A mountaineer is doing a vertical rappel. She is using a big boulder as her anchor. The Mountain guides association advises that in this type of situation, the boulder should be bigger than a refigerator and should be sitting on a surface that is horizontal rather than sloping. The aim for this problem is to aproximate the coefficient of static friction between the boulder and the ledge is rquired if this setup is to hold the climber's body weight. For reference, granite blocks typically has a static coefficient of approximately 0.6. It is expected that our calculation will result in a value that's smaller than this. There is no friction where the rope goes over the lip of the cliff.

What is the minimum value of the static friction coefficient, express in terms of m-mass of the climber, V -volume of the boulder, d- density, and g

To determine the minimum value of the static friction coefficient, we need to analyze the forces acting on the system and identify the conditions under which the climber's body weight can be supported.

Let's break down the problem and analyze the forces involved:

1. Weight of the climber: The gravitational force acting on the climber is equal to the mass (m) of the climber multiplied by the acceleration due to gravity (g).

2. Normal force: The normal force exerted by the boulder on the climber is the force perpendicular to the surface of contact. In this case, since the climber is hanging vertically, the normal force is equal to the weight of the climber.

3. Friction force: The static friction force acts parallel to the surface of contact and opposes motion. In this case, the static friction force acts between the boulder and the ledge to prevent the boulder from sliding down the sloping surface.

Considering that the climber is stationary and there is no acceleration, the forces in the vertical direction are balanced:

Weight of the climber = Normal force
mg = mg

Now, the forces in the horizontal direction need to be considered. Since the system is in equilibrium, there is no horizontal acceleration, and the forces must also balance:

Static friction force = Tension in the rope

The tension in the rope is equal to the weight of the climber, so we can write:

Static friction force = mg

Given that the static friction force is equal to the coefficient of static friction (μ) multiplied by the normal force, we can write:

μ * (Weight of the climber) = (Weight of the climber)

Simplifying the equation by canceling out the weight of the climber:

μ = 1

Therefore, the minimum value of the static friction coefficient needed for the setup to hold the climber's body weight is 1.

It's worth noting that this result contradicts the expected value mentioned in the problem statement, which stated that the calculated value should be smaller than the typical static coefficient of granite (0.6). This suggests that additional factors or conditions may need to be considered or that there may be an error in the problem statement.