Question

In Fig. 12-41, a climber with a weight of 533.8 N is held by a belay rope connected to her climbing harness and belay device; the force of the rope on her has a line of action through her center of mass. The indicated angles are $\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{40}\mathbf{.0}\mathbf{°}$and$\mathbf{}\mathit{\varphi }\mathbf{=}\mathbf{30}\mathbf{.0}\mathbf{°}$. If her feet are on the verge of sliding on the vertical wall, what is the coefficient of static friction between her climbing shoes and the wall?

Short answer

Coefficient of static friction between the wall and the shoesis 1.19.

Step-by-step solution

Understanding the given information

Weight of the climber,$W=533.8\text{\hspace{0.17em}N}$

$\theta ={40.0}^0$

$\varphi ={30.0}^0$

Concept and formula used in the given question 

To find coefficient of friction, first find the tension in the wire using equilibrium conditions. Then, find the friction force and normal force. Once you know friction force and normal force, you can find the coefficient of friction force using the basic definition of friction force. The equations used are given below.

$\begin{array}{c}\sum \tau =0\\ \sum {\text{F}}_{\text{x}}=0\\ \sum {\text{F}}_{\text{y}}=0\\ {\text{F}}_{\text{s}}=\mu {\text{F}}_{\text{n}}\end{array}$

Calculation for thecoefficient of friction between the shoes and the wall

Free body diagram:

Horizontal force:

$F_N=T\text{\hspace{0.17em}sin\hspace{0.17em}}\varphi$

Vertical force:

$\begin{array}{c}{F}_s+T\text{\hspace{0.17em}cos\hspace{0.17em}}\varphi =W\\ F_s=W-T\text{\hspace{0.17em}cos\hspace{0.17em}}\varphi \end{array}$

Torque:

$\begin{array}{c}WL\text{\hspace{0.17em}sin\hspace{0.17em}}\theta -TL\text{\hspace{0.17em}sin}(190-\theta -\varphi )\\ T=\frac{W\text{\hspace{0.17em}sin\hspace{0.17em}}\theta }{\sin (180-\theta -\varphi )}\\ =\frac{533.8\text{\hspace{0.17em}sin\hspace{0.17em}}40}{\sin (180-30-40)}\\ =365.14\text{\hspace{0.17em}N}\end{array}$

So,

$\begin{array}{c}{F}_N=365.14\text{\hspace{0.17em}sin\hspace{0.17em}}30\\ =182.6\text{\hspace{0.17em}N}\\ F_s=533.8-365.14\text{\hspace{0.17em}cos\hspace{0.17em}}30\\ =217.6\text{\hspace{0.17em}N}\end{array}$

Hence,

$\begin{array}{c}\mu =\frac{F_s}{F_N}\\ =\frac{217.6}{182.6}\\ =1.19\end{array}$

Hence, the coefficient of static friction between the wall and the shoes is 1.19.