1. The problem statement, all variables and given/known data So after determining the way an anchor setup will handle the high tension forces created when at different angles, here is the next step in the question. attached is a rough diagram that can be refered to for some clarity (hopefully). So assuming we have a 68kg climber going up a wall being belayed (belayer keeps the slack on the rope to a minimum as the climber ascends the wall, thus reduceing his fall distance) by a 70 kg person (not sure if this matters right now). If the climber were to slip off and fall a distance of 3m before the belayer arrests the fall ... at which point the belay device kicks in and stops the fall in time of roughly 1 second.. I get something like this Vf2 = Vi2 + 2(a)([tex]\Delta[/tex]y) giving a final velocity of 7.67 m/s then I believe I can use an impulse formula here for when the belay device arrests the fall... Fnet = [tex]\Delta[/tex]P / [tex]\Delta[/tex]t lets say there is a spring like effect and the climbers final velocity is then upward at 0.32 m/s after impact. so.. Fnet = ( (68Kg)(-0.32m/s) - (68Kg)(7.67m/s) ) / 1 second = 540 Newtons? I'm thinking there must be more than this because that is less force than the weight alone of the person hanging on the anchor... the anchor would experience a force = to the weight of the person PLUS whatever force due to acceleration from falling ? in a dynamic belay situation the person belaying will actually jump up right before the falling climber tensions the rope completely in effect increaseing the amount of time it takes to bring the falling climber to a complete stop thus reducing the impact... just curious how one would go about calculating these forces on the anchor etc.. it may be more advanced but not quite sure.. in the situation where the belayer is also lifted off the ground then the anchor must support the weight of both the belayer and the climber..plus whatever impact forces are involved im assuming, however maybe this is less than if the belayer were to statically absorb the fall? I am also not sure how the angle between the belayer and the climber(angle theta in diagram) will affect this? any thoughts?