What condition is not met?
And I'm sticking with easily answered. Does it fly? Yes? Conveyor and wheels would not change this no matter how you want to interpret the question
I think you and I agree that, in reality, the fact a plane is on a conveyor wouldn't affect its ability to take off. However, the question has this in it: "The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation." Assuming the question means that if the plane starts to move east and the wheels roll, then the surface of the conveyor must move west at exactly the speed of the wheels, then we have a problem as this can't happen. If the plane begins moving east at 1 mph, then the conveyor would start moving west, but when it reached 1 mph the planes wheels would then be moving 2 mph (even though the plane itself was still moving 1 mph). No matter the speed of the plane moving east, the speed of the wheels would be the speed of the plane plus the conveyor, which means the speed of the wheels would always exceed the speed of the conveyor and we have not met the condition of the original question. Most of us are arguing semantics and not physics. We understand the rolling of the wheels doesn't affect the planes ability to take off. The wheels can roll either direction on a conveyor, or even skid on ice without rolling at all and the plane can still take off.
I understand what you're saying. The conveyor could match airspeed (which would double wheel speed in relation to conveyor). I believe this example was the intended nature of the question designed to confuse people but as you said, not well written. What if it's meant to move in the opposite direction of rotation that the conveyor is moving with the plane and not against it. Conveyor matches airspeed of the plane and the plane's wheels do not turn at all (speed of wheel in relation to conveyor is the same, zero). And yes, if it was matching wheel speed it would just exponentially accelerate the wheel trying to achieve a goal it could never reach. Either way, I don't believe we are all arguing semantics as several have stated that the treadmill could stop forward movement of the plane. And it is that line of thinking I am disagreeing with. The plane flies as a conveyor under a plane is of minuscule influence no matter what it is doing.
Assuming there is indeed friction between the conveyor belt and the tires of the plane, it's vital to distinguish between two VERY distinct situations: 1. A conveyor belt moving at a constant speed will impart NO lateral, or translational, force to the axle of the wheels. Under this condition, the conveyor cannot balance/counter-act the forward thrust of the plane's engines ... and the plane will lift-off, for any SINGLE chosen speed of the belt. 2. However ... a conveyor belt that is accelerating must impart a lateral, or translational, force to the axle of the wheels. Under this condition, it is indeed possible (at least, theoretically) for the conveyor to balance/counter-act the forward thrust of the plane's engines ... thereby preventing the plane from lifting-off. The reasoning behind point #2 is this : an accelerating conveyor belt must exert a net force on the edge of the tire, in order for the rotational speed of the tire to increase (assuming no-slip between tire and conveyor). A single, net force acting on any body will always exert a translational force on the center-of-mass of the body (as well as, possibly, a rotational torque on the body if the force is applied off-center). Therefore, an accelerating conveyor must impart a lateral/translational force on the axle of the plane's wheels. If that force is chosen to match/balance the forward thrust of the engines, the plane will remain stationary (relative to surrounding air) ... but ONLY if the conveyor is accelerating. There's a WORLD of difference between inertial frames of reference, and non-inertial (accelerating) frames of reference. In a sense, an accelerating conveyor is essentially utilizing a non-inertial frame of reference to impart a translational (lateral) force on the wheel's axle (and thereby, on the plane itself) ... this is something that a constant-speed conveyor cannot do. In my view, any discussion of this topic must differentiate between a constant-speed conveyor belt, and the VERY different situation of a conveyor belt that is accelerating.
Doesn't matter. An accelerating conveyor could still only add additional drag of tires and wheel bearings. This would be minuscule if even measurable for the engine(s) to overcome.
That's right, and of course the plane still flies. Right, an impossibility Well, yeah. Some don't get it. Yes sir, that is correct if not necessarily meeting the possible interpretations of the original question.
I REPEAT - YOU NEED AIRSPEED Stick your head outside in the air. All through this experiment if the wind is calm at the beginning and at the end. You could tie the wings to the tarmac as there is no lift produced. Hell - crank the conveyer up to 10,000 mph and the plane will still have no airspeed. Is this really that hard to comprehend for some ???????
airspeed will remain at zero if you had plane tied down, but plane isn't tied down and it will gain airspeed to take off. as said 2,287 times wheels and treadmill have nothing to do with keeping the airspeed at zero, only to add a little more friction to gain airspeed.
Not true. A net force is required for acceleration of a body, whereas no net force is required for constant velocity ... that's a VERY fundamental difference. An accelerating conveyor applies a net force (as torque) to the tire's perimeter ... and therefore a net force (as translational) to the wheel's axle, and therefore to the plane itself. This does not happen with a conveyor operating at a constant speed. You simply can't accelerate the rotational speed of the tires without a net force ... and that net force also appears as translational to the wheel's axle, and to the plane itself.
Scott, we are on the same page. No need to repeat. Quote the post of just one person that said you -don't- need airspeed. What's mind boggling is that it's seemingly hard for some to understand that a conveyor can not hold an airplane stationary. True. As stated, whatever additional drag is created would not be enough to stop the plane from flying. Energy required to accelerate the tires, even to the point of explosion would still transfer a relatively small amount of energy to the airplane via the wheel bearings.
For those thinking a conveyor could somehow hold an airplane stationary, could you offer a technical explanation as to why? And do you just dismiss this example? https://www.youtube.com/watch?v=YORCk1BN7QY
Imagine, for a moment, the engines of the plane turned off. The question then becomes, can a conveyor belt exert a net, rearward, translational force on the plane? If the conveyor is moving rearward at a constant speed, the answer is NO. There is no net force required to rotate a plane's tires at a constant speed. Therefore, no net force is applied to the perimeter of the tires, and consequently no net translational force is applied to the axle (or plane). If the conveyor is accelerating, it's a VERY different scenario (an accelerating body is a fundamentally different scenario than a body moving at constant speed). An accelerating belt must accelerate the rotation of the plane's tires ... which means a net torque is applied to the perimeter of the tires (unlike the above scenario), which means a net translational force applied to the axle of the wheels. A rearward accelerating conveyor will absolutely accelerate the plane backwards. So ... in the first scenario (conveyor at constant speed), there is NO net translational force to be found. In the second scenario (accelerating conveyor), there IS a net translational force to be found. These two scenarios ... absence of a net translational force, versus presence of a net translational force ... are fundamentally very different. (note that, in both scenarios, we assume perfectly ideal, frictionless bearings at the wheel's axle ... but we do recognize friction at the tire/conveyor interface)
I already gave that example with a way to measure it. Put the plane on the conveyor. Tether the plane to a fixed point in front of the conveyor belt and accelerate the conveyor rearward. Strain gauge in line on tether. No one is saying it wouldn't have -any- measurable effect, but it would be minimal, would not hold the plane in one position and would not affect it's ability to fly. Did you watch the video? Do you take any issue with what you saw?
A rearward-accelerating conveyor exerts a net rearward, translational force on the plane ... this does not happen with a conveyor moving at a constant speed (assuming perfect, frictionless bearings at the axle in both cases). But ... if the plane's forward thrust of the engines is greater than the rearward translational force applied by the accelerating conveyor, the plane will (of course) move forward relative to still, surrounding air ... and the plane will take off. It's not sufficient to simply accelerate the conveyor rearward, arbitrarily ... the net translational force applied by the rearward-accelerating conveyor must balance the forward thrust of the engines. If the plane accelerates forward, or backward, relative to still air ... the balance has not been met.
That conveyor did not exactly match the speed of the wheels on that plane at all times, as is required in this problem. Read the problem. We don't need to give a technical explanation. The fact that the speed of the wheels on the plane are matched by the conveyor at all times is given. The conveyor has the same effect as leaving the parking brakes on. The plane cannot move forward, cannot create enough lift and cannot take off. The only way for the plane to defeat the conveyor is to leave it or sabotage it...or for the reader to ignore this important parameter of the problem. If you must experiment, take a toy plane and place it on a treadmill. Turn on the treadmill and the plane will roll off the back of the treadmill because it doesn't have an engine. Now place the toy plane back on the treadmill and use your hand to simulate thrust. You can push the plane forward, BUT THAT WOULD VIOLATE THE TERMS OF THE PROBLEM. THE TREADMILL MATCHES THE SPEED OF THE WHEELS, SO BY DEFINITION THE PLANE CANNOT MOVE FORWARD. Is there enough wind flowing over the wings of your toy plane to take off? Only if it creates enough thrust with its toy engine. You can simulate that by lifting the plane vertically off of the treadmill. I remember a friend's tale of sailing into San Francisco Bay under the Golden Gate Bridge as the tide was going out. Seems that no matter what he did, he could not make headway in his sailboat as the speed of the tide matched exactly the speed of his boat generated by the wind in his sails. Since the speed of the tide matched the hull speed of his boat, he stayed in one place. Although his speedo registered 6 knots, his speed over ground was 0. He was only able to get back to his slip because the tide started to reverse, and reduced speed and no longer matched his hull speed. Then and only then was he was able to move forward.
If you lock the brakes you can still thrust the plane forward by adding more thrust or skid the tires. Bad example.
dakharris, are you changing your answer to a conveyor cannot physically hold the plane in one spot to it violates the terms of the problem? Because this- "The conveyer belt is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation." Can be explained by the conveyor matching the planes airspeed while continually cancelling out any wheel movement. Therefore falling into the category as answered as asked, meeting all criteria. Wheel speed in relation to conveyor is matched at zero, plane of course flies.
Yes, of course thrust is an answer. But the plane does not move forward. It lifts off and flies. Some continue to insist that the plane can move forward over the conveyor. That violates the rule outlined in the problem that the conveyor matches the speed of the wheels. If the conveyor is matching the speed of the wheels, it is, by definition, not moving forward. In order to defeat the treadmill, we must assign attributes not included in the original problem. As I said, there can be no answer because we cannot all agree on the additional necessary parameters to definitively solve the problem. As written, there can be no right or wrong answer because each of us can impose his own limitations to defeat any proposed solution. What is fascinating is that people are willing to argue in circles for years. It has been fun playing this version of the Kobayashi Maru simulation again, but I've lost interest. Maybe we can resurrect this thread in another nine years.
Thanks Rob! From a democratic perspective I guess Art's right...and my answers is better than yours. There are possibly some tenacious people on FChat?
For an aircraft to fly the wing needs to produce lift. Lift = Lift co-efficient or lift angle x 1/2 p (density) v2 (velocity) S (speed relative to the air) With zero wind or when the aircraft is not moving relative to a fixed point on the earth S=0 Therefore the lift produced is zero. The fact are the aircraft will not fly if there is; no lift no true airspeed The speed of the wheels has nothing to do with producing lift. I cant understand why this is not obvious. The aircraft will not change speed relative to a zero wind condition. WAKE UP GUYS.
