Question #38469

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Question #38469

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Answer 1 · 2017-05-30T01:50:59

The amount of surface area in contact with the ground is far less when the man is walking than when he is standing still.

Explanation:

If we calculate pressure as the the force exerted per unit area

$a a a a a a a a a a a a a a Pressure = \frac{Force}{Area}$ $color(white)(aaaaaaaaaaaaaa)"Pressure"= ("Force")/("Area")$

and compare the pressure when walking (only the heel of 1 foot in contact, in this instance) vs. when standing still (both feet are in contact), taking the average weight of the man as $720 N$ $720" N"$ , then

$- - {Heel SA: 0.0064 m}^{2} and {SA of both feet: 0.0258m}^{2}$ $color(white)(--)"Heel SA: 0.0064 m"^2 and "SA of both feet: 0.0258m"^2$

(source: http://hypertextbook.com/facts/2003/JackGreen.shtml)

$a a$ $color(white)(aa)$

$when walking \Rightarrow P = \frac{720 N}{{0.0064 m}^{2}} = 112500 \frac{N}{m^{2}}$ $color(blue)"when walking"rArr"P"= ("720 N")/("0.0064 m"^2)=112500 N/m^2$

$- - vs$ $color(white)(--)"vs"$

$when standing still \Rightarrow P = \frac{720 N}{{0.0258 m}^{2}} = 27907 \frac{N}{m^{2}}$ $color(red)"when standing still"rArr"P"= ("720 N")/("0.0258 m"^2) = 27907 N/m^2$

we clearly see that the Pressure is $4 x$ $"4 x"$ greater when the person walks as opposed to when he stands still on both legs.

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Question #38469

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