A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h (< l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall

Correct answer: D. (h)/(l) = 3 3 16 , f = 16 3 3 N Force equation in x-direction, N _ 1 cos 30 ^ o - f = 0 ....(i) Force equation in y-direction, N _ 1 sin 30 ^ o + N _ 2 - mg = 0 ....(ii) Torque equation about O, mg(l)/(2)cos 60 ^ o - N _ 1 h cos 30 ^ o = 0 ...(iii) Also, given N _ 1 = N _ 2 .....(iv) [Note taking reaction from floor as normal reaction only] Solving (i), (ii), (iii) and (iv) we have (h)/(l) = 3 3 16 and f = 16 3 3

JEE Advanced 2016Paper 1

Exam: JEE Advanced
Year: 2016
Paper: Paper 1
Subject: Physics
Chapter: Laws of Motion

JEE Advanced 2016 · Paper 1 · Physics · Laws of Motion

A uniform wooden stick of mass 1.6 kg and length l rests in an...

A uniform wooden stick of mass 1.6 kg and length

l

rests in an inclined manner on a smooth, vertical wall of height h \left(< l\right)

such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of

{30}^{o}

with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/

and the frictional force f at the bottom of the stick are: \left(g = 10m{s}^{- 2}\right)

A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h

Options

A $\dfrac{h}{l} = \dfrac{\sqrt{3}}{16}, f = \dfrac{16\sqrt{3}}{3}N$
B $\dfrac{h}{l} = \dfrac{3}{16}, f = \dfrac{16\sqrt{3}}{3}N$
C $\dfrac{h}{l} = \dfrac{3\sqrt{3}}{16}, f = \dfrac{8\sqrt{3}}{3}N$
D $\dfrac{h}{l} = \dfrac{3\sqrt{3}}{16}, f = \dfrac{16\sqrt{3}}{3}N$

Correct answer

D. $\dfrac{h}{l} = \dfrac{3\sqrt{3}}{16}, f = \dfrac{16\sqrt{3}}{3}N$

Step-by-step solution

Force equation in x-direction,

{N}_{1}cos {30}^{o} - f = 0

....(i)
Force equation in y-direction,

{N}_{1}sin {30}^{o} + {N}_{2} - mg = 0

....(ii)
Torque equation about O,

mg\dfrac{l}{2}cos {60}^{o} - {N}_{1}\dfrac{h}{cos {30}^{o}} = 0

...(iii)
Also, given

{N}_{1} = {N}_{2}

.....(iv)
[Note taking reaction from floor as normal reaction only]
Solving (i), (ii), (iii) and (iv) we have

\dfrac{h}{l} = \dfrac{3\sqrt{3}}{16}

and

f = \dfrac{16\sqrt{3}}{3}

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