A 56 kg rock climber is climbing a "chimney" between two rock slabs as shown in the figure. The static coefficient of friction between her shoes and the rock is 1.02; between her back and the rock it is 0.70. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. What is her push against the rock?
Push against the rock= (mg)\(u1+u2)
To determine the push against the rock, we need to find the maximum force that can be exerted without slipping occurring at both the shoes and the back. We can start by analyzing the situation at the shoes and then at the back.
1. Analyzing the situation at the shoes:
Since the coefficient of friction between the shoes and the rock is given as 1.02, we can use this value to calculate the maximum force before slipping occurs.
The maximum frictional force (Ff) can be calculated using the equation:
Ff = μs * N
where μs is the static coefficient of friction and N is the normal force.
The normal force (N) exerted by the climber on the rock slab can be calculated as the weight of the climber:
N = mg
where m is the mass of the climber (56 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we have:
N = 56 kg * 9.8 m/s²
N = 548.8 N
Using the max frictional force equation, we can find Ff:
Ff = 1.02 * 548.8 N
Ff = 559.58 N (approximately)
Therefore, the maximum force the climber can exert at her shoes without slipping is approximately 559.58 N.
2. Analyzing the situation at the back:
Similarly, we can use the coefficient of friction between the back and the rock (0.70) to find the maximum force at her back.
Following the same steps, we find the normal force (N) as the weight:
N = 56 kg * 9.8 m/s²
N = 548.8 N
Using the max frictional force equation, we can find Ff:
Ff = 0.70 * 548.8 N
Ff = 384.16 N (approximately)
Therefore, the maximum force the climber can exert at her back without slipping is approximately 384.16 N.
In summary, the climber can exert a maximum force of approximately 559.58 N at her shoes and 384.16 N at her back without slipping occurring.
To determine the climber's push against the rock, we need to consider both the horizontal and vertical forces acting on her.
1. Vertical forces:
The only vertical force acting on the climber is her weight, which can be calculated using the formula:
Weight = mass * gravitational acceleration
Weight = 56 kg * 9.8 m/s^2
Weight = 548.8 N
2. Horizontal forces:
To find the climber's push against the rock, we need to consider the limiting friction forces between her back and the rock, as well as her shoes and the rock.
The limiting friction force can be calculated using the formula:
Friction force = coefficient of friction * normal force
a. Friction force between her shoes and the rock:
The normal force acting on her shoes can be calculated by balancing the vertical forces:
Normal force = Weight
Normal force = 548.8 N
The friction force between her shoes and the rock can be calculated as:
Friction force (shoes) = coefficient of friction (shoes) * Normal force
Friction force (shoes) = 1.02 * 548.8 N
Friction force (shoes) = 559.936 N
b. Friction force between her back and the rock:
To find the normal force acting on her back, we need to balance the vertical forces again, considering that she is on the verge of slipping.
The normal force acting on her back is equal to her weight minus the vertical component of the friction force between her shoes and the rock.
Vertical component of the friction force (shoes) = Friction force (shoes) * sin(θ)
[where θ is the angle between the back and the rock]
Since the climber is on the verge of slipping, the friction force between her back and the rock is at its maximum. Therefore, the horizontal component of the friction force between her back and the rock is equal to the horizontal component of the friction force between her shoes and the rock.
Friction force (back - horizontal component) = Friction force (shoes) * cos(θ)
Since the climber's back and shoes are both on the verge of slipping, we have:
Vertical component of friction force (shoes) = Friction force (shoes) * sin(θ)
Horizontal component of friction force (shoes) = Friction force (shoes) * cos(θ)
Equating these two components, we get:
Friction force (shoes) * sin(θ) = Friction force (shoes) * cos(θ)
tan(θ) = 1
θ = 45 degrees
Now we can calculate the normal force acting on her back:
Normal force (back) = Weight - Vertical component of friction force (shoes)
Normal force (back) = 548.8 N - (Friction force (shoes) * sin(θ))
Normal force (back) = 548.8 N - (559.936 N * sin(45))
Normal force (back) = 548.8 N - (559.936 N * 0.7071)
Normal force (back) ≈ 387.37 N
The friction force between her back and the rock can be calculated as:
Friction force (back) = coefficient of friction (back) * Normal force (back)
Friction force (back) = 0.70 * 387.37 N
Friction force (back) ≈ 271.16 N
Since the climber has reduced her push until her back and her shoes are on the verge of slipping:
Push against the rock = Friction force (shoes) + Friction force (back)
Push against the rock = 559.936 N + 271.16 N
Push against the rock = 831.096 N
Therefore, the climber's push against the rock is approximately 831.096 Newtons.