Question

The 70.0-kg climber in Fig. 4–72 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 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are ver- tical and that the static friction forces are both at their maximum. Ignore his grip on the rope.

Answer 1

Answer:

490 N

Explanation:

Please refer to the diagram below. Since the climber is currently at rest, we know that all forces must be equal.

∑F = 0

$\Sigma F_{x} = 0\\\Sigma F_{y} = 0$

The normal forces, N, exerted on the climber in x-direction, $N_{L}$ and $N_{R}$, are equal.

$\Sigma F_{x} = N_{R}-N_{L}=0\\N_{R}=N_{L}$ ----------------------------------- Eq. ( 1 )

The forces exerted on the climber in y-direction are equal to zero.

$\Sigma F_{y} = 0\\\Sigma F_{y} = F_{L}+F_{R} - mg = 0\\F_{L} + F_{R} = mg$

We know that Friction F, is equal to μN, where μ  is the coefficient of friction.

$\mu_{L} N_{L} + \mu_{R} N_{R} = mg$     -------------------------------- Eq. ( 2 )

(m is the mass of the climber and g is gravity)

Substituting Eq. ( 1 ) in Eq. ( 2 )

$\mu_{L} N_{R} + \mu_{R} N_{R} = mg\\N_{R} (\mu_{L} + \mu_{R}) = mg\\\\N_{R} = \frac{mg}{(\mu_{L} + \mu_{R})} \\\\N_{R} = \frac{ 70 \times 9.8} {0.80 + 0.60}\\\\N_{R} = \frac {686}{1.4}\\\\N_{R} = 490$

Since $N_{R} = N_{L}$ from equation 1, we know that the normal force exerted is 490N.