Externally? The weight of the plane is fully supported by a normal force exerted by the ground. Engine thrust is opposed by a friction force between the tire and belt/pavement. I think that much we agree on. We agree that this friction force kinematically couples the belt to the wheel; there is a no-slip condition at the tread/belt interface. We agree that any bearing losses, etc., will be transmitted through this interface. We agree that if thrust is not fully balanced by friction, the plane will accelerate and take off. Where we disagree is the magnitude of that friction force.
If the belt is moving at a constant velocity, you’re absolutely right. Takeoff thrust will be something on an order of magnitude greater than bearing friction; the forces are not balanced, and the airplane accelerates forward. F = ma and such.
However, if the belt is allowed to accelerate linearly, the wheels will undergo angular acceleration. If this is the case, the reaction force on the wheels will be the sum of bearing friction and force defined as Ialpha/r*, where I is the moment of inertia of the wheels about their rotational axis, alpha is angular acceleration of the wheels, and r is the wheel radius.
That force, imparting angular acceleration to the wheels, could counter engine thrust just as applying the brakes or tying a rope to the gear struts. It’s an incredibly impractical way to do it (hence “magic belt”), but the physics works.
So what happens when v of belt reaches infinity? By definition, it cannot accelerate any more, and therefore there will no longer be angular acceleration and force exerted on the plane. Therefore the plane will gain forward acceleration.
The thrust by the engines need not be infinite, because constant acceleration is needed to overcome even the slightest constant thrust.
Figures?
OK I hate to admit it but Cecil is correct, a plane can take off under those conditions, but his logic is wrong. The plane, however must have vertical takeoff capability to pull this off.
Why would the velocity of the belt ever reach infinity?
If you assume force is transferred to the plane via the angular acceleration of the wheels (as KeithT explains), then that requires a constant acceleration of the treadmill. The velocity constantly increases, yes, but it never reaches infinity.
If you assume massless wheels (as well as friction that doesn’t change with velocity and no limitations on treadmill acceleration and no relativity–again, all perfectly valid for a thought experiment), then the treadmill will instantly reach infinite speed, there will be insufficient force coupling between the treadmill and the plane, and the plane will take off.
Eh? Why? Assuming wheels with some mass, there must always be acceleration in order to maintain a force to counter the thrust, and so when t=infinity, v also has to be infinite. What would the limiting factor be?
Well, time of course. You’re asking what would happen after an infinite amount of time has passed, which is essentially meaningless. THere’s nothing “after” infinity.
We are in complete agreement that it might be possible to design a treadmill such that the plane does not take off. Read my first post through. (I did screw up in my last post and include the word acceleration.) Given the new problem of an designing accelerating treadmill, the question becomes, can the force of rolling friction be great enough to keep the plane in place - to exert sufficient force to accelerate the wheels sufficiently. This is a matter of picking the right materials.
In this case the coefficient of rolling friction is the parameter of interest. I would assume some material can be found such that the force of rolling friction is strong enough. (A plane with it’s brakes on experiences static friction, which is almost always greater than rolling friction.) Clearly not for whatever tarmacs are made of, as planes take off all the time, even though the wheels are accelerating, and are subject to exactly the relevent force of friction. Presumably there is some substance that will work. However, now you are solving a completely different problem, since the op explicitly states that you match the speed of the treadmill to the plane. In that case, the only way to accelerate the treadmill is if the plane is accelerating, in which case it will take off.
This may or may not be a nitpick but rolling friction is static friction becuase the tire is not moving in relation to the surface at the point of contact.
I think you are wrong, but since friction is poorly understood, certainly by me, who really knows? Presumably, static friction is stronger than rolling friction because of stronger bonding between surfaces, with the surfaces comingling to some extent. In this case, an atom in the tire and an atom on the treadmill are in contact for an infinitisimal amount of time, before they accelerate away from each other. I think it is analgous to two wheels rotating at the same speed with one on top of the other and each just touching the other. That would involve rolling friction, and this would too.
I have a problem with this logic. It is clear that if you apply a tangential force F to the wheels, the wheels will have an angular acceleration alpha, and there is relationship between the moment of inertia I of the wheel, alpha, and F, i.e. F = Ialpha/r*. But this force F is the frictional force from the belt, not some new reaction force. Besides, it is tangential, and so can only provide a torque to the wheels and make them spin faster. Acceleration of the belt has no impact on the plane.
No, if there’s a force on an object, there must either be : a) an equal force in the opposite direction which cancels the first force, or b) acceleration. If the plane stays still, the frictional force is transferred to the wheel hub. Take a look at this post in the other thread for a rather thorough explanation.
I apologize, zut already corrected me in the other thread and I found the flaw in my reasoning. So, an accelerating belt does create a force that opposes the movement of the plane. If I understood correctly, though, the frictional force on the wheel has nothing to do with this. The frictional force on the bearings, on the other hand, can impact the movement of the plane. But it is the force from the acceleration of the belt that is the key as zut showed.
You and Mr Infinity need to spend a little more time together and get to know one another better. Although I hear he’s kind of hard to reach. Well, OK, impossible.
Assume we have a magic treadmill with no friction in the mechanism, but there *is *friction between the whells and treadmill surface. When the propeller starts to turn the surface of the wheels are stationary relative to the treadmill surface. Since there is no friction in the treadmill mechanism, as the plane nudges forward the friction between the wheel and the treadmill will drag the treadmill in the direction of the plane. Since the wheels are stationary relative to the treadmill surface, the speed of the wheel and treadmill are the same. When the plane reaches takeoff speed it will lift from the ground with the wheels never having turned at all. This is the logical result if the surface of the wheel must match the treadmill speed.
If the interpretation is that the treadmill speed matches the speed of the wheel hub, then the treadmill is just moving a 2x the speed of takeoff and does not have enough drag to prevent the plane from lifting off.
Next question: if a zombie pterodactyl is on a treadmill, and the treadmill is going backwards as fast as it is running forwards, does the pterodactyl beat a T-Rex, assuming the T-Rex has a headache?
