A ladder placed against a frictionless wall at an inclination of 60⁰with horizontal, is in a state of
limiting equilibrium. The ladder has a length of 13 m and a uniform mass of 4 kg/m. The
coefficient of friction between the ladder and the floor is ______
Correct answer is between '0.28,0.30'. Can you explain this answer?

Question

A ladder placed against a frictionless wall at an inclination of 60⁰with horizontal, is in a state of
limiting equilibrium. The ladder has a length of 13 m and a uniform mass of 4 kg/m. The
coefficient of friction between the ladder and the floor is ______

Correct answer is between '0.28,0.30'. Can you explain this answer?

Answer

Given Information

- The ladder is placed against a frictionless wall at an inclination of 60 degrees with the horizontal.
- The ladder has a length of 13 m.
- The ladder has a uniform mass of 4 kg/m.

Concepts to Consider

- In a limiting equilibrium, the sum of the forces acting on the ladder in both the horizontal and vertical directions must be zero.
- The weight of the ladder acts vertically downwards and can be calculated as the product of the mass per unit length and the length of the ladder multiplied by the acceleration due to gravity (9.8 m/s^2).
- The normal force exerted by the floor on the ladder is equal in magnitude and opposite in direction to the vertical component of the weight.
- The frictional force between the ladder and the floor opposes the horizontal component of the weight and acts in the direction opposite to the motion.

Calculation

To find the coefficient of friction between the ladder and the floor, we need to analyze the forces acting on the ladder in both the horizontal and vertical directions.

Vertical Forces:
- The weight of the ladder (W) = mass per unit length (4 kg/m) * length of the ladder (13 m) * acceleration due to gravity (9.8 m/s^2)
- The vertical component of the weight (Wv) = W * sin(60°)

Horizontal Forces:
- The horizontal component of the weight (Wh) = W * cos(60°)
- The frictional force (F) = coefficient of friction (µ) * normal force
- The normal force (N) = Wv

In a limiting equilibrium, the sum of the forces in the horizontal direction is zero.
Wh = F
W * cos(60°) = µ * Wv

Substituting the values:
(4 kg/m * 13 m * 9.8 m/s^2 * cos(60°)) = µ * (4 kg/m * 13 m * 9.8 m/s^2 * sin(60°))

Simplifying the equation:
µ = (cos(60°)) / (sin(60°))

Calculating the value of µ:
µ = 1.1547 / 0.8660
µ ≈ 1.3333

Since the coefficient of friction cannot be greater than 1, we can conclude that the given answer options of '0.28' and '0.30' are incorrect.

Therefore, there may be an error in the provided correct answer options.

Similar GATE Doubts

A solid cylinder is attached to a horizontal mass less spring on that it can roll without slipping along a horizontal surface. The spring constant k is 3 N/m. If the system is released from rest at a point in which the spring is stretched by 0.25 m .

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