A uniform wooden stick of mass 1.6 ,kg and le | vedclass

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(D) Let the normal reaction of the floor and the wall be $N$. From vertical equilibrium: $N + N \sin 30^{\circ} = 1.6g$. $N(1 + 0.5) = 1.6 	imes 10 =

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the normal reaction of the floor and the wall be $N$. From vertical equilibrium: $N + N \sin 30^{\circ} = 1.6g$. $N(1 + 0.5) = 1.6 	imes 10 = 16 \Rightarrow 1.5N = 16 \Rightarrow N = \frac{32}{3} \,N$. From horizontal equilibrium: $f = N \cos 30^{\circ} = \frac{32}{3} 	imes \frac{\sqrt{3}}{2} = \frac{16 \sqrt{3}}{3} \,N$. Taking torque about the bottom point $A$: $1.6g 	imes (\frac{l}{2} \sin 30^{\circ}) = N 	imes x$, where $x$ is the distance from the bottom to the wall contact point. $16 	imes \frac{l}{4} = \frac{32}{3} 	imes x \Rightarrow 4l = \frac{32}{3} x \Rightarrow x = \frac{3l}{8}$. From geometry, $h = x \cos 30^{\circ} = \frac{3l}{8} 	imes \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3} l}{16}$. Thus, $\frac{h}{l} = \frac{3 \sqrt{3}}{16}$ and $f = \frac{16 \sqrt{3}}{3} \,N$.

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