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JEE Advanced · Physics · 5. Laws of Motion

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 μ1 and that between the floor and the ladder is μ2 . The normal reaction of the wall on the ladder is N1 and that of the floor is N2 If the ladder is about to slip, then

  1. A μ1=0 μ20 and N2tanθ=mg2
  2. B μ10 μ2=0 and N1tanθ=mg2
  3. C μ10 μ20 and N2=mg1+μ1μ2
  4. D μ1=0 μ10 and N1tanθ=mg2
Verified Solution

Answer & Solution

Correct Answer

(D) μ1=0 μ10 and N1tanθ=mg2

Step-by-step Solution

Detailed explanation

Condition of translational equilibrium
N1=μ2N2
N2+μ1N1=Mg
Solving N2=mg1+μ1μ2
N1=μ2mg1+μ1μ2
Applying torque equation about corner (left) point on the floor
mgl2cosθ=N1 lsinθ+μ1N1 lcosθ
Solving tanθ=2-μ1μ22μ2
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