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Arachnidman · 28 Dec 2006, 04:15

Ah good.

The plane WILL take off....

Imagine:

Conveyor belt is travelling at a speed of "infinity" miles per hour (stop being gay and ignore mass and time dilation etc)

You plonk a plane down on the conveyor belt.

The plane (of mass M) whizzes backwards.

The plane whizzes backwards past a stationary tree.

How much force needs to be applied in the opposite direction of the motion of the plane, to keep the plane stationary, level with the tree?

Simple, the same force required to de-couple the wheels from the plane.

What force is this I hear you ask?

F = uR

What is R? R is the reaction force caused by gravity. R = Mg. (g the gravitational field strength of 9.8 N/kg)

u is the co-efficient of friction between the planes wheels and the axel (between 0 and 1). Lets say this is a reasonable plane and the wheels are freely spinnable.. so 0.2

F = 0.2Mg

For a plane weighing 5000 Kg

F = 9800N

so, applying 9800 Newtons of force (from the engines) will keep a plane stationary, even when a conveyor belt is spinning the wheels backwards at an infinite speed (stop being gay about the wheels breaking).

What effect will a force of 10000N have?

Well, there is now a resultant force of 200N, and since F=Ma, this force will have the effect of accelerating the plane up the runway at:

0.04 metres per second, every second.

Wait enough seconds, and the speed of the plane will reach the required takeoff speed of 300 kph or whatever it is....

Come on!

28 Dec 2006, 04:15	#172
Arachnidman You love me really Join Date: Aug 2005 Posts: 342	Re: A Plane Problem Ah good. The plane WILL take off.... Imagine: Conveyor belt is travelling at a speed of "infinity" miles per hour (stop being gay and ignore mass and time dilation etc) You plonk a plane down on the conveyor belt. The plane (of mass M) whizzes backwards. The plane whizzes backwards past a stationary tree. How much force needs to be applied in the opposite direction of the motion of the plane, to keep the plane stationary, level with the tree? Simple, the same force required to de-couple the wheels from the plane. What force is this I hear you ask? F = uR What is R? R is the reaction force caused by gravity. R = Mg. (g the gravitational field strength of 9.8 N/kg) u is the co-efficient of friction between the planes wheels and the axel (between 0 and 1). Lets say this is a reasonable plane and the wheels are freely spinnable.. so 0.2 F = 0.2Mg For a plane weighing 5000 Kg F = 9800N so, applying 9800 Newtons of force (from the engines) will keep a plane stationary, even when a conveyor belt is spinning the wheels backwards at an infinite speed (stop being gay about the wheels breaking). What effect will a force of 10000N have? Well, there is now a resultant force of 200N, and since F=Ma, this force will have the effect of accelerating the plane up the runway at: 0.04 metres per second, every second. Wait enough seconds, and the speed of the plane will reach the required takeoff speed of 300 kph or whatever it is.... Come on! Last edited by Arachnidman; 28 Dec 2006 at 04:26.