Anybody ever heard of Rube Goldberg? ROTFLMAO!!!
As long as the treadmill is moving slowly, the plane would go backwards, assuming there actually is some wheel bearing friction. It would only take a touch of thrust to overcome this, though, and as the treadmill sped up, the engines could easily overcome the slight bit of backwards force the treadmill might exert on the plane. With wheels that are free to rotate, there’s no way a treadmill could exert the kind of force on the plane to seriously retard its motion, unless you bolted the whole plane to the treadmill.
The ideal situation is easier to picture, though. With zero wheel bearing friction, the treadmill exerts zero backwards force on the plane.
You’re basically asking why objects on the surface of the Earth don’t go flying off due to the Earth’s motion.
The answer is that the gravitational force of attraction the Earth exerts on you is sufficient to overcome the Earth’s orbital motion. However, at the Equator, where the linear velocity of the Earth’s orbital motion at a maximum, objects actually do weigh slightly less than those at the poles due to the resultant centrifugal force, which slightly counteracts the force of gravity. Because this force reduction is so minimal, it’s indicative of how minor the forces are that are created by the Earth’s motion.
These forces are so miniscule, that they are swamped by air resistance, wind, friction, etc. You only see their action on very large objects, such as hurricanes, or objects in free-fall over long distances, such as ballistic missiles.
WAG Y E S ! Did you expect it to sit still and spin it’s prop or turbines?
As I understand the OP, the question is actually “on a giant treadmill capable of infinite speed/acceleration, is it possible for a plane to turn on its engines and remain stationary with respect to the ground next to the treadmill?” I believe everyone agrees that if the plane is motionless with respect to the ground next to the treadmill, it won’t be able to generate any lift and take off.
I posit that this is possible, i.e. the plane won’t take off. The reason is this: at takeoff and at low speeds, the plane sits on its wheels. When the plane moves forward, the wheels roll over the ground smoothly. On an open tarmac (at low enough speeds to not have any lift), the velocity of the plane with respect to the ground is exactly proportional to how fast the wheels are spinning (v=r*omega), because the plane is rolling on them. Now, if the plane is moving forward over a treadmill whose surface is moving backward with respect to the ground at the precise relative speed of the plane with respect to the treadmill, it will remain stationary. It has to, if it is rolling on wheels.
An equivalent problem is a can of soda rolling down a gently sloping treadmill. Here gravity is taking the place of the airplane’s thrust. Assuming you can continuously accelerate the treadmill, is it possible to keep the can of soda at a fixed point on the slope? The answer, again, is yes.
I hesitate to jump into this thread, because it’s starting to look like another Monty Hall problem, but…
While we are given that the speed of the treadmill matches the speed of the wheels, we aren’t given that it influences the plane at all. What makes this problem tricky, I think, is that it has this magic treadmill - given frictionless bearings (why not) there is no way the treadmill can affect the speed of the plane. Since the plane is only influenced by engine thrust, it should be able to roll from one end of the conveyor belt to the other and take off. BUT, if it did that that would mean that it’s wheels would be rolling faster than the belt, and we are told that that doesn’t happen. So the plane can’t take off, not because of physics, but because of the constraints on the problem.
In real life, you would never be able to set up an experiment, because the engine would do what engines do and the plane would roll down the moving runway and take off, thus violating the constraint that the treadmill moves at the same speed as the wheels. As long as the speed of the treadmill matches the speed of the wheels, that plane ain’t going nowhere.
You know what? Now I get it. Thank you. I retract my earlier post, and nominate Valgard for an Igny Award, for the valorous fighting of my Ignyrance.
OK, I get that part, but say i expend additional effort to maintain the new speed of 101 mph, the treamill will compensate and any increase in ground speed relative to the world off the treadmill would be negated, correct?
Assuming an engine thrust sufficient to propell the aircraft forward at 150 knots and further assume that the treadmill is operating in the opposite direction at an equal 150 knots, in one hour, how much weight will be lost by those on board?
This has gone beyond ridiculous, I give up. (but I’m not buyin’ a ticket on that airline)
People move themselves by the action of their feet against the surface upon which they are standing; placing them on a moving treadmill means that they will have to move relative to the treadmill.
planes move themselves by the reaction to the action their engines thrusting mass (in the form of air and hot exhaust gases) in a rearward direction; except for the comparatively insignificant effects of friction experienced by the wheels, placing them on a moving treadmill doesn’t make any difference, because the normal forward motion of the plane is not imparted by any interaction with the ground (or treadmill); they will still thrust air and hot gases backward and be pushed forward in reaction. It’s that simple.
I (basically) disagree.
In the idealized situation, the can rolls with no friction. The treadmill can exert no force and thus has no influence on the can’s motion.
In the real world, the can has some friction, and thus when it rolls down a shallow slope, your treadmill can do as you claim. But if you steepen the slope enough, this no longer works. The critical angle will depend on the can’s mass (because “thrust” = mass * sin(angle)) and its friction (which will be proportional to mass * cos(angle) plus something for air resistance).
Back to the airplane (in this example, a small one): In the real world, it takes (say) 50 lbs of force to overcome the friction of the wheels and wheel bearings on a smooth paved surface. This is the force the treadmill can exert on the plane. But the propellor can easily generate 200 lbs of thrust. No way can the treadmill hold the plane still or indeed prevent it from accelerating to flying speed.
Rinse and repeat until clear.
Here’s a simple practical solution for anyone in this situation. This is true no matter which of the camps you are backing here. Apply thrust, and adjust flaps to make a turn. The plane will turn even if you don’t have enough lift to raise the plane into the air. You may taxi forward and fly away, when the plane is facing 90 degrees from the axis of belt travel. The track will not turn when you are 90 degrees from the axis of belt travel, since there is no vector of force applied in the conveyor’s direction of travel. You would be able to take off with a lesser degree of rotation from the belt axis, but you don’t know how much.
Regardless I am flying away and you are stuck in the Twilight Zone.
Don’t think about the treadmill for now. You roll something, e.g. a barrel, over the ground. If a force is acting on the barrel, it is going to accelerate and cover a certain amount of ground in a certain time. The movement of the barrel with respect to the ground is the same, regardless of whether the ground is itself accelerating in the opposite direction or remaining stationary. Unless it’s sliding, how can you not be able to compensate for the barrel’s relative motion with respect to the surface it is on by moving the surface itself in the opposite direction?
I think the problem is that you’re ascribing forces to the treadmill that it doesn’t have.
Let’s simplify the problem. Let’s pretend we’re talking about a guy on rollerblades with a rocket pack on his back. His rocket pack, when turned on, provides exactly enough force to overcome friction and air resistance and move at a constant 5 mph. Now, put him on top of a train which is moving 5 mph. How can he not stay motionless with respect to the ground next to him? How is this different from a plane on a magic treadmill? His rocket pack determines his speed relative to the ground beneath his feet. If that ground is itself moving, that doesn’t affect how the rocket pack works.
Here are my pretty pictures I drew to explain this case.
Ho boy, lets start back from the beginning:
The plane will take off with its wheels spinning twice as fast as normal but it will take more force than normal. First off let me say that it is impossible for a tire to rotate without an applied force at some point. Saying so is flat out wrong and someone saying so that has taught college physics is a bit frightening. It is true that at a constant velocity there is no force necessary to spin a wheel but we are talking about a sitation in which there is acceleration.
Lets first talk about the general case of a plane taking of on a normal runway. The plane at an instant in time has a certain velocity Vp and a certain acceleration Ap. Now, lets look at the tire. We know that there is a point of 0 velocity for an observer on the ground where the tire contacts the road. Since the tire has a foward linear velocity of Vp there must be a velocity caused by rotation at that point of -Vp. That means our tire is rotating at an angular velocity W=Vp/r where r=radius of the tire. Since the plane is also accelerating we know that there must be an angular acceleration of aplha (if you are looking at the attached picture its the symbol that looks like a little fishy). Otherwise there would be no longer a point of 0 velocity and in order to move forward with the new Vp the tire must begin sliding. Now our alpha relates to Va in the same way W did, alpha=Aa/r.
I am going to go into a brief explanation of moments and the moment of inertia so if you know what these are you can skip this paragraph. From experience you should realize that there are two ways to make something move. The first is linearly which means you just move something forward. The second is rotational such as when you close a door. For linear motion the equation for Force is F=massacceleration. The equation for rotational motion looks similar but it deals with moments and moments of inertia. A moment is caused by a force couple and is equal to Fdistance. Going to the door example since you are applying a force but the door does not move linearly there must be a force opposing yours. However, since that occurs at the door hinge there is an unbalanced moment with magnitude M=FD. The equation for rotational acceleration is M=AlphaI. I is what we call the moment of inertia and it corrosponds to mass with one difference. I includes the distance of the mass away from the axis of rotation. Experience should tell you that its harder to spin a long stick than a ball of the same mass.
Ok back to the explanation. Now, since there is an angular acceleration we know that there is a moment. The only place the moment comes from is friction becuase the force from the plane occurs through the center of rotation. We know M=Ialpha=Ffrictionradius. Now, since alpha is non-zero we know that Ffriction is non-zero. If there is an Ag, there is an alpha and therefore a Ffriction. If you look at the plane there are two forces in the x direction, Ffriction and the F from the engines (these two result in a moment but it is counteracted by an unequal distribution of Fweight on the wheels). Now M*Ap=Fengine-Ffriction so when Ffriction is greater Fengine must be larger to get the same acceleration.
Now onto the case, lets talk about the instant where the velocity of the plane is Vp and the accleration is Ap from an observer on the ground and corrospondingly the velocity of the treadmill is Vp and acceleration is Ap in the opposite direction. If you take the situation from the perspective of a guy on the treadmill it appears that the plane is moving forward with a velocity of 2Vp. Since it is going forward at Vp and since there is a point of 0 velocity for the rotating wheel we can determine w (angular velocity). We know that the velocity of the wheel due to rotation at the point of contact with the road is 2Vp to counteract the forward velocity of 2Vp mph. That means w=2Vp/r and alpha= 2Vp/r. Now the same equations apply in the first case so we get Ffrictionr=I*alpha. Since in this case alpha is twice the alpha in the first case Ffriction is twice as large.
Comparing the two cases, MA1=Fengine1-Ffriction1 and MA2=Fengine2-2*Ffriction1, we see that in order to have the same acceleration the second case requires the engine to produce more force in the amount of Ffriction. Now, in these cases the F due to friction is much smaller and can more or less be ignored. It is likely that our plane will be able to take off with no noticable difference between being on a treadmill or off.
Thats false, if the treadmill starts up with no other applied force on the wheel the wheel will not rotate and merely move along with the treadmill. Wheel bearing friction has nothing to do with this situation is you assume the wheel can rotate without slipping. All bearing friction will do is cause loss of energy.
This is an incorrect interpertation. If the friction in the bearings equaled the force from the jets it would only do so at very high rotational speeds. Since the wheels are rotating there must be some forward velocity. Now, it is possible that this forward velocity would not be sufficient to takeoff but that is moving from the theoretical aspect of the question in what would happen in the real world when you include this type of friction. I am 99.9% certain that a plane would take off before this happened.
In both situations the plane must apply some thrust to remain stationary as the treadmill starts up. If there are no forces then the wheel would not spin. Since it is spinning we know there is a friction force point backwards. For the plane to remain stationary there must be a force opposing that.
I think you need to draw a free body diagram (or look at mine) before you start telling people to do so. 
Wheel bearing friction is not the relevent concept in this point. The friction between the wheel and the treadmill. If there is friction there the plane accelerates backwards and the wheel rotates, if not then the wheel does not rotate and the treadmill slides underneath the plane.
The earth is not exactly a giant treadmill in this case. You are moving with the same velocity as the earth’s surface and the plane so they appear to be motionless.
“Gallilean” relativity allows us to treat all non-accelerating frames of reference as equivalent. But saying “regardless of whether the ground is itself accelerating in the opposite direction or remaining stationary” violates this - an accelerating frame of reference isn’t equivalent to a stationary one.
If you can move the surface at will, you can of course create any apparent barrel motion you wish - relative to that surface. But without exerting force on the barrel, you can’t influence its motion with respect to other frames of reference - the earth, for example.
Interesting - that’s sort of what I think you are doing. To be more precise, I believe that for things to work out as you say, the treadmill would have to generate forces that aren’t possible.
I’m saying that one way to understand the problem is to realize that the force the treadmill can exert is small. In the ideal case (zero friction) the force is nil. In the real-world case, it’s still far smaller than the airplane’s thrust.
Can we assume no wind (with respect to the earth) and that our rocketman is facing the direction opposite to the train’s motion? If so, 5mph of motion relative to the train would indeed make him stationary with respect to the earth. But you have stated that his rocket “provides exactly enough force to overcome friction and air resistance and move at a constant 5 mph.” If he’s stationary with respect to the earth, he feels no wind and thus has no air resistance. So his rocket pack, which was sufficient to move him at a constant 5mph taking air resistance into account, is good for 8mph in his current state. He thus moves at 3mph relative to the earth.
As noted, this isn’t compatible with “provides exactly enough force to overcome friction and air resistance and move at a constant 5 mph.” This may be clearer if you consider an extreme case: would the same rocket pack be able to move him at 5mph against a 60mph wind?
I thought about this the whole drive home, and I agree, there is a bit of a paradox in this problem that makes it break down.
First of all, the plane is going to take off. There is a net force on the plane due to its engines, so the plane will accelerate with respect to the ground, and eventually reach take-off speed.
There is a problem, though, as pointed out by Manduck and others. Let’s say we crank up our treadmill to 100 mph before starting the engines. Assuming frictionless wheel bearings, the plane will remain motionless with respect to the ground as its wheels rotate with a linear speed of 100 mph with respect to the axles. Now we start up the engines, creating a net force on the plane. The plane accelerates with respect to the ground, until a moment later it is travelling at 10 mph with respect to the ground. The plane’s wheels are now turning at 110 mph with respect to their axles. The treadmill, as dictated in the OP, speeds up to match this. This leads to a problem.
Free-body diagrams do not lie. If there is a net force acting on the plane due to its engines it will accelerate with respect to the ground. During this accelerated motion, the plane’s wheels will always be turning faster with each passing second. The treadmill will constantly be trying to catch up. This leads to a loop, I think, where the wheels and treadmill speed up infinitely.
God help us all: we’re taking this seriously! :smack:
OK. The engines generate thrust, right? Where are we going to get a counter thrust, assuming frictionless bearings etc? The only thing I can see is if the treadmill is constantly accelerating. This will constantly increase the angular momentum of the wheels which will compensate for the thrust of the engines. This does mean a constant throttle setting requires constant acceleration of the belt so the wheels are spinning like they have jet engines directly attached to their edges.
But I don’t think this is the same as the proposition that the speed of the treadmill is governed by the speed of the wheels, although I couldn’t say why.
You added “friction and air resistance” in to the problem such that the net force on rocketguy is zero. In this case, indeed, he will be motionless with respect to the ground.
What happens, though, when we crank up his rocket pack? Surely rocketguy’s rocket pack, when cranked up a bit more, can create a larger force than that of the retarding air resistance and wheel bearing friction. Now we have a net force on rocketguy, who will also accelerate with respect to the ground.
OK, now I think you’re nitpicking. Let me rephrase: we’ve got a guy with rollerblades and a jet pack on top of a train. He starts at one end, turns it on and accelerates down the length of the train. For the purpose of this example, there is no air resistance. I am asserting that there exists some possible function of acceleration vs. time that can be applied to the train which will keep the rocketman motionless with respect to the surrounding environment. The fact that figuring out and applying said acceleration to the train is unrealistically difficult is not the question, simply whether or not it’s possible to move the train in such a way that it exactly cancels out the rocket man’s acceleration.
If so, how is that different from the airplane problem? If not, why not?
As I just posted, pretend there is no air resistance. Also, let’s say his wheel bearings are frictionless. Is it possible for the train to move in such a way that the rocketman stays stationary with respect to the ground?