In the figure a 56 kg rock climber is in a lie-back climb along a fissure, with

Question

In the figure a 56 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 = 0.35 m, and the center of mass of the climber is a horizontal distance d = 0.45 m from the fissure. The coefficient of static friction between hands and rock is 1 = 0.50, and between boots and rock it is 2 = 1.10. The climber adjusts the vertical distance h between hands and feet until the (identical) pull by the hands and push by the feet is the least that keeps him from slipping down the fissure. (He is on the verge of sliding.) (a) What is the least horizontal pull by the hands and push by the feet that will keep the climber stable? (b) What is the value of h?

see the image http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c12/q02f.jpg

Explanation / Answer

draw an origin in the middle of the climber right where the cable wire touches her waist. make the y axis go straight up in the air vertically and make the x axis level with the horizonal
Positive y goes straight up in the air and positive x goes off to the right.
draw a free body diagram which is a dot with four arrows extending from the dot. One arrow labled F which stands for wall friction goes from the dot straight up along the +y axis.

arrow number 2 labled N which stands for wall normal goes from the dot straight to the right in the plus x direction

arrow 3 goes from the dot straight down towards the ground along the -y direction and its lable is W which stands for 'weight'

and then there is arrow number 4 the last arrow. it is at an angle. the previous 3 arrows were not at angles they were flush with an axis and needed no angle.

draw arrow 4 from the dot (which is the origin) upward and off towards the left so that it lies in quadrant four. this arrow is drawn parallel with the direction of the cable that holds the climber. all four arrows are straight lines, not curvy.

do the sum of the forces in the x axis and get an equation
N=Tsin27.5
do the sum in the y axis and get an equation
F+ Tcos(27.5)= (629.8)(98)
do the sum of torques about the origin and get an equation
Fsin(35)=Ncos(35)
do some algebra and solve for N and T and F. three equation and only 3 unknowns. when you get N go back to the sum in the y axis. instead of F+Tcos27.5=(629.8)(9.8)
say: uN + Tcos27.5=(629.8)(9.8) and solve for u.

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