 In mechanics, we must know three things about the any force;

(i) when does it act?

(ii) In what direction does it act?

(iii) What is its magnitude?

Normal Reaction is a contact force which acts on body if its motion is constrained is any particular direction. It acts in a direction opposite to the direction in which motion of body is constrained. The magnitude of normal reaction depends on the degree to which the motion of body is constrained. Let us take an example to understand this let us consider a block of mass 10 kg in four different situation.

(a) Body is under free fall and its motion is not at all constrained. It moves freely under gravity with acceleration of 9.8 m/s2. Since, the motion of the body is not at all constrained, The normal reaction will be zero.
(b) The body is at rest on the surface. The weight (10g) act in downward direction. Had it been free, it should have moved with acceleration g m/s2 downward direction but it is at rest.
So, we can say the motion of body is completely constrained in downward direction. Since, its motion is constrained a normal reaction act. In fact, it is this reaction exerted by the surface on the block which constrains the motion of body. So, the normal reaction N acts on the block in upward direction constraining the motion of body in downward direction.
Since, there is no acceleration of body in vertical direction. So, net force in upward direction is zero.
N – 10g = 0
So, normal reaction N = 100 Newton
The surface exert a normal reaction of 100 N to constrain the motion of body in downward direction. On the other hand the body will apply a reaction force of 100 N on the surface in opposite direction i.e. downward direction according to Newton’s third law.

(c) Here, 10 kg mass moves with acceleration 2 m/s2 is downward direction. Had it been free, it would have accelerated with 9.8 m/s2 in downward direction. So, its motion is also constrained in downward direction but less as compared to case (b). So, here also N will act in upward direction. The magnitude of N can be expected to be less compared to (b) as the motion of body in less constrained compared to previous case.

Applying Newton second law on 10 kg mass
10 gN = 10 × 2
N = 80 Newton

(d) In this case, motion of 10 kg mass is over constrained. It moves with acceleration of 2 m/s2 in upward direction opposite to direction in which it would have moved if it would have been free. We can expect N to be greatest in this case.
Applying Newton Second law
N – 10g = 10 × 2
N = 120 Newton
So, the magnitude of normal reaction acting on the body depends on degree to which motion of body is constrained. Let us take more examples on normal reaction.

Example 1 :
Two blocks of masses 10 kg and 5 kg are put one above the other as shown in figure. The whole system moves downward with an acceleration 2m/s2. Find the normal reaction between 10 kg and
5 kg and between 10 kg and the surface.

Solution :
Let us draw the free body diagram of each body.

In order to draw the free body diagram, you first analyse all the interactions it is subjected to
(i) FBD of 5 kg block
It is subjected to following interactions.
(a) It interacts with earth which attracts it with force 5g in downward direction.
(b) It interacts with 10 kg body which constrains its motion by exerting force N in upward direction.
(ii) FBD of 10 kg block
It is subjected to following interactions.
(a) It interacts with earth which attracts it with force 10g in downward direction.
(b) It interacts with 5 kg body by exerting a force N on it in upward direction. As a reaction, 5 kg body will apply a reaction force N on 10 kg block in downward direction.
(c) It interacts with surface which constrains its motion by exerting a force R on 10 kg block in upward direction.
Applying Newton’s second law on 10 kg and 5 kg blocks
10g + NR = 10a
5gN = 5a
Putting a = 2m/s2, we get N = 40 Newtons and R = 120 Newtons

Example 2 :
Two blocks of mass 5kg and 3kg are kept one above the other on the floor of lift. Which is accelerating in downward with an acceleration of 2m/s2. Find the normal reaction between two blocks.

Solution :
Let us draw the free body diagram of two blocks.

Both the blocks are accelerating with 2m/s2 in downward direction.
Applying Newton second Law.
3gN1 = 3a = 6
5g + N1N2 = 5a = 10
Solving above two equation
N1 = 24N
N2 = 64N
So, normal reaction between two blocks is N1= 24N

Example 3 :
Three blocks of masses 5kg, 3kg & 2kg are kept on the floor of a lift which moves in downward direction with acceleration 2m/s2. Find the force of interaction between lift and 5kg body.

Solution :
Let us draw the free body diagram of all the three blocks.
Block of mass 2kg.
Block of 2kg mass interacts 3kg mass and earth. So, there are two forces, N1 (interaction force between 3kg mass and 2kg mass) and 2g (interaction force between 2kg body and earth). Block of mass 3kg exert force N1 on 2kg, which constrains the motion of 2kg body in downward direction and hence acts in upward direction and 2g acts towards earth in downward direction.

Block of Mass 3kg
Block of 3 kg interacts with 2kg mass and 5kg mass and earth. So, there are three forces N1 (interaction force between 2kg mass and 3kg mass), N2 (interaction force between 3kg mass and 5kg mass) and 3g (interaction force between 3kg mass and earth)
2kg mass exerts reaction force N1 on 3kg mass in downward direction Motion of 3kg mass is constrained in downward direction hence, N2 is exerted on 3kg mass in upward direction by 5kg mass in upward direction. The weight 3g acts on 3kg mass in downward direction.

Block of Mass 5kg.
Block of mass 5kg interacts with 3kg mass and floor and earth. So, there are three forces : N3 (interaction force between 5kg mass and floor of lift), N2 (interaction force between 3kg mass and 5kg mass) and 5g (interaction force between 5kg mass and earth) Since, motion of 5kg mass is constrained in downward direction, so, floor exert normal reaction N3 on 5kg body in upward direction. 5kg mass exert N2 on 3kg mass, if this is action of 5kg body on 3kg body then 3kg body exerts reaction N2 on 5kg in downward direction according to Newton’s third law. The weight 5g acts on 5kg mass in downward direction towards earth.

The acceleration of all the three blocks is 2m/s2. Applying Newton’s second law
2gN1 = 2a = 4 ….(i)
3g + N1N2 = 3a = 6 ….(ii)
5g + N2N3 = 5a = 10 ….(iii)
Solving (i), (ii) & (iii)
N1 = 16N
N2 = 40N
N3 = 80N

Example 4 :
Two blocks of mass 4 kg and 2 kg are placed side by side on a smooth horizontal surface as shown in the figure. A horizontal force of 20 N is applied on 4 kg block. Find :
(a) the acceleration of each block
(b) the normal reaction between two blocks.

Solution :
Let us first draw free body diagram (FBD) of both bodies

Here, motion of 4 kg and 2 kg mass are constrained in downward direction. So, a normal reaction N1 and N2 act in upward direction. Also motion of body is constrained in forward direction due to presence of 2 kg mass. So, a reaction force R will act on 4 kg mass in direction opposite to the direction in which motion is constrained. So, R will act in backward direction. It is 2 kg mass which exert R on 4 kg mass which can be called action. So, 4 kg mass will exert a reaction force R on 2 kg mass in opposite direction.
Let a be the acceleration of bodies in right direction.
Applying Newton’s second law on 4 kg mass
There is no acceleration in 4 kg body in vertical direction
So, N1 – 4g = 0 … (i)
Let acceleration of 4 kg body in forward direction is a
20 – R = 4a … (ii)
Now, let us apply Newton’s second law on 2 kg mass
In this case also, there is no acceleration in vertical direction and acceleration in horizontal direction is a
N2 – 2g = 0 … (iii)
R = 2a … (iv)
From (i) and (iii),
N1 = 4g = 40 N
N2 = 2g = 20 N
From (ii) and (iv), $a=\dfrac{{10}}{3}m\text{/}{{s}^{2}}$

$R =\dfrac{{20}}{3}N$

Example 5 :
Two blocks of mass 3 kg and 5 kg are released from rest over a smooth inclined plane of inclination 30º as shown in figure. What is the normal force between the two blocks?

Solution :
Let us first draw free body diagram
Weight 5 g and 3 g will act on 5 kg and 3 kg mass in downward direction

Now, motion of 5 kg and 3 kg mass is constrained in direction perpendicular into the inclined plane. So, a normal reaction N1 and N2 will act in direction opposite to it

Now, the presence of 3 kg body will constrain the motion of 5 kg mass down the inclined plane. Its motion wont be as free as in absence of 3 kg mass. So, a normal force will act on 5 kg in direction upward along the inclined plane.

Since R is exerted by 3 kg block on 5 kg block, (treated as action), then according to Newton’s third law, 5 kg will exert an equal reaction force on 3 kg mass in opposite direction.

There is no motion of two bodies in y direction. So, net force in y direction on each body will be zero.

N1 – 5g cos 30º = 0  $\displaystyle \Rightarrow$  N1 = 43.3 N
N2 – 3g cos 30° = 0  $\displaystyle \Rightarrow$  N2 = 26 N

Let bodies are moving down the incline with acceleration a
For 5 kg mass
Net force in positive x direction
Fx = 5g sin 300° – R
Applying Newton second law Fx = max
$\displaystyle \Rightarrow$ 5g sin 300 – R = 5a … (i)

Similarly, for 3 kg mass
R + 3g sin 300 = 3a … (ii)
(i) + (ii),
8g sin 30° = 8a
a = 5m/s2
R = 3a – 3g sin 30° = 0
Significance of R = 0N.
If two masses 5 kg and 3 kg would have been alone, It would have come down with acceleration g sin 30° or 5 m/s2. So, presence of 3 kg mass is not constraining the motion of 5 kg mass. Hence, normal force between 5 kg and 3 kg mass is zero.