When an object moves through a fluid, it feels a resistance to its motion for two reasons: (1) It is dragging the fluid along with it and making the fluid slide along itself, and the fluid has an internal resistance to doing that (viscosity); (2) It is pushing the fluid in front of it, making the fluid in front of it go with the same speed that it is going.  To do that the object has to exert a force on the molecules of fluid in the direction of motion, so, by Newton's 3rd law, the fluid molecules exert a force back on the object, opposite to the direction of motion.  This second force is called drag or sometimes inertial drag.   An object moving in a fluid always feels both of these, but depending on the various parameters of the situation, one can be much more important than the other.  The ratio of the drag force to the viscous force is called the Reynolds number and tells which of the two forces you have to pay attention to in a given situation.

Though motion in a fluid is a very complex process, we can get a pretty good idea of what the drag force looks like with a simple model.  Indeed, let's pick the simplest possible model to calculate: a cylinder moving through a fluid as shown in the figure at the left.  We have a cylinder of radius, R, moving in a direction along its axis with a speed v through a mass of molecules of density, ρ (rho).

We'll model this situation with the simplest model we can create.  We know the molecules in the fluid are jittering around at a very high speed.  But on the average, as many are going in the opposite direction as in any given direction.  We'll assume that on the average all the molecular jiggling cancels out.  The cylinder will be pushing molecules in front of it.  It therefore has to add speed to those molecules, making them move (on the average) the same speed as the cylinder is going.  So how much force does it have to exert to add speed to those molecules? A little bit of fluid will be sped up. The force needed to do this is given by Newton's second law:

$$F_{cyl \rightarrow fluid} = m_{fluid moved} \frac{\Delta v_{fluid}}{\Delta t}$$

In a small amount of time, Δt, the cylinder will move a distance $Δx = v Δt$.  The front surface of the cylinder will therefore sweep out all the molecules in a thin volume equal to the area of the circle on top of the cylinder times $Δx$. So it sweeps out a small volume $ΔV = πR^2 Δx$.  The mass of molecules in this volume is this volume times the density, or $Δm = ρΔV$.  The change in velocity is from 0 to $v$ so $\Delta v = v$ for this small mass. To make this happen, the the cylinder has to exert a force of

$$F_{cyl \rightarrow fluid} = (\rho \Delta V) \frac{\Delta v}{\Delta t}$$

$$F_{cyl \rightarrow fluid} = (\rho \pi R^2\Delta x) \frac{v}{\Delta t}$$

Sliding the $\Delta x$ over the $\Delta t$ gives

$$F_{cyl \rightarrow fluid} = (\rho \pi R^2 v) \frac{\Delta x}{\Delta t}$$

But $\frac{\Delta x}{\Delta t} = v$ so this becomes

$$F_{cyl \rightarrow fluid} = \rho \pi R^2 v^2$$

Since $\pi R^2$ is the area of the cylinder facing into the fluid and pushing it, we might in general expect an equation that looks like

$$F_{cyl \rightarrow fluid} = \rho A v^2$$

where $A$ is the area of the object facing into the fluid.

By Newton's 3rd law, the force the fluid exerts back on the cylinder is equal in magnitude and opposite to this:

$$F_{fluid \rightarrow cylinder} = \rho A v^2$$

So we conclude that the drag force that a fluid exerts on an object moving through it

  • opposes the motion
  • is proportional to the density of the fluid.
  • is proportional to the area being pushed through the fluid.
  • is proportional to the square of the velocity of the object through the fluid.

Now our model was pretty simplified.  We ignored the fact that the molecules of the fluid were moving, and we ignored the fact that they in fact had to go someplace — they are being pushed into fluid ahead of the object so the fluid has to slide through itself. The result of all this is summarized by putting in a dimensionless coefficient $C_{drag}$ — the drag coefficient — which is determined by measurement. 

$$F^{drag}_{fluid \rightarrow cylinder} = \frac{1}{2} C_{drag} \rho A v^2$$

The factor of ½ has to do with the derivation of this law from momentum conservation and makes the interpretations of the coefficient more natural.*)

*Note that the drag coefficient can hide some complexity. But for a wide range of conditions and objects, it is in the neighborhood of "1" (that is, between 0.5 and 5.0). 

Joe Redish and Julia Gouvea 9/30/11


Article 377
Last Modified: January 31, 2019