rho g}} $$
Answer: $h = \sqrt{\frac{2\sigma(1 – \cos \theta)}{\rho g}}$