Name: Thomas R.
Date: October 2003
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.
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
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
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...
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