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A question of physics.....

a simple little question, or is it?

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WolstonFlyer12/10/2016 23:50:21
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Interesting isn't it!

WolstonFlyer13/10/2016 00:14:52
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OK because it is late and I am not cruel:


I agree that the problem is stupid and has no true solution, it is a trick question.

Statement A: if the treadmill moves in the opposite direction of the wheels, always at the exact same speed of the wheels, the wheels cannot ever move in reference to the treadmill.

Statement B: the engines generate an enormous force in the opposite direction of the plane, therefor, due to the law of equal and opposite reaction, the plane has an equally enormous force pushing it forwards.

Statement C: in a free body diagram of a frictionless system, there is no force opposing the force of the engines

Combining these statements shows that it is impossible for the plane to move in relation to the treadmill; however, there is no force stopping it from doing so.

These statements are contradictory and therefor the problem is void.

If you really want to have a read go here

Biggles' Elder Brother - Moderator13/10/2016 00:22:25
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Sorry no. The conveyer is merely a frictionless support. So a direct analogy would be imagine the aircraft was on a sheet of ice. Could it take off? Of course it could!
Second one - all the conveyer does is remove the effect of traction. But if an aircraft required traction to move forward it couldn't fly could it because when it's flying it's not in contact with the ground!
BEB
Martin Harris13/10/2016 00:22:44
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As there is no mention of the belt's coefficient of friction, why can't we simply assume the material used has a negligible value and take off with the wheels locked? Works for me!

Martin Harris13/10/2016 00:23:52
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Great minds (in your case BEB!) ?

Edited By Martin Harris on 13/10/2016 00:24:12

Biggles' Elder Brother - Moderator13/10/2016 00:24:58
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If there is no slip between wheels and belt then dynamically it is frictionless Martin
Biggles' Elder Brother - Moderator13/10/2016 00:26:35
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Indeed Martin!
john stones 113/10/2016 00:30:11
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Would it be possible for you to send me the two envelopes containing money W.F i promise not to open first and send one back wink

Martin Harris13/10/2016 00:39:33
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There must be more energy being applied to the rolling road (allowing for mechanical losses) than is being generated by the aircraft in order for the conditions specified to be met and maintain equilibrium - does this impact on your free body explanation (I don't know - just wondering).

There can be no doubt that if the wheel's rotation is matching the belt speed and there's no slip as per the original question, the aircraft isn't moving relative to the surrounding air - disregarding air movement generated by friction along the length of the belt.

As stated by WF - a trick question in reality!

WolstonFlyer13/10/2016 00:47:45
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OK

The engines apply thrust and the plane would roll forwards with the tyres touching the belt, we all get that, perfectly normal.

At the same exact instant the belt moves in the opposite direction because as the question states, it is "designed to exactly match the speed of the wheels, moving in the opposite direction". Is the plane moving forwards? How can it be, the wheels are rolling forwards at the same speed that the belt is moving backwards.

Then the engines power up more to push the plane forwards, the wheels are still in contact with the belt and the belt instantly matches that change in speed, the plane does not move forwards, it is sitting still with the wheels turning forwards at the same speed as the belt running backwards. For the plane to move forwards and get any airspeed to generate lift the wheels have to turn forwards faster than the belt is moving backwards.

The question is a paradox designed to create discussion

Edited By WolstonFlyer on 13/10/2016 00:49:31

WolstonFlyer13/10/2016 01:16:24
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Posted by Biggles' Elder Brother - Moderator on 13/10/2016 00:22:25:
Sorry no. The conveyer is merely a frictionless support. So a direct analogy would be imagine the aircraft was on a sheet of ice. Could it take off? Of course it could!
Second one - all the conveyer does is remove the effect of traction. But if an aircraft required traction to move forward it couldn't fly could it because when it's flying it's not in contact with the ground!
BEB

Of course an aircraft can take off from a sheet of ice, the surface of the ice is not moving in the opposite direction to the rotation of the wheels, so the wheels roll forwards relative to the ice, the wings have air speed and then lift off.

If the plane is on the huge conveyor belt and the belt started to move backwards slowly and the plane did nothing at all then the plane would travel backwards sat on the belt at the same speed as the belt - it is carried along on the belt just like shopping in a supermarket.

So the plane starts its engines to provide thrust and now the wheels start to turn in relation to the surface of the conveyor belt, the engines provide enough forward thrust to turn the wheels fast enough so that the plane is no longer moving backwards sat on top of the belt, it is stationary, held in place by the thrust of the engines pushing it forward in relation to the belt.

Now the plane increases its thrust to move forwards but instantly the conveyor belt increases its backwards speed to match the forward speed of the wheels, the plane is still not moving forwards in relation to the surface of the belt.

If the belt is designed so that it can always match the speed of the rotation of the wheels but in the opposite direction does the jet plane ever get any forwards air speed to generate lift and take off?

It's enough to make your brain melt !

Rich too13/10/2016 06:47:34
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no

Don Fry13/10/2016 07:04:06
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No, the plane breaks.

The belt travels at the rotational speed of the wheels. With jet engines trying to move the aircraft forward, and failing because the belt constantly speeds up, the rotational speed of the wheels becomes enormous, exceeding design conceptions and limits, and undercarriage failure is inevitable.

The belt survives, its designers built for this scenario.

Gary Manuel13/10/2016 07:09:18
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It's not a void or trick question. As your sub-heading says, it's a physics question.

Those who apply the laws of physics will see that the plane will take off. Those that don't apply the laws of physics will argue all day that it won't, but they would be wrong. That's what the question is really about - getting people to argue.

Martin Harris13/10/2016 07:28:50
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How can you apply the laws of physics to a flawed scenario? The laws of physics would also say that neither the wheels nor the belt could achieve infinite speed, which would need to happen instantaneously - which is what the question's conditions would require.

The wheels will also generate friction even though they are rotating in harmony with the belt surface - tyre cling, deformation, bearing friction, even friction between the air and the rotating periphery...

The physicist will tell you that it will take off whereas the engineer will tell you that something must break.

Edited By Martin Harris on 13/10/2016 07:53:33

Gary Manuel13/10/2016 07:50:49
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OK, here's my interpretation of the laws of physics to this scenario.

The thrust will push the plane forwards, overcoming drag until a point where the wing is generating enough lift for it to overcome the mass of the plane. At this point the it will be capable of flight. Let's say this speed is 150mph.

The wheel will accellerate at the same speed as the plane. i.e. from 0 to 150mph.

The conveyor will accelerate BACKWARDS matching the speed of the wheel. i.e. from 0 to 150mph.

The only effect the conveyer has is to make the wheels rotate at twice the speed that they would do on a stationary runway, i.e. 0 to 300mph as measured on a wheel mounted speedometer, but the plane would still take off at 150 mph ground speed.

Gary Manuel13/10/2016 08:38:25
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Oh - and as an Engineer, heres how I'd build a working system.

The high tensile steel rope attaches to the wheel on the plane, passes around a pully at the rear end of the conveyor, runs the full length of the conveyor ans attaches to an anchor on the top surface of the belt.

The belt will match the speed of the wheels but in the opposite direction.

design.jpg

Andy Meade13/10/2016 08:59:08
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Posted by Phil Cooke on 12/10/2016 21:11:50:

It really doesn't matter how fast the wheels go round, no forward motion of the airframe = no aerodynamic lift = no take off.

I'm with Phil!

Bryan Anderson 113/10/2016 09:50:33
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Posted by WolstonFlyer on 12/10/2016 21:06:43:

This simple little question has caused a lot of healthy discussion on Facebook today.

747.jpg

Edited By WolstonFlyer on 12/10/2016 21:07:49

This is extremely badly phrased. "Exactly match the speed of the wheels". What is the speed of a rotating wheel? A wheel typically has two speeds: one translational and the other rotational. I think it means there is no slip between the tyre contact patch and the belt.

Short answer: the plane will take off if the engines provide normal take-off thrust.

Long answer.

Start with no motion and no engine thrust. The situation is stable and nothing will move, ever. There is no problem.

If, somehow, the initial conditions are such that the wheels are spinning and the belt moving such that the plane as a whole is stationary, then. in the absence of friction, that state will persist for ever. With friction, what happens depends on the inertia in the wheels /plane and the conveyor belt systems. Unless matched magically, I would expect some movement.

Now begin again and start the engines. If there is no friction between the contact area of the tyres and the belt, the wheels will not rotate and just slide contradicting the conditions implied in the question. The plane will accelerate and take off.

With friction, there will be a force in the backwards direction trying to rotate the wheel. This force must be supplied by the belt. As the plane accelerates, the wheel rotational speed has to increase so a force must persist. The belt system has to supply rotational energy to the wheel. Normally, the Earth does that.

Now consider a very large wheel. If large enough, the contact area becomes almost indistinguishable from a caterpillar track in contact with the ground and with no relative motions. (Alternatively, zoom in on this area). The backwards velocity of the tyre at the contact patch relative to the hub exactly matches the forward velocity of the wheel hub, the plane velocity, So there is no problem. The belt does not need to move at all though it must be constrained. The Earth doesn't when a plane takes off (at least to an observer with poor sensors standing on the Earth).

John F13/10/2016 10:06:45
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Posted by WolstonFlyer on 13/10/2016 01:16:24:

If the belt is designed so that it can always match the speed of the rotation of the wheels but in the opposite direction does the jet plane ever get any forwards air speed to generate lift and take off?

It's enough to make your brain melt !

Not really.

This reminds me of a discussion that ended up with a few members of a motorbike forum being sin binned for a few days as they pedantically argued whether a piston actually stops at TDC.

Pedantic, and inconsequential, stuff about friction, roller bearings not taking the strain, the tyre not the wheels is moving, etc, etc, ad inifinitum et nauseum, to one side, the question is simple.

The forward speed of the aircraft is negated by the reverse speed of the treadmill. Therefore, if there is no forward movement can the aircraft take off?

Take off energy is supplied by the lift coefficient of air passing over the wing. No air moving over the wing = no lift.

(VTOL capable aircraft are obviously excluded from this scenario!)

Edited By John F on 13/10/2016 10:10:52

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