We've ignored air resistance in all scenarios previously, so it's finally time to face it. But first, a bit of explanation as to why I put this in the fluids unit. Actually, I'll let you see for yourself with the interactive demo below.

Air Resistance Demo

Speed: 8

Particle Rate: 10ms

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You see, air resistance is really due to the collision of flowing air particles with an object! This does sound an awful lot like how air pressure works. Moreover, air resistance and drag forces really only exist if the object is going through a fluid. Therefore, it makes perfect sense to have air resistance here, even though it does concern itself more with forces.

Air resistance is proportional to the relative velocity of the air and the object moving through it. In other words, the force of the wind pushing on you is air resistance, as is the wind that blows in your hair when you stick your head out of the car window. There are actually two common equations for air resistance. One is directly proportional to this relative velocity, while the other is proportional to the square of it.

I'll show the general forms of the equations first, with an arbitrary constant in front to indicate the strength of air resistance.

$$ F_d = - bv $$ $$ F_d = - cv^2 $$
The constants $b$ and $c$ are just arbitrary constants that describe how strongly the air resistance affects the object. The negative sign indicates that the force is in the opposite direction of the relative velocity, which makes sense as air resistance is a dissipative force. This is a good conceptual understanding, but how do we know when to use each formula?

There is no exact relationship, but the first formula is generally used when we have small objects moving relatively slowly through a thin medium or objects moving through a very viscous medium. The second is used when we have relatively large ojects moving at medium to high speeds through a medium. No exact formula can tell you which drag equation to use, but most of the time it should be obvious or given to you.

Now, I want to explore the actual drag equation. This one is a quadratic model, meaning it's closer to the second equation I gave above.

$$ F_d = \dfrac12 C_d A \rho v^2 $$
That's a lot of terms! Don't worry, though, I'll explain what each of these are. First, $v$ is obviously the relative velocity between the object and the air, and $\rho$ is the density of the air. The next variable $A$ represets the cross-sectional area of the surface moving through the air. The smaller this area, the better, which is why many airplanes have sleek airframes. Finally, $C_d$ is the drag coefficient, similar to the coefficient of friction but for objects moving in fluids. It has to be experimentally measured, and depends to some degree on what shape the object is. There are lots of models for drag, but we'll only focus the simplest one. This one is directly proportional to the relative velocity between the fluid and the object. It's relative velocity because air resistance doesn't care if you're moving or the air is moving. Think about how a breeze is essentially the same as the wind that blows when you open the window on a moving car. The simplest model uses a constant $b$ to describe air resistance.

$$ F_d = - b v $$
The constant $b$ is just an arbitrary constant that describes how strongly the air affects the object. This is related to things like the area of the object (larger areas mean larger air resistance, since there's more surface for air particles to hit), the density of the air (denser fluid means more particles hitting the object), and the nature and characteristics of the fluid itself. The negative sign indicates that the air resistance tends to resist relative motion, just like the other dissipative force we know about: friction.

I want to test your conceptual understanding of how air resistance works here before we move on. There's not going to be much calculation in this lesson, as air resistance is mainly a conceptual thing at the level we're approaching it. (We're not designing planes here.)


That last question leads us into our new concept: the terminal velocity. Basically, as something falls under gravity, it experiences an increasing drag force on it as it accelerates. However, at some point the drag force will equal the gravitational force pulling the object downwards, and the object will no longer accelerate. This velocity is called the terminal velocity.

Your terminal velocity can change! This is what parachutes do: they increase the drag coefficient on you by making a much larger area for the air to hit, which causes your terminal velocity to be less. However, when you reach that new terminal velocity the gravitational force is still equal to the drag force! This is a common misconception and trips up countless people, because it is unintuitive at first thought.

We can actually calculate what the terminal velocity would be. We're going to use both simplified models, though I will say that the quadratic model would likely be more accurate for a skydiver. It's a simple matter of force balance.

First, we'll start with the linear model.

$$ mg - bv_t = 0 $$ $$ v_t = \dfrac{mg}{b} $$
We use the same approach for the quadratic model.

$$ mg - c{v_t}^2 = 0 $$ $$ v_t = \sqrt{\dfrac{mg}{c} } $$
Simple enough. Now, you should be able to use this general approach to calculate the terminal velocity given any model for drag.

Okay, so now we know what happens at the end of the falling motion when there is drag. But what about the in-between parts? Those are important as well. It is possible to get an exact formula for what happens, but this is often unnecessary and it is easier to think conceptually. We start off with the moment that the object is dropped, and we consider the motion to "end" when we reach the terminal velocity.

Fortunately for you calculus students, we can actually use calculus to derive the exact equations of motion for an object that falls under air resistance. We will use the linear drag model for simplicity, and we can simply use our integration skills to get through this problem. First, we want to write the forces on the falling object:

$$ mg - bv = ma $$ $$ g -\dfrac{b}{m} v = a $$
Now, we want to recall that $a = \frac{dv}{dt}$, which lets us set this up as a differential equation:

$$ g -\dfrac{b}{m} v = \dfrac{dv}{dt} $$
We want to separate the variables here, so we can write this as:

$$ \dfrac{dv}{g -\dfrac{b}{m} v} = dt $$
Now, we can integrate both sides. The left side is a bit tricky, but we can use u-substitution. I won't show it here since it's not a complex substitution, but you can use $u = g - \frac{b}{m} v$ if you have trouble.

$$ \int_0^v \dfrac{dv}{g -\dfrac{b}{m} v} = \int_0^t dt $$ $$ -\dfrac{m}{b} \ln\left|g -\dfrac{b}{m} v\right| \bigg|_0^v = t $$ $$ -\dfrac{m}{b} \left( \ln\left|g -\dfrac{b}{m} v\right| - \ln(g) \right) = t $$
We can use a property of logarithms to combine the two logarithms:

$$ -\dfrac{m}{b} \ln\left(\dfrac{g -\dfrac{b}{m} v}{g}\right) = t $$ $$ \dfrac{g -\dfrac{b}{m} v}{g} = e^{-\dfrac{b}{m} t} $$ $$ g -\dfrac{b}{m} v = g e^{-\dfrac{b}{m} t} $$ $$ v = -\dfrac{mg}{b} \left( e^{-\frac{b}{m} t} - 1 \right) $$
We can use this to find the position and acceleration as well, though I won't make you do that. This is the exact equation of motion for an object falling under linear drag. It is a bit complicated, but it does make sense. It's easier to understand if we look at it conceptually instead of focusing on the equation.

When the object is initially dropped, it has very little air resistance acting on it because it is not moving very fast. THis means that the initial acceleration should be roughly linear and equal to $g$. As the object falls, the acceleration will gradually decrease as the force of air resistance increases.

As the object approaches the terminal velocity, the velocity with respect to time should asymptote to a fixed value, and the acceleration should asymptote to zero. The initial slope of the velocity versus time graph would look roughly linear, and the slope would decrease with time as the acceleration decreased. With this information, we can draw a rough graph of the motion. When the object approaches the terminal velocity, the acceleration should drop to zero as the forces get balanced, and the velocity will approach the terminal velocity value. This information is enough to generate a rough graph of the motion of velocity versus time.

Take a look at the graph below. It has all of the characteristics that the hypothetical graph we talked about above should have.

That's it for our exploration of air resistance! This topic can get pretty complex, but the introductory physics and the general concepts behind it are quite simple. This is also the end of our fluids unit, and the end of mechanics. I hope you enjoyed your journey with us thus far.