Follow your lead

Classical Mechanics Level 4

One technique of rock climbing is called "lead climbing". A lead climber starts out with a rope which is held at one end by a partner. As the climber goes up the wall, their friend lets out more rope, and the climber clips the rope into metal loops that are attached to the wall. In this way, if the climber loses their grip, they will fall a distance equal to twice the length of rope between them and the last loop they clipped into (the last loop is point $A$ in the diagram). The less of a distance the climber falls, the less force they experience on their body, and the better the chance of escaping injury.

Suppose a climber is $h$ m above the last clip $A$ , which is $d$ m above the ground, falls and feels an average force $F_d$ from the rope. How does $F_d$ compare to the force they feel when falling from the same distance $h$ m above their last clip when the last clip is $2d$ m above the ground?

Assumptions

Pretend the rope acts like an elastic solid with Young's modulus $E$
Assume your friend holds their end of the rope with infinite grip

F_{2d} > F_d

F_{2d} < F_d

F_{2d} = F_d

Cannot tell from the information provided

1 solution

Snehal Shekatkar
Jan 11, 2015

I will post the better solution later but the main point is that when distance is $2d$ instead of $d$ , more rope is available for the same stretch and hence less is the force.

Follow your lead

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