Question

In the figure, a climber with a weight of 390 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 θ = 50˚ and φ = 25˚. 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?

Answer 1

Given:

W = weight of the climber = 390 N

θ = angle between climber and wall = 50°

φ = angle between rope and wall = 25°

Let:

L = distance from her feet to her center of mass

T = tension in the rope

Thee angle between the rope and a line perpendicular to the climber is:

α = 90° - θ - φ

α = 15°

The angle of the rope to horizontal is:

β = 90° - φ

β = 65°

Summing up the torque about her feet, you get:

L×W×sin(θ) = L×T×cos(α)

T = W×sin(θ)/cos(α)

T = 309.3 N

The horizontal component of the reaction of the wall on her feet is:

Rx = T×cos(β)

Rx = 130.7 N

The vertical component of the reaction of the wall on her feet is:

Ry = W - T×sin(β)

Ry = 109.7 N

The coefficient of friction between her feet and the wall is

µ = Ry / Rx

µ = 0.839

Kindly upvote :)