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The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}}$
For upward motion, $F_{\mathrm{up}}=m g(\sin \theta+\mu \cos \theta)$
For downward motion, $F_{\text {down }}=m g(\sin \theta-\mu \cos \theta)$
...(ii)
According to the question,
$$
\begin{aligned}
& F_{\text {up }}=2 F_{\text {down }} \\
& m g(\sin \theta+\mu \cos \theta)=2 m g(\sin \theta-\mu \cos \theta) \\
& \sin \theta+\mu \cos \theta=2 \sin \theta-2 \mu \cos \theta \\
& 3 \mu \cos \theta=\sin \theta \\
& \mu=\frac{1}{3} \tan \theta \\
& \mu=\frac{1}{3} \times \tan 60^{\circ} \\
& \Rightarrow \quad \mu=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}} \\
&
\end{aligned}
$$
For downward motion, $F_{\text {down }}=m g(\sin \theta-\mu \cos \theta)$
...(ii)
According to the question,
$$
\begin{aligned}
& F_{\text {up }}=2 F_{\text {down }} \\
& m g(\sin \theta+\mu \cos \theta)=2 m g(\sin \theta-\mu \cos \theta) \\
& \sin \theta+\mu \cos \theta=2 \sin \theta-2 \mu \cos \theta \\
& 3 \mu \cos \theta=\sin \theta \\
& \mu=\frac{1}{3} \tan \theta \\
& \mu=\frac{1}{3} \times \tan 60^{\circ} \\
& \Rightarrow \quad \mu=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}} \\
&
\end{aligned}
$$
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