What Does Air Resistance Do

Have yous ever had a feeling that something is trying to ho-hum you down when you ride a bicycle? When you move in the forwards direction, the frictional force exerted by the air tends to reduce your speed. The frictional forcefulness acts on your face and body in the opposite direction of the bicycle's motion. The air resistance force increases proportionally to the speed. Crouching down on the bicycle allows you to decrease the effect of air resistance force and motility faster.

Y'all may now think of the air resistance force as something negative and preventing motion, but actually, it turns out to be quite useful in our everyday lives. For instance, when a skydiver jumps out of an airplane and opens the parachute, the air slows down the fall. The speed of the skydiver decreases as the ground is approached, due to the resistance provided by air. As a outcome, the person reaches land safely and smoothly - all because of the resistive force. In this article, we will discuss the science backside air resistance in more detail.

Air Resistance Definition

Thus far, in nigh physics problems involving movement, it's explicitly stated that air resistance is negligible. In existent life that is not the case as all objects feel some level of resistance as they pass through the air.

Air resistance or drag forcefulness is a type of friction that occurs betwixt an object and the air surrounding it.

These drag forces crusade the object to move more slowly by acting in the direction of the incoming period and are proportional to the velocity. It is a type of not-conservative forces since it makes the energy misemploy.

Let's say we take a ball with mass $thousand$. We drop it and every bit information technology falls, it's going to experience a resistive force. The resistive force mathematically is equal to

$$ \vec{F}_{\mathrm{r}} = - k \vec{v} $$

where $thousand$ is a positive abiding, and $five$ is the velocity of the object relative to the medium. The negative sign indicates that the resistive force is in the contrary direction to the velocity.

At this stage in your learning, knowing this version of the resistive strength equation is sufficient, however, a more precise and realistic representation of air resistance would be given by $\vec{F}_{\mathrm{r}} = - k \vec{v}^ii$. Read further about information technology in the deep dive!

In literature, y'all volition nearly likely see a modified version of this equation with the velocity term squared

$$ \vec{F}_{\mathrm{r}} = - k \vec{v}^2.$$

That's because the resistance depends on the type of flow. Turbulent flow is known to be fast and requires the employ of $\vec{5}^2$, meanwhile laminar menstruum is slow and uses $\vec{v}$. Because the terms "slow" and "fast" are relative, a dimensionless quantity known as the Reynolds number has to be considered, where low values correlate with laminar menstruum, and high values with turbulent menses. Real-life examples, such as skydiving and claret flowing in our arteries, are events of high-speed menses, and therefore would require the employ of $\vec{v}^two$. Unfortunately, such an in-depth analysis of air resistance is beyond the AP Physics level, so we will be considering air resistance linear in air speed.

Air Resistance Coefficient

As discussed earlier, $k$ is a constant of proportionality. Its value is determined by the properties of the medium and the unique characteristics of the object. The main contributing factors are the density of the medium, the surface area of the object, and a dimensionless quantity known as the drag coefficient. In a real-life example involving a skydiver, the medium would be the air and the surface surface area would refer to either the skydiver or the parachute.

Now nosotros can explain the effectiveness of a parachute when it comes to slowing downward a skydiver. As the surface area $A$ of the object falling increases,

$$ A_{\mathrm{skydiver}} \ll A_{\mathrm{parachute}},$$

$k$ increases, so the magnitude of the resistive forcefulness increases as well, therefore slowing the object down.

The total expression used to summate the resistive forcefulness is

$$\vec{F}_\mathrm{r} = \frac{ane}{2} D \rho A \vec{v}^2$$

where $D$ is the drag coefficient, $\rho$ is the density of the medium, $A$ is the surface area of the object, and $\vec{v}$ is the velocity.

Allow'southward look at a free-body diagram to understand its motion better.

Air Resistance Free Body Diagram

What happens to an object every bit it gets dropped and is falling down? It experiences a downwardly force in the course of weight and a resistive forcefulness in the opposite direction of the move due to air resistance, both of which are visualized in the free-torso diagram visible below.

Air Resistance An image of an object falling, with drag force pointing upward and weight pointing downwards StudySmarter Fig. 1 - As the object falls, the resistive force acts upwards on information technology, meanwhile the weight pulls it downwards.

Co-ordinate to Newton's 2nd constabulary, the cyberspace force acting on an object $\vec{F}_{\mathrm{net}}$ is equal to the mass $m$ of the object times its dispatch $\vec{a}$. Then knowing all that, we can obtain the following expression

$$ m\vec{g} - k\vec{v} = m\vec{a}.$$

When we commencement the movement at $t=0$, its initial velocity is $\vec{5}_0=0$, therefore, the initial air resistance forcefulness is also zero. As fourth dimension passes and the object starts moving, somewhen it will reach a constant velocity, which is chosen concluding velocity $\vec{v}_\mathrm{T}$. Because the velocity is constant, the acceleration volition be aught. The correct-hand side of the expression becomes naught, and nosotros can rearrange the remaining terms

$$ thousand\vec{g} = k\vec{5}_\mathrm{T} $$

to find the equation for terminal velocity

$$ \vec{five}_\mathrm{T}= \frac{yard\vec{g}}{chiliad}. $$

Terminal velocity is the maximum speed accomplished by an object moving under the influence of a constant force and a resistive force that is exerted on the object in contrary directions.

Terminal velocity is reached when there is no net forcefulness applied to the object, meaning that the acceleration is zero. Let's wait at an example problem involving terminal velocity.

Air Resistance Formula

Let's now find the velocity as a role of time. To achieve that, we need to convert Newton's second law into a differential equation. Dispatch is the commencement derivative of velocity, so $\vec{a}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$. So we can write

$$ m\frac{\mathrm{d}\vec{5}}{\mathrm{d}t}=k\vec{1000}-m\vec{5}. $$

Let's split up our variables:

$$ \frac{\mathrm{d}v}{mg- kv}=\frac{\mathrm{d}t}{thousand}.$$

To perform all the necessary mathematical operations, for now, we'll await at one dimension only and regard the vector quantities equally scalars.

Here, information technology is important to set the integration limits. The fourth dimension goes from zilch to time $t_{\mathrm{f}}$. When time is equal to nada, our initial velocity is zero also, and as time goes to $t_{\mathrm{f}}$, our velocity becomes velocity $v_{\mathrm{f}}$.

The reason we don't set up the upper limit as the terminal velocity is considering nosotros are trying to find the velocity as a function of time!

$$\int_{0}^{v_\mathrm{f}} \frac{\mathrm{d}v}{mg-kv} = \int_{0}^{t_{\mathrm{f}}} \frac{\mathrm{d}t}{g}$$

If nosotros take the antiderivative, nosotros will obtain a natural logarithm

$$\left.\frac{\ln(mg-kv)}{-k}\right|_0^{v_\mathrm{f}} = \left.\frac{t}{yard}\correct|_0^{t_\mathrm{f}}$$

Now permit's apply the limits

$$ \begin{align} \frac{\ln(mg-kv_{\mathrm{f}})}{-k} - \frac{\ln(mg)}{-thousand} & = \frac{t_{\mathrm{f}}}{m}, \\ \ln \left ( \frac{mg-kv_{\mathrm{f}}}{mg} \right ) & = \frac{-kt_{\mathrm{f}}}{chiliad}. \end{marshal} $$

Finally, get rid of the natural logarithm:

$$ \begin{align} \mathrm{e}^{\ln \left ( \frac{mg- kv_{\mathrm{f}}}{mg} \right )} &= \mathrm{east}^{\frac{-kt_{\mathrm{f}}}{m}} \\ \frac{mg-kv_{\mathrm{f}}}{mg} &= \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ one- \frac{kv_{\mathrm{f}}}{mg}&= \mathrm{east}^{\frac{-kt_{\mathrm{f}}}{m}} \\ \frac{kv_{\mathrm{f}}}{mg} & = 1- \mathrm{eastward}^{\frac{-kt_{\mathrm{f}}}{1000}} \\ v_{\mathrm{f}} &= \frac{mg}{k} \left ( 1-\mathrm{east}^{\frac{-kt_{\mathrm{f}}}{m}} \right ). \finish{align} $$

The final version of the equation including all the vector values is equally follows

$$ \vec{v_{\mathrm{f}}}=\vec{5}_\mathrm{T} \, (1-\mathrm{due east}^{-\frac{t_{\mathrm{f}}}{T}}) $$

where $T$ is the fourth dimension constant and equal to $\frac{thou}{k}$.

And that's how nosotros derive the velocity expression every bit a time function! The final equation confirms our previous conclusions about the concluding velocity. If the value of $t_{\mathrm{f}}$ is set to cypher, $\vec{v_{\mathrm{f}}}$ also will be null, meanwhile if $t_{\mathrm{f}}$ is set to something huge, let's say infinity, we will be left with $\vec{v_{\mathrm{f}}} = \vec{v_\mathrm{T}}$.

What would happen though if the initial velocity was non cipher?

Permit'southward say we have a car with an initial velocity $\vec{v}_0$ against some resistive force $\vec{F}_\mathrm{r}$ that is again equal to $-thousand\vec{five}$. When nosotros draw a complimentary-trunk diagram of the car, the weight is downward, the normal strength is upward, and the air resistance force is in the opposite direction of the motion.

In this example, the final velocity will exist cypher, and the car will stop. The simply force acting on the object in the direction of the motility is the resistive force, and so information technology will be our net force. Then we can write

$$ grand\vec{a} = -k\vec{v}.$$

We are going to echo the aforementioned process every bit previously since this becomes a differential equation when we write acceleration as $\vec{a}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$ and obtain

$$ \brainstorm{align} m \frac{\mathrm{d}\vec{five}}{\mathrm{d}t} & = - one thousand\vec{v} \\ \frac{\mathrm{d}v}{v} & =\frac{-k}{thou} \mathrm{d}t. \end{align}$$

Once again, for the calculations, nosotros'll consider the scalar version of the equation. Hither nosotros have to take integrals of both sides, but first, we need to decide on the limits. Time one time again goes from goose egg to $t$. Still, at present we accept an initial velocity, so our velocity limit is from $v_0$ to $five$

$$\int_{v_0}^{v_{\mathrm{f}}} \frac{\mathrm{d}five}{five} = \int_{0}^{t_{\mathrm{f}}} \frac{-k}{m} \mathrm{d}t. $$

Over again, have the derivative to have a natural logarithm, apply the limits and obtain the following expression

$$ \ln \left ( \frac{v_{\mathrm{f}}}{v_0} \correct ) = \frac {-kt_{\mathrm{f}}}{m}.$$

We can rewrite this as:

$$ \begin{align} \mathrm{e}^{\ln \left (\frac{v_{\mathrm{f}}}{v_0} \correct )} & = \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{1000}} \\ \frac{v_{\mathrm{f}}}{v_0} & =\mathrm{east}^{\frac{-kt_{\mathrm{f}}}{m}} \end{align}$$

where the final expression including all the vector quantities becomes

$$ \vec{v_{\mathrm{f}}} = \vec{five}_0 \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}}.$$

Air Resistance Instance

Let's look at an case problem involving the same skydiver mentioned earlier, to check our knowledge!

A skydiver is falling with the initial speed $\vec{v}_0$ through the air. At that moment ($t = 0$), they open up the parachute and experience the force of air resistance whose strength is given by the equation $\vec{F} = -k\vec{five}$, where the variables are the same as defined earlier. The total mass of the skydiver and the equipment is $yard$.

Determine the expression for the skydiver's acceleration, terminal speed, and make a graph of velocity equally a function of time.

Solution

Nosotros know that

$$ \vec{F}_{\mathrm{net}} = \vec{F}_\mathrm{1000} - \vec{F}_\mathrm{r} $$

so considering the free body diagram explained earlier, nosotros tin can detect the expression for the acceleration

$$ \begin{marshal} m\vec{a} & = grand\vec{grand} - k\vec{v}, \\ \vec{a} & = \frac{m\vec{thousand} - k\vec{v}}{m}.\terminate{align}$$

Based on the definition from before, the skydiver will reach their terminal velocity, when the velocity is abiding ($\vec{v} = \vec{v}_\mathrm{T}$). That means that the dispatch becomes zero

$$ 0 = \frac{m\vec{g} - thousand\vec{five}_\mathrm{T}}{m} $$

which rearranges into

$$ \vec{5}_\mathrm{T} = \frac{thousand\vec{g}}{one thousand}.$$

At present let's use this expression to plot the velocity-time graph.

Air Resistance A velocity time plot showcasing the changes in velocity of a skydiver as they start from inital velocity and approach the terminal velocity. The gradient of this plot represents acceleration, which isn't linear StudySmarter Fig. 3 - The changes in velocity from the initial descent of the skydiver until they approach the last velocity over time. The gradient of this plot represents the acceleration of the skydiver.

Initially, the skydiver is descending at the velocity $\vec{five}_0$ and accelerating at roughly the gravitational acceleration $\vec{m}$. Every bit the parachute is released, the skydiver is subjected to considerable resistive force - air resistance. The acceleration from the drag force results in an upwardly acceleration, so the downwards velocity decreases. The slope of our velocity versus time plot represents the acceleration. Based on the previous observations, it volition not be constant, just rather will approach zero as the velocity reaches the concluding velocity $\vec{five}_\mathrm{T}$. Every bit a result, the plot isn't linear.

Some other examples of air resistance in our everyday lives would exist

Walking in a tempest makes walking challenging quite frequently. A significant amount of resistance is experienced by the individual walking confronting the wind, making it difficult to walk forward. The aforementioned reason makes information technology challenging to concord an umbrella in your manus when at that place is a stiff air current present.
A feather falling to the ground has a tendency to bladder and move slowly, rather than autumn within seconds similar other objects, of slightly bigger mass. The gravitational force pulls the feather towards the earth; nevertheless, the air resistance force prevents the feather from falling or moving while in move.
Newspaper planes, if built correctly, wing effortlessly in the air. To attain this, the front surface of the newspaper plane is sharpened. Every bit a result, the paper plane cuts through the air and escapes the air resistance strength just plenty to keep it in the air for longer.
A real plane's engine, wings, and propellers are all congenital to provide enough thrust to help the plane overcome the force of air resistance. Turbulence is also caused by the friction that the air creates. Spacecrafts, notwithstanding, just take to worry near air resistance during launching and landing, as there is no air in space.

Air Resistance - Key takeaways

The forces that oppose an object's relative move as it moves through the air are referred to equally air resistance.
These drag forces cause the object to move more slowly by interim in the direction of the incoming flow and are proportional to the velocity.
The mathematical expression for air resistance is $ \vec{F}_\mathrm{r} = - thou \vec{v}$, where the negative sign indicates the opposite direction of the motion.
Terminal velocity is defined equally the maximum speed achieved by an object moving under the influence of a constant forcefulness and a resistive strength that is exerted on the object in opposite directions.
When no internet force is applied to the object, meaning that the dispatch is zero, the concluding condition is reached.
Some air resistance examples include walking in the storm, a plume falling to the ground, a newspaper plane, an airplane, a skydiver using a parachute, and riding a bicycle.