Newton’s Third Law of Motion
Newton’s third law of motion states that to every action, there is always an equal and opposite reaction i.e. the force of action and reaction are always equal and opposite.
Explanation of Newton’s Third Law of Motion
Here, the term action means the force exerted by one body on the other body, while the term reaction means the force exerted by the second body on the first. If $F_{AB}$ is the force exerted on body A by body B (i.e., action) and $F_{BA}$ is the force exerted on body B by body A (i.e., reaction), then according to Newton’s third law of motion:
$$ F_{AB} = -F_{BA} \quad $$
To prove this law, consider an isolated system consisting of two bodies A and B moving along the same straight line.
Let them collide so that they mutually act and react upon each other. Due to this, their velocities will change, and hence their linear momenta will also change.
Let $\Delta \overrightarrow{P_1}$ and $\Delta \overrightarrow{P_2}$ be the change in linear momentum of body A and B respectively. According to the law of conservation of linear momentum, the net change in linear momentum of the system is zero, i.e.,
$$ \Delta \overrightarrow{P_1} + \Delta \overrightarrow{P_2} = 0 \quad $$
Dividing both sides by $\Delta t$ and taking the limit as $\Delta t \rightarrow 0$, we get
$$ \lim_{\Delta t \to 0} \frac{\Delta \overrightarrow{P_1}}{\Delta t} = – \lim_{\Delta t \to 0} \frac{\Delta \overrightarrow{P_2}}{\Delta t} $$
or
$$ \frac{d \overrightarrow{P_1}}{dt} = – \frac{d \overrightarrow{P_2}}{dt} \quad$$
i.e. Rate of change of linear momentum of B = – rate of change of linear momentum of A
or Force on B $=$ Force on A
$$ F_{BA} = -F_{AB} $$
Hence, the forces of action and reaction are equal and opposite, which proves Newton’s third law of motion.
Important Notes:
1. Newton’s third law signifies that forces in nature always occur in pairs. A single isolated force is not possible. Any agent applying a force (called direction) experiences an equal force in the opposite direction (called reaction).
2. Forces of action and reaction act always on different bodies. Hence, the action and reaction are not balanced forces, as each force is acting on a different body. Therefore, these forces cannot cancel each other.
3. The action and reaction do not act simultaneously, but they are always in action.
4. Newton’s third law is applicable whether the bodies are at rest or they are in motion. The third law is applicable for all types of forces like gravitational, electric, magnetic, mechanical, muscular, or magnetic forces, etc.
Illustrations of Newton’s Third Law
Some of the examples of Newton’s third law of motion are given below:
1. Book kept on a table. A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action. The table supports the book by exerting an equal force upwards. This is the force of reaction. As the system is at rest, net force on it is zero. Therefore, forces of action and reaction must be equal and opposite.
2. Walking. While walking, a person presses the ground in the backward direction (action) by his feet. The ground pushes the person forward with an equal force (reaction). The magnitude of reaction in the horizontal direction compels the person to move forward.
3. Swimming. A swimmer pushes the water backwards (action). The water pushes the swimmer forward (reaction) with the same force. Hence, the swimmer swims.
4. Firing from a gun. When a gun is fired, the bullet moves forward (action). The gun recoils backwards (reaction).
5. Flight of jet planes and rockets. The burnt fuel which appears in the form of hot and highly compressed gases escapes through a nozzle (action) in the backward direction. The escaping gases push the jet plane or rocket forward (reaction) with the same force. Hence the jet or rocket moves.
6. Rebounding of a rubber ball. When a rubber ball is struck against a wall or floor, it exerts a force on the wall (action). The wall exerts an equal force (reaction) exerted by the wall or floor on the ball.
7. It is difficult to walk on sand or ice. This is because on pushing, sand gets displaced and reaction from sandy ground is very little. In the case of ice, force of reaction is small because friction between ice and feet is very little.
8. Driving a nail into a wooden block without holding the block is difficult. This is because, when the wooden block is not resting against a support, the block and nail both move forward on being hit with a hammer. However, when the block is held firmly against a support, and the nail is hit, an equal reaction of the support drives the nail into the block.
Horse and Cart Problem
The various forces acting on a system of horse and cart at rest are shown as Figure 3(a).7.
Here, the weight $(W_1)$ of the cart $C$ is balanced by the reaction $(R_1)$ of the ground on the cart. The weight $W_2$ of the horse $H$ is balanced by the reaction $R_2$ of the ground on the horse.
The horse pulls the cart with a force $F$ in the forward direction. The cart, in turn, pulls the horse with the same force $F$ in the backward direction. These two forces are also balanced.
While pulling the cart, the horse pushes the ground backwards with its foot by a force $F \sin \theta$ at an angle $\theta$ with the horizontal. As a reaction, the ground exerts force $R$ on the horse equal and opposite to $F$.
$ \displaystyle R\sin \theta $ vertically upward and $ \displaystyle R\cos \theta $ along the horizontal.
The component $R \cos \theta$ tends to move the cart forward. This motion is opposed by the force of friction $f$ between the cart and the ground. The cart will move only when $R \cos \theta > f$.
Example 1:
A 40 kg shell is flying at a speed of 72 km/h. It explodes into two pieces, one piece of mass 15 kg stops. Calculate the velocity of the other piece.
Solution: Here,
$$ m_1 = 15 \, \text{kg}, \,\, m_2 = 25 \, \text{kg}, \,\, v_1 = 0 $$
Let the velocity of the other piece be $v_2$.
Using the principle of conservation of momentum,
$$ m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2 $$
$$ \text{or} \, 0 + 25 \cdot v_2 = 40 \cdot \frac{72 \, \text{km/h}}{18 \, \text{m/s}} $$
$$ \therefore v_2 = \frac{800}{25} = 32 \, \text{m/s} $$
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Example 2:
A machine gun has a mass of 10 kg. It fires 5 bullets of mass $35 \, \text{g}$ each in a second with a speed of $400 \, \text{m/s}$. What force must be applied to keep the gun in position?
Solution:
Mass of each bullet, $m = 5 \times 35 \, \text{g} = 0.035 \, \text{kg}$
Number of bullets fired per second $= 5$
$$ v = 400 \, \text{m/s} $$
Number of bullets/sec $= n \, v = 5 \times 0.035 \times 400 = 70 \, \text{kg m/s} $$
Therefore,
$$ F = \frac{m \cdot v}{t} = 70 \, \text{N} $$
Apparent Weight of a Man in a Lift/Elevator
Suppose a person of mass $m$ is standing on a weighing machine placed in an elevator/lift. The actual weight of the person $= mg$. This acts on the weighing machine, which offers a reaction $R$ given by the reading of the weighing machine. This reaction exerted by the surface of contact on the person is the apparent weight of the person. We shall discuss how $R$ is related to $mg$ in the following different situations:
(i) When the elevator is at rest
Acceleration of the person $= 0$
Net force on the person $= 0$
i.e. $$ R – mg = 0 $$
or
$$ R = mg $$
i.e. apparent weight is equal to the actual weight of the person.
(ii) When the elevator is moving uniformly in the upward/downward direction
Again, in uniform motion, acceleration of the person = 0. Since, $$ R = mg $$ i.e.,
apparent weight = actual weight of the person.
(iii) When the elevator is accelerating downwards
Suppose uniform downward acceleration of the person in the lift $= a$
Therefore, net downward force on the person,
$$ f = ma $$
As is clear from Fig. (c),
$$ R_2 = m(g – a) \quad $$
Thus, $R_2 < mg$
Hence apparent weight of the person becomes less than the actual weight when the elevator is accelerating downwards.
(v) In free fall of a body under gravity, $a = g$:
$$ R_2 = m(g – g) = 0 $$
i.e. apparent weight of the body becomes zero, or the body becomes weightless.
(vi) When downward acceleration is greater than $g$
i.e. $a > g$
$$ R_2 = m(g – a) $$
$ \displaystyle R_2$ becomes negative i.e. apparent weight of the person becomes negative. In that event, the person will rise from the floor of the lift and stick to the ceiling of the lift.
