Derivation of formula for capillary action

[jd12345]

the height upto which liquid rises in capillary tube is 2S cosθ / ρgR
I need a derivation for this - please
I know that a combination of adhesive force , surface tension and weight creates the contact angle θ but how to proceed after that?
To calculate height h i need to know the force by which tube is pulling the liquid(adhesive force) but how can i know that?

Its really important for me to know the derivation so please help me

 

[jd12345]

Cmon - please, even a single reply would help me a lot

 

[whoyameye]

http://hyperphysics.phy-astr.gsu.edu/hbase/surten2.html
hope this helps.. too lazy to type the whole thing.. also read articles on wikipedia baout surface tension and capillary action.. you should be able to figure out..

 

[Jabuz]

A short search on google came up with this:

http://www.schoolphysics.co.uk/age1...face tension/text/Capillary_action/index.html

 

[pmacias]

Sure, I can provide a derivation for the formula for capillary action. First, let's define the variables involved:

h = height of liquid rise in capillary tube
S = surface tension of liquid
θ = contact angle between liquid and capillary tube
ρ = density of liquid
g = acceleration due to gravity
R = radius of capillary tube

Now, let's consider a capillary tube immersed in a liquid as shown in the figure below:

[insert image of capillary tube immersed in liquid]

The liquid rises up the tube to a certain height h due to the combined effect of adhesive force, surface tension, and weight.

The adhesive force is the force of attraction between the liquid molecules and the molecules of the capillary tube. This force is responsible for the liquid rising up the tube.

The surface tension is the force per unit length acting at the interface between the liquid and the tube. It is responsible for the formation of a meniscus at the liquid-air interface.

The weight of the liquid column above the point of measurement also contributes to the liquid rise.

Now, let's consider the forces acting on a small section of the liquid column inside the tube, as shown in the figure below:

[insert image of forces acting on small section of liquid column]

The forces acting on this section of liquid are the adhesive force F_a, the surface tension force F_st, and the weight of the liquid column W.

Using trigonometry, we can resolve the surface tension force into its components as shown in the figure below:

[insert image of resolved surface tension force]

The vertical component of the surface tension force balances the weight of the liquid column, while the horizontal component contributes to the adhesive force.

Now, let's consider the equilibrium of forces acting on the liquid column. The sum of all vertical forces must be equal to zero, since the liquid column is not moving up or down. Therefore, we can write:

F_a + F_st cosθ - W = 0

Solving for F_a, we get:

F_a = W - F_st cosθ

Substituting the value of weight W = mg, where m is the mass of the liquid column and g is the acceleration due to gravity, we get:

F_a = mg - F_st cosθ

We can also express the weight of the liquid column as the product of its volume and density (m = ρV). Substituting this value, we get: