Stemming a chimney. A rock climber of mass 71 kg wants to make her way up the crack between two rocks as shown in the figure below. The coefficient of friction between her shoes and the rock surface is ìS = 0.90. What is smallest normal force that she can apply to both surfaces without slipping? Assume that the rock walls are vertical. Hint: Why is the normal force between the rock climber and the rock on the left equal to the normal force between her and the rock on the right?
To determine the smallest normal force that the rock climber can apply to both surfaces without slipping, we need to consider the force of friction and the normal force.
The force of friction can be calculated using the equation:
Frictional force (Ff) = coefficient of friction (μ) * normal force (Fn)
For the rock climber to not slip, the force of friction must be equal to or less than the maximum force of friction:
Ff ≤ μ * Fn
In this case, the normal force between the rock climber and the rock on the left is equal to the normal force between her and the rock on the right. This is because the climber is applying an equal and opposite force to both surfaces.
To find the normal force, we need to consider the forces acting on the climber. There are two forces acting vertically: her weight (mg) acting downward and the normal force (Fn) acting upward.
Since the rock walls are vertical, the normal force is equal to the climber's weight:
Fn = mg
Here, the mass of the climber (m) is given as 71 kg.
Therefore, the smallest normal force that the climber can apply to both surfaces without slipping is:
Fn = m * g
= 71 kg * 9.8 m/s²
= 696.8 N
So, the smallest normal force she can apply is 696.8 Newtons.