In the figure, a 55 kg rock climber is in a lie-back climb along a fissure, with

Question

In the figure, a 55 kg rock climber is in a lie-back climb along a fissure, with hands pulling on one side of the fissure and feet pressed against the opposite side. The fissure has width w = 0.17 m, and the center of mass of the climber is a horizontal distance d = 0.39 m from the fissure. The coefficient of static friction between hands and rock, ?hands, is 0.38, and between boots and rock, ?boots is 1.20.
What is the least horizontal pull by the hands and push by the feet that will keep him stable?

Explanation / Answer

moment:

(d+w)*mg = w*F*u + F*h

(0.17+0.39)*55*9.8 = 0.17*0.38*F + F*h

F*u1 + F*u2 = mg

=> F = 55*9.8/(0.38 + 1.20) = 341.139241 N ANSWER

puttin in 1:

(0.17+0.39)*55*9.8 = 0.17*0.38*341.139241 + 341.139241*h

=> h= 0.8201999987 m ANSWER

moment:

(d+w)*mg = w*F*u + F*h

(0.17+0.39)*55*9.8 = 0.17*0.21*F + F*h

F*u1 + F*u2 = mg

=> F = 55*9.8/(0.80 + 0.21) = 533.66336634 N

puttin in 1:

(0.17+0.39)*55*9.8 = 0.17*0.21*533.66336634 + 533.66336634*h

=> h= 0.52989999999 m

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