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

Correct answer: C. μ _ 1 ≠ 0 μ _ 2 ≠ 0 and N _ 2 = mg 1 + μ _ 1 μ _ 2 Condition of translational equilibrium N _ 1 = μ _ 2 N _ 2 N _ 2 + μ _ 1 N _ 1 = Mg Solving N _ 2 = mg 1 + μ _ 1 μ _ 2 N _ 1 = μ _ 2 mg 1 + μ _ 1 μ _ 2 Applying torque equation about corner (left) point on the floor mg(l)/(2)cos θ = N _ 1 lsin θ + μ _ 1 N _ 1 lcos θ Solving tan θ = 2 - μ _ 1 μ _ 2 2 μ _ 2

JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Physics
Chapter: Laws of Motion

JEE Advanced 2014 · Paper 1 · Physics · Laws of Motion

In the figure, a ladder of mass m is shown leaning against a wall....

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is ${\mu}_{1}$ and that between the floor and the ladder is ${\mu}_{2}$ . 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

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with

Options

A ${\mu}_{1} = 0{\mu}_{2} \ne 0$ and ${N}_{2}tan \theta = \dfrac{mg}{2}$
B ${\mu}_{1} \ne 0{\mu}_{2} = 0$ and ${N}_{1}tan \theta = \dfrac{mg}{2}$
C ${\mu}_{1} \ne 0{\mu}_{2} \ne 0$ and ${N}_{2} = \dfrac{mg}{1 + {\mu}_{1}{\mu}_{2}}$
D ${\mu}_{1} = 0{\mu}_{1} \ne 0$ and ${N}_{1}tan \theta = \dfrac{mg}{2}$

Correct answer

C. ${\mu}_{1} \ne 0{\mu}_{2} \ne 0$ and ${N}_{2} = \dfrac{mg}{1 + {\mu}_{1}{\mu}_{2}}$

Step-by-step solution

Condition of translational equilibrium

{N}_{1} = {\mu}_{2}{N}_{2}

{N}_{2} + {\mu}_{1}{N}_{1} = Mg

Solving

{N}_{2} = \dfrac{mg}{1 + {\mu}_{1}{\mu}_{2}}

{N}_{1} = \dfrac{{\mu}_{2}mg}{1 + {\mu}_{1}{\mu}_{2}}

Applying torque equation about corner (left) point on the floor

mg\dfrac{l}{2}cos \theta = {N}_{1}lsin \theta + {\mu}_{1}{N}_{1}lcos \theta

Solving

tan \theta = \dfrac{2 - {\mu}_{1}{\mu}_{2}}{2{\mu}_{2}}

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