# Viscous friction (force): coefficient and examples

The **viscous friction** arises when a solid object is moved through a fluid or gas – liquid – a. It can be modeled as a force proportional to the negative of the object’s velocity or its square.

The use of either model depends on certain conditions, such as the type of fluid in which the object is moving and whether it is very fast or not. The first model is known as *linear resistance* , and in it the magnitude of the viscous _{friction} F is given by:

F _{rub} = γv

Here γ is the proportionality constant or viscous friction coefficient and v is the object’s velocity. It is applicable to bodies moving at low speeds in fluids with a laminar regime.

In the second model, known as *quadratic resistance* or Rayleigh’s law, the magnitude of the friction force is calculated according to:

F _{rub} = ½ ρ.AC _{d} .v ^{2}

Where ρ is the density of the fluid, A is the cross-sectional area of the object and C _{d} is the drag coefficient.

The product ½ ρ.AC _{d} is an aerodynamic constant called D, whose SI units are kg / m, therefore:

F _{rub} = Dv ^{2}

This model is most appropriate when the velocity of objects is medium or high, as the motion produces turbulence or eddies as it passes through the fluid.

A moving tennis ball and cars on the road are examples of objects on which this model works very well.

The viscous force arises because the solid must separate layers of fluid to move through it. The existence of several models is due to the fact that this force depends on multiple factors, such as fluid viscosity, velocity and object shape.

There are more aerodynamic objects than others and many are precisely designed so that the resistance of the medium minimizes its speed.

__Examples of viscous friction__

__Examples of viscous friction__

Any person or object that moves in a fluid necessarily experiences resistance from the environment, but these effects are often overlooked in simple applications such as free fall.

The statements of almost every free fall problem show that the effects of air resistance are overlooked. This is because air is a very “thin” fluid and therefore we expect that the friction it offers will not be significant.

But there are other movements where viscous friction has a more decisive influence, let’s see some examples:

**Stones falling into water and pollen grains**

-A rock that is thrown vertically into an oil-filled tube experiences a force that opposes its descent, thanks to the resistance of the fluid.

-Pollen grains are very small; therefore, for them, air resistance is not negligible, because, thanks to this force, they can stay afloat for a long time, causing seasonal allergies.

**Swimmers and Cyclists**

-In the case of swimmers, they wear a cap and shave completely so that the resistance of the water doesn’t slow them down.

-Like swimmers, time trial cyclists experience air resistance; consequently, helmets have aerodynamic designs to improve efficiency.

Likewise, the position of the cyclist within a competing group is relevant. Whoever leads the march evidently receives the greatest resistance from the air, while for whoever closes the gear this is almost nil.

**parachutists**

-Once a skydiver opens the parachute, it is exposed to the viscous friction of the air, the most suitable model is the one with the square of velocity. In this way, it reduces its speed and, as friction opposes the fall, it reaches a constant limit value.

**Automobiles**

For automobiles, the coefficient of aerodynamic resistance, a constant determined experimentally and the surface that it presents against the wind, are the determining factors for reducing air resistance and reducing consumption. That’s why they are designed with slanted windshields.

**Millikan Oil Drop Experiment**

-In the Millikan oil drop experiment, physicist Robert Millikan studied the movement of oil droplets in the middle of a uniform electric field, concluding that any electric charge is a multiple of the electron charge.

For this, it was necessary to know the radius of the drops, which could not be determined by direct measurement, due to their small size. But in this case, the viscous friction was significant and the drops ended up slowing down. This fact made it possible to determine the radius of the drops and subsequently their electrical charge.

__Exercises__

__Exercises__

**– Exercise 1**

In the equation of the viscous friction force at low speed:

*F _{rub} = γv*

a) What dimensions should the viscous friction coefficient γ have?

b) What are the units of γ in the International System of Units?

**Solution for**

Unlike static friction or kinetic friction coefficients, the viscous friction coefficient has dimensions that should be:

strength / speed

Force has dimensions of mass x length/time ^{2} , while those of velocity are length/time. Denoting them as follows:

-Mass: M

-Length: L

-Time: T

The dimensions of the viscous friction coefficient γ are:

[ML / T ^{2} ] / [L / T] = [MLT / LT ^{2} ] = M / T

**Solution b**

In SI, the units of γ are kg / s

**– Exercise 2**

Taking the water resistance into consideration, find an expression for the terminal velocity of a metallic ball that is dropped vertically into an oil-filled tube, in these cases:

a) low speed

b) high speed

**Solution for**

The figure shows the diagram of the free body, showing the two forces acting on the sphere: the downward weight and the upward velocity-proportional fluid resistance. Newton’s second law for this motion states the following:

γv _{t} – mg = 0

Where v _{t} is the terminal speed, given by:

v _{t} = mg / γ

**Solution b**

If we assume medium to high speeds, the appropriate model is the one with the speed squared:

F _{rub} = ½ ρ.AC _{d} .v ^{2}

Thus:

½ ρ.AC _{d} .v ^{2} – mg = 0

Dv ^{2} – mg = 0

v = √ [mg / D]

In both situations, the greater the object’s mass, the greater its terminal velocity.