In a Coutte flow, the applied force on one of the plates can be expressed by $F = \eta A \frac{dv}{dz}$. F here can also be defined using Newton's second law of motion, $F = m \frac{dv}{dt}$.

The two forces are equal here. That said, there has to be a way to derive $m \frac{dv}{dt}$ from $\eta A \frac{dv}{dz}$ or vice versa. But can't quite figure it out. What am I missing here?

(I can "sense" the similarity between the second law of motion, $F = m \frac{dv}{dt}$ and the viscosity, $\eta$, coming from $F = \eta A \frac{dv}{dz}$, but can't derive $m \frac{dv}{dt}$ from $\eta A \frac{dv}{dz}$).

Any help/comment would be appreciated.


closed as unclear what you're asking by Kyle Kanos, ACuriousMind, Ryan Unger, user10851, Gert Mar 18 '16 at 2:10

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    $\begingroup$ Can you make the analogy between two scenarios more clearer? so that we know what you know and what you don't know and then see if we can help you with things you don't know. Thanks, $\endgroup$ – Benjamin Mar 17 '16 at 18:36
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    $\begingroup$ The similarity you "sense" is entirely superficial and based on the appearance of two similar looking derivatives: $\frac{dv}{dt}$ and $\frac{dv}{dz}$. There's no connection. One's a derivative to time, the other's a spatial gradient. $\endgroup$ – Gert Mar 17 '16 at 18:58
  • $\begingroup$ @Benjamin, considering Couette flow, the force deriving the viscosity equation should be the same force in Newton's second law. That said, there is no reason why a conversion from one to another won't be possible as stated in my op. $\endgroup$ – gamahuri Mar 17 '16 at 19:44
  • $\begingroup$ You're comparing apples and oranges. Newton's law of viscosity gives the contact forces between various parcels of deforming fluid. These sum, together with body forces, to give the net force on a parcel, which is then related by Newton's 2nd law of motion to the acceleration of the parcel. So viscous forces are contact forces and Newton's 2nd law relates to net force. $\endgroup$ – Chet Miller Mar 18 '16 at 14:24
  • $\begingroup$ @Chester, viscous force is the reaction force here. Fluid is a continuum here. That makes the "parcel" and "sum" consideration invalid. $\endgroup$ – gamahuri Mar 19 '16 at 0:05

The proportional relation between force and velocity we sometimes refer to as a viscous friction factor is a highly idealized model that may fit actual physical behavior sometimes, but in general, not.

Although we might be able to fudge the units and attempt to wedge viscosity in, the result doesn't stand as any general law in connection with Newton's second law.

So the answer is no.

  • $\begingroup$ "[...] highly idealized model that may fit actual physical behavior sometimes, but in general, not." It fits non-Newtonian fluids rather well, actually. $\endgroup$ – Gert Mar 17 '16 at 22:46
  • $\begingroup$ @Gert not blood $\endgroup$ – docscience Mar 17 '16 at 22:48
  • $\begingroup$ And blood's your typical fluid? $\endgroup$ – Gert Mar 17 '16 at 22:49
  • $\begingroup$ I don't regard the viscous behavior of a Newtonian fluid as a "highly idealized model" by any stretch of the imagination. The Newtonian Fluid relationship between the stress tensor and the rate of deformation tensor involving viscosity as a proportionality constant provides the fundamental basis of modern fluid mechanics. $\endgroup$ – Chet Miller Mar 18 '16 at 14:20
  • $\begingroup$ @ChesterMiller What I'm trying to illustrate is that if you discount relativistic effects, the force of inertia $F=ma$ is a very clean model that adheres closely to physical observations. On the other hand 'Newtonian' fluids are an idealistic concept. Different fluids have different behavior but mass (inertia) is simply inertia no matter how you cut it. Thus the force due to viscous frictional loss, $F=Bv$ is not a very clean model. It is (more) idealistic than the second law. $\endgroup$ – docscience Mar 18 '16 at 14:54

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