HISSIN50 said:
Because of how this is worded, we assume the plane has no movement on the X-axis, correct? It matters not how power is delivered (wheels or jets) - if the wheel starts to turn, the treadmill negates the forward progress. So when an observer looks at it from the tower, there is no net horizontal displacement.
E.g., if a dude is on rollerskates and is on a this treadmill. I am standing on terra firma to the side of him and push his back to propogate forward movement - as soon as his little wheelies start to move, the treadmill goes the otherway and my pushing on his back does not make him change his absolute X-displacement.
I dont know much about airplane dynamics. There might be some Bernouli effect because of the turbulence caused by the plane's engine, though none from cutting the ambient air. I also dont know about the dynamic needed for take off - if the little bit of self-inflicted turbulence is enough to allow a slight lift, then with enough theoretical thrust/power, the plane could start to lift off, it would seem.
If we say it needs lift from cutting the air (via X-axis movement), I say it wont take off.
The horrible wording/phrasiology of the question (not by Stangster - I think it's intentionally ambiguous) makes it so that I kinda wonder if there is a correct answer - there is enough left open to interpretation.
I love this one. It always gets people.
I agree with you about the first part JT:
#1 - The plane must move as observed in relation to a fixed object off of the belt in order to create lift. With out the planes movement, it will not take off. You need air to flow underneath the planes wings to create lift. The planes speed creates this air flow, which allows take off. Period.
Here is where I disagree.
#2 - If the belt is spinning backwards, it will actually turn the planes wheels forward. If the belt spins forward, it will spin the tires backwards. A person on roller skates on a treadmill will move forward, not backward. Regardless, there is friction between the bearings on the roller skates, which will propell the skater forward, which means he will have to hold the side rails to stay in one place. Only if the bearings were somehow magically (frictionless), then the skater would remain stationary. The skaters weight multiplies the coefficient of static/dynamic friction present in the bearings.
i.e. If the static coefficient of each wheel was .09, then you would use the equation
View attachment 466969
Where F = force due to friction, u = coefficient and N = the force normal to the contact surface. (normal means the amount of force applied perpendicular to the contact surface. Since gravity points straight down, in this case, the normal force is always the weight of the object that is on the bearings.
Which means the greater the weight of the object, the more force is applied to lateral movement.
#3 - Whatever way/speed the tires spin, it has little affect on the planes movement. It is attached to bearings, so the tires spin and the plane does not stay in the same spot on the belt. (Because the plane weighs as much as it does, the friction that exists on the bearings will apply a slight force to the plane's axles, and it may move, but not very much). IMO, I think the friction is too small overcome the plane's inertia.
#4 - The only reason a plane's wheels spin forward is because the ground/tire contact has friction, which makes the tires spin forward. And that only happens because the ground is stationary.
#5 - If the belt is moving forward (making tires spin backwards), The plane is still on bearings, and the axle inside the bearing which is mounted to the frame of the airplane does not turn. It only has a tiny amount of friction applied from the bearings, which the planes jet engines can easily overcome, thus the plane will still move forward, and take off.
I think the force of friction will still be very small compared to the force that the jet engines apply, thus it will be able to overcome the force friction applies, and the jet will take off. It will use more runway length to get up to speed, but that's the extent of it.
Scott