Static Equilibrium Climber Problem

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SUMMARY

The Static Equilibrium Climber Problem involves calculating the coefficient of static friction for a climber weighing 510 N, with angles θ = 41˚ and φ = 26˚. The forces acting on the climber include the normal force (FN), tension from the rope (T), and static friction (Fs). The relationship Fs = μFN is crucial, where μ represents the coefficient of static friction. The solution requires resolving the tension into its horizontal and vertical components and ensuring that the sum of forces and torques equals zero.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of force resolution in physics
  • Familiarity with torque calculations
  • Basic concepts of friction, specifically static friction
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  • Study the principles of static equilibrium in detail
  • Learn how to resolve forces into components in physics problems
  • Explore torque calculations and their applications in mechanics
  • Research the factors affecting static friction and its calculations
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Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as educators looking for practical examples in teaching these concepts.

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Homework Statement



A climber with a weight of 510 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 = 41˚ and φ = 26˚. 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?

Homework Equations



Sum of all forces = 0
Sum of all torques = 0

The Attempt at a Solution



[URL=http://img7.imageshack.us/my.php?image=physicsproblem.jpg][PLAIN]http://img7.imageshack.us/img7/5139/physicsproblem.th.jpg[/URL][/PLAIN]

FN - Normal force from the wall
T - Tension from rope
Fs - Static friction

What am I doing wrong?
 
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One has

FN - Normal force from the wall
Fs - Static friction

but Fs = μFN and must act vertically, opposite the weight of the climber.

T must be resolved into x (horizontal) and y (vertical) components,

and one must consider the torques, which must equal zero.

The climber's moment is the length of rope, but the moment opposing is the distance from from the rope at the wall to the feet.
 

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