Why don't we include the adhesive and cohesive force while calculating rise in a capillary tube?

Question

The contact angle of a liquid solid interface is explained by saying that the liquid surface must be perpendicular to the resultant of adhesive cohesive and gravitational forces acting on it, since it cannot sustain shear stresses.

However, once the contact angle is determined, the cohesive and adhesive forces are always omitted from the discussion. For instance, one way of calculating the rise of water in a capillary tube is to equate the force due to the surface tension to the weight of the liquid risen.

$$2\pi R T \cos \theta = \pi R^2 \rho g h $$

which gives

$$h = \frac{2T\cos \theta}{R\rho g}$$

where $R$ is the radius of the tube, $T$ the surface tension and $\theta$ the contact angle. However, why are adhesive and cohesive forces excluded from this discussion. As far as I'm aware, the adhesive forces is the main reason the liquid rises (or falls) in the capillary and not the surface tension.

Answer 1

Adhesive forces are accounted for when calculating capillary height.

My guess is that you think they are not because you read, somewhere, a discussion in which adhesive forces were used to calculate a contact angle, then the contact angle was used to calculate the height. In that case, adhesive forces are being use the calculate the height. They are simply being used through the intermediary of the contact angle.

If you want, you can do the calculation like this:

The water height will rise so that the energy of the system is minimized. Let us assume that the shape of the surface of the water in the capillary is fixed and focus only on the height of the column.

There is gravitational energy to account for as well as surface energy between the water and the column.

If we raise the water height by an amount $\mathrm{d}h$, we have increased the gravitational energy by $\rho g A h \mathrm{d}h$, where $h$ is the height of the bottom of the surface of the water, A is the cross-sectional area of the capillary, $\rho$ is the density of the water, and $g$ is gravitational acceleration.

If the surface energy per unit area of contact between the water and the capillary is $-\gamma$, we reduce the energy by $2\pi r \gamma \mathrm{d}h$ when raising the height by $\mathrm{d}h$.

The energy is minimized when these two are equal,

$$\rho g \pi r^2 h = 2\pi r \gamma$$

or

$$h = \frac{2 \gamma}{\rho g r}$$

If we take $\gamma = \cos\theta T$ we reproduce your expression.