Question

A block kept on a rough surface starts sliding when the inclination of the surface is $\mathrm{\theta }$ with respect to the horizontal. The coefficient of static friction between the block and the surface is

• A. $\sec \mathrm{\theta }$
• B. $\sin \mathrm{\theta }$
• C. $\tan \mathrm{\theta }$
• D. $\cos \mathrm{\theta }$

Answer 1

The correct option is C

$\tan \mathrm{\theta }$

Step 1: Given data and free body diagram

Let mass of block$=\mathrm{m}$

Inclination of the surface with respect to horizontal$=\mathrm{\theta }$

Acceleration due to gravity$=\mathrm{g}=10{\mathrm{ms}}^{-2}$

The free body diagram is given below

Step 2: Calculating the angle of inclination

Balancing the forces which are normal to the block we get'

$\mathrm{N}=\mathrm{mgcos}\left(\mathrm{\theta }\right)$

And balancing the forces which are parallel to the inclined surface we get,

$\mathrm{f}=\mathrm{mgsin}\left(\mathrm{\theta }\right)$

We know that when the body begins to slide the frictional force acting on the body is given by:

$$\begin{gathered} \mathrm{f}=\mathrm{\mu N}(\mathrm{\mu }=\mathrm{coefficient}\mathrm{of}\mathrm{friction}) \\ \Rightarrow \mathrm{mgsin}\left(\mathrm{\theta }\right)=\mathrm{\mu mgcos}\left(\mathrm{\theta }\right) \\ \Rightarrow \sin \left(\mathrm{\theta }\right)=\mathrm{\mu cos}\left(\mathrm{\theta }\right) \\ \Rightarrow \mathrm{\mu }=\frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)} \\ \Rightarrow \mathrm{\mu }=\tan \left(\mathrm{\theta }\right) \end{gathered}$$

Therefore, the coefficient of static friction is equal to $\tan \left(\mathrm{\theta }\right)$.

Hence option(C) is the correct answer.