Static Equilibrium Climber Problem

[craig22]

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

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F_N - Normal force from the wall
T - Tension from rope
F_s - Static friction

What am I doing wrong?

 

[Astronuc]

One has

F_N - Normal force from the wall
F_s - Static friction

but F_s = μF_N 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.

 

[nlightner]

Your approach seems to be on the right track, but there are a few things that need to be clarified and corrected in order to solve this problem correctly.

First, it's important to note that the climber is in a state of static equilibrium, meaning that all forces and torques acting on her are balanced and she is not accelerating. This is indicated in the problem statement by the fact that her feet are on the verge of sliding, but not actually sliding. This means that the sum of all forces and torques must be equal to zero.

Next, it's important to properly label all forces and torques in the problem. In your diagram, you have correctly labeled the normal force FN and the tension T from the rope. However, you have not included the weight of the climber, which should be labeled as mg, where m is the mass of the climber and g is the acceleration due to gravity. This weight force should be acting straight down through the climber's center of mass, as indicated in the problem statement.

In addition, you have labeled the static friction force as Fs, but you have not included its direction or magnitude in your diagram. The static friction force always acts in the opposite direction of motion or the tendency for motion, so in this case it should be acting upwards along the wall. Its magnitude is what we are trying to solve for in this problem.

Now, let's look at your equations. The sum of all forces should be equal to zero, so we can write:

ΣF = FN + T + Fs + mg = 0

Since we are looking for the coefficient of static friction, we can rewrite this equation as:

Fs = μsFN

where μs is the coefficient of static friction. This equation tells us that the static friction force is equal to the coefficient of static friction multiplied by the normal force. We can then substitute this into our original equation to get:

FN + T + μsFN + mg = 0

Next, we need to consider the sum of all torques. In this case, we can choose any point to calculate the torques about, but it is most convenient to choose the point where the rope is attached to the climber's harness, since we know that the tension force T is acting at this point and will not contribute to any torque. Using the equation for torque (τ = rFsinθ), we can write:

Στ = (r1)(T