A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during th

NCERT Exemplar · Q3

Q. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

(a) constant and equal to

mg

in magnitude.

(b) constant and greater than

mg

in magnitude.

mg

(d) at first greater than

mg

, and later becomes equal to

mg

Punjab PsebMCQ· 1mImportance★★★★★est

46% · 38/83 Questions

✓ Free question

Concept understanding — Newton Third Law Pairs

Newton's Third Law Pairs

The Intuition: Forces Never Act Alone

Imagine you are standing on a skateboard and you push against a wall. The wall does not move, but you roll backward. You pushed the wall, and the wall pushed back on you — at the very same instant, with the very same strength.

This is the core idea behind Newton's Third Law: whenever one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first. These two forces are called an action-reaction pair (or "Third Law pair").

Note

The two forces in a Third Law pair act at the same instant — there is no delay between "action" and "reaction." They are two sides of a single interaction, not a cause followed by an effect.

The Precise Statement

Newton's Third Law: If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}

Both forces have the same magnitude and act along the same line, but point in opposite directions — and, crucially, they act on two different objects.

The Key Rule for Spotting a Third Law Pair

A genuine Third Law pair must satisfy two conditions:

The two forces act on different objects (never on the same object).
They arise from the same interaction (the same pair of bodies in contact, or the same gravitational/electric attraction).

If a pair of forces acts on the same object, they are not a Third Law pair — even if they happen to be equal in magnitude and opposite in direction.

Why Equal and Opposite Forces Don't Cancel

This confuses many students: if the two forces are always equal and opposite, why does anything ever move?

Because the two forces act on different objects — they never appear together in the same free-body diagram. When you push against a wall, the wall pushes back on you. That reaction force acts on you, not on the wall, so it can accelerate you backward even though the wall itself does not move (the wall is anchored to the Earth, which is far too massive to notice the force).

A Careful Worked Example: A Book on a Table

A book sits at rest on a table. Two different force pairs are involved here, and it is important not to mix them up:

Pair 1 (gravitational interaction): Earth pulls the book downward with a force equal to the book's weight, $mg$ . By the Third Law, the book pulls the Earth upward with the same magnitude, $mg$ . This pair acts on two different bodies — the book and the Earth.
Pair 2 (contact interaction): The table pushes up on the book with a normal force $N$ . By the Third Law, the book pushes down on the table with a force of the same magnitude, $N$ . This pair also acts on two different bodies — the book and the table.

Watch out

A very common mistake is to say "the normal force equals the weight because of Newton's Third Law." This is wrong. The book's weight and the table's normal force both act on the same object (the book), so they cannot be a Third Law pair — they merely happen to be equal in magnitude here because the book is in equilibrium (Newton's second law with zero acceleration, $N - mg = 0$ ). The true Third Law partner of the book's weight is the pull the book exerts on the Earth; the true Third Law partner of the normal force is the push the book exerts on the table.

More Everyday Examples

Walking: Your foot pushes backward on the ground; the ground pushes forward on your foot. That forward push is what propels you ahead.
Rocket propulsion: The rocket pushes exhaust gases backward; the gases push the rocket forward. This works even in the vacuum of space, since no air is needed — only the exhaust and the rocket exerting forces on each other.
Swimming: A swimmer pushes water backward with their hands and feet; the water pushes the swimmer forward.

How to Use This in Problems

For every force you draw on an object, ask: "What is the Third Law partner of this force, and on what other object does it act?"
Draw a separate free-body diagram for each object. A Third Law pair will show up as one arrow in each of two different diagrams — same length, opposite direction, same type of force (both gravitational, or both contact, or both electric).
Never draw both members of a Third Law pair on the same free-body diagram — they belong to different objects.

The Deeper Reason

Newton's Third Law is closely tied to the conservation of momentum. In any isolated interaction between two objects, whatever momentum one gains, the other loses — because the forces they exert on each other are equal and opposite at every instant. This is why the law holds for every kind of fundamental interaction: gravitational, electromagnetic, and even contact forces, which are ultimately electromagnetic in origin.

Students preparing for boards often pair a search for "Newton Third Law Pairs class 11 physics" with "NCERT Physics syllabus" — Newton Third Law Pairs is a syllabus-aligned topic under Laws of Motion in NCERT Class 11 Physics, making it a natural fit for both board exams and JEE/NEET practice sets. Working through the worked examples above alongside the official NCERT Physics textbook is the most reliable way to turn this understanding into exam-ready recall.

The ground reaction force must first exceed $mg$ to accelerate the man upward from rest, then return to $mg$ once he moves at constant speed (or stops). The correct option is (D).

The key is to think about Newton’s second law — not just the third law. When the man squats and then stands up, his centre of mass does not move at constant velocity. It starts at rest, accelerates upward, then decelerates to rest again at the top. The ground reaction force is the upward normal force $N$ from the floor. The man’s weight $mg$ acts downward. The net force on the man is $N - mg$ , and this equals $ma$ , where $a$ is the acceleration of his centre of mass.

If the man simply stood still, $N = mg$ . But during the act of standing, his centre of mass must gain upward speed, so there must be a period of upward acceleration. That requires $N > mg$ . Later, as he approaches the upright position, he must slow down (decelerate upward), which means $N < mg$ briefly. However, the question’s options only mention “greater than $mg$ ” and “equal to $mg$ ”, so the simplest correct description is that $N$ is first greater than $mg$ , then becomes equal to $mg$ once he is stationary.

Let’s walk through the phases.

Initial state (squatting, at rest)

The man is stationary on the ground. His centre of mass has zero velocity. The net force is zero, so $N = mg$ . But this is only the starting point — the process hasn’t begun yet.
Beginning to stand — upward acceleration

To start moving upward, the man must push harder against the ground. By Newton’s third law, the ground pushes back with an equal and opposite force. So $N$ becomes greater than $mg$ . The net upward force $N - mg$ provides the upward acceleration $a$ .

N - mg = ma \quad \Rightarrow \quad N = m(g + a) > mg

Middle of the motion — possible constant speed

If the man rises at constant speed for a while, then $a = 0$ and $N = mg$ during that interval. But in a natural squat-to-stand movement, the acceleration phase is short and followed by deceleration. The question’s options don’t mention a period where $N < mg$ , so the simplest match is that after the initial acceleration, $N$ returns to $mg$ as the man becomes stationary.
Nearing the top — deceleration

To stop at the upright position, the man must have a downward acceleration (i.e., upward deceleration). That would require $N < mg$ . However, the options given do not include “less than $mg$ ” at any stage. So the intended answer focuses on the fact that $N$ is first greater than $mg$ (to start the motion) and later equal to $mg$ (when at rest or moving at constant speed).

Watch out

A common mistake is to think that because the man is “pushing” on the ground, the reaction is always greater than $mg$ . But once he is moving at constant speed or is stationary, the net force is zero, so $N = mg$ . The extra force is only needed to change his speed.

Tip

Think of standing up as a controlled upward throw of your own body. To throw something upward, you must exert a force greater than its weight initially. Once it’s moving, you can ease off.

Thus, the reaction force is not constant — it varies — and it is greater than $mg$ only during the upward acceleration phase, then equal to $mg$ when the man is at rest (or moving uniformly).

✓Final answer

mg

←Previous questionQ2 — A proton is kept at rest. A positively charged particle is released fr…→Next questionQ4 — A bicyclist comes to a skidding stop in 10 m. During this process, the…

Q.A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is