Question

The 64.5-kg climber in is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.64, respectively. Assume the walls are vertical and that the static friction forces are both at their maximum. Ignore his grip on the rope.
Determine the minimum normal force he must exert.

Answer 1

Answer:

The minimum force the climber must exert is about 439N.

Explanation:

We use the relationship between friction and normal force to answer this question:

$F_{friction} = \mu_{static} \cdot F_{normal}\implies F_{normal}=\frac{F_{friction}}{\mu_{static}}$

We are given the static coefficients of friction but need to determine the friction force. To do that we consider the totality of forces acting on this hapless gentleman stuck in a chimney. There is the gravity acting downward (+), then there are two friction forces acting upward (-), namely through his shoes and his back. The horizontal force exerted by the climber on both walls of the chimney is the same and is met with equally opposing normal force. Since the climber is not falling the net force in the vertical direction is zero:

$F_{net} = 0 = F_g - F_{shoes}-F_{back}= mg - \mu_{shoes}F_{norm}-\mu_{back}F_{norm}\\F_{norm}=\frac{mg}{\mu_{shoes}+\mu_{back}}=\frac{64.5kg\cdot 9.8\frac{m}{s^2}}{0.8+0.64}\approx 438.96N$

The normal force in this equilibrium is about 439N and  because we are told that the static friction forces are both at their maximum, this value is at the same time the minimum force needed for the climber to avoid starting slipping down the chimney.