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A $2 \mathrm{~kg}$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with respect to horizontal axis. The co-efficient of static friction between their surfaces is:
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Verified Answer
The correct answer is:
$1$
$\begin{aligned}
& \mathrm{mg} \sin 45=\mathrm{f}_{\mathrm{L}} \\
& \mathrm{mg} \cos 45=\mathrm{N} \\
& \mathrm{f}_{\mathrm{L}}=\mu_{\mathrm{s}} N \\
& \mu_{\mathrm{s}}=\tan 45=1
\end{aligned}$
or
$\tan \theta=\mu_s$ ( $\theta$ is angle of repose)
$\tan 45=\mu_{\mathrm{s}}=1$
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