Angle of Contact

When a liquid's surface comes into contact with a solid surface (for example, the wall of a narrow tube dipped into the liquid), it forms a meniscus — a curved surface that can be either convex (bulging upward, as with mercury against glass) or concave (dipping downward like a bowl, as with water against glass). The angle of contact θ, between a given liquid and a given solid surface, is defined as the angle, measured within the liquid, between the tangent drawn to the liquid's free surface and the tangent drawn to the solid's surface, both taken at their common point of contact. (In practice, when reading the level of water in a narrow capillary, one reads the tangent to the meniscus from inside the water; when reading the level of mercury, one reads the tangent to the meniscus from just above the mercury column.)

i) Concave meniscus — acute angle of contact. Consider a molecule A on a liquid's surface very close to a vertical container wall — kerosene in a glass bottle, say. Molecule A experiences both a net adhesive force $\vec{AP}$, drawn purely horizontal because the wall is vertical, and a net cohesive force $\vec{AC}$, directed into the liquid at very nearly 45° to the two surfaces. If the magnitude of the adhesive force is large enough, the resultant $\vec{AR}$ of these two forces is directed into the solid wall itself. For the liquid surface to remain in stable equilibrium, however, the net force $\vec{AR}$ acting on every surface molecule must be exactly normal (perpendicular) to the liquid surface everywhere — so, for $\vec{AR}$ to end up normal to the tangent AT drawn to the surface, the liquid must actually pile up higher against the solid wall, which is exactly what produces a concave meniscus. Such a liquid is said to wet the solid surface.

ii) Convex meniscus — obtuse angle of contact. Now consider a molecule A on a liquid's surface near a vertical wall for a liquid like mercury in a glass bottle. Again the net adhesive force $\vec{AP}$ is horizontal, but here the magnitude of the cohesive force is so large that the resultant $\vec{AR}$ is instead directed into the liquid itself. By the same equilibrium requirement, the liquid must creep away from (downward against) the solid boundary so that AR again ends up normal to the surface — this produces a convex meniscus, and such a liquid is said not to wet that solid surface.

iii) Zero angle of contact. For a liquid that completely wets a solid — pure water on very clean glass, for instance — the number of liquid molecules actually near the contact region is so small that the cohesive force is essentially negligible, $\vec{AC} \approx 0$, and the net adhesive force is itself the whole resultant force, $\vec{AP} = \vec{AR}$. In this limiting case the tangent AT runs flush along the wall, inside the liquid, and the angle of contact comes out essentially zero.

iv) Angle of contact = 90° and the acute/obtuse condition. For a hypothetical liquid whose angle of contact with a given container happens to be exactly 90°, the net cohesive force $\vec{AC}$ sits at exactly 45° to both surfaces and the resultant $\vec{AR}$ works out to be exactly vertical, running along the solid surface. For this special case to hold, $AP = AC/\sqrt{2}$, where AC is the magnitude of the net cohesive force. From this boundary condition, the general rule follows directly: the angle of contact is acute (giving a concave meniscus) whenever $AP > AC/\sqrt{2}$, and obtuse (giving a convex meniscus) whenever $AP < AC/\sqrt{2}$.

b) Shape of liquid drops on a solid surface. The very same competition of forces decides whether a small liquid drop placed on a flat solid surface spreads out or beads up. Let θ be the angle of contact for a given solid-liquid pair, and let $T_1$ be the force due to surface tension at the liquid-solid interface, $T_2$ the force due to surface tension at the air-solid interface, and $T_3$ the force due to surface tension at the air-liquid interface — each acting tangentially along its own interface. For the drop's edge to be in equilibrium:

$T_2 = T_1 + T_3\cos\theta, \qquad \cos\theta = \dfrac{T_2 - T_1}{T_3} \qquad \text{--- (2.18)}$ …