When are stairs NOT an inclined plane?

AI Thread Summary
The discussion revolves around the mechanical advantage (MA) of a staircase, calculated as the length divided by the height, yielding an MA of 2.2 for a staircase that is 10 meters long and 5 meters high. Participants question whether stairs truly function as inclined planes, noting that the configuration of steps and the mechanics of climbing may not provide a reduction in the force required to lift one's body. They highlight that while the formula suggests a mechanical advantage, the actual experience of climbing stairs involves bending and straightening legs, which affects the center of mass movement. The conversation emphasizes the complexity of how force is applied while climbing stairs compared to a traditional inclined plane. Ultimately, the participants seek clarity on the relationship between the formula and the physical experience of using stairs.
Student800
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Homework Statement


What is the mechanical advantage of a staircase that is 10 meters long (horizontally) and 5 meters high (vertically)?


Homework Equations



MA=Length/Height


The Attempt at a Solution



MA=2.2

I'm certain that this is the answer my teacher is looking for, but it does not make sense to me. It seems to me that stairs are not always the same as an inclined plane, but that this may depend on the configuration of the steps and how they are climbed.

For example, what if the stairs have 2 equally spaced steps (each 2.5, meters high) or 4 equally spaced 1.25 meter high steps. I don't see how those configurations provide any mechanical advantage -- I have to climb each high step and then walk over the flat distance between steps. There is no reduction in the force I have to use to life my body.

In the more typical configuration, if I move my body horizontally over the flat part of the steps and vertically at the edge of the steps, how do the stairs provide any mechanical advantage? Again, there is no reduction in the force I have to use to lift my body. Furthermore, even if I were to move my body at a constant slope upward equal to the average slope of the stairs (as if I were a block sliding on a flat inclined plane), isn't it really me that preserves that slope? In other words, couldn't I climb a vertical ladder and essentially zigzag body back and forth as I climb so that my body center of mass is always moving up at any given slope? By my teacher's formula, the ladder has an MA=1.

I can only imagine that I'm missing something important here because my teacher said he's been using this same problem set since before I was born. Can someone please help me figure out where my reasoning has gone wrong?
 
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Student800 said:
In the more typical configuration, if I move my body horizontally over the flat part of the steps and vertically at the edge of the steps, how do the stairs provide any mechanical advantage?
But that is not how one actually walks up steps. The legs bend and straighten in such a way that the mass centre moves in a reasonably straight line.
 
Thank you for your reply!

Yes, I see how walking up certain stairs in a certain way may provide for your center of mass to make a smooth path up the stairs. On a ladder I can make my center of mass approximate any path you might like despite what the formula might give for the mechanical advantage of a ladder.

I guess I was just stuck on:
1) the how important it is for the stair-walker to use their legs to make their center of mass track the incline path in order for the formula to work, and
2) how the force in the direction of the incline is only a component of the force your feet have to apply to move up the steps while for a traditional inclined plane analysis, the force in the direction of the incline is the only force applied to the weight.

Thanks again!
 
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