Newton’s Third Law of Motion

Newton's Third Law of Motion is a key concept in physics. It says that for every action, there's an equal and opposite reaction. In simple terms, when one object applies a force on another, the second object pushes back with the same force, just in the opposite direction. These forces always come in pairs, and they help us understand many everyday actions, like how a rocket launches or how we walk. Knowing this law is crucial for solving problems about movement, force, and momentum in physics.

1.0Statement of Newton's Third Law of Motion

• If a body A exerts a force $\vec{F}$ on another body B, then B exerts a force $\vec{F}$ on A, the two forces acting along the line joining the bodies.

• Newton's Third Law states that every action has an equal and opposite reaction, also known as the action-reaction law.
• Here ${\vec{F}}_{12}$ (force on first body due to second body) is equal in magnitude and opposite in direction to (force on second body due to first body).

$\vec{F}12=-{\vec{F}}_{21}$

2.0Properties of Action – Reaction Pairs

Newton's Third Law of Motion: For every action, there is an equal and opposite reaction.

3.0Key Points About  Newton’s Third Law of Motion

•  A single isolated force cannot exist in nature; forces are always produced in equal and opposite action-reaction pairs.

1. Simultaneous Action and Reaction
   There is no time delay between action and reaction. Neither force is the cause or effect—they occur at the same time.
2. Applicable in All States of Motion
   Newton’s Third Law holds true whether the bodies are at rest or in motion.
3. Acts Without Physical Contact
   The law is valid even for non-contact forces such as gravity and electrostatic interactions.
4. Universal Applicability
   It applies to all types of forces, including gravitational, electrostatic, magnetic, tension, friction, and viscous forces.
5. Forces Act on Different Bodies
   Action and reaction forces never cancel each other out because they act on different objects.

Examples Of Newton’s Third Law of Motion

1.Book Kept on a Table

• When a book rests on a table, it pushes down due to its weight (action), and the table pushes up with an equal and opposite force (reaction). This illustrates Newton’s Third Law: every action has an equal and opposite reaction.

2.While walking we press the ground (action) with our feet slightly slanted in the backward direction.

The ground exerts an equal and opposite reaction force on us. Its vertical component supports our body weight, while the horizontal component propels us forward, enabling us to walk.

4.0Horse and Cart Problems

• A cart connected to a horse by string.The horse while pulling the kart produces a tension T in the string in the forward direction(action).The cart in turn pulls the horse by an equal force T in the opposite direction.
• Initially the horse presses the ground with a force F in an inclined direction.The reaction R of the ground acts on the horse in the opposite direction.
• The horse exerts an inclined force F on the ground. In response, the ground applies an equal and opposite reaction force R on the horse. This reaction force R can be resolved into two perpendicular components.

1.Vertical component V which balances the weight W of the horse

2.Horizontal component H which helps the horse to move forward 

Let f be the frictional force; the horse advances forward, H>T

Net force acting on the horse=H-T

H-T=ma………..(1)

The cart moves forward if T>f

Net force acting on the cart=

The cart's weight is balanced by an equal and opposite normal force from the ground,acceleration acting on the cart is also a,if mass M is the mass of the cart,then

T-f=ma………..(2)

adding(1) and (2)

$H-f=(M+m)a$

$a=\frac{H-f}{M+m}$

If a is positive if H-f is positive or if H>f

The system moves if H>f

5.0Conservation of Linear Momentum from Newton’s Third Law

During collision the body A exerts a force ${\vec{F}}_{BA}$ on body B from Newton’s Third Law ,the body B also exerts a force ${\vec{F}}_{AB}$ on body A such that,

$$\begin{gathered} {\vec{F}}_{AB}=-{\vec{F}}_{BA} \\ \text{Impulse of}{\vec{F}}_{AB}={\vec{F}}_{AB}\Delta t=\text{change in momentum of A}=m_1{\vec{v}}_1-m_1{\vec{u}}_1 \\ \text{Impulse of}{\vec{F}}_{BA}={\vec{F}}_{BA}\Delta t=\text{change in momentum of B}=m_2{\vec{v}}_2-m_2{\vec{u}}_2 \\ {\vec{F}}_{AB}\Delta t=-{\vec{F}}_{BA}\Delta t \\ {m}_1{\vec{v}}_1-m_1{\vec{u}}_1=-(m_2{\vec{v}}_2-m_2{\vec{u}}_2) \\ {m}_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2 \end{gathered}$$

Total linear momentum before collision=Total linear momentum after collision

6.0Linear Momentum Conservation via Newton’s Third Law

• Two object of masses m1 and m2 moving along a straight line and colliding against each other.The velocities and hence momentum of two bodies undergoes a change,p1 and p2 be the change in momentum produced in time t
• The net change in linear momentum in the absence of external force is zero.

$$\begin{gathered} \Delta p_1+\Delta p_2=0 \\ \Delta p_2=-\Delta p_1 \\ \frac{\Delta p_2}{\Delta t}=-\frac{\Delta p_1}{\Delta t} \\ \text{Rate of change of momentum of}{m}_2=\text{- Rate of change of momentum of }{m}_1 \\ \text{Force on}{m}_2=\text{- Force on}{m}_1 \\ \text{Action =- Reaction} \end{gathered}$$

This proves Newton’s Third Law of Motion

7.0FAQ On Newton's Third Law of Motion

Q-1.Show the normal reaction forces in the following figures?

Solution: Here $N_1\text{and}{N}_2$ normal reaction forces perpendicular to the well defined surface towards the centre of the body.

Q-2.Draw the free body diagrams and find tension in each string and acceleration of the system.

Solution:

$\text{Acceleration}=\frac{\text{Net pulling force}}{\text{Total mass to be pulled}}$

$a=\frac{30}{10}=3{\text{m/s}}^2$

(a) FBD of 3 kg

$$\begin{gathered} T_1=3a \\ {N}_3-3g=3\times 0 \\ {N}_3=30\text{N} \end{gathered}$$

(b) FBD of 2 kg

$$\begin{gathered} T_2-T_1=2a \\ {N}_2-2g=2\times 0 \\ {N}_2=20N \end{gathered}$$

(c) FBD of 5 kg

$$\begin{gathered} 30-T_2=5a \\ {N}_5-5g=5\times 0 \\ {N}_5=50N \end{gathered}$$

Note: Here, $N_{AB}\text{and}{N}_{BA}$ are the action–reaction pair (Newton’s third law).

Q-3.A book of 1 kg is held against a wall by applying a perpendicular force F. If ${\mu }_s=0.2$, then what is the minimum value of F ? $(Takeg=9.8m/{\mathrm{s}}^2)$

Solution: The forces acting on the book are –

For book to be at rest it is essential that $mg=f_s$

$$\begin{gathered} f_s^{\text{max}}={\mu }_{s}N\text{and }N=F \\ mg={\mu }_{s}F \\ F=\frac{mg}{{\mu }_s}=\frac{1\times 9.8}{0.2}=49N \end{gathered}$$