A block is between two surfaces as shown in the figure. Find the normal reaction at both surfaces. [Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

• Correct answer
  A

  $N_1=37.2 \mathrm{~N}$ and $N_2=9.6 \mathrm{~N}$
• B

  $N_1=38.2 \mathrm{~N}$ and $N_2=8.6 \mathrm{~N}$
• C

  $N_1=40 \mathrm{~N}$ and $N_2=4 \mathrm{~N}$
• D

  $N_1=37.5 \mathrm{~N}$ and $N_2=9.9 \mathrm{~N}$

Solution

The given situation is shown in the figure. Resolving applied force, we have following forces on block
Given, $\tan \theta=\frac{3}{4}, \sin \theta=\frac{3}{5}$ and $\cos \theta=\frac{4}{5}$
So, horizontal component, $F_1=12 \times \cos \theta=12 \times \frac{4}{5}=\frac{48}{5} \mathrm{~N}$ and vertical component, $F_2=12 \times \sin \theta=12 \times \frac{3}{5}=\frac{36}{5} \mathrm{~N}$ So, total applied force on ground is $10+\frac{36}{5}+2 \times 10=37.2 \mathrm{~N}$ So, reaction $N_1$ of ground $=37.2 \mathrm{~N}$ upwards Also, total force horizontally on wall is $F_1=\frac{48}{5}=9.6 \mathrm{~N}$ So, reaction $N_2$ of wall $=9.6 \mathrm{~N}$ towards left

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