Question

In Figure 12-43, a climber leans out against a vertical ice wall that has negligible friction. Distance ais $\mathbf{0}\mathbf{.914}\mathbf{\text{\hspace{0.17em}m}}$ and distance Lis$\mathbf{2}\mathbf{.10}\mathbf{\text{\hspace{0.17em}m}}$. His center of mass is distance $\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.940}\mathbf{\text{\hspace{0.17em}m}}$from the feet–ground contact point. If he is on the verge of sliding, what is the coefficient of static friction between feet and ground?

Short answer

The coefficient of static friction between the feet and the groundis$0.216$ .

Step-by-step solution

Understanding the given information 

i) Distance from the ice wall to the feet, $a=0.914\text{\hspace{0.17em}m}$.

ii) Length of the climber, $L=2.10\text{\hspace{0.17em}m}$.

iii) Distance of center of mass from the bottom, $d=0.940\text{\hspace{0.17em}m}$.

Concept and formula used in the given question

Using the concept of static equilibrium, you can write the equation for torque in terms of force and distances. Using these equations, you can find the frictional force. From the frictional force value, it is possible to find the value of the coefficient of friction.

Calculation for thecoefficient of static friction between feet and ground 

A free body diagram can be drawn as;

From the diagram, we can conclude that:

Forces acting in the x direction can be written as,

$F_{N2}-F_s=0$ (1)

Forces acting in the y direction can be written as,

$\begin{array}{c}{F}_{N1}-mg=0\\ F_{N1}=mg\end{array}$ (2)

Torque can be calculated as,

$\begin{array}{c}mgd\cos \theta -F_{N2}L\sin \theta =0\\ F_{N2}=\frac{mgd\cot \theta }{L}\end{array}$

Substituting values from equation 2 in the above equation and we get,

$F_{N2}=\frac{F_{N1}d\cot \theta }{L}$

Substituting values from equation 1 in the above equation and we get,

$F_s=\frac{F_{N1}\cot \theta d}{L}$ (3)

We know that,

$F_s=\mu F_{N1}$ (4)

And

$\cot \theta =\frac{a}{\sqrt{L^2-a^2}}$ (5)

Comparing equations 3 and 4, we get,

$\mu =\frac{d\cot \theta }{L}$

Substitute values from equation 5 in the above expression, and we get,

$\mu =\frac{d}{L}\frac{a}{\sqrt{L^2-a^2}}$

Substitute values in the above expression, and we get,

$\begin{array}{c}\mu =\frac{0.940}{2.10}\frac{0.914}{\sqrt{{2.10}^2-{0.914}^2}}\\ =0.4476\times 0.4834\\ =0.2163\end{array}$

Thus, the coefficient of static friction between the feet and the ground is $0.216$.