PhysicsScience

The Physics Philes, lesson 10: Resistance is Fluid

In which fluid is defined, terminal speed is reached, and I still can’t do calculus.

The last couple of weeks I’ve discussed friction forces and how they are applied through a few example problems. However, I only discussed three friction forces: kinetic, static, and rolling. There is one more aspect of friction that we need to look into, and that’s fluid resistance and terminal speed.

You must forgive me for neglecting these concepts. There’s – gulp – calculus. But the concepts are simple enough, so I thought I would power through. Smarter people than I can let me know where I goofed it up in the comments.

Fluid resistance is the force that a fluid exerts on a body moving through it. When I think of a fluid my mind automatically thinks of a liquid, but that’s not how it’s defined in physics. A fluid for our purposes can be a liquid or a gas. According to Newton’s third law of motion, when a body passes through a fluid it exerts a force on that fluid and the fluid exerts and equal and opposite force on the body. The direction of the fluid resistance force always acts in the opposite direction of the body’s velocity relative to the fluid and the magnitude of the force usually increases with the body’s speed. (This is different from kinetic friction, which is independent of speed.)

There are two different equations we use to determine fluid resistance, one for very low speeds and one for high speeds. For very low speeds, the magnitude of the fluid resistance force is approximately proportional to the body’s speed multiplied by a proportionality constant. Or, in math-speak:

f = kv

The proportionality constant depends on the size and shape of the body and the properties of the fluid. The units for k are N · s/m  or kg/s. For objects moving at the speed of a tossed tennis ball or faster, we have to deal with air drag. The resisting force is approximately proportional to v^2. So for higher speeds, our equation is:

f = Dv^2

Because of the dependence on v^2, air drag increases rapidly with increasing speed. In this equation, f still equals the magnitude of the fluid resistance force, v equals the velocity of the moving body, and the value of D depends on the size and shape of the body and the density of the air. The units we use for D are N · s^2/m^2 or kg/m.

An object falling through a fluid does not have a constant acceleration because of the effects of fluid resistance. This makes sense if you’ve ever dropped anything in a pond or a bucket of water. But this is important because we can’t use the constant acceleration equations to describe the motion. Instead, we have to use Newton’s second law of motion.

Let’s say that we drop a rock in the water. According to Newton’s second law, ∑F = ma. According to the superposition of forces, we could also write it like this:

∑F = mg + (-kv) = ma

When the rock first starts to move, it’s v = 0, so the resisting force is 0 and the initial acceleration equals the force of gravity (g). But as the speed increases, so does the resisting force until it equals the magnitude of the weight. When this happens, the equation above will equal 0 and there is no further increase in speed. This final speed (v sub t) is called the terminal speed. The terminal speed equation for bodies moving at very low speed is:

v sub t = mg / k

OK, up until this point I think I understand. But now we run into terminal speed for bodies moving at a faster speed, and this is where I start to fall apart. I know that it has something to do with derivatives, which I haven’t quite figured out. And by “haven’t quite figured out” I mean “am still totally lost on.”

But, while I can’t explain the math (which assumes that I’ve adequately explained any math in these posts), I can at least show you how it’s applied. For an object falling through the air at high speeds, the terminal speed is reached when Dv^2 equals mg (otherwise known as weight). So the equation for terminal speed in this situation is:

v sub t = √mg / D

(Style note: v sub t should equal the square root of mg / D. not just the square root of mg. Understand? Good.)

This equation explains why more massive objects will fall faster than lighter objects of the same physical size. If we drop a ping pong ball and a ping pong sized lead ball from the roof, the value of D will be the same, but the value of m will be different. The same concept explains why a crumpled up piece of paper will fall faster than a flat piece of paper. The mass will the same, but the smaller size of the crumpled up piece will make the drag (D) smaller and the final speed larger.

When you think about this, it all makes sense. We’ve all dropped something that sank to bottom of the swimming pool. We’ve all stuck our hands out the car window as it was speeding down the street. Whenever we do this, we’re experiencing fluid resistance. This is just the math behind what we already intuitively know.

(As a final note, I want to thank commenter IBY for trying to explain fluid resistance. It was helpful!)

Featured image credit: A_of_DooM

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

3 Comments

  1. August 26, 2012 at 2:40 am —

    Unfortunately, in that long comment, I used the angled bracket, and so it activated the html thingy and now a chunk of my explanation is useless. 🙁

    Also, the stuff I did there won’t apply for raindrops and people. They fall in the v^2 domain. It will apply, though, with throwing stuff in syrup. So I don’t know why I said “in most cases” when I talked about the v^1 domain.

    BTW, you already got the terminal velocity for the v^2 domain. You don’t need calculus for that, as you have demonstrated up there. But you do need calculus to know how long it will take to nearly be there. In fact, you need to know how to solve something called “differential equation”. So yeah, this one is way beyond you. And I don’t know how to explain it with less calculus.

    • August 26, 2012 at 2:54 am —

      But, I can tell you the equation for v^2, where v0 is terminal velocity:

      v=v0*(e[gt/v0]+e[-gt/v0])/(e[gt/v0]-e[-gt/v0])

      See why I don’t know how to reduce the calculus? What a pain.

    • August 26, 2012 at 10:28 am —

      Aarg! I hate when things are way beyond me! But! I am planning on taking trig and calc from an actual university next spring and fall. I WILL NOT BE DEFEATED.

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