Quote:
Originally posted by JeffW:


This is not an aerospace engineering problem. An assumption is that the plane is already properly engineered. So we are back to Physics 101. Regardless, I wouldn't trust an aerospace engineer who has no grasp of basic physics (like the one you quoted).Please refer him to the physics professor quoted earlier.

It is disappointing that you have chosen to fanatically support the incorrect answer instead of taking a few minutes to understand this high-school level problem.

Please read the basic physics:

[QUOTE]Originally posted by JeffW:
[qb]Relavent information:

A plane is standing on a runway that can move (like a giant conveyor
belt). This conveyor has a control system that tracks the plane's
speed and tunes the speed of the conveyor to be exactly the same (but
in the opposite direction).

Will the plane be able to take off?

Let's agree on what assumptions are sensible then. All of the assumptions below apply to virtually all aircraft.

Plane is powered by engines that push air (props or jets)
Plane's speed is measured by a windspeed meter as well as GPS
Plane's wheels spin freely
Wheel friction is negligible when compared to thrust
Newtonian physics apply



Coefficient of dynamic friction:

Force of friction:

F(f) = -uN

(negative because it opposes motion)

u = coefficient of static friction

N = weight of plane

Notice velocity is not included!

That means that the velocity of the conveyor belt is irrelevant for all practical purposes.

F(t) is force of thrust

a = (F(t)+F(f))/m

You won't find a scenario where |F(f)| is greater than |F(t)|. Therefore, in ALL cases the plane moves with respect to the atmosphere, thus achieving lift.

Possible scenario:

t = 0 :

Plane 0 mph
Conveyor 0 mph
Wheels 0 mph

t = 15 :

Plane 60 mph
Conveyor 60 mph
Wheels spin @ 120 mph

t = 55:

Plane 160 mph
Conveyor 160 mph
Wheels spin @ 320 mph

....

The plane takes off.

Either learn why it is correct or refute it....

..scientifically.
how do assume that thrust is 20% of the normal force? yes your simple fbd shows that f(t) is greater than the f(f), but you also need the f(l) to be greater than the f(g) for the plane to become ariborne, correct? how do we obtain a f(l) greater than f(g)? there needs to be a forward speed great enough to create the f(l) to overcome the f(g). you said that velocity does not factor into this problem, but i think it does.

lets take a practical example. forget the conveyor for a minute. a plane moves at 20mph down the runway. the f(t) is greater than the f(f). will the plane take off? it is moving forward, so by your reasoning it will fly, but it will not fly. why is that? because there is not eoungh lift created f(l) to overcome the f(g).

so back to the original problem, any thrust created by the plane to move it forward is counteracted by the conveyor, therefore there is no forward motion, therefore the plane does not fly.

jeffw, just out of curiosity, why wouldn't you trust the aerospace engineer? i'm sure that he has taken several physics courses past the grade 9 level. are you an engineer yourself?