All right, then, I’m going to repost something from the other thread about force transferrence. First of all, let me point out again that the answer to this question depends on the assumptions you’re making about what’s happening in the problem.
There’s nothing wrong with interpreting “wheel speed” as “hub speed relative to the ground” or “tire diameter speed.” There’s nothing wrong with assuming no friction or massless tires (or high friction or high mass). There’s nothing wrong with pointing out that you doubt a physical system like this could be fabricated, for a 747 at full thrust. But for heaven’s sake, state your assumptions.
Now, with that over. I see that a number of people (including Cecil, I might add) are forgetting that rotational acceleration of the tires will produce a force on the airplane. Nothing wrong with assuming massless wheels, but you’ve gotta say so. So let’s go to the basics and talk about forces and force transfer. Let’s talk about forces in the context of, say, a wheel:
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The wheel is round, and it has some mass, m, with a center of mass in the center of the wheel at “O”. It’s also got rotational inertia, I. One way to think of mass is as the resistance of the wheel to motion when a force is applied. In the same way, rotational inertia can be thought of as the resistance of the wheel to rotation when a torque is applied. With me so far?
In addition, let’s assume that the wheel has a bearing in the center, at the “O”. The other side of the bearing is connected to Something. We haven’t defined what Something is, yet, but we will. Be patient. Let’s make this simple and assume the bearing is frictionless. We can add friction in later if we want, but let’s keep it simple for now. Now, the bearing is a pin joint–it allows both horizontal and vertical forces between the wheel and the Something that it’s connected to, but does not allow torques to be transferred.
One last thing. Let’s make this wheel and the associated forces as simple as possible. So let’s ignore gravity and everything else. No gravity, no treadmill, no Earth, no nothing. This is, right now, just a wheel in space connected to Something through the bearing. OK. Got it?
Now let’s change the story.
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* O *<----F
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I’ve added a force F on the wheel; this force F passes through the center of mass. This force is, as drawn, unbalanced, and it isn’t yet the whole story. From here on out, a few different things could happen.
Suppose the wheel isn’t connected to anything at the bearing (the Something is Nothing, in other words). In this case, there’s no way the force F can be resisted, and the entire wheel will accelerate to the left. The acceleration, in fact, will obey Newton’s law, F = ma.
OK, but what if the wheel is actually connected to Something? What if that Something was the ground, and the wheel is prevented from actually moving to the left? In that case, there will be a reaction force at the bearing. The wheel will push on the ground, through the bearing, and the ground will push back:
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* R---->O *<----F
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The ground will push back, in fact, so that R exactly equals F. This is the key: because the wheel does not move, the forces must be add to zero. If they didn’t add to zero, the wheel would accelerate to the left or right. But it doesn’t. So the forces must be balanced, and the force F is transmitted to the ground.
Another possible thing that could happen would be that the Something that the wheel is connected to is something very massive, but moveable, which has a mass M. In that case, as soon as the force F is applied, everything accelerates–both the wheel and the Something–at an acceleration of F = (M+m)a. In this case, there will still be a reaction R on the bearing, but the reaction R will be less than F. How much less is left as an exercise to the reader.
All right, got all that? Now let’s do something a little different. Let’s move the force:
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*******<----F
Now the force F is no longer a the center of mass. How does that change things?
First or all, let’s go back to the assumption that the wheel is just free in space–nothing is connected to the bearing. In this case, the wheel will still accelerate to the left in accordance with F = Ma. The force F is still unbalanced, and the unbalanced force must result in an acceleration. However, the wheel will also rotationally accelerate. The force F is not through the center of mass, and that creates a torque T = Fr. This torque rotationally accelerates the wheel through the rotational equivalent of Newton’s law: T = I[symbol]a[/symbol] ([symbol]a[/symbol] being rotational acceleration). So the wheel will begin to move to the left, and spin clockwise.
Now let’s assume, instead, that the wheel is connected to the ground at the bearing–the Something is the ground again. This gives a case like so:
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* R---->O *
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*******<----F
Here’s the key question: does the center of mass of the wheel accelerate to the left? No! it’s connected to the ground! It can’t move to the left. And because it can’t move to the left, the forces must add to zero, and the force R equals the force F.
Of course, the difference between this case and the one above where the force F was pointed toward the center of mass is that now the wheel will spin. And we can calculate exactly how much it will spin: the two forces set up a torque T = Fr, which rotationally accelerates the wheel through the rotational equivalent of Newton’s law: T = I[symbol]a[/symbol].
So all that leads to the real question: how does this apply to an airplane? The point is, to keep an airplane from moving, you need to apply a force F to the plane that is equal and opposite the engine thrust. Assuming the plane won’t flip end-for-end or anything silly like that (adding back gravity and so forth), it doesn’t matter if that force is applied to the nose of the airplane, or the tail, or the landing gear.
It doesn’t even matter if the force is applied at the bottom of the tires. That force will still be transmitted to the plane and oppose the engine force. Given that this is a thought experiment, it is possible to apply an opposing force to keep the plane in place via the tires. An accelerating treadmill will accelerate the wheels, which will transmit the force to the plane. And, to counteract a 747 at full thrust, that acceleration is apt to be very large. A rubber-band powered toy? Not so large.