Lifeline Design ```Name: Thomas R. Status: Educator Age: 60s Location: N/A Country: N/A Date: October 2003 ``` Question: If an object attached to a 6 foot life line weighing 200 pounds falls 6 feet what would be the weight felt by the life line at the end of the fall be? Please show the formula so I can use it in my fall protection class. Replies: The force depends entirely on how much the life line stretches. If it did not stretch at all so the object stopped instantaneously, its acceleration would be infinite and so the force would be infinite. Remember F = ma (Force = mass times acceleration). If the rope exerted a constant force while it was stretching, it would be easy to calculate the (constant) force exerted on the body. The result is: F = mg(h+d)/d + mg Here mg is the weight of the object (say 200 lb), h is the distance it falls before the life line starts slowing its fall (say 6 ft) and d is the distance it falls while the life line is slowing its fall (say 1 ft). The answer is then eight (8) times its weight, or 1600 lb. More likely, however, the rope stretches like an ideal spring with the retarding force increasing like F = kx, where x is the amount the rope has stretched. The answer is then: F = 2mg(h+d)/d +mg With the same numbers, the force is 15 times the weight of the person, or 3,000 lb. I hope this is clear and answers your question fully. If you would like to see the derivation of these equations, please write again and I will be happy to provide the derivation, which is quite simple. Best, Dick Plano... Click here to return to the Engineering Archives

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