Can the Plane Take-Off?

[dlavery]

Ken is right. The airplane will take off. The only thing that is different than a "normal" take off is that the wheels will be freewheeling at twice the lift-off speed.

The problem is that you have to remember that the wheels are free spinning, and DO NOT provide motive force (other than the minute amount of drag due to bearing friction, also accounted for by Ken). The acceleration forces for the airplane are due to the thrust of the engines against the surrounding atmosphere - NOT by the wheels transmitting force to the ground.

Originally Posted by Elgin Clock

Fn - Force of Conveyer pushing up at the plane (+Y) (the force comes from the contact of the conveyor to the plane's wheels) and it it is a staple in every force equation which you have apparently forgot about and left out. Also known as Normal Force.
Fg - Force by gravity holding the plane down (-Y) on the conveyer
Ft - For this case, thrust of airplanes jets going in the reverse direction (+X) of:
Fcb - Force being countered by the conveyer belt (via really wicked cool sensors - which I would like to see in the kit next year ) in the equal and opposite direction of the airplane's thrust. (-X)

So.. in this equation.. and using Newton's third law, which states "For every action, there is an equal and opposite reaction."

Fn + -Fg = 0 = no movement in Y direction
Ft + -Fcb = 0 = no movement in X direction

In Elgin's analysis, the assumption is made that Ft is equal to Fcb. This is where the mistake creeps in. The problem statement does NOT state that the counter-rotating conveyor belt is applying a force to the airplane that is equal and opposite to the engine thrust. It simply states that the conveyor runs at a velocity inverse to the (ground) speed of the plane. The actual force transmitted to the airplane is limited to the relatively minute amount of drag due to friction in the wheel bearings (under the presumption that the pilot is not standing on the brakes). This drag force is a very tiny fraction of the forward thrust force applied by the airplane engines. In other words, Fcb is defined as the wheel bearing drag, and Ft >Fcb. And since Ft >Fcb, then Ft +-Fcb>0 => movement in X direction.

If you still think that the plane will not take off, then think through this: imagine exactly the same stating conditions. The airplane engines ramp up to full thrust. If you are assuming that the airplane (relative to an off-conveyor observing position) is "standing still" the explain the following: the pilot quickly retracts the landing gear, so the airplane is no longer in contact with the conveyor belt, the engines are still at full thrust, and the airplane is at zero velocity relative to the ground. What happens next? When you try to reconcile this self-conflicting condition, you will realize that in the original problem, the airplane takes off.

-dave