



brunoschull
Mar 3, 2011, 11:28 AM
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Registered: Nov 10, 2004
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Hi, I've been playing around with force calculations a bit, using the equation from rgold, and so on. Interesting stuff. Before I began to calculate forces with climbers and belayers, dynamic ropes, and so on, I wanted to have some baseline numbers from a simple model, to put everything else into perspective. For example, I wanted to calculate the force generated by a falling object. Let's say, a 100 kg mass falling 5 meters. Not a climbing situation really, just a falling mass calculation. I think I worked my way through it, following information on different web sites, but I'm no expert. Can anybody check my numbers? To calculate the force of a falling object Mass = 100 kg Height = 5kg 1Find the velocity Energy before = Energy after Energy before = gravitational potential energy Energy after = kinetic energy PE = mgh KE = 1/2mv^2 mgh = 1/2mv^2 V = (2gh)/2 g = 9.8 m/s^2 h = 5 m v = (2 X 9.8 X 5)/2 = 9.8 m/s 2Find the KE KE = ½mv^2 KE = (½ X 100 X (9.8)2 = 4802 J 3Find the force From work energy principle Average impact force X distance traveled = Change in Kinetic energy Average impact force = Change in kinetic energy/distance traveled Object comes to rest Change in kinetic energy =  KE Force = KE/distance traveled on impact before stopping Distance traveled on impact before stopping = 0.1 m Force = 4802/0.1 = 48,020 newtons Force = 48 kN Does this look right? Thanks!





shockabuku
Mar 3, 2011, 11:56 AM
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Registered: May 20, 2006
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You can skip a step and calculate the KE directly as mgh, then divide by the stopping length to get the average force. Of course it doesn't tell you much  it's probably like dropping a weight on a steel cable. The impact force will be very high, and will necessarily underestimate the peak force.





hafilax
Mar 3, 2011, 12:04 PM
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Registered: Dec 11, 2007
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You've got an extra step in there and you have some rounding error as well. You don't need to calculate the velocity from the kinetic energy and then go back to kinetic energy again. You can see that mgh=4900J. For the average force approximation from the work you would get 49 kN but there you are ignoring the extra potential energy lost from dropping the last 10 cm. This probably isn't a big deal for a small rope stretch but becomes more important as the stretch increases. If the person actually falls 5.1 m then h=5.1 m and the average force would be closer to 50 kN. People are usually more interested in the peak impact force rather than an average impact force. If you want to take a conservation of energy approach with the rope modeled as an undamped spring then you can do it in 1 calculation. The energy starting the fall is equal to the energy at the bottom of the fall. Initially all of the energy is gravitational potential energy since your person starts with zero velocity and the spring is unstretched E0=mgh You have to define the zero of your coordinate system at this point. If you set the gravitational energy to be zero at the bottom of the motion with the spring fully stretched then at the bottom the person will have zero velocity again and all of the energy will be stored in spring potential energy. With the 10 cm of stretch and a fall distance of 5 m before the rope starts to act you will have that h=5.1 m and mgh=1/2k(0.1)^2 For the person to stop in 10 cm the spring constant would have to be k=2mgh/(0.1)^2 and the peak force would be where the spring is maximally stretched F=k(0.1)=2mgh/(0.1)=100 kN This isn't as useful as rgold's approach because the spring constant is solved for depending on the fall length and not the other way around.





brunoschull
Mar 3, 2011, 12:35 PM
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Registered: Nov 10, 2004
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Thanks for the replies, I was sure that I was making some basic mistakes, and that there were more elegant and accurate ways to perform the calculation. Before I get to my follow up question, the practical lesson I draw from this, is, wow! The difference between static fall (can we call it that?) and an actual climbing fall with a dynamic rope, belay, and so on, is enormous! Of course, I knew that the force would be high, but I am still surprised by the number...obviously much higher than the ratings for much climbing gear...food for thought. In any case, now I am intrigued, and I have a follow up question: As you suggest, peak force would be more informative for my simple comparisons than average force... Would anybody mind explaining why these two things, peak force and and average force, are different? Is it that the forces do not happen at a single instant, but are instead averaged over a small time interval? Or it it that, even in a rigid system, forces are averaged over a whole mass? Basically, what is averaged? Many thanks, and please forgive my ignorance (I'm a biologist not a physicist!) B





hafilax
Mar 3, 2011, 12:54 PM
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The first approximation to the behaviour of a rope is to treat it like a spring. If you put a mass M on a length of rope it will stretch y, 2M it will stretch 2y, etc. For this model, as a rope stretches the force exerted is not constant but increases. If you treat the force exerted by the rope as constant for all amounts of stretch as is the case with your first model, you will overestimate the force for the small part of the stretch and underestimate it for the large part of the stretch. Putting it another way, your rope has a constant restoring force no matter how far you pull it. It's behaving more like friction than a spring.





dynosore
Mar 3, 2011, 1:20 PM
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Registered: Jul 29, 2004
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brunoschull wrote: Of course, I knew that the force would be high, but I am still surprised by the number...obviously much higher than the ratings for much climbing gear...food for thought. In any case, now I am intrigued, and I have a follow up question: As you suggest, peak force would be more informative for my simple comparisons than average force... Would anybody mind explaining why these two things, peak force and and average force, are different? Is it that the forces do not happen at a single instant, but are instead averaged over a small time interval? Or it it that, even in a rigid system, forces are averaged over a whole mass? Basically, what is averaged? Many thanks, and please forgive my ignorance (I'm a biologist not a physicist!) B Think of a bell curve. You're plotting force v. time. Is the peak the same as the average, across the width of the curve? Of course not. Average is not that useful for climbing. Peak is what matters. If your gear holds 13kn, which theoretical fall would you rather take: a fall with a peak of 10kn and an average of 8kn, or a fall with an average of only 4kn but a peak force of 16?








