Registered: 23/08/01
Posts: 4757
Loc: Mt. Zion, IL
Quote:
Originally posted by JeffW: Obviously people who have never taken a physics class find this tuff hard to grasp.
Please prove this answer wrong: (since you claim it is incorrect)
Quote:
Originally posted by JeffW: 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.
Guess those who took physics didn't take spelling.
And in your formula wait someone else posted that. It their formula it clearly states as long as the thrust is enough to overcome the friction then the plane takes off. I agree.
I just don't think the engines will with the counter action of the belt. That’s me and my opinion.