JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Rotational Mechanics

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 on the stick is perpendicular to the stick. The stick makes an angle of 30° 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 / l

and the frictional force

f

at the bottom of the stick are

(g = 10 m s^{- 2})

(A) $\frac{h}{l} = \frac{3}{16}, f = \frac{16 3}{3} N$
(B) $\frac{h}{l} = \frac{3}{16}, f = \frac{16 3}{3} N$
(C) $\frac{h}{l} = \frac{3 3}{16}, f = \frac{8 3}{3} N$
(D) $\frac{h}{l} = \frac{3 3}{16}, f = \frac{16 3}{3} N$ ✓ Correct

Answer: (D)

Solution

Let

N_{w}

be the normal reaction from the wall and

N_{f}

be the normal reaction from the floor. Given

N_{w} = N_{f} = N

.

For horizontal equilibrium:

N sin 30° = f

, so

f = N /2

.

For vertical equilibrium:

N cos 30° + f = m g

, giving

N (3 /2 + 1/2) = 16

, so

N = \frac{32}{3 + 1}

.

Taking moments about the bottom of the stick and using the geometry with the stick making 30° with the wall, we can determine the position where the wall contacts the stick. Through torque balance:

\frac{h}{l} = \frac{3 3}{16}

and

f = \frac{16 3}{3}

N

Note: The source solution indicates this problem may have issues with the given constraints, but option (D) is the stated answer.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Rotational Mechanics

Solution

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