02GF74
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| posted on 10/6/11 at 06:58 AM |
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Friday's homework - convert momentum to a "force"
Ok, now from my school days momentum = mv (where m is mass and v is velocity) and Force = ma (where m is mas and a is acceleration).
So how do you convert from momentum to force?
Here is the problem.
Let's say I have attached a spring to my foot and jump off a chair. I know poundage of spring, height of chair and my wieght. I want to know
how much the spring is compressed when I hit the ground.
Let's keep it simple by igniring air resitance, friction, length of spring etc.
My acceleration is due to gravity so my velocity will be given by v2 = u2 + 2as (I think that is right, s in this case is height of chair).
So what is the force on on the spring when I land?
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mcerd1
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| posted on 10/6/11 at 07:08 AM |
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you want to search for 'Impulse' calculations 
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jabs
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| posted on 10/6/11 at 07:11 AM |
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Gleened from Yahoo answers, well beyond me
The way to do this is by recognising that delta-P = Impulse
ie. Change in momentum = impulse = Force x time
(actually, integral (force dt), but this reduces to Force x time for a constant force, which we need to assume for simplicity)
When the object hits the ground, it has a momentum P (work this out using m x v, where v is the velocity with which it hits the ground).
Assume that it stops in.. say... 0.1 seconds. Given that all the momentum is gone in 0.1 seconds, we can calculate the force by:
F = P/t = P/0.1
(I don't have a calculator handy and I'm not good with imperial units.. so let's do a rough approximation:
Distance = 0.3 metres, m = 14 kg,
v^2 = 2as = 2 x 10 x 0.3 = 6.
therefore v = sqrt(6)... = 2.4 (OK.. I lied about having a calculator)
(of course, you can also get the speed by solving 1/2mv^2 = Kinetic energy = potential energy... it should, naturally, work out the same)
P = mv = 14 x 2.4 = 34
therefore F = 34/0.1 = 340 kg m/s^2 ... APPROXIMATELY).
[Edited on 10/6/11 by jabs]
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02GF74
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| posted on 10/6/11 at 07:21 AM |
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ok, so not that simple due to not kowing tme to stop.
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BigLee
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| posted on 10/6/11 at 07:28 AM |
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Have you done a full risk assesment for this 'stunt'? 
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britishtrident
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| posted on 10/6/11 at 08:03 AM |
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Several possible approaches I would use the Energy method with the added twist you have to add the compression of spring due to your static
weight.
Remember to work in standard units ie Newtons & Metres
Dynamic part of problem.
Find your kinetic energy at the instant where the spring touches the ground --- which is the same as your potential energy before you jump -----
which is the same as the energy which will be stored in the spring.
As : energy = force * distance and spring rate = force / distance, from basic calculus we can see spring rate is the the gradient of the
force-distance graph and energy is the area under the force-distance graph for the spring. As we know the energy and we know the rate of the
spring the rest is using calculus to get the compression of the spring.
However the compression of the spring due to static weight must be added.
[Edited on 10/6/11 by britishtrident]
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JoelP
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| posted on 10/6/11 at 08:12 AM |
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momentum isnt force. If you had a body of known mass experiencing a known force, ie drag, then from the time to stop you could work out how much
momentum it had and thus its mass.
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hughpinder
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| posted on 10/6/11 at 10:38 AM |
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In addition to britishtridents post - you need to consider time.
You have an undamped spring.
If you ignore air resistance, energy turned to heat in the spring etc, you will bounce up and down on the spring forever. At the lowest point of your
travel you will generate the highest force on the floor, at the top the lowest force, possibly 0 if you leave the ground! (ignoring all energy
absorbed by friction, bending the spring etc, you would bounce back to where you started from).
You would need to estimate how 'elastic' your legs are - if you keep your legs straight and locked, the impulse at maximum compression
will be more than if you allow your knees to bend etc.
iirc, a military parachute landing is the equivalent of jumping off a 10m building. You bend your legs to absorb the shock of landing or you break
your legs, so lets say you stop in 1m approx. plug in the maths and you get about 10g deceleration from that jump. So if you can take up to 10g, the
force on the spring will be 10* your body weight and will compress that much at its maximum point (so if you weigh 10 stone and its a 140lb/inch
spring it'll shorten 10 inches maximum).
So my definative answer is that the force will between 0 and 10 times your body weight depending on the exact time you take the reading at, so the
load on the spring will be between 0 and 10 times your body weight.
To do the calculation, you need to decide either the distace you want to stop in, or the maximum deceleration you want to take and work back from the
speed you will actually reach when jumping from whatever height you start at.
Regards
Hugh
[Edited on 10/6/11 by hughpinder]
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speedstar
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| posted on 10/6/11 at 11:51 AM |
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I agree with the energy conservation method.
When you're stoof on the chair, you have x Joules of potential energy
The energy is changed to kinetic energy as you fall, then back to potential as the spring is compressed.
Assuming ideal conditions, you have no losses to air resistance, heating the spring, etc, you can ignore all the caluclations in the middle - you PE
on the chair = your PE when the spring is at maximum deflection.
play aroudn with your spring theory for a little while (not got calcs to hand and not done it in a few years) but you get a displacement (possibly for
work done...?). Work out the equivalent force to give you that same displacement and bobs
your uncle  ... i think
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blakep82
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| posted on 10/6/11 at 11:54 AM |
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wtf?! why would anyone need to know that? 
high speed camera and a ruler?
[Edited on 10/6/11 by blakep82]
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02GF74
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| posted on 10/6/11 at 12:00 PM |
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quote:
Originally posted by blakep82
wtf?! why would anyone need to know that?
high speed camera and a ruler?
I was keeping it simple, hence the chair, no damping etc. but actual question is from what height do I jump off so that the rear spring gets coil
bound on my full suspension mountain bike.
Conclusion is that it is not so simple.
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cliftyhanger
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| posted on 10/6/11 at 12:41 PM |
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quote:
Originally posted by 02GF74
quote:
Originally posted by blakep82
wtf?! why would anyone need to know that?
high speed camera and a ruler?
I was keeping it simple, hence the chair, no damping etc. but actual question is from what height do I jump off so that the rear spring gets coil
bound on my full suspension mountain bike.
Trial and error approach
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