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 t

Question

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 $(g = 10 {m s}^{- 2})$ . [2016]

Smart Achievers · Accepted Answer

(4)

$By vertical equilibrium,$

$N + N \sin 30 ° = 1.6 g$

$\Rightarrow N = \frac{3.2 g}{3} . . . (i)$

$By horizontal equilibrium,$

$f = N \cos 30 °$

$= \frac{\sqrt{3}}{2} N = \frac{16 \sqrt{3}}{3}$

$Torque about A,$

$1.6 g \times A B = N \times x$

$1.6 g \times \frac{ℓ}{2} \cos 60 ° = \frac{3.2 g}{3} \times x \Rightarrow \frac{3 ℓ}{8} = x . . . (i)$

$But \cos 30 ° = \frac{h}{x} \Rightarrow x = \frac{h}{\cos 30 °} . . . (i i)$

$From eq. (i) and (ii),$

$\frac{h}{\cos 30 °} = \frac{3 ℓ}{8} \Rightarrow \frac{h}{ℓ} = \frac{3 \sqrt{3}}{16}$

Solution

1 $\frac{h}{l} = \frac{\sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

2 $\frac{h}{l} = \frac{3}{16}, f = \frac{16 \sqrt{3}}{3} N$

3 $\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{8 \sqrt{3}}{3} N$

4 $\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

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Solution

1 hl=316, f=1633 N

2 hl=316, f=1633 N

3 hl=3316, f=833 N

4 hl=3316, f=1633 N

Ans. (4) By vertical equilibrium, N+Nsin30°=1.6g ⇒N=3.2g3 ...(i) By horizontal equilibrium, f=Ncos30° =32N=1633 Torque about A, 1.6g×AB=N×x 1.6g×ℓ2cos60°=3.2g3×x⇒3ℓ8=x ...(i) But cos30°=hx⇒x=hcos30° ...(ii) From eq. (i) and (ii), hcos30°=3ℓ8⇒hℓ=3316

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1 $\frac{h}{l} = \frac{\sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$

2 $\frac{h}{l} = \frac{3}{16}, f = \frac{16 \sqrt{3}}{3} N$

3 $\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{8 \sqrt{3}}{3} N$

4 $\frac{h}{l} = \frac{3 \sqrt{3}}{16}, f = \frac{16 \sqrt{3}}{3} N$