5.0 Normal Force

  Laws of Motion

It is a force, which always acts along the common normal between the two bodies at the point of contact.

It is represented by a $N$.

Common normal: Common normal is perpendicular to the line of contact of block and ground as shown in the figure.

Note:

• The normal force always acts along the common normal
• Block exerts a downward normal force on a ground as shown in the figure.
• So, as a reaction, ground also exerts an upward normal force on block.
• If a normal force is greater than zero $(N>0)$, it means that the bodies are in contact
• If a normal force is zero $(N=0)$, it means that the bodies are not in contact
• If a normal force is less than zero $(N<0)$, it means that the bodies are in contact but the direction of normal force is in opposite direction

The diagram belows helps us to understand the normal force $(N)$ better.

Question 1. Find the force $F$, so that the block looses contact with the ground.

Solution: For equilibrium in the vertical direction, $$N + F = mg\quad ...(i)$$

Condition for block to loose contact with the ground, $$N = 0\quad ...(ii)$$

Therefore, from equation $(i)$ & $(ii)$ we get, $$F = mg$$

Question 2. A man of mass $M$ is standing on a weighing machine placed on ground. Calculate the machine reading $(W)$.

Solution: From FBD of man, $${N_1} = Mg\quad ...(i)$$

From FBD of weighing machine, $${N_2} = {N_1}\quad ...(ii)$$

From equation $(i)$ and $(ii)$ we get,

$${N_2} = Mg\quad $$

Note:

• $N_2$ is the force which is provided by ground to balance $N_1$
• $N_1$ is the force exerted by the man on the weighing machine