The 70.0-kg climber in Fig. 4–72 is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that the static friction forces are both at their maximum. Ignore his grip on the rope.

In this problem, the walls are vertical; so the normal forces will act in the horizontal direction.

Let us assume that the rope is not under tension force and does not apply any force.

Given data:

The mass of the climber is $m=70\mathrm{kg}$.

The static coefficient of friction between the shoes and the wall is ${\mu }_s=0.80$.

The static coefficient of friction between the back and the wall is ${\mu }_s\text{'}=0.60$.

The free body diagram of the man is as follows:

The relation to calculate the forces in the x-direction is given by:

$\begin{array}{c}\Sigma F=0\\ F_{NL}-F_{NR}=0\\ F_{NL}=F_{NR}\end{array}$

Here, $F_{NL}$and $F_{NR}$are the normal forces on the left and right sides.

The relation to calculate the forces in the y-direction is given by:

$\begin{array}{c}\Sigma F\text{'}=0\\ F_{frL}+F_{frR}=mg\\ {\mu }_s{F}_{NL}+{\mu }_s\text{'}{F}_{NR}=mg\end{array}$

Here, $F_{frL}$and $F_{frR}$are the frictional forces on the left and right sides, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{c}{\mu }_s{F}_{NL}+{\mu }_s\text{'}{F}_{NL}=mg\\ F_{NL}=\left[\frac{mg}{{\mu }_s+{\mu }_s\text{'}}\right]\\ F_{NL}=\left[\frac{70\mathrm{kg}\times 9.81\mathrm{m}/{\mathrm{s}}^2}{0.80+0.60}\right]\\ F_{NL}=490.5\mathrm{N}\end{array}$

Thus, $F_{NL}=490.5\mathrm{N}$is the force exerted by the climber.