If I could embed a video in a paper-based test question, this would be on there.

And here is the original page with that video (and extra info from Boeing).

So, here is the test question: What is the mass of the brakes? Alternative question: what is the coefficient of friction between the tires and the runway? I would have asked about the final temperature of the brakes, but the video estimates that at 1400 C. The other important info from the video:

They tested the 747-8 with a take off mass of 443,181 kg.

The 747 had a speed of 200 mph before using just the brakes to stop.

The brakes were carbon brakes. I guess that means they were made of carbon?

The plane took about 200 meters to stop.

Estimating the temperature ————————–

Ok, this is really just a quick check on the temperature. If you consider the brakes to be a blackbody (which is an object that emits light mostly due to its temperature), then you can estimate the temperature of that object based on the color. Here is PhET's fairly awesome blackbody simulation. Also, here is an image of the super-hot brakes from the video:

Now, I will just adjust the temperature in the PhET simulator until it seems to match the color of those brakes. This is what I get:

Actually, that shows the brakes as well as the output from PhET and a list of colors for each temperature that I found online. It seems that PhET and this other site don't completely agree. However, the image of the brakes isn't that great anyway. So, it seems plausible that the brakes are around 1600 K.

Estimating the brake mass ————————-

When you have a question like this, the key thing to ask yourself is: what principle should I start with? In this case, I can use the temperature to get a measure of the thermal energy of the brakes (if I knew the mass). I can also get an estimate for the thermal energy by thinking about the reason the brakes get hot in the first place. So, this means I need to use the work-energy principle. It says:

Before applying the work-energy principle, I need to choose my system. In this case, I will use the real system that includes the 747 (brakes and all). This means that I can have things like thermal energy in the system. If, I had chosen the 747 as a point system, I couldn't have thermal energy. Also, by choosing this 747 as my real system, there is no work done on the object. There are essentially only 3 forces acting on the plane: friction, gravity and the runway pushing up. None of these do any work. The friction force is from runway, but since the runway doesn't move, there is no displacement and no work. This means my equation becomes:

If I take my initial kinetic energy to be the plane moving at 200 mph and the final to be at rest, then I can write:

Here, there are two masses: m p is the mass of the plane and m b is the mass of the stuff that increases in temperature (the brakes). There is also the constant c. This is the specific heat of the brake material. If I use carbon (as the video says), then I can say carbon has a specific heat capacity of 0.509 J/(g*K). Actually, that is value for diamond carbon from chemicool.com.

All that is left is to plug in my values. Doing this, I get:

Notice that I used a change in temperature of 1,400 K (not the final temperature). Ok, the other point - this mass looks very high. However, this is for all the breaks. Take a look at this image of the landing gear:

It's an 18-wheeler! So, assuming all the wheels are the same, this would put the mass of each brake at 136 kg (300 pounds). I think I could believe that. These suckers are huge.

Coefficient of Friction ———————–

Now for part 2 - let me go ahead and draw a force diagram for this plane while it is stopping.

Here, F N is the force the ground pushes up on the plane (the N stands for 'normal' - you know, perpendicular). F f is the frictional force. The most common model for friction is to say the magnitude of the friction force is:

Where μ is the coefficient of friction. For this situation, the plane really uses static friction to stop (which is technically different than friction where the two surfaces are rubbing against each other). I am assuming the wheels don't slide so that static friction would be appropriate to use.

Either way, I need to find the force the ground pushes up on the plane. This is pretty easy. Since the plane does not accelerate in the vertical direction (I will call this the y direction), then the net force in the vertical direction must also be zero. I can write this as:

So, the force the ground pushes up is equal to the weight of the plane. Before you say "duh", let me point out how many times I have seen students make mistakes on this part. Sure, in this case F N = mg. That doesn't mean it is always true. You should always do the above step if you don't want to get yourself into trouble. For instance, if the plane was going down a hill, F N would NOT equal the weight. Ok.

What about the frictional force? I could find the acceleration in the x-direction and use the following:

But I am not going to do that. First, that is what you would expect me to do. So, by not doing it, it will be harder for someone to ambush me. Second, it is actually easier and cooler to do it with the work-energy principle. If I go back to the work-energy equation and think of the plane as a point-object (which it clearly is not), then the only type of energy the object can have is kinetic energy. This makes my work-energy equation:

&Delta r is the displacement of the plane and &theta is the angle between the force doing the work and the displacement of the plane. Clearly, cos 180 = -1. This means that:

Combining this with the other equation (forces in the y-direction), I get:

Now, I just need to plug in my values:

And this is wrong. Typically, the coefficient of friction is less that 1.0. My mistake was that I used Δ r as 200 meters. In fact, the video said the 747-8 stopped 200 meters short of the target distance.

Here is the new plan. I searched for quite some time, but my google-skills failed me. I could not find the required rejected takeoff distance for a 747-8 (or for any 747). So, I will estimate the coefficient of friction and use that to calculate the minimum stopping distance. The only change above is to solve for Δ r instead of μ. Hypertexbook.com has a list of coefficients of friction between rubber and concrete. Several are listed, but let me go with 0.8. Using this, I get a stopping distance of

Half a kilometer seems like a reasonable value to stop a plane moving 200 mph. Ok, I feel better now.

Update ——

Thanks to David for finding this document for me.

In this document, Boeing lists the RTO (Rejected Takeoff) distance of 11,900 feet (3600 m). If this is true, then my coefficient of friction must be too high (since I had the plane stopping in 500 meters.

See Also: