Question

In the figure below, a climber with a weight of 699.8 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 42.0° and ф-31.0. 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? 288.1

Answer 1

As shown in the free-body diagram, the forces on the climber consist of T from the rope, normal force FN on her feet, upward static frictional force f,, and downward gravitational force mg. L mg Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton's second law to the vertical and horizontal directions, we have net,y In addition, the net torque about O (contact point between her feet and the wall) must also vanish: From the torque equation, we obtain T- mg sin6/sin(180°-e-ø)

Substituting the expression into the force equations, and noting that the coefficient of static friction to be 4FN, we find fs _1ng-T cosφ _mg_lng sin θ cos φ/sin(180°-e-の ng sin θ sin φ / sin(180°-θーの 1-sin θ cos φ / sin(180°-θーの sin θ sin φ / sin(1 800-9-φ) With e-:42° and φ-31°, the result is 1-Sin θ cos gs / sin(180°-t-の-1-sin 42° cos 31°/sin(180°-42°-31°) sin θ sin φ/sin(1800-8-φ) sin 42'sin 31° / sin(180°-42°-31°) 45 s S111 S111 () Sin 二10.667

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