JEE Advanced2014PhysicsLaws of MotionActual
In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ 1 and that between the floor and the ladder is μ 2 . The normal reaction of the wall on the ladder is N 1 and that of the floor is N 2 If the ladder is about to slip, then
Options
- Aμ 1 = 0 μ 2 ≠ 0 and N 2 tan θ = m g 2
- Bμ 1 ≠ 0 μ 2 = 0 and N 1 tan θ = m g 2
- Cμ 1 ≠ 0 μ 2 ≠ 0 and N 2 = m g 1 + μ 1 μ 2
- Dμ 1 = 0 μ 1 ≠ 0 and N 1 tan θ = m g 2
Correct answer
C. μ 1 ≠ 0 μ 2 ≠ 0 and N 2 = m g 1 + μ 1 μ 2
Step-by-step solution
Condition of translational equilibrium N 1 = μ 2 N 2 N 2 + μ 1 N 1 = M g Solving N 2 = m g 1 + μ 1 μ 2 N 1 = μ 2 m g 1 + μ 1 μ 2 Applying torque equation about corner (left) point on the floor m g l 2 cos θ = N 1 l sin θ + μ 1 N 1 l cos θ Solving tan θ = 2 - μ 1 μ 2 2 μ 2