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

(C) $\mu_{1} \neq 0 \quad \mu_{2} \neq 0$ and $N_{2}=\frac{m g}{1+\mu_{1} \mu_{2}}$; (D) $\mu_{1}=0 \quad \mu_{2} \neq 0 \operatorname{and} N_{1} \tan \theta=\frac{m g}{2}$

(A) $\mu_{1}=0 \quad \mu_{2} \neq 0 \operatorname{and} N_{2} \tan \theta=\frac{m g}{2}$

(B) $\mu_{1} \neq 0 \quad \mu_{2}=0 \operatorname{and} N_{1} \tan \theta=\frac{m g}{2}$

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Laws of Motion

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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

(A) $μ_{1} = 0 μ_{2} \neq = 0 and N_{2} tan θ = \frac{m g}{2}$
(B) $μ_{1} \neq = 0 μ_{2} = 0 and N_{1} tan θ = \frac{m g}{2}$
(C) $μ_{1} \neq = 0 μ_{2} \neq = 0$ and $N_{2} = \frac{m g}{1 + μ _{1} μ _{2}}$ ✓ Correct
(D) $μ_{1} = 0 μ_{2} \neq = 0 and N_{1} tan θ = \frac{m g}{2}$ ✓ Correct

Answer: (C), (D)

Solution

Forces:

N_{1}

(wall normal, horizontal),

μ_{1} N_{1}

(wall friction, downward),

N_{2}

(floor normal, upward),

μ_{2} N_{2}

(floor friction, horizontal),

m g

(weight, downward).

Horizontal equilibrium:

N_{1} = μ_{2} N_{2}

.
Vertical equilibrium:

N_{2} + μ_{1} N_{1} = m g

.

Substituting:

N_{2} + μ_{1} μ_{2} N_{2} = m g

, so

N_{2} = \frac{m g}{1 + μ _{1} μ _{2}}

. This is option (C).

Torque about bottom (length

ℓ

N_{1} ℓ sin θ = m g \frac{ℓ}{2} cos θ

, giving

N_{1} tan θ = \frac{m g}{2}

.

If

μ_{1} = 0

N_{2} = m g

and

N_{1} tan θ = \frac{m g}{2}

still holds. This is option (D).

Option (A) states

N_{2} tan θ = \frac{m g}{2}

, which is incorrect (should be

N_{1}

).
Option (B) requires

μ_{2} = 0

, but then

N_{1} = 0

and the ladder cannot be in equilibrium against the wall.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Laws of Motion

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