Question

In the given figure, a mass of $4 kg$ rests on a horizontal plane. The inclination of plane is gradually increased. The mass just begins to slide at $\theta=15^{\circ} .$ What is the coefficient of the static friction between the block and the surface?

• A. 0.50
• B. 0.27
• C. 0.85
• D. 0.60

Answer 1

Answer: (B)

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 $f$, opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown.
We have $m g \sin \theta=f_{s}, m g \cos \theta=N $
$\Rightarrow \tan \theta=\frac{f_{s}}{N}$
As $\theta$ increases, the self-adjusting frictional force $f$ increases until at $\theta=\theta_{\max }, f _{s}$
achieves its maximum value,
$\left(f_{s}\right)_{\max }=\mu_{s} N $
$\Rightarrow \left(f_{s}\right)_{\max } / N=\mu_{s}$
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 is independent of the mass of the block.
For $\theta_{\max }=15^{\circ}$
$ \Rightarrow \mu_{s}=\tan 15^{\circ}=0.27$