Question

In the given figure, a mass of $4kg$ 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 $mg\sin \theta =f_s,mg\cos \theta =N$
$\Rightarrow \tan \theta =\frac{f_s}{N}$
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 ${\theta }_{\max }=15^{\circ }$
$\Rightarrow {\mu }_s=\tan 15^{\circ }=0.27$