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adrenaline_smack
Feb 4, 2006, 5:34 PM
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ok heres a question. physics dictates for equilibrium to occur, all forces must cancel, however on a top rope setup, a 70kg climber can be belayed by a 50kg belayer. that is to say, the 70kg climber can be hanging in dead air caused by the normal force exerted by a 50kg belayer. this is not possible, 50kg*9.8m/s^2 does not equal 70kg*9.8m/s^2. but since obviously it does happen, my guess is that the coefficient of static friction (dynamic as well) on the caribeaners that the rope feeds through the anchor is much greater than i thought, its all i can think of to account for the difference of 196N. my knowledge of physics and statics/ dynamics is still in the "learning" phase. i suppose because friction is directly proportional to the force normal, the greater the magnitude force exerted by the rope on the beaners ( by the hanging climber), the greater the force of friction. my question to those who have a DEGREE in the related field or actually know, please tell me what accounts for that 196N of unresolved force to reach static equilibrium
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altelis
Feb 4, 2006, 5:39 PM
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beaners have have as much friction as you initially supposed. biners on the other hand...... :lol: (can't help but use the "jerk off" emoticon....) sorry, end of obnoxious reply, and end of post because i really can't answer your question...
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dudemanbu
Feb 4, 2006, 5:44 PM
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1. friction at the carabiners is greater than you think. 2. In a dynamic rope energy is disspiated via heat as it stretches. I think your estimates of a 50kg person belaying a 70kg person is slightly flawed. my partner and i weight just about that and i always get sucked up when he falls.
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daithi
Feb 4, 2006, 5:47 PM
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You're right. Friction accounts for the other force to obtain equilibrium. A generally accepted rule of thumb for the effect of friction on a carabiner is that the average tension ratio is 52% with a rope making a 180° bend, according to R&I Jul-Aug '95. So according to that a 70 kg climber should be able to be statically held by a 37 kg belayer! You wouldn't want to fall though! :D
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adrenaline_smack
Feb 4, 2006, 6:17 PM
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so heres another question whats the function for the stretch of dynamic rope with respect to force? obviously a 4kN force will stretch a dynamic rope more than a .5kN force would. ah and a question for people with a biomedical engr or kiniseisology background, what magnitude of force is required for injuries such as whiplash, broken bones from impacts of the leg ( a situation might be decking onto slab), slamming into a wall (jaming/ breaking bones in the wrist by trying to brace yourself with your arms before you hit), etc... i have zero knowledge in this field, so if you could offer ranges of force with a scenario where such injuries could occur would be helpful.
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curtis_g
Feb 4, 2006, 6:32 PM
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for your second question...that depends on the spicific rope. there is a k value so that F = xk (where F is force and x = ditance of the stretch) that is spicific to the 'spring' called the coefficient of elasticity. It is a constant (assuming a perfect spring). I havn't asked enough questions about a rope manufacturer's tests or units or how they measure their rated dynamic/static elasticity attributes that are measured and listed with the UIAA fall rating specs and other specs on a climbing rope but I'd think that you could figure out a rope's coefficient of elasticity from those specs.
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daithi
Feb 4, 2006, 6:33 PM
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In reply to: so heres another question whats the function for the stretch of dynamic rope with respect to force? A more useful quantity is to relate it to fall factor (f). The percentage increase is delta L / (100 * L) = 1/k [1 + sqrt(1 + 2*k*f)], where k is E*A/(m*g) (Young's modulus, cross sectional area, mass and acceleration due to gravity) and f is the fall factor, which is d/L (where d is the falling distance and L is the length of rope out). It is derived from energy conservation and is quite easy to do. You should give it a go. I'll start you off! The elastic energy is defined as 0.5 * EA/L * (delta x)^2.
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greenketch
Feb 4, 2006, 6:36 PM
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It's my understanding that a way old US Air Force study found that human bones break at close to 12 Kn of force. This was a major reason that modern dynamic climbing ropes moderate impact forces to less thatn 8Kn.
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scottquig
Feb 4, 2006, 7:15 PM
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In reply to: for your second question...that depends on the spicific rope. there is a k value so that F = xk (where F is force and x = ditance of the stretch) that is spicific to the 'spring' called the coefficient of elasticity. It is a constant (assuming a perfect spring). I havn't asked enough questions about a rope manufacturer's tests or units or how they measure their rated dynamic/static elasticity attributes that are measured and listed with the UIAA fall rating specs and other specs on a climbing rope but I'd think that you could figure out a rope's coefficient of elasticity from those specs. This would be true of a linear spring. However, I don't think a climbing rope falls under that catecory, though. You could probably model a climbing rope as a linear spring for small displacements (i.e. 1% or less). Climbing ropes, though, often experience elongation of 10 to 30%. I feel like the effects of damping are very significant also.
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112
Feb 4, 2006, 8:50 PM
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You might also want to look at the friction caused by a flat belt over a drum. Equation: T2 = T1*exp(mu*beta) where T1 = Tension felt by belayer T2 = Tension felt by Climber mu = coeffitient of friction (static or dynamic) between rope and carribeaner beta = angle of contact between rope and beaner (approximatley 180 degrees, or PI radians) - angle MUST be in RADIANS Proof is here
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c22
Feb 6, 2006, 11:32 PM
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double post :oops:
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c22
Feb 6, 2006, 11:36 PM
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it doesn't matter if the rope stretches or not (at least for static equillibrium), as once the rope stretches to it's full length you would still have (in given example) 686N on one side and 490 on the other. The only remaining force is friction which must have a coefficient of .4. u=f/N, where f is the force of friction (necissarily 196 newtons), N is normal force (remember this is the weight of the hanging climber, as the weight of the first is counteracted by the normal force from the ground) and u is the coefficient of friction. *as a side note, if the lighter climber is lifted of the ground the normal force on the caribiner is increased as it begins to hold the belayer's weight as well.
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c22
Feb 6, 2006, 11:42 PM
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after a little more thought, the problem is much harder than this, though the easy answer is "static friction of the biner" the why is harder, as you have a variable force exerted by the smaller climber.
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brent_e
Feb 7, 2006, 12:12 AM
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In reply to: 1. friction at the carabiners is greater than you think. 2. In a dynamic rope energy is disspiated via heat as it stretches. yeah, have you ever felt one after a fall??? she's hot! oh, and the falling climber loses some energy from that huge scream he expells during the fall. :D Brent
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brent_e
Feb 7, 2006, 12:17 AM
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In reply to: it doesn't matter if the rope stretches or not (at least for static equillibrium), as once the rope stretches to it's full length you would still have (in given example) 686N on one side and 490 on the other. The only remaining force is friction which must have a coefficient of .4. u= f/N, where f is the force of friction (necissarily 196 newtons), N is normal force (remember this is the weight of the hanging climber, as the weight of the first is counteracted by the normal force from the ground) and u is the coefficient of friction. *as a side note, if the lighter climber is lifted of the ground the normal force on the caribiner is increased as it begins to hold the belayer's weight as well. pretty sure this is wrong. the normal force would be the one experienced by the carabiner at the top. this would not be equal to the climber, as the biner experiences force from both climbers. So, you'd have to add the forces from each part of the rope to get the biner to get the normal, thus your coefficient o friction is less and this occurs even before the belayer gets lifted, or else the climber would just keep falling, right? and the force isn't static. none of these forces, AFAIK, are static. it's kinetic friction and kinetic forces. Everything here is moving. Brent
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