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Viscous stress is proportional to $\frac{du}{dy}$, so they only arise from relative motion between different layers of fluid. That makes it somewhat similar to kinetic friction, which requires relative motion between two objects.

But what if there is no relative motion between the different layers of fluid? Is there an equivalent of static friction in fluid dynamics?

In particular, I'm considering this hypothetical static scenario for fluids. Say you place a small fluid droplet on a gentle incline, such that it remains in static equilibrium without sliding down the incline. How exactly is this equilibrium achieved?

Viscosity definitely doesn't play a role here, because everything is stationary, so there is no velocity differential and thus no stress due to viscous effects.

I could see how the droplet might end up getting deformed in a peculiar manner in order for this equilibrium to be achieved, such that the force due to surface tension will have sufficient component upward along the incline to balance the downward component of its weight along the incline. So, surface tension could be what enables it to be at equilibrium.

But could there be any other forces at play? Something doesn't feel too right about the claim that surface tension does all of this.

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  • $\begingroup$ There will be an adhesive force between the fluid droplet and slope. If you have a drop of mercury on an inclined surface it will "roll" down the slope assuming that mercury does not adhere to the slope. $\endgroup$
    – Farcher
    Oct 23, 2021 at 7:01
  • $\begingroup$ @Farcher I see. What is known about the nature of this adhesive force? E.g. is it proportional to the normal force on the fluid, just like static friction? I'm trying to find out more about how this adhesion is modelled... $\endgroup$
    – chris97ong
    Oct 23, 2021 at 7:08
  • $\begingroup$ Adhesion $\endgroup$
    – Farcher
    Oct 23, 2021 at 7:12

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Viscous stress is proportional to $\frac{du}{dy}$

This is actually only partly true; not all fluids are Newtonian fluids and non-Newtonian fluids behave differently from Newtonian fluids.

There are many models for viscosity as a function of shear:

  • Newtonian: $\tau=\mu\frac{du}{dy}$
  • Bingham: $\tau = \tau_0+\mu_{\infty}\frac{du}{dy}$ for $\tau>\tau_0$ else $\frac{du}{dy}=0$
  • Ostwald–de Waele: $\tau = K(\frac{du}{dy})^n$
  • others like pseudoplastic, carreau-yasuda, etc

Basically in general (very simplified): $$\tau = \tau_0 + K(\frac{du}{dy})^n$$ for $\tau>\tau_0$ else $\frac{du}{dy}=0$

Here $\tau_0$ is known as the yield stress and it is basically the equivalent of static friction. It is the stress that must be applied before the fluid even starts to move and only then will viscosity (or dynamic friction) play a roll.

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    $\begingroup$ I would add a statement up front that not all fluids are Newtonian fluids, and that non-Newtonian fluids behave differently from Newtonian fluids. $\endgroup$ Oct 23, 2021 at 11:56
  • $\begingroup$ Looking at Bingham's model, it does seem like only specific types of fluid follow such a relation for viscous stress. But surely, there should be a form of static friction involved even for Newtonian fluids like water right? $\endgroup$
    – chris97ong
    Oct 23, 2021 at 23:37
  • $\begingroup$ @chris97ong, the yield stress is generally due to molecular interactions for e.g. long-chain polymers which require to be elongated somewhat by a stress before they start flowing. For newtonian fluids the molecules are considered to be almost ideal and little effects from molecular interactions (akin to a ideal gas assumption) so yield stress for newtonian fluids is possible just so small it is considered negligble. $\endgroup$
    – nluigi
    Oct 24, 2021 at 13:46

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