A.P. French, Mechanics Problem

  • #1
Dorothy Weglend
247
2
I am using A.P. French, Mechanics, to supplement my Physics studies (i.e., on my own). I'm stumbling over problem 6-3 (b): A man of 80 kg jumps down from a window ledge 1.5 m above ground, bending his knees so that his center of gravity descends an additional distance h after his feet touch the ground, what must h be so that the average force exerted on him by the ground is only three times his normal weight?

So, obviously, a=3 m/s^2, since we want F=(80 kg)(3 m/s^2). I use v^2 = 2g(1.5) to get the velocity at the point that the knees start to bend, so v^2 = 29.4 (m^2/s^2).

Since I want a=3 m/s^2, I reason that all I need is to use 0 = v^2 - 2ah, with a=3 and v^2=29.4, to find the stopping distance at that acceleration, which should be h.

This gives me 4.9 m, which seems excessive, and French gives h = 0.75 m, and I can't see how he ended up with that. It seems to me this would correspond to a final velocity of 2.12 m/s, which is way too low for a 1.5 m drop.

Well, thank you for any suggestions or pointers.

Dot
 

Answers and Replies

  • #2
Pengwuino
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... AP french? what's that have to do with physics
 
  • #3
quantumdude
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Pengwuino said:
... AP french? what's that have to do with physics

A. P. French is the author of the book, ding-a-ling! :wink:

Dorothy said:
So, obviously, a=3 m/s^2, since we want F=(80 kg)(3 m/s^2).

You mean a=3g=29.4 m/s2, don't you?
 
  • #4
Pengwuino
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Tom Mattson said:
A. P. French is the author of the book, ding-a-ling! :wink:

:rofl: :rofl: :rofl: :rofl: :rofl:
 
  • #5
Dorothy Weglend
247
2
You mean a=3g=29.4 m/s2, don't you?[/QUOTE]

Thanks Tom. What a gracious way to point out my foolish error.

However, this would give h=0.5 m, not h = 0.75 m. So I am still puzzling...

Thanks again,
Dot
 
  • #6
quantumdude
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I haven't worked out this problem yet, but here is one other thing:

Dorothy Weglend said:
Since I want a=3 m/s^2, I reason that all I need is to use 0 = v^2 - 2ah, with a=3 and v^2=29.4, to find the stopping distance at that acceleration, which should be h.

That would be true if the acceleration were constant, which it almost certainly is not.

I'm not familiar with French, so let me ask you something: Do you at this point have energy methods (work-energy theorem, conservation of energy) at your disposal? How about impulse?
 
  • #7
balakrishnan_v
50
0
The force exerted by the ground is 3mg and the downward gravity is mg
Hence the deceleration is 3g-g=2g

hence we have
v^2=2(2g)(h)

or
h=v^2/(4g)
and v^2=2g.1.5
So
h=2*1.5/4=3/4=0.75 metres or 75 centimetres
 
  • #8
Dorothy Weglend
247
2
balakrishnan_v said:
The force exerted by the ground is 3mg and the downward gravity is mg
Hence the deceleration is 3g-g=2g

Thank you. I find this simple problem very confusing, and I appreciate both of you helping me very much.

Thanks again,
Dot
 

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