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Torrere · 12-08-2006, 12:47 PM

Can the treadmill possibly exert enough force on the plane to counteract the force of the thrust?

1) Pie is correct that any force exerted on the wheels is subsequently exerted on the plane (although LabRat's drawing is better).

2) By SteveDallas' example, we know that
Fthrust => Ftreadmill
otherwise the plane would be pushed backwards. If Fthrust > Ftreadmill, the plane must accelerate forward and then eventually take off.

3) The force exerted by the treadmill on the wheels is a friction force, and therefore limited to:

Ftreadmill <= μR * Weight of plane

Where μR is the coefficient of rolling friction.

There are three different coefficients of friction that we could use: static, rolling, and kinetic. Static means that the plane is not moving at all w/r to the treadmill. With kinetic friction, the plane and the wheels are sliding forward, as in Maggie's story of brakes on ice.

μK < μR < μS

I'm not sure what the coefficient of rolling friction is for a 747, but the largest μR listed on Wikipedia is 0.03, and that's for a bus on asphalt. I assume that μR for a plane would be much smaller, but I'll use 0.03 for effect. According to Boeing's site, for a 747-400, the maximum takeoff weight of the plane is 3886 kN. Each of the 4 engines produces a maximum of 281 kN of thrust, for a total of 1124 kN of thrust.

So

Ftreadmill <= 3886 kN * 0.03 = 116.6 kN
Fthrust = 1124 kN

1124 kN > 116.6 kN

I'm not entirely sure, but it looks like this plane is going to move forward, and the treadmill can't go fast enough to stop it, because the plane will just start sliding. Since the plane is moving forward, air goes over the wings, and the plane takes off.

Can anyone get μR for a plane wheel?

12-08-2006, 12:47 PM	#201
Torrere a real smartass Join Date: Dec 2001 Location: Kirkland, WA Posts: 1,121	Can the treadmill possibly exert enough force on the plane to counteract the force of the thrust? 1) Pie is correct that any force exerted on the wheels is subsequently exerted on the plane (although LabRat's drawing is better). 2) By SteveDallas' example, we know that Fthrust => Ftreadmill otherwise the plane would be pushed backwards. If Fthrust > Ftreadmill, the plane must accelerate forward and then eventually take off. 3) The force exerted by the treadmill on the wheels is a friction force, and therefore limited to: *Ftreadmill <= μR Weight of plane Where μR is the coefficient of rolling friction. There are three different coefficients of friction that we could use: static, rolling, and kinetic. Static means that the plane is not moving at all w/r to the treadmill. With kinetic friction, the plane and the wheels are sliding forward, as in Maggie's story of brakes on ice. μK < μR < μS** I'm not sure what the coefficient of rolling friction is for a 747, but the largest μR listed on Wikipedia is 0.03, and that's for a bus on asphalt. I assume that μR for a plane would be much smaller, but I'll use 0.03 for effect. According to Boeing's site, for a 747-400, the maximum takeoff weight of the plane is 3886 kN. Each of the 4 engines produces a maximum of 281 kN of thrust, for a total of 1124 kN of thrust. So Ftreadmill <= 3886 kN * 0.03 = 116.6 kN Fthrust = 1124 kN 1124 kN > 116.6 kN I'm not entirely sure, but it looks like this plane is going to move forward, and the treadmill can't go fast enough to stop it, because the plane will just start sliding. Since the plane is moving forward, air goes over the wings, and the plane takes off. Can anyone get μR for a plane wheel?