A 67.5 kg rock climber clings to protrusions in a wall. The climber\'s center of

Question

A 67.5 kg rock climber clings to protrusions in a wall. The climber's center of gravity is 0.360 m from the wall, and his hands are 1.57 m from his feet. What horizontal force is required at his hands to keep from falling backwards? Hint: Use the feet of the climber as the center of rotation when calculating torque. What is the horizontal force exerted by the climber's feet? What is the total vertical force exerted by the climber's hands and feet? The horizontal force depends on the distance of the climber's center of gravity from the wall. What should that distance be if the climber wants to reduce the required horizontal force from hands by 50%? (Use DNE if no solution is possible.) What should the distance of center of gravity from the wall be if the climber wants to reduce the vertical force of her hands and feet by 50%? (Use DNE if no solution is possible.)

Explanation / Answer

let us consider centre of rotation at the feet

here the forces exerted by rock on feet and hands little bit inclined . so these forces can be resolved as vertical and horizontal and vertical components

here the torque due to horizontal component of force exerted on feet about the same point =0

according yo princicile of moments

torque due to horizontal force exerted at hand +torque due to horizontal force at feet=torque due to gravity

F horizontal at hand(1.57)+0=mg(.36)=67.5(9.8)(.36)

F=151.68N

the vertical components of the two forces should be balanced by his weight

sum of the vertical forces =mg=67.5(9.8)661.5N

hence

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