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
[tex]\Sigma F_{x} = 0\\\Sigma F_{y} = 0\\[/tex]
The normal forces, N, exerted on the climber in x-direction, [tex]N_{L}[/tex] and [tex]N_{R}[/tex], are equal.
[tex]\Sigma F_{x} = N_{R}-N_{L}=0\\N_{R}=N_{L}[/tex] ----------------------------------- Eq. ( 1 )
The forces exerted on the climber in y-direction are equal to zero.
[tex]\Sigma F_{y} = 0\\\Sigma F_{y} = F_{L}+F_{R} - mg = 0\\F_{L} + F_{R} = mg\\[/tex]
We know that Friction F, is equal to μN, where μ is the coefficient of friction.
[tex]\mu_{L} N_{L} + \mu_{R} N_{R} = mg\\[/tex] -------------------------------- Eq. ( 2 )
(m is the mass of the climber and g is gravity)
Substituting Eq. ( 1 ) in Eq. ( 2 )
[tex]\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[/tex]
Since [tex]N_{R} = N_{L}[/tex] from equation 1, we know that the normal force exerted is 490N.
