(c12p24) In the figure, a 55 kg rock climber is in a lie-back climb along a fiss

Question

(c12p24) 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 = 0.25 m, and the center of mass of the climber is a horizontal distance d = 0.41 m from the fissure. The coefficient of static friction between hands and rock, hands, is 0.42, and between boots and rock, boots is 1.25.
What is the least horizontal pull by the hands and push by the feet that will keep him stable?
3.23×102 N

For this horizontal pull, what must be the vertical distance h between hands and feet?
9.97×10-1 m

If the climber encounters wet rock; so that The coefficient of static friction between hands and rock, hands, is 0.21, and between boots and rock, boots is 0.90. What is the least horizontal pull by the hands and push by the feet that will keep him stable?

For this horizontal pull, what m

Explanation / Answer

a)
use of vhorizonta forces = 0

Fx = 0

Net force on Nhand = Nboots = N

again from the sum of vertica force Fy = 0

mg = N*(hand) + N*(boots)

N = mg/(hand + boots)

N = 55*9.8/(0.42+1.25)

N = 322.75 N

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b) mg(d+w) - N*hand*(w) - N*h = 0

55*9.8*(0.41+0.25) - 322.75*0.42*(0.25) - 322.75 *h = 0

355.74 - 31.81601 - 374.306*h = 0

h = 0.997 m
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c)

Fx = 0

then Nhand = Nboots = N

Fy = 0

then

mg = N*(hand) + N*(boots)

N = mg/(hand + boots)

N = 55*9.8/(0.21+0.90)

N = 485.58 N

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d.

mg(d+w) - N*hand*(w) - N*h = 0

55*9.8*(0.41+0.25) - 485.85 *0.21*(0.25) - 485.85 *h = 0

355.74 - 25.5 -485.85*h = 0

h = 0.68 m

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