A 52.0 kg rock climber clings to protrusions in a wall. The climber\'s center of
ID: 1362136 • Letter: A
Question
A 52.0 kg rock climber clings to protrusions in a wall. The climber's center of gravity is 0.380 m from the wall, and her hands are 1.58 m from her feet.
What horizontal force is required at her hands to keep from falling backwards? Hint: Use the feet of the climber as the center of rotation when calculating torque. N?
What is the horizontal force exerted by the climber's feet? N ?
What is the total vertical force exerted by the climber's hands and feet? N?
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.) M?
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.) M?
Explanation / Answer
net torque about the feet = 0
torque due to weight = T1 = W*x = +52*9.8*0.38 = 193.648 Nm
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torque due horizantal force = T2 = F*y = Fx*1.58
in equilibrium T2 = T1
Fx = 122.56 N
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Fv + Fg = 0
Fg = -w
Fv = W = 52*9.8 = 509.6 N
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if Fx1 = 122.56/2
W*x1 = Fx1*y
x1 = 0.189 m <<<--------------answer
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DNE
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