In the figure, a 52 kg rock climber is climbing a “chimney” between two rock slabs. The coefficient of static friction between her shoes and the rock is 1.20; between her back and the rock it is 0.707. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. (a) What is the magnitude of each of her forces of push against the two columns of rock? (b) What fraction of her weight is supported by the frictional force on her shoes?
14.6
To find the magnitude of the forces of push against the two columns of rock, we can use the equation:
Force of push = coefficient of static friction * normal force
(a) For her back:
Force of push on her back = coefficient of static friction between her back and the rock * her weight
Force of push on her back = 0.707 * (52 kg * 9.8 m/s^2) (assuming g = 9.8 m/s^2)
Force of push on her back = 357.24 N
For her shoes:
Force of push on her shoes = coefficient of static friction between her shoes and the rock * her weight
Force of push on her shoes = 1.20 * (52 kg * 9.8 m/s^2)
Force of push on her shoes = 616.32 N
(b) To find the fraction of her weight supported by the frictional force on her shoes, we compare the frictional force with her weight:
Fraction of weight supported = frictional force on her shoes / her weight
Fraction of weight supported = (1.20 * (52 kg * 9.8 m/s^2)) / (52 kg * 9.8 m/s^2)
Fraction of weight supported = 1.20
Therefore, the fraction of her weight supported by the frictional force on her shoes is 1.20.
To determine the magnitude of the forces of push against the two columns of rock and the fraction of her weight supported by the frictional force on her shoes, we need to analyze the forces involved.
(a) To find the magnitude of each of her forces of push against the two columns of rock, we need to consider the forces acting on the climber. There are three vertical forces involved: her weight (mg), the normal force on her shoes (N_shoes), and the normal force on her back (N_back).
Since the climber is on the verge of slipping, the force of static friction on her shoes is at its maximum value. The magnitude of the force of static friction on her shoes (F_shoes) is given by: F_shoes = μ_shoes * N_shoes, where μ_shoes is the coefficient of static friction between her shoes and the rock.
Similarly, the magnitude of the force of static friction on her back (F_back) is given by: F_back = μ_back * N_back, where μ_back is the coefficient of static friction between her back and the rock.
Since the climber is on the verge of slipping, the forces of push against the two columns of rock must equal the forces of static friction on her shoes and back. Therefore, we have:
F_shoes = Force of push against the shoe column
F_back = Force of push against the back column
(b) To find the fraction of her weight supported by the frictional force on her shoes, we need to compare the force of static friction on her shoes (F_shoes) to her weight (mg).
To solve these two problems, we first need to calculate the normal forces acting on her shoes and back columns.
To find N_shoes, we have the equation: N_shoes - mg = 0 since there is no vertical acceleration in this case. Solving for N_shoes, we get: N_shoes = mg.
To find N_back, we have the equation: N_back - mg = 0. Solving for N_back, we get: N_back = mg.
Substituting these values into the equations for the forces of static friction, we get:
F_shoes = μ_shoes * mg
F_back = μ_back * mg
To find the magnitude of each force of push against the two columns of rock, we equate them to the corresponding forces of static friction:
F_shoes = Force of push against the shoe column
F_back = Force of push against the back column
To find the fraction of her weight supported by the frictional force on her shoes, we divide the magnitude of the force of static friction on her shoes (F_shoes) by her weight (mg):
Fraction of weight supported = F_shoes / (mg)
Now, we can substitute the given values:
- The mass of the climber (m) = 52 kg
- The coefficient of static friction between her shoes and the rock (μ_shoes) = 1.20
- The coefficient of static friction between her back and the rock (μ_back) = 0.707
Using these values, we can now calculate the answers to both parts (a) and (b) of the question.