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Maths/Physics question 4-Aug-2005 At 6:03:48 PM Ronny

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*assume that the force is applied equally over the entire acceleration period* (that's probably incorrect - but if you don't assume that it doesn't work, and the maximum force could be anything. This assumption will give you the lowest possible maximum for the process)

also assume that the acceleration takes place over 1m of arm movement, and then there's one metre of body movement during the dyno.

Forget for a moment that the arms are part of the body and so it effectively weighs less when you're pulling. (the arms arn't being accelerated in the same manner)

let:
(v1) = final velocity on hold = 0m/s
(u1) = velocity at time that start hold is released = ?
(s1) = distance from letting go to finish hold = 1m
m = mass = 100,000g
(u2) = initial velocity = 0m/s
(s2) = distance body moves while pulling = 1m
a = constant acceleration while pulling = ?
F = constant force on hold while pulling = ?
acceleration due to gravity = -9.8m/s-2 (up is positive)

so:

(v1)*(v1) = (u1)*(u1) + 2a(s1)
0 = (u1)*(u1) + 2*-9.8*1
(u1)= square root of 19.6 = 4.43

and
(u1)*(u1) = (u2)*(u2) + 2a(s2)
19.6=0 + 2*a*1
a = 9.8

(This of course makes sense - same distance, same difference in velocity, uniform acceleration on both - acceleration has to be the same)

so F due to gravity on climber while at rest = ma = 100,000 * -9.8 = -980,000N
force on hold therefore = -980,000 N while at rest.
To accelerate climber upwards at 9.8m/s-2, (+)980,000 N required.
Therefore force required on climber while pulling = force on hold = 1960kn

If the same speed was to be reached over 1m of acceleration without uniform acceleration, at some point the force would have to be greater than this, so what we know is that for a 100kg climber, the minimum force that a hold must be albe to withstand in order for the dyno to take place is 1960kn.

I think this makes sense - but its prob no different to what's already been said, i just didn't understand.
Anyway - don't blame me, i'm a lawyer not a physicist.
J