Question: 12: In the diagram shown, the normal reaction force between 2 kg and 1 kg is
(Consider the surface, to be smooth):
Given g = 10 \,\mathrm{m\,s^{-2}}
(1) 25 N
(2) 39 N
(3) 6 N
(4) 10 N
Answer: Option (1)
Explanation:
Since the surface is smooth, only normal reactions and applied forces act between the blocks.
All three blocks lie on an inclined plane of angle \theta = 30^\circ.
The component of weight of each block along the incline is given by
mg\sin\thetaFor the 3 kg block, a force of 60 N is applied up the incline.
For the 1 kg block, a force of 18 N is applied down the incline.
Let us first find the acceleration of the system. Total mass of the system is
M = 3 + 2 + 1 = 6\ \text{kg}Total force acting along the incline is
F_{\text{net}} = 60 - 18 - (6 \times 10 \times \sin 30^\circ) F_{\text{net}} = 60 - 18 - 30 = 12\ \text{N}Hence, acceleration of the system is
a = \frac{12}{6} = 2\ \text{m s}^{-2}Now consider the 1 kg block alone. Forces acting on it along the incline are:
1. Normal reaction N from the 2 kg block upward along the incline
2. Applied force of 18 N downward along the incline
3. Component of its weight downward along the incline
Applying Newton’s second law along the incline for the 1 kg block:
N - 18 - (1 \times 10 \times \sin 30^\circ) = 1 \times a N - 18 - 5 = 2 N = 25\ \text{N}Therefore, the normal reaction force between the 2 kg and 1 kg blocks is 25 N.
Hence, the correct option is (1).