A helicopter of mass 2000kg rises with a vertical acceleration of $15$ m s$^{-2}$. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the ($g = 10$ m s$^{-2}$) (a) force on the floor of the helicopter by the crew and passengers. (b) action of the rotor of the helicopter on the surrounding air. (c) force on the helicopter due to the surrounding air.

A helicopter of mass 2000kg rises with a vertical acceleration of 15 m s⁻². The total mass of the crew and pas

NCERT Exemplar · Q42

15

(a) force on the floor of the helicopter by the crew and passengers.

(b) action of the rotor of the helicopter on the surrounding air.

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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. …

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Applying Newton's second law to the passengers and then to the whole helicopter, and using Newton's third law for the action-reaction pairs: the force on the floor by the crew is 12500 N downward; the rotor's action on the air is 62500 N downward; the force on the helicopter by the air is 62500 N upward.

Why Newton's Third Law Pairs Are Central

The problem asks for three forces, each of which is an action or reaction in a different interaction. The trick is to isolate the correct system for each part and apply $F_{\text{net}} = ma$ to that system. Then, by Newton's third law, the force that system exerts on something else is equal in magnitude and opposite in direction to the force exerted on it.

We take $g = 10\ \text{m s}^{-2}$ upward as positive.

Step-by-step solution

1. Force on the floor by the crew and passengers

Consider the crew and passengers as a system. Their total mass is $m_p = 500\ \text{kg}$ . They move upward with the same acceleration $a = 15\ \text{m s}^{-2}$ as the helicopter.

Two forces act on them:

Their weight $m_p g$ downward.
The normal reaction $N$ from the floor, upward.

Newton's second law for the passengers (taking upward positive):

N - m_p g = m_p a

Substitute $m_p = 500$ , $g = 10$ , $a = 15$ :

N - 500 \times 10 = 500 \times 15

N = 5000 + 7500 = 12500\ \text{N (upward)}

This $N$ is the force the floor exerts on the passengers. By Newton's third law, the force the passengers exert on the floor is equal in magnitude and opposite in direction — i.e., 12500 N downward.

Watch out

A common mistake is to forget that the floor pushes up on the passengers with $N$ , and then to give $N$ itself as the answer. The question asks for the force on the floor by the crew, which is the reaction — so it points downward.

2. Action of the rotor on the surrounding air

The rotor blades push air downward. That push is the "action" referred to in the question. To find its magnitude, we first find the upward thrust the air exerts on the rotor (the reaction), then use Newton's third law.

Consider the entire helicopter (including crew and passengers) as one system. Total mass:

M = 2000 + 500 = 2500\ \text{kg}

Forces on this system:

Total weight $Mg$ downward.
Upward thrust $T$ from the air on the rotor.

The system accelerates upward at $a = 15\ \text{m s}^{-2}$ . Newton's second law:

T - Mg = M a

T - 2500 \times 10 = 2500 \times 15

$T = 25000 + 37500 = 62500\ \text{N (upward)}$ …

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Q.A helicopter of mass 2000kg rises with a vertical acceleration of $15$ m s $^{-2}$ . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the ( $g = 10$ m s $^{-2}$ )

Concept understanding — Newton Third Law Pairs

Newton's Third Law Pairs

The Intuition: Forces Never Act Alone

The Precise Statement

The Key Rule for Spotting a Third Law Pair

Why Equal and Opposite Forces Don't Cancel

A Careful Worked Example: A Book on a Table

More Everyday Examples

Why Newton's Third Law Pairs Are Central

Step-by-step solution

1. Force on the floor by the crew and passengers

2. Action of the rotor on the surrounding air

Concept: Newton's Second and Third Laws for Connected Systems

Q.A helicopter of mass 2000kg rises with a vertical acceleration of 151515 m s−2^{-2}−2. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (g=10g = 10g=10 m s−2^{-2}−2)

Why Newton's Third Law Pairs Are Central

Step-by-step solution

1. Force on the floor by the crew and passengers

2. Action of the rotor on the surrounding air

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Q.A helicopter of mass 2000kg rises with a vertical acceleration of $15$ m s $^{-2}$ . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the ( $g = 10$ m s $^{-2}$ )

Unlock everything free for 14 days