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In Fig. 12-39, arock 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 = 0.20 m,and the center of mass of the climber is a horizontal distance d = 0.40 mfrom the fissure. The coefficient of static friction betweenhands and rock is,μ1=0.40and between boots and rock it isμ2=1.2. (a) What is the least horizontal pull by the hands and push by the feet that will keep the climber stable? (b) For the horizontal pull of (a), what must be the vertical distance h between hands and feet? If the climber encounters wet rock, so thatμ1andμ2are reduced, what happens to (c) the answer to (a) and (d) the answer to (b)?

Short Answer

Expert verified

Answer:

a)The least horizontal pull by the hands and push by the feet that will keep the climber stableis.3.4×102N

b)For the horizontal pull, the vertical distance h between hands and feetis

c)Would increases

d)h should decrease if the coefficients decrease

Step by step solution

01

Listing the given quantities

Rock climberof mass55.0 Kg

Fissure has width, = 0.20 m

Horizontal distance d = 0.40 m from the fissure

Coefficient of static friction between hands and rock is,μ1=0.40

Coefficient of static friction between boots and rockμ1=1.2

02

Understanding the concept the force   

Here,The problem asks for the person's pull (his force exerted on the rock) but since we are examining forces and torques on the person, we solve for the reaction force FN1 (exerted leftward on the hands by the rock) _ At that point, there is also an upward force of static friction on his hands f1, which we will take to be at its maximum valueμ1FN1. We note that equilibrium of horizontal forces requiresFN1=FN2(the force exerted leftward on his feet); on his feet there is also an upward static friction force of magnitudeμ2FN2we have to calculate horizontal and vertical component of the force.

Formula:

f1+f2-mg=0

03

Calculation of the least horizontal pull by the hands and push by the feet that will keep the climber stable 

(a)

Equilibrium of vertical forces givesf1+f2-mg=0

fN1=mgμ1+μ2=3.4×102N

The least horizontal pull by the hands and push by the feet that will keep the climber stable is 3.4×102N.

04

Calculation of the vertical distance h between hands and feet

(b)

Computing torques about the point where his feet come in contact with the rock, we find

mg(d+w)-f1w-FN1h=0h=mg(d+w)-μ1FN1wFN1=0.88 m

For the horizontal pull, the vertical distance between hands and feet is

05

 Effect of coefficient of friction 

(c)

Both intuitively and mathematically (since both coefficients of friction are in the denominator) we see from part (a) that FN1would increase in such a case.

06

Explanation

As for part (b) it helps to plug part (a) into part (b) and simplify:

h=d+wμ2+dμ1

from which it becomes apparent that h should decrease if the coefficients decrease.

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