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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. If the angle of inclination of the plane is $60^{\circ}$, then the coefficient of friction is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{3}}$
According to the question, angle of inclination $\theta=60^{\circ}$.
Now, force of friction, $f=\mu N=\mu m g \cos \theta$ and net retarding force, $\left(F_1\right)=m g \sin \theta+f$ $\therefore$ Net accelerating force down the inclined plane is
$\therefore$ External force needed (up the inclined plane) to maintain sliding motion is (net retarding force)
From Eqs. (i),(ii) and (iii) we get
$m g \sin \theta+\mu m g \cos \theta=2(m g \sin \theta-\mu m g \cos \theta)$
or
$3 \mu m g \cos \theta=m g \sin \theta$
or
$$
\frac{\sin \theta}{\cos \theta}=3 \mu
$$
or
$$
\tan \theta=3 \mu \quad\left(\because \theta=60^{\circ}\right)
$$
or
$$
\tan 60^{\circ}=3 \mu
$$
$$
\begin{aligned}
\sqrt{3} & =3 \mu \\
\mu & =\frac{1}{\sqrt{3}}
\end{aligned}
$$
When inclination of plane is $60^{\circ}$ then the coefficient of friction, $\mu=\frac{1}{\sqrt{3}}$.
Now, force of friction, $f=\mu N=\mu m g \cos \theta$ and net retarding force, $\left(F_1\right)=m g \sin \theta+f$ $\therefore$ Net accelerating force down the inclined plane is
$\therefore$ External force needed (up the inclined plane) to maintain sliding motion is (net retarding force)
From Eqs. (i),(ii) and (iii) we get
$m g \sin \theta+\mu m g \cos \theta=2(m g \sin \theta-\mu m g \cos \theta)$
or
$3 \mu m g \cos \theta=m g \sin \theta$
or
$$
\frac{\sin \theta}{\cos \theta}=3 \mu
$$
or
$$
\tan \theta=3 \mu \quad\left(\because \theta=60^{\circ}\right)
$$
or
$$
\tan 60^{\circ}=3 \mu
$$
$$
\begin{aligned}
\sqrt{3} & =3 \mu \\
\mu & =\frac{1}{\sqrt{3}}
\end{aligned}
$$
When inclination of plane is $60^{\circ}$ then the coefficient of friction, $\mu=\frac{1}{\sqrt{3}}$.
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