A 83 kg person is hanging off the edge of a cliff, holding a rope. A 101 kg person is on the top of the cliff holding the other end of the rope. Find the minimum coefficient of static friction that would prevent the 83 kg person from falling.
You have to use F=ma and draw a force diagram for each person.
For the person hanging down
F = ma
83*g - T = 83*0
(you don't want it to accelerate)
(T = tension in the string)
For the person standing on top.
F = ma
T - Ff = 101*0
(Ff = force of friction = (mu)*R
where mu is the Greek letter representing the coefficient of friction and R is the force exerted by the ground on the person standing on the ground. R = mg = 101*g for the person on top)
You get two equations:
T = 83*g (from the 1st equation)
T = (mu)*101*g (from 2nd equation)
Therefore:
83*g = (mu)*101*g
83 = 101*mu
mu = 83/101
Hope that helps.
the velocity of a car A relative to a car B is 15km/h in a direction
To find the minimum coefficient of static friction needed to prevent the 83 kg person from falling, we need to consider the forces acting on the system.
Let's denote the mass of the 83 kg person as M1 and the mass of the 101 kg person as M2. We also know the accelerations for both persons will be zero since they are not moving.
First, let's consider the forces acting on the 83 kg person (M1) hanging off the edge of the cliff:
1. The weight of M1: This force is equal to M1 multiplied by the acceleration due to gravity (9.8 m/s²), directed downward.
So the weight of M1 = M1 * g = 83 kg * 9.8 m/s² = 813.4 N (newtons) directed downward.
2. The tension in the rope: This force is transmitted directly to M1 from the 101 kg person (M2) holding the other end of the rope. The tension force will be directed upwards.
Next, let's consider the forces acting on the 101 kg person (M2) on top of the cliff:
1. The weight of M2: This force is equal to M2 multiplied by the acceleration due to gravity (9.8 m/s²), directed downward.
So the weight of M2 = M2 * g = 101 kg * 9.8 m/s² = 990.8 N (newtons) directed downward.
2. The tension in the rope: Again, this force is transmitted directly to M2 from the 83 kg person (M1) hanging off the edge of the cliff. The tension force will be directed downwards.
Now, to find the minimum coefficient of static friction that would prevent the 83 kg person from falling, we need to consider the conditions for equilibrium. In other words, the sum of the forces in the horizontal (x) and vertical (y) directions must be zero:
In the vertical direction:
- The tension force acting upwards on M1 must balance the weight force acting downwards: Tension - Weight of M1 = 0.
- Tension = Weight of M1 = 813.4 N.
In the horizontal direction:
- The force of static friction acting at the edge of the cliff on M1 is the force that prevents M1 from sliding and falling.
- The maximum static friction force (F_friction) can be calculated by multiplying the coefficient of static friction (μ) by the normal force (N).
- In this case, the normal force (N) is equal to the weight of M2: N = Weight of M2 = 990.8 N.
- The maximum static friction force (F_friction) is given by F_friction = μ * N.
Since we want to find the minimum coefficient of static friction that prevents M1 from falling, we need to consider the worst-case scenario when the static friction force is at its minimum possible value (just enough to prevent sliding):
- The maximum static friction force (F_friction) acting horizontally on M1 must equal the tension in the rope: F_friction = Tension.
- Therefore, μ * N = Tension.
Substituting the values:
μ * 990.8 N = 813.4 N.
Now we can solve for the minimum coefficient of static friction (μ):
μ = Tension / N = 813.4 N / 990.8 N ≈ 0.822.
So the minimum coefficient of static friction that would prevent the 83 kg person from falling is approximately 0.822.