1
$\begingroup$

Edit:(4/1/2016) Okay, I get how the required equation arises from force balance. All forces are acting on the same length, so the length cancels out of the equation and what we've got is essentially the net force per unit length.

But, could someone provide a thermodynamic derivation of the equation? $$\frac{\sin \phi_{\alpha\theta}}{\gamma_{\alpha\theta}} = \frac{\sin \phi_{\alpha\beta}}{\gamma_{\alpha\beta}} = \frac{\sin \phi_{\beta\theta}}{\gamma_{\beta\theta}} $$

Original Question:

Why does net force per unit length have to be zero for equilibrium? Specifically, I'm looking at angles at the triple line with 3 phases $\alpha, \beta, \theta$.

This wikipedia article says

In equilibrium, the net force per unit length acting along the boundary line between the three phases must be zero

Why is this the case? I though net force, not force per unit length had to be zero in equilibrium. I would appreciate any derivation of the wanted equation, $$\frac{\sin \phi_{\alpha\theta}}{\gamma_{\alpha\theta}} = \frac{\sin \phi_{\alpha\beta}}{\gamma_{\alpha\beta}} = \frac{\sin \phi_{\beta\theta}}{\gamma_{\beta\theta}} $$

using thermodynamics, or force balance. I can find no derivation of this relation (called Young's Equation) online.

$\endgroup$
  • 1
    $\begingroup$ For the triple point to remain stationary, the surface tensions of all the boundaries must sum to zero at the triple point. Otherwise, there will be a net force to move the triple point and it is not in equilibrium. The rest is geometry and is left as an exercise for the reader. The real question is, do you understand surface tensions and contact angles? $\endgroup$ – Jon Custer Jan 4 '16 at 17:53
  • $\begingroup$ @JonCuster And I'm interested in a derivation using thermodynamics. $\endgroup$ – Aritra Das Jan 4 '16 at 18:02
  • $\begingroup$ @JonCuster Would you mind having a look now? $\endgroup$ – Aritra Das Jan 4 '16 at 18:30
  • $\begingroup$ Assume a planar triple junction (i.e. infinite in z-direction). Write down the energy of the junction as a function of the surface energies and angles. This is the free energy of the system. Differentiate and set equal to zero. This is the equilibrium point. $\endgroup$ – Jon Custer Jan 4 '16 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.