91影视

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 ${渭}_s=0.80$.

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

The free body diagram of the man is as follows:

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

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

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:

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

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:

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

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