“Imagine a plane is sitting on a massive conveyor belt, as wide and as long as a runway. The conveyer belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?"
My initial reaction was to think that it could, but the more I think about it (and draw some good ol' force diagrams) the more I'm inclined to think that it doesn't.
However, as some people in various comments have pointed out the problem is a bit vague so I figure I should explain some of my assumptions (some other comments I read also talked about wheels bursting into flames etc... and IMO those suggestions ignore the question of interest):
- The airplane's engines apply force purely in the horizontal direction
- The plane can only take off when the lift generated by the wings is greater than the downward force (i.e. its weight) that it's exerting on the treadmill (L>W).
- Wings can only generate lift as a function of moving in the horizontal plane (L=f(v))
- All the airplane's engines are being applied equally (no rotation from either side pushing more or less)
- The wheels, treadmill etc... are all indestructible and all bearings etc... are perfectly frictionless
- The only friction is that between the wheels and the treadmill (so wheels can't magically slide off)
- This is a closed system with no external forces (no wind etc...). If it's not the jet or the treadmill it doesn't count (no nudging the plane)
Even though it's true that the engine is pushing against the air it can't make the plane start to lift off until the the velocity of the plane is great enough that the wings produce lift. If the plane had strong enough engines that could lift its entire weight and those engines were mounted such that they could move or apply force away from the horizontal then it could take off (like a Harrier), but so long as the engines can only provide thrust in the horizontal then all the force that the engines generate goes into the plane (i.e. they're pushing/pulling the plane) and into the landing gear and into whatever the airplane is sitting on. Until a plane reaches take off speed it is still basically an engine (or engines) pushing on parts of a vehicle in order to make the wheels turn upon a surface. If that surface is matching the wheels' speeds in the opposite direction then the plane can't move forward, therefore it can't get up enough speed to lift off (this is why a car can sit on a dynamo with its drive wheels spinning at top speed and it won't go anywhere).
Arguing that it doesn't matter what the wheels are doing in this problem would imply that a plane could take off if the wheels (and the treadmill) were standing still (and that doesn't seem to make sense). Until it takes off a plane's velocity is a function of its wheel's speeds (vw) minus the speed of the surface (vt) it's sitting on (i.e. v=vw-vt=0). Until that speed is greater or equal to the take-off speed (vto) the plane won't take off. You could re-conceptualize the problem as putting a plane on an infinitely long treadmill that's going at half of its take off speed in the opposite direction (0.5vto). If the plane can only move forward it would have to move at 1.5x its take off speed (vw=1.5vto) in order to overcome the speed of the treadmill (v=vw-vt=1.5vto-0.5vto=vto)and if its wings could produce life in either direction the plane would only need to accelerate under its own power to 1/2 of its take off speed in to lift off (v=vw+vt=0.5vto+0.5vt=vto) in reverse (momentarily ignoring the obvious aerodynamic issues)
A real world example of this problem would be seaplanes. Seaplanes can't take off if they're trying to take off against sufficiently fast moving water, however if they are turned around and they move with the water then they can take off much more easily (theoretically with no power if the water they were sitting in was moving fast enough).
So, any thoughts? Does that make sense or did I make an embarrassing 2+2=5 kind of error?