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vicum


Nov 28, 2001, 12:22 AM
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I was reading the thread about placment of pro over on the trad board and I'm confused about a few things. I've had calc III a little physics so you can speak a little technobabble to me:). Could somebody whose knowledgeble about both climbing and physics answere these questions?

On earth accerleation is a=9.8m/s^2 and we know
f=m*a

your mass always stays the same, like 80 kg.

so the force is a 784newtons, right?

What I don't understand is how you get a force of like 2000 kilonewtons on the system? where does that come from?

The only thing I could think of is it's something to do with momentum.
p=m*a
but that's not a force unit (kg*m/s)
If somebody could explain that to me I'd appriciate it. Not trying to pick a fight but I'd really like to know. Thanks ~Arnold


vicum


Nov 28, 2001, 12:46 AM
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Ok, after thinking and reading my old physics book a bit more...(assuming you wiegh 80kg)

is it that the acceleration used in f=m*a
is the deceleration. So you fall 15m.

x=Vot+.5at^2
t=(30/9.^.5=1.75 sec
v=Vo+9.8*t
v=17.1m/s

so be the end of your fall you going 17 meters per second and you must be slowed to 0m/s.

I came across one equation rearanged with momentum and force which looked like

F=delta P/delta t

intitial P=(17m/s)*80=1371 kgm/s

delta P=- 1371 kgm/s

delta t must be where the rope stretch comes into play. The more it strechtes, the longer it will take to strecth which makes t great and the force less. If you on a static line, t is close to 0 so F is huge.

F=-1371/t.
So that would be the force applied to you at your harness. I think the force at the last peice of pro is the same but I'm not sure. So would that force be what's given on the gear ratings or what? I'm close or totally screwed up:)? Thanks ~Arnold


newbieclimber


Nov 28, 2001, 1:30 AM
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hi vicum,

the force on a free falling body is constant in a vacuum:

f=ma

or

weight=mg with g=9.8m/s*s on earth

the potential energy of a falling body is given by the equation

PE = mgh

which you may recognize. h=height. PE represents a force(mg) applied over a distance(h). since energy must be conserved the distance it takes you to decelerate multiplied by the decelerating force must equal PE from above. but things are complicated because in the real world there is some potential energy that escapes in the form of heat and the rope is dynamic and acts somewhat like a spring storing some of the potential energy from the fall.

sorry i dont know enough physics to explain it better but that is the general idea.

as for the force on your gear it goes like this. at the gear the climber is exerting 175 lbs of force from her weight but the belayer is pulling in the opposite direction with 175 lbs of force so the gear is subjected to twice the force. it actually isnt twice the force because the friction of the rope running through the caribiner clipped to the gear lessens the amount of force the belayer needs to exert to hold the climber. so the gear experiences about 190% of the weight of the climber.

i hope that helps.

[ This Message was edited by: newbieclimber on 2001-11-28 01:59 ]


wigglestick


Nov 28, 2001, 7:57 AM
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Not quite right newbie. For the conservation of energy to work the potential energy you speak of must equal the kinetic energy that your falling carcass possesses.

PE=mgh (you go that one right)
Conservation of energy states that PE=KE
and KE=.5*m*v^2
By cancelling the mass out of both equations:
gh=.5*v^2
Since g=9.8
we can calculate your velocity by entering the height from which you fell say 15 meters
9.8*15*2=v^2
v=17.1 m/s (amazing how vicum and I came up with the same answer by doing it 2 different ways, isn't it)
Now this is in a vacuum and does not account for air drag etc.
So in vacuum if you fell from 7.5 meters above your last piece (15 meters total) you would be going 17.1 m/s when the rope caught you. Now you have to do some work calculations (work=force*distance) to determine what force must be applied to your body by the rope in order to stop you. And you need to through in some spring equations (f=kx) and some other things that I can't remember right now.

[ This Message was edited by: wigglestick on 2001-11-28 08:48 ]


coach


Nov 28, 2001, 8:38 AM
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Vicum,
I'm not into physics or any of that math stuff but I'm confused what you mean by 2000 Kn on the system. Most of our gear is rated to about 22-24 Kn so if you ever took a fall that exerted 2000 Kn you can kiss your ass goodbye. Even if the system held, your body wouldn't. That would be somewhere in the area of 450,000 lbs of force!

Climb On


talons05


Nov 28, 2001, 10:00 AM
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I think he is confused... Probably talking lbs. or Kg. here.

AW


daisuke


Nov 28, 2001, 10:26 AM
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if a force of 2000 KN was applied to you you'd instantly turn into a red and white mist of flesh and bone fragments.
In short, gear is ideally designed to take more force than you could ever put on it in a fall, overkill rules!

D


eclarke98


Nov 28, 2001, 10:50 AM
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These posts have just reminded me how much I have forgotten in the 4 years since I took my last physics class.


vicum


Nov 28, 2001, 11:36 AM
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Yeah, sorry about the 2000kn. I was kinda confused:-) But thanks for the responeses. So the rating is actually the maximum force the peice of pro is able to generate(decelerating the climber)? Thanks ~Arnold


newbieclimber


Nov 28, 2001, 11:52 AM
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hi again vicum,

the rope is going to be decelerating the climber and as it does the rope will exert force on the gear. the rating of a piece of gear is the force up to which it has a 99.9% chance of retaining its integrity. in statistical terms the rated strength is three standard deviations(three sigma) below the mean breaking strength. beware that not all gear has a three sigma rating. some ratings are just the mean breaking strength which means if the gear experiences force as high as the rated strength it has a 50% chance of failing.


[ This Message was edited by: newbieclimber on 2001-11-28 11:59 ]


Partner pianomahnn


Nov 28, 2001, 1:06 PM
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Vicum, your equations are right, but for an instant stop. You're neglecting the fact that rope stretches. That right there will make your equation quite different. You would need to factor in Hooks Constant (I think), into that, and then you could figure out the actual force placed on your piece of protection.

In real world situations, I wouldn't assume ANY fall would result in pro failing. Now, if you were on a wire of some sort, and took a HUGE whipper, then chances are something might break.


coach


Nov 28, 2001, 1:14 PM
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One thing about all the gear used in climbing; if it has not been abused, the rock is good and your system is correct there is no reason to believe that the equipment will fail. It is all made to take loads far in excess of what your body can take. Most failures are due to old or damaged/abused equipment, poor rock or improper system setup.

Climb On


awkward


Nov 28, 2001, 3:57 PM
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I was unable to read all the above posts, but wouldn't impulse come into effect here? Like if you hit something with a hammer it moves, but if you press the hammer with the same force it does not move. (crappy explaination, but the point is there).

-Bryan


nayjay


Dec 5, 2001, 2:43 AM
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All i know is when you let go, YOU FALL


stigonrock


Dec 5, 2001, 12:10 PM
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Wow - You guys.

I haven't picked up a maths/ physics book in years!, ... reading this forum I almost feel like an idiot ... as I can give no input on this.

But hey, what the heck...nice to know there are some knowledgeable people out there.

Angela


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