A ladder of length 5m has mass of 25 Kg. the ladder is leaned against a frictionless vertical wall at an angle of 100 with the vertical. A repairman with a mass of 82Kg needs to stand on an upper rung of the ladder that is 1m from the end. What is the minimum coefficient of friction between the ladder and the floor such that the ladder doesn’t slip?
To determine the minimum coefficient of friction between the ladder and the floor, we need to analyze the forces acting on the ladder.
First, let's identify the forces present:
1. Weight of the ladder (mg): The ladder's weight is given as 25 kg. Therefore, the weight can be calculated using the equation: weight = mass * acceleration due to gravity ( g = 9.8 m/s^2).
So, the weight of the ladder is 25 kg * 9.8 m/s^2.
2. Tension in the ladder (T): Since the ladder is leaning against a vertical wall, there is tension in the ladder. The tension provides the necessary support to prevent the ladder from slipping.
3. Normal force (N): The normal force acts perpendicular to the plane of contact, which in this case is the floor. The normal force counterbalances the vertical components of the weight and the tension.
4. Frictional force (Ff): The frictional force acts parallel to the floor's surface and opposes the motion of the ladder.
Now, let's analyze the forces acting on the ladder:
1. Vertical direction:
The vertical components of the ladder's weight (mg) and the tension (T) must be balanced by the normal force (N).
The equation for the vertical forces is as follows: N - mg cos(θ) - T cos(θ) = 0, where θ is the angle between the ladder and the floor.
2. Horizontal direction:
The horizontal components of the ladder's weight (mg) and the tension (T) must be balanced by the frictional force (Ff).
The equation for the horizontal forces is as follows: Ff - mg sin(θ) - T sin(θ) = 0.
However, we want to find the minimum coefficient of friction, so we need to determine the maximum value for the frictional force (Ff). This happens when the ladder is on the verge of slipping, meaning the frictional force reaches its maximum value, which is the product of the normal force and the coefficient of friction (Ff = μN).
Substituting Ff = μN and rearranging the equations, we get:
1. Vertical direction: N = mg cos(θ) + T cos(θ)
2. Horizontal direction: μN = mg sin(θ) + T sin(θ)
Now, let's calculate the values of mg, T, and θ:
1. mg = 25 kg * 9.8 m/s^2 (as given in the problem statement)
2. T: The tension can be determined by taking moments about the lower end of the ladder. Since the distance from the lower end to the upper rung is 1 m, the moment caused by the repairman is (82 kg * 9.8 m/s^2) * 1 m.
Therefore, T = (82 kg * 9.8 m/s^2) * 1 m
With the values of mg and T, we can substitute them into the equations above:
1. Vertical direction: N = (25 kg * 9.8 m/s^2) * cos(θ) + (82 kg * 9.8 m/s^2) * cos(θ)
2. Horizontal direction: μ[(25 kg * 9.8 m/s^2) * cos(θ) + (82 kg * 9.8 m/s^2) * cos(θ)] = (25 kg * 9.8 m/s^2) * sin(θ) + (82 kg * 9.8 m/s^2) * sin(θ)
Combining the two equations, we can solve for the minimum coefficient of friction (μ) by dividing both sides by [(25 kg * 9.8 m/s^2) * cos(θ) + (82 kg * 9.8 m/s^2) * cos(θ)]:
μ = [(25 kg * 9.8 m/s^2) * sin(θ) + (82 kg * 9.8 m/s^2) * sin(θ)] / [(25 kg * 9.8 m/s^2) * cos(θ) + (82 kg * 9.8 m/s^2) * cos(θ)]
Now, you can plug in the value of θ (100 degrees) into the equation and calculate the minimum coefficient of friction (μ).