Young's equation Can somebody derive the Young's equation for the contact angle and surface energy of a droplet.

$$\gamma_{la}Cos\theta+\gamma_{ls}+\gamma_{sa}$$
I've scratched my head for days finding how it exist.but couldn't find it anywhere on internet.please help:|
 A: To try and make this clearer I've drawn a three dimensional picture of the drop:

The surface tension is the force per unit length acting normal to a line. Consider the small segment of the perimeter of the drop I have marked as $\ell$. We'll assume $\ell$ is small enough that it can be considered as straight. If the surface-vapour surface tension is $\gamma_{sv}$ then this produces a force on our line $\ell$ of $F = \gamma_{sv} \ell$ as shown by by the red line. So the total outwards force is:
$$ F_\text{out} = \gamma_{sv} \ell $$
Likewise inside the drop the surface tension between the solid and the liquid is pulling the drop inwards with a force $F = \gamma_{ls} \ell$. This is shown by the blue line.
And finally the surface tension at the liquid-vapour interface is pulling our line element with a force $F = \gamma_{lv} \ell$ and it is pulling upwards at an angle $\theta$ where $\theta$ is the contact angle. So the horizontal component of this force is pulling our line inwards with a force $F = \gamma_{lv} \ell \cos\theta$. This is shown by the green line. So the total inwards forces (the blue and green arrows) are:
$$ F_\text{in} = \gamma_{sl} \ell + \gamma_{lv} \ell \cos\theta $$
And the last step is just to say that if the drop is in equilibrium, i.e. it is neither spreading out not rolling up, then the inwards and outwards forces must be the same:
$$ \gamma_{sv} \ell = \gamma_{sl} \ell + \gamma_{lv} \ell \cos\theta $$
And just divide through by $\ell$ to get Young's equation:
$$ \gamma_{sv} = \gamma_{sl} + \gamma_{lv} \cos\theta $$
