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When a person walks, the only force acting on him is the force of friction between him and the ground (neglecting air resistance and all). The magnitude of acceleration due to this force is independent of the mass of the object (longer legs have more mass). Hence the person should move with with a velocity independent of the length of his legs.

But I have heard (also observed) that people with longer legs walk faster than ones with shorter legs. If that is true, then why?

One can argue that the torque about the pivot due to friction is more in case of longer legs, But then the torque due to gravity (when one raises his leg to move), which opposes the frictional torque, is also more for longer legs. And why would these torques make a difference anyway, as they have no effect on the acceleration of the center of mass?

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It is a relevant question, as students often don't discriminate between maximum static friction and actual static friction. Unless it is duplicate, leave it as it is. – Pygmalion Apr 17 '12 at 17:03

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Think about the limiting cases. An ant-sized marching band would take a long time to march the length of a football field. The reason they take so long has nothing to with friction - it's just that their legs are smaller and so each stride moves them a shorter distance.

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Yes, but their legs weight less and the torque required to moev them is much less, so why can't they simply move their legs faster? – John Rennie Apr 18 '12 at 8:02
From their perspective their bodies are proportioned exactly as ours, and so the timing of their strides are the same as ours. Naturally this is an extreme example - it is just meant to indicate that someone with shorter legs (say, a child) will walk slower compared to someone with longer legs. They could, of course, hasten the timing of their strides, but that would require a noticeable extra input of effort – kleingordon Apr 18 '12 at 18:38
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Note, in your problem you have static friction, and for static friction

$$F_\text{sf} \le F_\text{sf,max} = \mu_\text{sf} N = \mu_\text{sf} m g.$$

Note "less or equal" in the equation above, which means that actual static friction is less or equal than maximum static friction. So when you are walking, most of the time "less" sign applies, which means that mass of the body isn't relevant for the friction!

(Example: if you are standing still, $F_\text{sf} = 0$, despite the fact that $\mu_\text{sf} m g > 0$.)

People with longer legs walk faster simply because they are making bigger steps. I guess period between two steps for casual walking is some kind of a constant for all people... maybe has something to do with our heart beats.

Dynamic friction is different, there is always "equal" sign:

$$F_\text{df} = F_\text{df,max} = \mu_\text{df} N = \mu_\text{df} m g.$$

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Oh right! I don't know why I neglected that(maximum and actual static friction acting). – Megha Apr 17 '12 at 17:08
No problem. Students of traffic sciences have problem with this difference, and it is a very very important issue for them. – Pygmalion Apr 17 '12 at 17:15
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Interesting question. I had a Google around and came across http://silver.neep.wisc.edu/~lakes/BME315ScalingWalk.html, which seems a reasonable discussion of the mechanics (very simplified). The conclusion is that the walking speed is proportional to the square root of leg length, so taller people do walk faster but the square root dependance means it's not not much faster.

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Hmmm it is nice. Thank You:) – Megha Apr 17 '12 at 17:07
From 12 to 14 I was very athletic, and excelled particularly in the long jump, my style chosen as an example for other students! Then the growth spurt came and I realized I could never compete since other girls grew much longer legs and could start jumping up to 5 meters, whereas I, with my short legs, good style or not, could only reach 4m or so :( . I then concentrated in mathematics and physics :). – anna v Apr 17 '12 at 17:18
But you could outperform them by jumping over pole, leaving your center of pass under pole ;) – Pygmalion Apr 17 '12 at 17:23
The simple pendulum mechanism works well if the person is walking casually but if we consider an athlete running a race, factors like muscle power have to be taken into account. – Megha Apr 18 '12 at 14:52
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I think the simplest model that may be useful here is to treat the legs as simple pendula. In "steady state" comfortable walking, it is reasonable to assume that the legs oscillate close to their natural frequency. That is, the forward contacting leg lifts allowing the rear to swing forward freely over the stride. For a (simple) pendulum with:

$$\omega = \sqrt{\frac{g}{l}}$$ the velocity along the ground will be: $$v \propto l\omega = \sqrt{lg}$$

Note that this result is independent of the mass of the walker and the ground contact forces.

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