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,$\mathit{\mu }\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$and between boots and rock it is$\mathit{\mu }\mathbf{2}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}$. (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 that$\mathit{\mu }\mathbf{1}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\mu }\mathbf{2}$are reduced, what happens to (c) the answer to (a) and (d) the answer to (b)?

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,$\mu 1=0.40$

Coefficient of static friction between boots and rock$\mu 1=1.2$

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 value$\mathit{\mu }\mathbf{1}\mathit{F}\mathit{N}\mathbf{1}$. We note that equilibrium of horizontal forces requires$\mathit{F}{\mathit{N}}_{\mathbf{1}}\mathbf{=}\mathit{F}{\mathit{N}}_{\mathbf{2}}$(the force exerted leftward on his feet); on his feet there is also an upward static friction force of magnitude$\mathit{\mu }\mathbf{2}\mathit{F}\mathit{N}\mathbf{2}$we have to calculate horizontal and vertical component of the force.

Formula:

$\mathit{f}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathit{f}\mathbf{2}\mathbf{}\mathbf{-}\mathbf{}\mathit{m}\mathit{g}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

(a)

Equilibrium of vertical forces gives$f1+f2-mg=0$

$$\begin{gathered} fN1=\frac{mg}{\mu 1+\mu 2} \\ =3.4\times {10}^{2}N \end{gathered}$$

The least horizontal pull by the hands and push by the feet that will keep the climber stable is $3.4\times {10}^{2}N$.

(b)

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

$$\begin{gathered} mg(d+w)-f1w-FN1h=0 \\ h=\frac{mg(d+w)-\mu 1FN1w}{FN1} \\ =0.88 m \end{gathered}$$

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

(c)

Both intuitively and mathematically (since both coefficients of friction are in the denominator) we see from part (a) that $F{N}_1$would increase in such a case.

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

$h=\left(d+w\right)\mu 2+d\mu 1$

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