A rock climber of mass $m$ is ascending a vertical dihedral corner formed by two mutually perpendicu

Question

A rock climber of mass $m$ is ascending a vertical dihedral corner formed by two mutually perpendicular vertical rock walls. To hold position, the climber presses their two hands against the two walls symmetrically without using their feet. Let $\mu$ be the coefficient of static friction between the climber's hands and the rock surfaces, and $g$ be the acceleration due to gravity. Which of the following statements is/are CORRECT for the minimum force required to maintain static equilibrium?

Professor PC Thomas Classes · Answer

Let Wall 1 be along $y=0$ and Wall 2 along $x=0$. By symmetry, normal forces $N_1 = N_2 = N$. Horizontal equilibrium requires the horizontal friction from Wall 1 to balance $N_2$ and vice versa, so $f_{1x} = N$ and $f_{2y} = N$. Vertical equilibrium requires $f_{1z} + f_{2z} = mg \implies f_z = \frac{mg}{2}$. Total friction at each hand is $f_{	ext{fric}} = \sqrt{N^2 + \left(\frac{mg}{2}ight)^2}$. Using $f_{	ext{fric}} \le \mu N$, the limiting condition yields $(\mu^2 - 1)N^2 = \left(\frac{mg}{2}ight)^2 \implies N = \frac{mg}{2\sqrt{\mu^2 - 1}}$, which requires $\mu > 1$. Total magnitude of force exerted by each hand is $F = \sqrt{N^2 + f_{	ext{fric}}^2} = N\sqrt{1+\mu^2} = \frac{mg}{2}\sqrt{\frac{\mu^2 + 1}{\mu^2 - 1}}$.