# Understanding Drag Force against a non-linear Force

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1. Jun 26, 2017

### Kiwigami

1. The problem statement, all variables and given/known data
I'd like to understand Drag Force better; but school always ignores it. Thus, I'm asking this purely out of obssession. I'm picturing a scenario where a non-constant force is pushing an object horizontally, ignoring friction. But, I'd like to understand how Drag Force influences the object's velocity.

My questions are: 1) Do the steps that I'm taking in my math make sense? 2) How do I interpret the math?

2. Relevant equations
Let F(x) be a function of Force with respect to Displacement: F(x) = 200000 - 100x
Force is decreasing linearly as displacement increases from 0 meters to 2000 meters.

A quick google search led me to this formula: FD(x) = 0.5 * Drag Coefficient * Frontal Area of Object * Air Density * V2

Suppose I know the factors involved, and I get an overall constant of 0.18: FD(x) = 0.18v2

Let's assume the object has a mass of 70 kg.

3. The attempt at playing with math
1. Finding Work: W(d) = ∫ from 0 to d of (200000-100x) dx = -50(d-4000)d
2. Kinetic Energy: KE(d) = -50(d-4000)d = 0.5(70)v2
3. Solve for Velocity: v(d) = sqrt((2/70) * -50(d-4000)d)
4. Plug velocity function into the Drag Force: FD(d) = 0.18(sqrt((2/70) * -50(d-4000)d))2
5. Net Force: F(d) - FD(x) = FNET(x) = [200000 - 100d] - [0.18(sqrt((2/70) * -50(d-4000)d))2]
Here is where I feel like I did something wrong. When I integrate the Net Force to find Work and find the velocity function with respect to displacement all over again, the velocity decreases down to 0 at about 376 meters and continues to decrease into the negative numbers.

However, originally the force acting on the object is a 2 kilometers influence. I don't think the object would suddenly go backwards at the 376 meter mark.

2. Jun 26, 2017

### TSny

Are you assuming that your formula for F(x) applies for x greater than 2000 m as well as x < 2000 m, or are you assuming F(x) remains zero for x > 2000 m?

These steps ignore the drag force, so they don't give you the correct velocity. If you set up Newton's second law, you will obtain a second order, nonlinear differential equation for the position as a function of time, x(t). I think the equation is too difficult to solve analytically, but you could use a computer to find a numerical solution.

3. Jun 26, 2017

### Kiwigami

Yeah, I am assuming x < 2000 m

4. Jun 26, 2017

### TSny

OK

5. Jun 26, 2017

### Kiwigami

I have zero background in Differential Equations, but is this what you mean?

Setting up Net Force: Net Force = [200000-100x] - [0.18v^2]
Work = Integral from 0 to d of [200000-100x] - [0.18v^2] dx = -50 d (d + 0.0036 v^2 - 4000) according to an online calculator.
Kinetic Energy: Then I set -50 d (d + 0.0036 v^2 - 4000) = 0.5 mv^2
Solve for v, which according to Wolfram Alpha is: V(d) ± (50 sqrt(-(d - 4000) d))/sqrt(9 d + 1750)

6. Jun 26, 2017

### TSny

In doing the integral, you have treated the velocity v as a constant. So, this isn't correct.

I don't see a way to solve for the velocity as a function of t or as a function of x by using work/energy concepts. To find the work done by the drag force, you would need to know how the velocity varies with x. But you don't know v(x) at the outset.

Newton's second law is $F_{\rm net} = m \frac{d^2x}{dt^2}$. So,

$200000 - 100x(t) - .18x'(t)^2 = 70 x''(t)$

This is the differential equation for the position as a function of time. The drag force term with the square of $x'(t)$ makes the equation difficult to solve analytically.

7. Jun 26, 2017

### Ray Vickson

No: the situation is much more complicated (and difficult) than you seem to think. The drag force has the form
$$F_d(v) = \begin{cases} -0.18 v^2 & \text{if} \; v > 0 \\ +0.18 v^2 & \text{if} \; v < 0 \end{cases}$$
This is because the drag force acts in the direction opposite to velocity, so when velocity is >0 the drag acts backwards, but when velocity < 0 the drag force acts forward (to oppose the velocity).

So, your force has the form $F(x,v) = 200,000 - 100 x + F_d(v).$

You say you have not yet taken material on differential equations, but in this case it does not matter: the problem does not have an "analytical" solution, but must be tackled numerically (using a numerical differential equation solver). For example, when the mass is $m = 70$ (kg) and the initial position and initial velocity are both = 0, the Maple package gives a solution that "turns around" at about time $t = 4.2523$ (seconds) at position $x(4.2523) = 2194.4168$(meters) and velocity very near 0.

Do not bother trying to write formulas for the solution; nobody can do it! You cannot just use "conservation of energy" because that does not apply to your particle in this case: the drag force invalidates the simple energy-conservation law! Of course, total energy is conserved, but some of the kinetic energy of the particle is transferred to the air (via friction), serving to heat the air and perhaps form energetic vortices, etc. However, the energy of the particle all by itself is not conserved.

Last edited: Jun 26, 2017
8. Jun 26, 2017

### TSny

Yes. The switch in sign of the drag force when the object turns around is important. This means that the differential equation that I wrote is only valid as long as the particle is moving in the positive x direction. That's why I wanted to know if the applied force formula is still in effect once the particle reaches x = 2000 m. If the applied force remains zero once the object reaches 2000 m, then we don't need to worry about the object turning around.

Here's a graph of the position (blue) and velocity (orange) as functions of time for the case where both the the applied force and drag force act for 0 < x < 2000 m. For x > 2000 m, the applied force is assumed to be zero and only the drag force is acting.

The horizontal axis is time in seconds. The vertical axis is distance in meters for x and velocity in meters per second for v.

9. Jun 26, 2017

### Ray Vickson

I am more interested in the (physically more realistic) case where the form of $F$ applies also for $x > 2000$ m. In the absence of drag, the resulting dynamics would be a simple harmonic motion about the point $x = 2000$. For example, with mass $m = 70$ (kg) and starting from $x(0) =0$ with velocity $v(0)=0$, the motion is periodic with amplitude 2000 (m) and period $p = \pi \sqrt{70}/7$ sec, which is about 5.257 sec..

The effect of drag is most easily seen by looking at the new "displacement variable" $y = x - 2000$, because in the absence of drag the values of $y(t)$ oscilllate with amplitude 2000 about the point $y = 0$. With drag given as above, the nonlinear second-order differential equation satisfied by $y = y(t)$ is
$$70 y'' = - 100 y - 0.18\; \text{Sign}(y')\; y'^2$$
Solving this in Maple (using a numerical method) gives the following solution.

The blue curve is $y(t)$ while the red curve is $v(t) = y'(t)$. We started from $x=0$ with initial velocity 0, so the initial value of $y$ is $-2000$.
We can see the damping effects of drag, with the amplitudes of the oscillations lessening as time passes, and the velocities (speeds) also slowing over time.

Last edited: Jun 26, 2017
10. Jun 26, 2017

### TSny

That's nice. As a check, I get the same graphs (Mathematica).

#### Attached Files:

• ###### Drag Oscillate.gif
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11. Jun 26, 2017

### Ray Vickson

Note: I wrote $y''$ in the posting, but the actual left-hand-side should be $70 y''$. The Maple solution used the correct "70" form. I have corrected the posting in an edit.