# Are drag coefficients negative?

• amolv06
So I suppose the only thing I need to do now is figure out how to solve for my constants so that I can use this equation for real world problems.In summary, the conversation discusses finding an equation that describes a body in free-fall in air in a differential equations class. The equation is derived using the assumption that the drag force is always opposite to the motion. After plugging in a particular solution, it is found that the equation does not make sense and asserts that the body is falling up. It is clarified that the sign of the drag force should be negative and the particular solution is incorrect. The correct particular solution should be proportional to t^2 and the direction of the y-axis does not affect the solution. The conversation concludes with
amolv06

## Homework Statement

Find an equation describing a body in free-fall in air. This is not really homework. I'm in a differential equations class write now, and I have fun finding real world applications for such things.

## Homework Equations

$$m\frac{d^{2}y}{dt^{2}} = B\frac{dy}{dt}-mg$$

## The Attempt at a Solution

Rewriting the above:

$$\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = -g$$

The corresponding homogeneous equation is:

$$\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = 0$$

Two solutions to the homogeneous equations are:

$$y_{1} = C_{1}$$ and $$y_{2} = C_{2}e^{\frac{b}{m}t}$$

And as a particular solution:

$$y_{p} = \frac{gm}{b}t$$

Therefore by the superposition principle, we have a general equation as follows:

$$y(t) = y_{1}(t) + y_{2}(t) + y_{p}(t) = C_{1} + C_{2}e^{\frac{b}{m}t} + \frac{gm}{b}t$$

So here's where my question arises. I believe my assumption (the initial differential equation) should be a reasonable approximation of the real world. And I plugged my solution into the original differential equation. I seem to have answered it correctly. If both of these assumptions are true, then this equation doesn't make sense. It asserts that we are falling up! The only we I can think that this equation can still hold is if B were negative. I think after that, everything should make sense. Is this a correct assumption, or did I just make a mistake in the mathematics above?

Thanks in advance for your time and any help.

Last edited:
amolv06 said:

## Homework Statement

Find an equation describing a body in free-fall in air. This is not really homework. I'm in a differential equations class write now, and I have fun finding real world applications for such things.

## Homework Equations

$$m\frac{d^{2}y}{dt^{2}} = B\frac{dy}{dt}-mg$$
The drag force is $$- B\frac{dy}{dt}$$ since it is always opposite to th emotion (where B is assumed to be a positive constant).

## The Attempt at a Solution

Rewriting the above:

$$\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = -g$$

The corresponding homogeneous equation is:

$$\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = 0$$

Two solutions to the homogeneous equations are:

$$y_{1} = C_{1}$$ and $$y_{2} = C_{2}e^{\frac{b}{m}t}$$

And as a particular solution:

$$y_{p} = \frac{gm}{b}t$$

Therefore by the superposition principle, we have a general equation as follows:

$$y(t) = y_{1}(t) + y_{2}(t) + y_{p}(t) = C_{1} + C_{2}e^{\frac{b}{m}t} + \frac{gm}{b}t$$

So here's where my question arises. I believe my assumption (the initial differential equation) should be a reasonable approximation of the real world. And I plugged my solution into the original differential equation. I seem to have answered it correctly. If both of these assumptions are true, then this equation doesn't make sense. It asserts that we are falling up! The only we I can think that this equation can still hold is if B were negative. I think after that, everything should make sense. Is this a correct assumption, or did I just make a mistake in the mathematics above?

Thanks in advance for your time and any help.

Your particular solution is incorrect, it should be proportional to t^2
(and of course, if you set B=0 you should recover the usual free fall equation)

Sorry for the confusion, I'm using g as 9.8 rather than -9.8. I should have clarified. But I believe that should clear up the confusion with the drag force sign. Basically I set the drag force to be positive whereas the gravitational force is negative.

I'm not sure why my particular solution does not work.

$$\frac{d^{2}y_{p}}{dt^{2}} = 0$$

$$\frac{dy_{p}}{dt} = \frac{gm}{b}$$

Plugging this into the original differential equation I have

$$-\frac{b}{m}\frac{gm}{b} = -g$$ which seems to be what I need. I must be missing something.

amolv06 said:
Sorry for the confusion, I'm using g as 9.8 rather than -9.8. I should have clarified. But I believe that should clear up the confusion with the drag force sign. Basically I set the drag force to be positive whereas the gravitational force is negative.

I'm not sure why my particular solution does not work.

$$\frac{d^{2}y_{p}}{dt^{2}} = 0$$

$$\frac{dy_{p}}{dt} = \frac{gm}{b}$$

Plugging this into the original differential equation I have

$$-\frac{b}{m}\frac{gm}{b} = -g$$ which seems to be what I need. I must be missing something.

But it does not matter whether you use +9.8 or -9.8 (i.e. whether your y-axis points upward or downward) because the drag force is always opposite to the motion so no matter what your choice is for your direction of the y axis, the force will be -B dy/dt

Ahh, yes! How could I have missed that.

Thanks a lot. That clears up a lot.

## 1. What is a drag coefficient?

A drag coefficient is a dimensionless quantity that is used to quantify the drag or resistance of an object in a fluid environment, such as air or water. It is a measure of how much the fluid resists the motion of the object.

## 2. Why would a drag coefficient be negative?

A drag coefficient can be negative if the direction of the drag force is opposite to the direction of motion of the object. This can occur in certain situations, such as when an object is moving upward and the drag force is acting downward.

## 3. Can a negative drag coefficient be considered a good thing?

In most cases, a negative drag coefficient would not be considered a good thing. It would mean that the object is experiencing a drag force in the opposite direction of its intended motion, which could be detrimental to its performance.

## 4. How is a drag coefficient determined?

A drag coefficient is typically determined through experimentation and testing. The object in question is placed in a fluid environment and its motion and the forces acting on it are measured. The drag coefficient is then calculated by dividing the drag force by the product of the fluid density, velocity, and surface area of the object.

## 5. Are there any real-world examples of negative drag coefficients?

Yes, there are some real-world examples of negative drag coefficients. One example is the motion of a rocket in the upper atmosphere, where the drag from the thin air can be overcome by the thrust of the rocket engine, resulting in a negative drag coefficient. Another example is the aerodynamics of certain types of aircraft, such as the F-117 Nighthawk, where the shape of the aircraft creates a negative drag coefficient at certain angles of attack.

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
246
• Introductory Physics Homework Help
Replies
10
Views
342
• Introductory Physics Homework Help
Replies
3
Views
933
• Introductory Physics Homework Help
Replies
6
Views
304
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
13
Views
840
• Introductory Physics Homework Help
Replies
30
Views
590
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
463