Plane on a Conveyor Belt Interview Question
I saw this physics interview question pop up on the internet, and thought it might be worth discussing:
A plane is standing on a runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction). Can the plane take off?
There are a number of forces here that apply: gravity, the force of the engines / airspeed / lift, and the drag of the wheels against the conveyor / the speed of conveyor. However, the wheels provide essentially a frictionless boundary between plane and ground; unlike car wheels, the wheels of an airplane spin freely in place. So, as the conveyor belt speeds up, the airplane stays in place, but its wheels spin at the same velocity. Furthermore, the lift of the airplane is relative to its airspeed, and its engines push against air. So, the airplane will accelerate forward and take off as normal.
The commenter “Max” has a nice summary as well: “A comparable example in my mind would be a car on a treadmill. If the car is being pulled along by a winch and the wheels are turning freely then the car is going to be pulled at an identical rate whether or not the treadmill is there or not (assuming as you did that the treadmill’s speed is inverse to that of the car).”
This entry was posted on Saturday, February 9th, 2008 at 10:15 pm and is tagged with conveyor belt, car wheels, plane speed, question pop, interview question, airspeed, winch, treadmill, airplane, velocity, gravity, tunes, physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback.
13 Responses to 'Plane on a Conveyor Belt Interview Question'
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on February 9th, 2008 at 10:27 pm
The issue would be if the wheel bearings could handle the increased speeds. If the plane hits the ground before reaching lifting off speeds, then I guess flight wouldn’t happen. Neat to think about though!
on February 10th, 2008 at 1:34 am
Isn’t the lift of the plane generated by the uneven flow of air over the wings? So, if the plane is standing in place, it’s not going to be able to take off.
Now, I suppose the question is whether the plane is standing in place. After all, the forward momentum isn’t coming from the wheels and friction (like in a car), it’s coming from an engine’s thrust against air. Which means that the plane actually is going to move forward, and the conveyor belt is just going to cause the wheels to spin faster (without actually detracting air flow over the wings).
So, I suppose, yes, the plane will take off.
I started off disagreeing with the post and got to agreeing with it. That should count for something in an interview, right?
on February 11th, 2008 at 1:44 am
Mythbusters did this one. Basically the plane takes off because the propeller pulls it forward which allows for the airflow. The wheels on the plane do not propel it in any way and that is what confuses people (so they say).
I’m sure you could find a youtube video if you wanted to.
on February 15th, 2008 at 7:44 pm
An airplane with mass M sits on a conveyor belt. The airplane’s jet engine is turned OFF. The conveyor belt starts moving backward at a constant velocity V_conveyor. (note that all velocities will be quoted relative to the ground referential)
Which direction does the plane go? Backward, with velocity V_conveyor.
Intuitively it is reasonable to expect that the plane will have to overcome this backward velocity before taking off.
At time T=0, the jet engine is turned on, generating a thrust, F by pressing against (we all agree) the atmosphere. As a result, the plane experiences a forward acceleration A = F/M. Note that the conveyor belt provides no acceleration, just constant velocity. [As an aside, also note that acceleration is independent of any fixed frame of reference].
We are allowed to decompose the velocity of the plane relative to ground (V_total) into the velocity component due to the conveyor belt (V_conveyor) and the velocity resulting from thrust acceleration (V_plane):
V_total = V_plane - V_conveyor
As velocity is the integral of acceleration, at the instant in time T=0 its (forward) velocity component due to the thrust force is still V_plane = 0: the plane is still moving backward relative to earth with velocity V_conveyor - V_plane = V_conveyor.
Time, T, continues to pass. Thrust F guarantees constant acceleration A=F/M, and the equation for the velocity of the plane due to thrust is V_plane=TA.
Thus V_total = TA - V_conveyor. The plane’s ground speed V_total will be zero only when
T_null = V_conveyor/A
Until this time the plane will continue to move backward. As more time passes, the plane’s forward velocity component due to thrust will continue to increase linearly with time, whereas the backward velocity component due to the conveyor belt motion will remain constant, and so the plane will progressively pick up forward velocity (V_total = TA - V_conveyor > 0 ), develop lift and take off.
Thus if the conveyor belt has constant velocity the plane will, of course, teventually take off. If the conveyor belt’s velocity increases linearly with time at a rate A, then the plane will not develop forward velocity and will not take off.
on February 15th, 2008 at 10:26 pm
you go off track at “where does the plane go”–it actually does not move!
on February 17th, 2008 at 6:15 am
The plane does take off. The thing that confuses people, including me at first, is that the movement of the conveyor does not keep the plane stationary, it just spins the wheels twice as fast as normal. Because the wheels are spinning faster the wheel mechanism will generate more friction than normal, but this would not be enough to prevent the plane from taking off.
So basically, the plane needs its normal take off thrust plus some additional thrust to over come wheel mechanism friction in order to take off. If you assume the system is frictionless then only normal thrust is needed.
on February 19th, 2008 at 10:38 am
The plane rises because of airlift, not wheel speed silly.
So the plane will not rise at all.
on February 19th, 2008 at 10:46 am
The plane rises because of airlift, not wheel speed silly.
So the plane will not rise at all.
Whats the matter with you people? Do you think planes take off cause their WHEELS spin? No, they rise because the difference in the airflow over the wing, from under the wing. Thats all that matters, not the wheels.
on February 20th, 2008 at 1:01 pm
it’s not gonna take off..
there is no lift generated because of the groundspeed / airspeed.
so the airplane wheels will only spin very hard, but for the rest, nothing happens
on February 29th, 2008 at 3:04 pm
Sweet jesus, is science education that bad in the US? Of course the plane doesn’t life off.
on March 2nd, 2008 at 2:48 pm
It lifts off; here’s a hint–the conveyor belt, no matter how fast it moves, imparts no “backwards” force on the plane. Why? Because it just makes the wheels spins backwards really fast…
on April 14th, 2008 at 11:37 am
How do planes go from take off speed to cruising speed?
After lift off they are not in contact with the runway.
The answer to that is also how they go from standing still to take off speed.
And it is independant of the runway.
Therefore the runway only holds the plane up till it takes off and the conveyor makes no difference.
on April 19th, 2008 at 5:59 pm
I like what Stephen just wrote. I also think a key point was made by Elliott Back who opines that the conveyor belt imparts no retardung force on the plane–actually maybe a wee bit of force but almost none–and thus the operation of the belt will not prevent liftoff. One thing is certain: it is important that the riddle be correctly stated: The experiment does NOT involve an apparatus that keeps the ground speed of the plane at zero. The experiment involves an apparatus that matches belt speed to wheel speed.
As I am sure everyone here knows, this subject has about 40 entries on the Discovery Channel Mythbusters wiki. Lots of confusing comments all around in that mix. Some funny, some sad.
This is an interesting problem for students in physics to contemplate.
Reminds me of the Monte Hall Problem in some ways….