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*/*s*^{2}. 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 (10*g*) act in downward direction. Had it been free, it should have moved with acceleration g *m*/*s*^{2} 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* – 10*g* = 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*/*s*^{2} is downward direction. Had it been free, it would have accelerated with 9.8 *m*/*s*^{2} 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 *g* – *N* = 10 × 2* N* = 80 Newton

(d) In this case, motion of 10 *kg* mass is over constrained. It moves with acceleration of 2 *m*/*s*^{2} 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* – 10*g* = 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 2*m*/*s*^{2}. 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) F**BD of 5 kg block**

It is subjected to following interactions.

(a) It interacts with earth which attracts it with force 5*g* 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 10*g* 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

10*g* + *N* – *R* = 10*a*

5*g* – *N* = 5*a*

Putting *a* = 2*m*/s^{2}, we get *N* = 40 Newtons and *R* = 120 Newtons

**Example 2 :**

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

**Solution :**

Let us draw the free body diagram of two blocks.

Both the blocks are accelerating with 2*m*/*s*^{2} in downward direction.

Applying Newton second Law.

3*g* – *N*_{1} = 3*a* = 6

5*g* + *N*_{1} – *N*_{2} = 5*a* = 10

Solving above two equation

*N*_{1} = 24*N*

*N*_{2} = 64*N*

So, normal reaction between two blocks is *N*_{1}= 24*N*

**Example 3 :**

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

**Solution : **

Let us draw the free body diagram of all the three blocks.**Block of mass 2kg.**

Block of 2*kg* mass interacts 3*kg* mass and earth. So, there are two forces, *N*_{1} (interaction force between 3*kg* mass and 2*kg* mass) and 2*g* (interaction force between 2*kg* body and earth). Block of mass 3*kg* exert force *N*_{1} on 2*kg*, which constrains the motion of 2*kg* body in downward direction and hence acts in upward direction and 2*g* acts towards earth in downward direction.

**Block of Mass 3 kg**

Block of 3

*kg*interacts with 2

*kg*mass and 5

*kg*mass and earth. So, there are three forces

*N*

_{1}(interaction force between 2

*kg*mass and 3

*kg*mass),

*N*

_{2}(interaction force between 3

*kg*mass and 5

*kg*mass) and 3

*g*(interaction force between 3

*kg*mass and earth)

2

*kg*mass exerts reaction force

*N*

_{1}on 3

*kg*mass in downward direction Motion of 3

*kg*mass is constrained in downward direction hence,

*N*

_{2}is exerted on 3

*kg*mass in upward direction by 5

*kg*mass in upward direction. The weight 3

*g*acts on 3

*kg*mass in downward direction.

**Block of Mass 5kg.**

Block of mass 5*kg* interacts with 3*kg* mass and floor and earth. So, there are three forces : *N*_{3} (interaction force between 5*kg* mass and floor of lift), *N*_{2} (interaction force between 3*kg* mass and 5*kg* mass) and 5*g* (interaction force between 5*kg* mass and earth) Since, motion of 5*kg* mass is constrained in downward direction, so, floor exert normal reaction *N*_{3} on 5*kg* body in upward direction. 5*kg* mass exert *N*_{2} on 3*kg* mass, if this is action of 5*kg* body on 3*kg* body then 3*kg* body exerts reaction *N*_{2} on 5*kg* in downward direction according to Newton’s third law. The weight 5*g* acts on 5*kg* mass in downward direction towards earth.

The acceleration of all the three blocks is 2*m*/*s*^{2}. Applying Newton’s second law

2*g* – *N*_{1} = 2*a* = 4 ….(i)

3*g* + *N*_{1} – *N*_{2} = 3*a* = 6 ….(ii)

5*g* + *N*_{2} – *N*_{3} = 5*a* = 10 ….(iii)

Solving (i), (ii) & (iii)

*N*_{1} = 16*N*

*N*_{2} = 40*N*

*N*_{3} = 80*N*

**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 *N*_{1} and *N*_{2} 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, *N*_{1} – 4*g* = 0 … (i)

Let acceleration of 4 *kg* body in forward direction is a

20 – *R* = 4*a* … (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

*N*_{2} – 2*g* = 0 … (iii)

*R* = 2*a* … (iv)

From (i) and (iii),

*N*_{1} = 4*g* = 40 *N*

*N*_{2} = 2*g* = 20 *N*

From (ii) and (iv), a=\dfrac{{10}}{3}m\text{/}{{s}^{2}}

**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 *N*_{1} and *N*_{2} 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.

*N*_{1} – 5*g* cos 30º = 0 \displaystyle \Rightarrow *N*_{1} = 43.3 *N**N*_{2} – 3*g* cos 30° = 0 \displaystyle \Rightarrow *N*_{2} = 26 *N*

Let bodies are moving down the incline with acceleration a

For 5 *kg* mass

Net force in positive *x* direction

*F _{x}* = 5

*g*sin 300° –

*R*

Applying Newton second law

*F*=

_{x}*ma*

_{x}\displaystyle \Rightarrow 5

*g*sin 300 –

*R*= 5

*a*… (i)

Similarly, for 3 *kg* mass

*R* + 3*g* sin 300 = 3*a* … (ii)

(i) + (ii),

8*g* sin 30° = 8*a*

*a* = 5*m*/*s*^{2}

*R* = 3*a* – 3*g* 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*/

*s*

^{2}. 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.