# Is it possible to do the stair climber without using any energy?

In gyms I sometimes work out on a stair climber machine. This machine has stairs that move down and you have to climb upward to stay on the machine. I've taken a high school physics class, and I was thinking, since your body stays in the same position, only your legs have to move down, is it possible to cheat the stair climber and do it using little energy? I know this wouldn't be possible for a human, but would it be possible for a machine? Without leaning on the rails.

• Not only is it possible, but people do it all the time. You see them leaning on the handrails. The more weight you put on the handrails, the less energy you spend. Of course in the limit where you put all your weight on the rails, you wouldn't be spending any energy with your legs, since they wouldn't even need to touch the stairs. Nov 11 '19 at 21:09
• without using the rails I mean Nov 11 '19 at 22:11
• I think this comes down to the definition of "do it", as in "do it using little energy". If "do it" means use it in the way it's supposed to be used, then by design it's supposed to be used in a way that requires expenditure of a lot of energy. If on the other hand it allows use in a non-standard way, I suppose you could just stand on one of the foot pedals and not use any energy. But if it entails moving the foot pedals up and down in the presence of friction, that motion will always require the expenditure of some energy. Nov 11 '19 at 22:34
• Yes, I mean use the stair climber the proper way. And I think you misunderstood what a stair climber is. Nov 11 '19 at 23:01
• Pay attention to alephzero's comments on Ben Crowell's answer. If you push on a thing, and if you and the thing are moving away from each other, then you are doing work. If part or all of your weight is on a pedal, and if the pedal is moving away from your center of mass while your weight is on it, then you are doing work. Nov 11 '19 at 23:25

Suppose there are two conveyor belts, A and B. You're standing on A. You notice that B is moving relative to A.

Question 1: Is it possible to let B go by without doing any mechanical work?

Answer: Yes, just let B go by without touching it.

Note that it doesn't matter whether A is really moving or standing still, or whether B is really moving or standard still, or both/neither. We can't say whether things are "really" moving.

Question 2: Is it possible to be in this situation, and make a force on B in the direction of B's motion relative to A, without doing work?

Answer: No. Work equals force times distance.

If you replace "conveyor belt A" with "the rails," then your question is question 2.

• Ok but you don't actually have to exert a force on the stairs, you just have to let your leg move down while you keep your body in the same position. Nov 11 '19 at 23:11
• @dandan That is not correct. You are exerting a force on the stairs equal to your own weight. And because the stairs are moving down, that force is doing work (= force x distance moved). Nov 11 '19 at 23:14
• When you're standing you also exert a force equal to your weight right? When you normally move up stairs you have to exert a force more than your weight. Nov 11 '19 at 23:17
• To stand you don't have to use energy. Nov 11 '19 at 23:19
• You only have to exert a force more than your weight to start moving upwards. Ignoring air resistance, etc, you only need a force equal to your weight to keep moving at constant speed. Nov 11 '19 at 23:19

Assume a stair climbing machine of the sort that resembles a short, pointless escalator. Such a machine is, perforce, getting more or less constant force applied to it. Because the stairs constantly go down, that force is exerted over a distance (unless you cheat and hang on the rails).

So, no.

The sort of "stair climber" that features a pair of steps that are geared so that one rises as the other falls could theoretically work so that no energy is absorbed -- but only if the machine is specifically designed not to absorb energy (i.e., designed without drag, so it always pushes up on the rising tread with the same force as is being exerted on the falling tread).

Again (probably) no.