eklhad@ihnet.UUCP (K. A. Dahlke) (06/04/85)
< I seem to get no where > Suppose we have two men, X and Y. X is climbing a long flight of stairs, while Y is climbing a down escalator. X has already begun his ascent, and is no longer accelerating. Y also has a constant velocity, namely 0 (relative to the building). As Y climbs up, the escalator moves down. The question, who is doing more work? I did not submit this question to net.physics, because I already know the physics answer. X is traversing a distance, against a force, line integrals F.dS etc etc, and so X is doing work while Y is lazy. However, X and Y seem to be making the same movements, against the same force. Who is burning more calories? Who is working harder biologically? If X is working harder, what is he doing differently? It seems like, somewhere in each step, X must pay a price. If the effort expended is comparable, better move to the stairs, at least you get somewhere that way. Any help appreciated. -- Karl Dahlke ihnp4!ihnet!eklhad
hes@ecsvax.UUCP (Henry Schaffer) (06/05/85)
They are both moving their legs in the same manner, but the forces involved are not the same (assuming they both are "climbing" at a steady pace.) X, the stair climber, has to lift his body up each step. Y, on the escalator, just lets his leg, which took a step up, ride down with the tread. Y has to transfer weight between his two legs, but doesn't have to use his leg muscles to lift his body. (Y could spend more energy by quickly taking a step up, then riding one step worth down, take another quick step up, ... This should be approximately equal to the stair climber.) --henry schaffer n c state univ
jpexg@mit-hermes.ARPA (John Purbrick) (06/06/85)
> Is it harder work climbing stairs or going up a down escalator? > Karl Dahlke ihnp4!ihnet!eklhad This belongs in net.puzzles. Imagine a sealed box put over the person on the escalator. It would be impossible to tell if one was climbing or not, only that the stairs were always moving under one's feet. Hence, it's intuitive that the amount of work done is the same. But where does the energy go? It goes to backdrive the escalator's motor--in other words, it will draw less power (or even, if it's designed by a physicist, return power!) from the electric supply. Or imagine the escalator replaced by a smooth, moving belt. To stop oneself slipping down, one would have to exert a force up the belt, and one would have to expend that force through a distance in order to hold one's position. Mechanical work can be represented as (force * distance_through_which_force_is_ moved). QED. John Purbrick jpexg@mit-hermes.ARPA {...decvax!genrad! ...allegra!mit-vax!} mit-eddie!mit-hermes!jpexg
alexis@reed.UUCP (Alexis Dimitriadis) (06/08/85)
In article <237@ihnet.UUCP> eklhad@ihnet.UUCP (K. A. Dahlke) writes: > Suppose we have two men, X and Y. X is climbing a long flight > of stairs, while Y is climbing a down escalator. > X has already begun his ascent, and is no longer accelerating. > Y also has a constant velocity, namely 0 (relative to the building). > As Y climbs up, the escalator moves down. > The question, who is doing more work? > I did not submit this question to net.physics, because I already > know the physics answer. X is traversing a distance, against a force, > line integrals F.dS etc etc, and so X is doing work while Y is lazy. > However, X and Y seem to be making the same movements, against the same force. Actually, physics does have the answer: If you use, in each case, the steps as a reference frame, you find that both men are doing the same amount of work. To use a "stationary" reference frame in the case of the escalator is misleading, since an immobile man in the escalator is losing potential energy in a stationary reference frame, but not in the reference frame of the escalator. Bot men are doing the same amount of work, it is the work done by the escalator that causes the difference. Alexis Dimitriadis -- _______________________________________________ As soon as I get a full time job, the opinions expressed above will attach themselves to my employer, who will never be rid of them again. alexis @ reed ...teneron! \ ...seismo!ihnp4! - tektronix! - reed.UUCP ...decvax! /
buck@shell.UUCP (06/11/85)
I have never read so many consistently incorrect replys on this network!
If all you armchair physicists would go out and actually test out your
theories, you would *very* quickly discover that it takes much, much
more effort to go up the down escalator than to go up stairs.
I say this as someone who has run up down escalators on numerous
occasions, and as a professional physicist.
Now, why is the escalator so much more work?
Power expended is Force<dot>Velocity. Integrate to get kinetic
energy , .5mv**2 . So, for example, in accelerating something
from rest to v=2, only one quarter of the energy is expending
from v=0 to v=1, and three quarters is spent from v=1 to v=2.
Well, then it ought to get easier to accelerate things as they
speed up because the velocity is going up in Force<dot>Velocity, right?
Wrong - it just practically becomes *much* more difficult to keep
the same force on something that is moving away from you.
Exactly the same thing is occuring with the foot trying to apply
a force to the escalator. Just do a simple thought experiment -
speed up the escalator enough and the man can make no net progress,
he just can't move his leg (impulsively) down fast enough to get
any reaction force from the escalator, so he can't make any net progress.
It is left as an excercise for the student to fill in the gaps
in the above and make a rigorous proof, or just step over to your
nearest stair/escalator combination...
A. Lester Buck @ Shell Development Co.
{ihnp4, pur-ee, ut-sally}!shell!buck