A block of mass $4 \mathrm{~kg}$ at rest on a rough inclined plane making an angle of $\theta$ with the horizontal. The coefficient of static friction between the block and plane is 0.5 and the frictional force on the block is $14.14 \mathrm{~N}$, find the value of $\theta$ ?

• A

  $30^{\circ}$
• Correct answer
  B

  $45^{\circ}$
• C

  $60^{\circ}$
• D

  $15^{\circ}$

Solution

Given, $m=4 \mathrm{~kg}, \mu_s=0.5$ $f r_k=14.14 \mathrm{~N}=10 \sqrt{2} \mathrm{~N}$ Normal reaction force, $N=m g \cos \theta=4 \times 10 \times \cos \theta=40 \cos \theta$ and kinetic friction, $\begin{array}{lrl} \Rightarrow & f r_k=\mu_s N \Rightarrow 10 \sqrt{2}=0.5 \times 40 \cos \theta \\ \Rightarrow & 10 \sqrt{2}=\frac{1}{2} \times 40 \cos \theta \\ \Rightarrow & 10 \sqrt{2}=20 \cos \theta \Rightarrow \cos \theta=\frac{10 \sqrt{2}}{20} \\ \Rightarrow & \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ} \end{array}$

Hence, the correct answer is option: 'B'.