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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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{mg} \sin 45=\mathrm{f}_{\mathrm{L}} \\ & \mathrm{mg} \cos 45=\mathrm{N} \\ & \mathrm{f}_{\mathrm{L}}=\mu_{\mathrm{s}} N \\ & \mu_{\mathrm{s}}=\tan 45=1 \end{aligned}$ or $\tan \theta=\mu_s$ ( $\theta$ is angle of repose) $\tan 45=\mu_{\mathrm{s}}=1$

A $2 \mathrm{~kg}$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with…

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