# Force, acceleration problems

force, acceleration problems!!!!

## Homework Statement

hi, im new to this and this is my first question, so any help would be great!!im a bit of a physics novice, but i am trying to work out the force exerted by the foot while walking. I know force=weight/area and so from this i can work out force exerted when standing. how do I work out force exerted while walking? the speed the person is walking is 130cm/s, this is constant for 5 minutes. can i use the F=mass x acceleration in this instance if the speed is constant?? i hope theres enough information there if i have left anything out let me know!thanks emily xx

## The Attempt at a Solution

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daniel_i_l
Gold Member
Welcome!
1) pressure = weight/area , not force.
2) If the speed is constant then there's no acceleration. But when a person is walking he stops and starts all the time so he's really accelerating.

hi, thanks for your reply! not sure why i put force sorry, typo error!!so do u think its possible to work out force exerted on the foot while walking? thanks again for your help!

cepheid
Staff Emeritus
Gold Member
Hi Emily,

I know force=weight/area
I'm afraid that's not quite correct. Weight is a type of force. It's pressure that is equal to force/area.

As for the force exerted on the foot while walking, here are my two cents. I don't think that the walking speed is necessarily the most useful information, because it is determined by the length of a stride and the frequency of those strides (how often they occur), but not by how hard the foot comes down on the ground. Furthermore, if the initial velocity of the foot is known and we can assume that the accelaration is constant (it is slowed to a stop by the ground at a constant rate), then the force exerted on the foot by the ground is constant and we can in principle use Newton's 2nd law with this constant acceleration to solve for it.

However, this assumption is unlikely to be valid. Taking a step is a very dynamic process, especially since the foot deforms upon impact, so that it seems more likely that the contact force goes from zero, increases rapidly (and not necessarily linearly) to some maximum value, and then falls off as the foot begins to rise again but is still in contact with the ground. In this case, the rate of change of velocity (i.e. acceleration) is not constant and therefore you need calculus just to do your calculation (assuming you had some way of measuring or modelling this acceleration as a function of time). Speaking of deforming, that brings me to another point: most of the physics you are learning at the introductory level is the mechanics of single particles (that we can consider to have fairly localized mass at a single point in space e.g. a projectile) or rigid bodies (they do have some finite extent in space, but they don't deform). The human body is neither of these idealizations, and the foot is connected to all kinds of other things that exert forces on it. How do we take into account the fact that the leg initially pushes down and then later starts to pull up, and this change occurs not abruptly but all in a fluid and continous way. What mass do we use in Newton's 2nd law? The foot? The entire leg? A larger portion of the body?

I don't mean to be discouraging, I just meant to point out that you have chosen to tackle a very tough problem!

Is this a homework assignment?

nrqed
Homework Helper
Gold Member
hi, thanks for your reply! not sure why i put force sorry, typo error!!so do u think its possible to work out force exerted on the foot while walking? thanks again for your help!
Hi. welcome to the forums.

This is a difficult question and it's not easy to calculate. In practice, one would have to attach so measuring device under the feet of the person to measure that, there are probably sites where this type of measurement is done.

Notice that what you need is how the vertical velocity of the foot changes with time. What you providedis the horizontal velocity of the person, which is roughly constant. But this is unrelated to the force on the feet. You would need to know the vertical velocity as a function of time of the foot and take the derivative of that. Or, if one coudl approcimate the acceleration as constant, what you would need is the vertical speed of the foot just before it reaches the ground and then you would need to knwo the time it takes for the foot to come to rest. Then the acceleration (the average) would be the speed before it reaches the ground divided by the time interval.

Thanks so so much for your help, its part of my final year master project/dissertation!i am enjoying writing it, but i have come up against some impossible problems!! ok so getting a value of force exerted while walking is looking unlikely, so could i just state that the force exerted on the foot while walking is greater than while standing?? Thanks again for your help, greatly appreciated!!

nrqed
Homework Helper
Gold Member
Thanks so so much for your help, its part of my final year master project/dissertation!i am enjoying writing it, but i have come up against some impossible problems!! ok so getting a value of force exerted while walking is looking unlikely, so could i just state that the force exerted on the foot while walking is greater than while standing?? Thanks again for your help, greatly appreciated!!
Well, how important is this t your thesis? Yes, it is clear that the force is greater than while standing. But to be more quantitative would require quite abit of work. It would depend on the type of shoe worn by the person and also on the way the person walks (you know how some people make much more noise while walking barefeet on a hardwood floor than others because they hit hard the surface). So it's very complex. But I am sure there are some sites with relevant data somewhere. Have you tried a google search?

hi, its not an integeral part of my thesis, more of a personal part of the project!!i have found some data on walking pressures, but the pressure seemed a little high!!on the front area of the foot, in 3inch heels the pressure measured was 58PSI, 2999mmHg(the lady weighed 125lbs) and when walkin in flat shoes the pressure was 31PSI. So if she weighs 125lbs and surface area of her foot was 11 square inches, pressure exerted is 11.36 PSI or 587 mmHg, does that seem reasonable? cant believe how helpful this site is!!i have some other queries regarding my project, could i post them aswell? thanks again

nrqed
Homework Helper
Gold Member
Thanks so so much for your help, its part of my final year master project/dissertation!i am enjoying writing it, but i have come up against some impossible problems!! ok so getting a value of force exerted while walking is looking unlikely, so could i just state that the force exerted on the foot while walking is greater than while standing?? Thanks again for your help, greatly appreciated!!
Do you have access to American Journal of Physics? Here's an article that could provide some data (I could look it up when I go back to my office)

no i dont have access and i dont think my university does either!thats so very kind of you to offer, but i dont want to trouble you. youve helped so much already! thanks :)

nrqed
Homework Helper
Gold Member
hi, its not an integeral part of my thesis, more of a personal part of the project!!i have found some data on walking pressures, but the pressure seemed a little high!!on the front area of the foot, in 3inch heels the pressure measured was 58PSI, 2999mmHg(the lady weighed 125lbs) and when walkin in flat shoes the pressure was 31PSI. So if she weighs 125lbs and surface area of her foot was 11 square inches, pressure exerted is 11.36 PSI or 587 mmHg, does that seem reasonable? cant believe how helpful this site is!!i have some other queries regarding my project, could i post them aswell? thanks again
I am not surprised that the pressure would go so high while walking. There is a fairly large acceleration when the foot is in contact with the floor (the change of speedis not large but it occurs in a very small time interval, which leads to a large impact force). So those numbers seem very reasonable to me !

When you calculated 11.36 PSI, did you divide by two to take into account the fact that the force is spread over both feet? I am assuming that the 11 square inches is for one foot alone...)

Of course you may ask as many questions as you want!!! There are people very very knowledgeable in all fields of physics here so some of them may be much more helpful than I am.

And you may want to post your next questions in the "General Physics" forum because the present forum is usually reserved to assignment problems.

Patrick