A $\sqrt{34} \mathrm{~m}$ long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If $\mathrm{F}_{\mathrm{f}}$ and $\mathrm{F}_{\mathrm{w}}$ are the reaction forces of the floor and the wall, then ratio of $\mathrm{F}_{\mathrm{w}} / \mathrm{F}_{\mathrm{f}}$ will be: (Use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Laws of Motion

34 m

long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If

F_{f}

and

F_{w}

are the reaction forces of the floor and the wall, then ratio of

F_{w} / F_{f}

will be:
(Use

g = 10 m / s^{2}

)

(A) $\frac{6}{110}$
(B) $\frac{3}{113}$
(C) $\frac{3}{109}$ ✓ Correct
(D) $\frac{2}{109}$

Answer: (C)

Solution

The ladder has length

ℓ = 34

m and base distance 3 m, so height

= 34 - 9 = 5

m. Thus

cot θ = 3/5

. Taking torque about the base:

N_{2} ℓ sin θ = m g \frac{ℓ}{2} cos θ

, giving

N_{2} = \frac{m g}{2} cot θ

. The floor reaction

F_{f} = (m g)^{2} + N_{2}^{2}

. The ratio

\frac{F _{w}}{F _{f}} = \frac{N _{2}}{( m g ) ^{2} + N _{2}^{2}} = \frac{1}{1 + \frac{4}{c o t ^{2} θ}} = \frac{3}{109}

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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