A mass of $4\; kg$ rests on a horizontal plane. The plane is gradually inclined until at an angle $\theta= 15^o$ with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
A mass of $4\; kg$ rests on a horizontal plane. The plane is gradually inclined until at an angle $\theta= 15^o$ with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
Answer The forces acting on a block of mass $m$ at rest on an inclined plane are $(i)$ the weight mg acting vertically downwards $(ii)$ the normal force $N$ of the plane on the block, and $(iii)$ the static frictional force $f_{ s }$ opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight $m g$ along the two directions shown, we have
$m g \sin \theta=f_{s}, \quad m g \cos \theta=N$
As $\theta$ increases, the self-adjusting frictional force
$f_{ s }$ increases until at $\theta=\theta_{\max ^{\prime}} f_{ s }$ achieves its
maximum value, $\left(f_{s}\right)_{\max }=\mu_{s} N$
Therefore.
$\tan \theta_{\max }=\mu_{s}$ or $\theta_{\max }=\tan ^{-1} \mu_{s}$
When $\theta$ becomes Just a little more than $\theta_{\max }$. there is a small net force on the block and it begins to slide. Note that $\theta_{\max }$ depends only on
$\mu_{ s }$ and $1 s$ independent of the mass of the block.
For $\quad \theta_{\max }=15^{\circ}$
$\mu_{s}=\tan 15^{\circ}$
$=0.27$
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