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ExplorationCh 6 How Forces Affect Motion

How Forces Affect Motion: Class 9 Science · Chapter 6

Exploration · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 95 min
  • 1 simulation
  • 30 practice questions
  • 23 MCQs

What you will learn in How Forces Affect Motion

  • Describe a force as a push or pull with magnitude and direction, and measure it in newtons with a spring balance
  • Find the net force when forces act along one line, and tell balanced forces from unbalanced ones
  • Explain how friction opposes sliding and depends on the surfaces in contact, and identify weight and the normal force on an object
  • State Newton's first law, link it to inertia, and draw position–time and velocity–time graphs for zero net force
  • Use F = ma and F = mg with the equations of motion to solve problems, and explain why a longer stopping time means a smaller force
  • State Newton's third law, identify the two forces of a pair acting on different objects, and explain walking, rowing, recoil and rockets
  • Treat joined objects as one system and find its acceleration from the external force alone

Original companion text aligned to NCERT Science Class 9, Chapter 6, pp. 94–115. Not affiliated with or endorsed by NCERT.

In Chapter 4 you described motion with position, velocity and acceleration, but you never asked what makes an object start, stop, speed up or turn. Free-wheel a bicycle down a slope and it gains speed without any pedalling. On a flat road it slowly loses speed and stops. Turn the handle and it changes direction. Each time, something is acting on it.

That something is a force. In this chapter we look closely at forces, see how several of them combine, and meet the three laws of motion with which Isaac Newton connected force to motion.

6.1 The Concept of Force

In earlier classes you met a force as a push or a pull. A force can do four kinds of things to an object. It can set the object moving from rest, change its speed, change its direction, or change its shape. Flick a carrom striker and a coin at rest shoots off. Squeeze the brakes of a bicycle and it slows down. A goalkeeper's palm sends a football off at a new angle. A rolling pin flattens a ball of dough into a roti. Fig. 6.1 shows all four effects on one rubber ball.

3/4A push from the side changes its direction of motion.

Fig. 6.1A force can start motion, change speed, change direction or change shape.NCERT Fig. 6.1 · p. 94

A rubber ball on a floor seen from above, in four steps. Step 1: the ball is at rest and a push from the left sets it moving to the right. Step 2: a second push from behind makes the moving ball go faster, shown by a longer velocity arrow. Step 3: a push from the side turns the ball, so its velocity arrow now points diagonally down and to the right. Step 4: two palms squeeze the ball from both sides and it becomes a tall oval.

Suppose a friend tells you, 'I pushed the box with 50 N.' Can you say where the box went? Not yet: you also need to know which way the push pointed. Every force you know has a direction built into it. The Earth's pull points down. Water pushes a floating bottle up. Friction always points against sliding. So a force is described fully only by its magnitude (how strong it is) together with its direction, just like displacement, velocity and acceleration in Chapter 4.

Definition
Force

A push or a pull on an object that comes from its interaction with another object. A force has a magnitude and a direction; its SI unit is the newton (N).

ExampleThe Earth's pull on a falling mango: about 5 N, towards the centre of the Earth.

Forces are measured in newtons (symbol N), the SI unit named after Isaac Newton. Written out in full, the unit begins with a small letter: newton, not Newton.

6.1.1 Measuring the magnitude of a force

How strong is a push or a pull? To measure it we use a spring balance. Pull on its hook and the spring inside stretches: the harder the pull, the longer the stretch. A pointer moving along a scale marked in newtons shows how strong the pull is. Hang a bag of vegetables from the hook, and the pull being measured is the Earth's pull on the bag, its weight. Pull the hook with your hand, a rubber band or a toy car instead, and the balance reads that pull just as happily (Fig. 6.2).

2/3The Earth pulls the bag down with 4.9 N, its weight.

Fig. 6.2A spring balance reads the weight of an object, or any other pull on its hook.NCERT Fig. 6.2 · p. 95

A vertical spring balance with a scale marked 0 N, 5 N and 10 N. Step 1: nothing hangs from the hook, the spring is short and the pointer reads 0 N. Step 2: a cloth bag of 500 g of dal hangs from the hook, the spring stretches and the pointer reads 4.9 N, the weight of the dal. Step 3: instead of the bag, a hand pulls the hook down and the pointer moves further, to 8 N.

Quick check
1 of 3
A batter hits a fast ball straight back past the bowler. Which effect of the bat's force is this mainly an example of?
Show answer

Answer (B) Changing the direction of motion

The ball was already moving towards the batter; the bat sends it the other way, so its direction of motion changes. The ball does squash a little for an instant, but that is not the main effect.

6.2 Balanced and Unbalanced Forces

Look at a bicycle standing on its stand. The Earth pulls it down, yet it does not sink into the ground, because the ground pushes up on its tyres and stand. Two forces act on it at once, and they cancel out. Most objects around you are like this, with several forces acting on them together. What, then, decides how such an object moves?

Picture a supermarket trolley with one friend pushing it from the front and another from the back. If both push equally hard, the trolley stays where it is. Two forces that are equal in size but opposite in direction cancel each other: they are balanced forces (संतुलित बल). If one friend pushes harder, the forces become unbalanced, and the trolley rolls the way the bigger push points (Fig. 6.3).

2/330 N against 20 N: unbalanced, net force 10 N towards the bigger push.

Fig. 6.3Opposite forces subtract, forces in the same direction add.NCERT Fig. 6.4 · p. 96

A trolley on a floor with force arrows drawn to scale. Step 1: a 20 N push from the left and a 20 N push from the right; the net force is zero and the trolley does not move. Step 2: 30 N from the left and 20 N from the right; the net force is 10 N to the right and the trolley moves right. Step 3: 30 N and 20 N both push from the left; the forces add to a net force of 50 N to the right and the trolley moves further.

The single force that has the same effect as all the forces together is the net force (परिणामी बल). For forces along one straight line, there are two cases. When two forces act in opposite directions, the net force equals the difference of their magnitudes and points along the bigger force. When they act in the same direction, the net force is their sum. Think of one person pulling a loaded handcart from the front while another pushes it from behind: their efforts add up.

Definition
Net force

The single force that has the same effect as all the forces acting on an object together. Along one line: add the forces pointing one way and subtract those pointing the other way.

ExamplePulls of 18 N east and 12 N west give a net force of 6 N east.

Definition
Balanced forces

Forces on an object whose net force is zero. They do not change the object's state of rest or motion.

ExampleTwo friends pushing a trolley from opposite ends with 20 N each.

Example 6.1NCERT Example 6.1 · p. 96

Two children, one carton of books

Meera pulls a carton of books with a force of 18 N. Kabir applies 12 N to the same carton. The floor is very smooth, so ignore friction. Find the magnitude and direction of the net force when:

  • (a) both forces act towards the east;
  • (b) Meera pulls east and Kabir pulls west;
  • (c) Meera pulls west and Kabir pulls east.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Pause and ponder 1–2: questions with answers

  1. Short answerEasyNCERT Pause and Ponder Q1

    A flower pot hangs from a hook in the ceiling by a chain and does not move.

    1. Name two forces acting on the pot.

      Show solution

      The Earth pulls the pot down (its weight). The chain pulls the pot up (the tension in the chain).

    2. Are these forces balanced? How do you know?

      Show solution

      Yes. The pot stays at rest, so the net force on it must be zero: the upward pull of the chain equals the pot's weight.

  2. Short answerMediumNCERT Pause and Ponder Q2

    Ravi and his friend push a wooden stool from opposite sides. The stool starts to slide towards Ravi.

    1. Are the forces on the stool balanced?

      Show answer

      Answer (B) No, unbalanced

      The stool starts moving, so its velocity changes: a net force must act on it.
    2. Who pushes harder?

      Show answer

      Answer (B) His friend

      The friend pushes the stool towards Ravi, and it moves that way. So the friend's push is larger than Ravi's push (and friction together).
Quick check
1 of 3
Forces of 25 N towards the east and 40 N towards the west act on a cart. What is the size of the net force?
N
Show answer

Answer 15 N

The forces are opposite, so subtract: 40−25=15 N, pointing west, along the larger force. Adding them (65 N) would be right only if both pointed the same way.

6.3 The Force of Friction: Often Overlooked but Always Present

Why does a heavy sack of rice refuse to budge when you give it a gentle push? Where the sack rests on the floor, tiny bumps on the two surfaces catch on each other. This gives rise to friction (घर्षण), a force that always points against sliding. Push gently and friction pushes back just as hard, so nothing happens. Push harder and friction grows too, but only up to a limit. Beyond that limit your push wins, a net force acts forwards, and the sack starts to slide.

Definition
Friction

The force that acts along two surfaces in contact and opposes their sliding over each other.

Draw every force on a crate being pulled along the floor and you find four, not two (Fig. 6.4). Two act along the floor: the rope's pull and friction. The other two act up and down. The crate's weight points down, and the floor answers with an upward push at right angles to its surface, the normal force (अभिलंब बल). The crate neither sinks into the floor nor jumps off it, so these two must balance. Air drag is a fifth force, but on a slow-moving crate it is tiny, so we leave it out.

3/3The pull is bigger than friction, so a net force acts forwards.

Fig. 6.4Weight and the normal force balance; the pull and friction decide whether the crate speeds up.NCERT Fig. 6.11 · p. 97

A wooden crate rests on a floor with a rope tied to its right side. Step 1: an arrow down from the crate is labelled weight and an equal arrow up is labelled normal force; these vertical forces are balanced. Step 2: an arrow along the rope to the right is labelled pull, and a dashed arrow along the floor to the left is labelled friction, shorter than the pull. Step 3: a green arrow to the right is labelled net force, because the pull is bigger than friction.

Definition
Normal force

The push of a surface on an object resting on it, acting perpendicular (at right angles) to the surface.

A carrom coin after a good strike, a toy car you let go of, a football rolling over grass: each one slows down and stops on its own. It is tempting to conclude that motion needs a force to keep it going. In fact, each of them is being slowed by friction. To keep something moving steadily on Earth, you must push just hard enough to cancel friction. Take the push away, and friction is left alone to bring the object to rest.

Why does the eraser first speed up and then slow down? Count the horizontal forces at each stage (Fig. 6.5). Held at the mark, it feels the band pulling it forwards and your finger pushing it back, and these cancel. Lift your finger and only the band and friction remain; the band is stronger, so the net force points forwards and the eraser speeds up. A moment later the band straightens and loses touch with the eraser. Friction is now the only horizontal force. It points backwards, so the eraser's speed falls to zero.

2/4The band's push is bigger than friction: a net force forwards, so the eraser speeds up.

Fig. 6.5During the launch the band beats friction; afterwards, friction alone slows the eraser to a stop.NCERT Fig. 6.13 · p. 99

A view from above of a table with two pencils and a rubber band between them. Step 1: the eraser is pulled back so the band forms a V; arrows show the band pulling it forwards and a finger holding it back with an equal force, so it stays still. Step 2: released, the band pushes the eraser forwards with a long arrow while a shorter dashed friction arrow points backwards; the eraser speeds up. Step 3: the eraser has left the band; only a dashed friction arrow acts backwards and its velocity arrow is short, so it slows down. Step 4: the eraser is at rest farther along, with the distance from the band marked.

3/3It can never reach that height, so it keeps rolling for ever.

Fig. 6.6If there were no friction, a ball on a flat track would never stop.

A ball starts high on a ramp on the left; a dashed line marks its starting height across the picture, and a note says to imagine no friction. Step 1: the second ramp is steep, and the ball stops on it at the same height it started from. Step 2: the second ramp is gentler; the ball rolls farther along it but still stops at the same height. Step 3: the second surface is flat; the ball can never reach the starting height, so it keeps rolling, shown by a long arrow.

Quick check
1 of 3
A box is sliding to the left across a floor. Which way does friction from the floor act on the box?
Show answer

Answer (C) To the right

Friction opposes the sliding, so it points opposite to the motion: to the right. Pointing left, with the motion, would speed the box up.

6.4 Newton’s First Law of Motion

Galileo's idea became the first of Newton's three laws.

LawNewton's first law of motion

Unless a net force acts on it, an object at rest stays at rest, and a moving object keeps moving with the same speed in the same direction.

Read the law backwards and it gives a quick test. Whenever an object's velocity is not changing, whether it is parked or cruising straight and steady, the net force on it is zero, and so is its acceleration. A parked scooter stays parked. A train coasting on a perfectly straight, level and frictionless track would keep the same speed in the same direction for ever. Because it describes inertia, the first law is also called the law of inertia.

Definition
Inertia

The property of an object by which it resists any change in its velocity: it stays at rest, or keeps moving uniformly in a straight line, unless a net force acts.

ExamplePassengers in a bus lurch forwards when it brakes suddenly: their bodies tend to keep moving.

Example 6.2NCERT Example 6.2 · p. 101

A handcart that keeps rolling

A vegetable seller is pushing his moving handcart along a level road. His forward push is exactly equal to the friction on the cart. Will the cart keep moving, or will it gradually stop?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.3NCERT Example 6.3 · p. 101

Graphs when the net force is zero

No net force acts on either of two scooters on a straight road. Scooter A is parked at the roadside. Scooter B moves at a steady 10 m s−1. Sketch the position–time and velocity–time graphs of each.

Try it yourself first. Then reveal the solution one step at a time.

2/2Position rises steadily; velocity stays the same.

Fig. 6.7Zero net force: velocity constant, so the velocity–time graph is flat.NCERT Fig. 6.16 · p. 101

Two pairs of axes side by side: position against time on the left, velocity against time on the right. Step 1, a parked scooter: the position–time graph is a flat horizontal line and the velocity–time graph runs along the time axis at zero. Step 2, a scooter at a steady velocity: the position–time graph is a straight line sloping upwards and the velocity–time graph is a flat line above the time axis.

Practice · 3 questions

Pause and ponder 3–5: questions with answers

  1. Quick answerEasyNCERT Pause and Ponder Q3

    Close to the ground, a raindrop falls straight down at a steady speed. What is the net force on it?

    Show answer

    Answer (A) Zero

    Its velocity is constant, so its acceleration and the net force are zero. Its weight acts down, but the air's drag acts up with an equal force and cancels it.
  2. Quick answerMediumNCERT Pause and Ponder Q4

    No net force acts on a hockey puck. Which of these could be happening? Choose all that apply.

    Select all that apply
    Show answer
    Answer
    • (A) It stays at rest
    • (B) It slides in a straight line at a constant speed
    With zero net force the velocity cannot change, so the puck either stays at rest or keeps a constant velocity. Speeding up changes the speed, and moving round a circle changes the direction, so both need a net force.
  3. Short answerMediumNCERT Pause and Ponder Q5

    Forces act on nearly everything around us, yet the net force on an object can still be zero. Give one example, and name the forces that cancel.

    Show solution

    A ceiling lamp hanging still: its weight acts downwards, and the pull of its wire acts upwards with an equal force.

    Another example: a car cruising at a steady speed on a straight, level road. The forward push of the road on its driving wheels balances the backward drag of the air and the friction that resists motion.

Quick check
1 of 3
A ball rolls across a frictionless floor at 2 m s−1, and no horizontal force acts on it. What is its speed 10 s later?
Show answer

Answer (C) 2 m s−1

With no net force its velocity cannot change, so it is still 2 m s−1. Answering zero assumes motion dies out by itself; it only does so when friction acts.

6.5 Newton’s Second Law of Motion

Starting, stopping, speeding up, slowing down and turning are all changes in velocity, and every change in velocity is an acceleration. So forces cause accelerations. But how exactly are the net force and the acceleration linked?

DerivationFrom two times to two accelerations

In both runs the truck starts from rest (u=0) and covers the same distance s with a constant acceleration. So s=ut+12at2 gives, for the two runs:

s=12 a1T12s=12 a2T22

The right-hand sides are equal, so:

a1T12=a2T22a2a1=T12T22

For example, with T1=2.0 s and T2=1.5 s, the ratio is 4.0/2.25≈1.8. Doubling the pull nearly doubled the acceleration.

Both activities point to one rule: the second of Newton's laws, and one of the most useful ideas in all of physics.

LawNewton's second law of motion

An object acted on by a net force accelerates along the direction of that net force. Its acceleration is directly proportional to the net force and inversely proportional to the object's mass.

a=Fm
(6.1)
where
a
acceleration of the object
F
net force on it
m
its mass
Newton's second law
F=ma
(6.2)

Here F is always the net force, not any single force, and the acceleration points the same way as the net force. With mass in kilograms and acceleration in  m s−2, the force comes out in newtons. Put m=1 kg and a=1 m s−2 into Eq. (6.2):

F=1 kg×1 m s−2=1 kg m s−2=1 N
Definition
Newton (N)

The SI unit of force. One newton is the net force that gives a mass of 1 kg an acceleration of 1 m s−2.

Let go of a stone and it falls faster and faster: the Earth's pull gives it an acceleration. This particular acceleration has its own name, the acceleration due to gravity, and its own symbol, g (unit  m s−2). Put a=g into Eq. (6.2) and you get the size of the Earth's pull on any mass m, which is its weight:

weight
F=mg
(6.3)

Measured anywhere near the ground, g is about 9.8 m s−2; it changes only slightly from place to place. In rough calculations, 10 m s−2 is close enough.

The second law also explains why soft landings hurt less. To stop a moving object, you must bring its velocity down to zero. Doing that in a very short time means a large deceleration and so, by F=ma, a large force. Stretching out the stopping time makes the deceleration, and the force, much smaller (Fig. 6.9).

3/3Stopped in about 0.1 s, it needs only about 2.4 N and survives.

Fig. 6.9Same change in velocity, different stopping times: the quicker the stop, the larger the force.NCERT Fig. 6.18 · p. 104

Two eggs fall side by side at 4 metres per second, one above a stone floor on the left and one above a pillow on the right. Step 1: both eggs are falling at the same speed. Step 2: the left egg hits the stone floor, stops in about 0.004 seconds, feels an average force of about 60 newtons and cracks. Step 3: the right egg sinks into the pillow, stops in about 0.1 seconds, feels an average force of only about 2.4 newtons and stays whole.

Check the numbers: each egg (0.06 kg) loses 4 m s−1. Over 0.004 s the deceleration is 1000 m s−2, which needs about 0.06×1000=60 N. Over 0.1 s it is only 40 m s−2, about 2.4 N. (These are average forces, ignoring the egg's small weight.)

This is why you pull your hands back as you catch a fast throw. Catch a water balloon at Holi with stiff arms and it bursts. Let your hands move back with it, and the same balloon slows down over a longer time, needs a smaller force, and survives. A fielder's 'giving' hands work the same way and also protect the fingers.

Sometimes we want a large force. A hammer head striking a nail stops in a tiny fraction of a second, so it presses on the nail with a very large force and drives it into the wood. The quicker the stop, the bigger the force.

Example 6.4NCERT Example 6.4 · p. 105

A porter's load

A railway porter balances a suitcase of mass 24 kg on his head and stands still. How large an upward force does his head apply to the suitcase? Take g=9.8 m s−2.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.5NCERT Example 6.5 · p. 105

Will the almirah move?

Asha pushes a steel almirah of mass 40 kg that stands on a level floor. The friction on it can be at most 120 N, and it stays 120 N once the almirah slides. How far does the almirah move in 3 s if she pushes steadily with (a) 120 N (b) 140 N?

Try it yourself first. Then reveal the solution one step at a time.

InteractiveForce lab: push a crate

A crate on a floor with three arrows: your push in blue, friction in orange and the net force in green, each labelled with its size in newtons. A slider sets the push from 0 to 60 newtons; buttons choose a floor with no friction, smooth tiles or a rough floor, and a mass of 5, 10 or 20 kilograms. Play runs ten seconds of motion, and a button switches the push on or off during the run. Cards show the friction, net force, acceleration, velocity, distance and time, a status line explains which law is at work, and a velocity–time graph is drawn as the crate moves.

Fig. 6.10Set the push, the surface and the mass, then press Play. Switch the push off during a run to see what friction does on its own.

1/3Speeding up: positive slope, so a forward net force.

Fig. 6.11The slope of a velocity–time graph gives the acceleration; multiply by the mass for the net force.NCERT Fig. 6.21 · p. 106

A velocity–time graph with time from 0 to 14 seconds and velocity from 0 to 8 metres per second. The line rises straight from 0 to 6 metres per second between 0 and 4 seconds, stays flat at 6 metres per second until 10 seconds, then falls straight to zero at 13 seconds. Step 1 highlights the rising part: acceleration 1.5 metres per second squared and a force of 900 newtons towards the north. Step 2 highlights the flat part: acceleration zero and net force zero. Step 3 highlights the falling part: acceleration minus 2 metres per second squared and a force of 1200 newtons towards the south.

Example 6.6NCERT Example 6.6 · p. 106

Reading forces from a velocity–time graph

An e-rickshaw with its passengers has a total mass of 600 kg. It moves north along a straight road, and Fig. 6.11 is its velocity–time graph. Find the net force on it during (a) 0 to 4 s (b) 4 to 10 s (c) 10 to 13 s.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Pause and ponder 6–8: questions with answers

  1. Quick answerEasyNCERT Pause and Ponder Q6

    A toy train of mass 250 g runs along a straight track at a steady 0.8 m s−1. What is the net force on it?

    N
    Show answer

    Answer 0 N

    Constant velocity means a=0, so F=ma=0. The mass and the speed are not needed: no acceleration, no net force.
  2. Short answerEasyNCERT Pause and Ponder Q7

    You want to give an empty shopping trolley and a fully loaded one the same acceleration. Which needs the larger push, and why?

    Show solution

    The loaded trolley. By F=ma, the same acceleration needs a force in proportion to the mass. If the loaded trolley has three times the mass of the empty one, it needs three times the net force.

  3. Short answerMediumNCERT Pause and Ponder Q8

    A new television comes packed in a carton with thick blocks of thermocol all round it. How does the thermocol protect the TV if the carton is dropped?

    Show solution

    When the carton hits the floor, the thermocol squashes, so the TV comes to rest over a longer time instead of stopping almost instantly. A longer stopping time means a smaller deceleration and, by F=ma, a smaller force on the TV, so it is less likely to crack.

Quick check
1 of 4
A net force of 30 N acts on a 12 kg trolley. What is its acceleration?
m s⁻²
Show answer

Answer 2.5 m s⁻²

a=F/m=30/12=2.5 m s−2. Multiplying instead (30×12) mixes up the formula.

6.6 Newton’s Third Law of Motion

So far we have followed one object at a time. But a force always comes from an interaction between two objects. Slap a table and your palm stings: while your hand pushes the table, the table pushes back on your hand. What happens to both objects in such an interaction?

A skateboarder kicks the road backwards with one foot and shoots forwards. Your own walking is the same trick, done more gently (Fig. 6.12). At each step your sole presses back against the ground. The ground presses forwards on your sole with an equal force, and that forward force is what carries you along. Notice what kind of force it is: friction. We usually think of friction as the force that stops things; in walking, it is the force that gets you going.

2/3The ground pushes your foot forwards, through friction.

Fig. 6.12Walking is a third-law pair made possible by friction.NCERT Fig. 6.25 · p. 107

Side view of a leg and shoe on the ground. Step 1: an arrow along the ground points backwards, labelled foot on ground: the foot pushes the ground back. Step 2: a second arrow at the sole points forwards, labelled ground on foot: the ground pushes the foot forwards with friction, the force that moves the walker. Step 3: the floor is covered with oil; there is hardly any friction, the forward push disappears and the foot slides backwards, away from a dashed outline showing where it was.

Experiments like these, repeated in endless variations, lead to Newton's third law.

LawNewton's third law of motion

If object A pushes or pulls on object B, then at the same instant B pushes or pulls on A with a force that is equal in size and opposite in direction.

Climbing a rope in a gym uses the same pairing: your hands and feet press down on the rope, and the rope presses you up. And now the shikara puzzle has an answer. The boatman's paddle shoves water towards the stern; the water shoves back on the paddle towards the bow, and the paddle is held by the boatman sitting in the boat. One force acts on the water, the other on the paddle, so there is nothing to cancel. Sweep harder, and the water's push on the paddle grows, so the boat gains speed faster. Load four tourists, and the same push must accelerate a much larger mass, so the speed rises more slowly (Section 6.5).

A rocket uses the same idea (Fig. 6.13). Its engines burn fuel and force huge amounts of hot gas downwards out of the nozzles. The gas pushes the rocket upwards with an equal force. When this upward push is bigger than the rocket's weight, the net force points up and the rocket lifts off the launch pad, as in the launches you may have watched from Sriharikota.

2/3The gas pushes the rocket upwards with an equal force; the weight acts down.

Fig. 6.13The engine pushes gas down; the gas pushes the rocket up.NCERT Fig. 6.30 · p. 109

A rocket stands upright on the ground with a flame below its nozzle. Step 1: two arrows point down from the nozzle, labelled engine pushes gas down. Step 2: a long green arrow points up beside the rocket, labelled gas pushes rocket up, and a shorter arrow points down from it, labelled weight. Step 3: the rocket has risen off the ground; a note says the upward push is bigger than the weight.

Compare

Third-law pair or balanced forces?

Third-law pair

  • Act on two different objects
  • Always equal in size and opposite in direction
  • Never cancel each other
  • Example: foot pushes ground back, ground pushes foot forwards

Balanced forces

  • Act on the same object
  • Equal and opposite only when they happen to balance
  • Cancel: the net force is zero
  • Example: a book's weight and the table's push on the book
Practice · 2 questions

Pause and ponder 9–10: questions with answers

  1. Short answerMediumNCERT Pause and Ponder Q9

    A rotating lawn sprinkler has bent arms that squirt water out sideways. Why do the arms spin round?

    Show solution

    Each nozzle pushes water out in one direction. By the third law, the water pushes the nozzle the opposite way with an equal force. The nozzles point sideways at the ends of the arms, so these backward pushes turn the arms round and round.

  2. Short answerHardNCERT Pause and Ponder Q10

    An astronaut floating outside a space station has lost her grip, and no safety line is attached. She is holding a heavy spanner. How can she get back to the station?

    Show solution

    She should throw the spanner hard in the direction away from the station. As she pushes the spanner, the spanner pushes her with an equal and opposite force, so she starts moving towards the station. With nothing to slow her down, she keeps drifting towards it at a constant velocity (first law). A spacecraft changes its velocity in the same way, by pushing out gas from its engines.

Must the two objects touch? No: the third law holds for non-contact forces too. Two magnets pull, or push, each other equally. A comb charged by rubbing it in dry hair attracts a scrap of paper, and the paper attracts the comb just as strongly. A coconut dropping from a palm is pulled down by the Earth, while the coconut pulls the Earth up with exactly the same force. Explore the magnets in Fig. 6.14.

SimulationEqual pulls between unequal magnets

Two bar magnets lie side by side with field lines around them; the right magnet is much stronger than the left. Arrows drawn on both magnets show the force each one feels from the other. You can drag either magnet closer or farther, and flip either one end for end to switch between attraction and repulsion; the two force arrows always stay equal in length and opposite in direction.

Fig. 6.14Drag either magnet, or flip one with its button. The arrows show the force each magnet feels.NCERT Fig. 6.31 · p. 110
Try this
  • The right magnet is much stronger. Compare the two force arrows: is either one longer?
  • Flip one magnet so that matching poles face each other. What happens to both arrows?
  • Drag the magnets closer together. Do the arrows stay equal as they grow?
Example 6.7NCERT Example 6.7 · p. 110

Why the Earth doesn't rush up to a coconut

A coconut of mass 1.5 kg drops from a palm. The Earth pulls the coconut down with a force of 14.7 N, and the coconut pulls the Earth up with 14.7 N. The Earth's mass is about 6.0×1024 kg. Find the acceleration of each, and explain why only the coconut seems to move.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.8NCERT Example 6.8 · p. 110

Two skaters push apart

Aman (60 kg) and Riya (40 kg) stand face to face on roller skates on a smooth floor. They push on each other's palms, and the push on each of them is 120 N. Find the size of each one's acceleration at that moment.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A book rests on a table, and the Earth pulls the book down. What is the third-law partner of this force?
Show answer

Answer (C) The book pulling the Earth up

The partner force acts on the other object in the same interaction (book and Earth), so it is the book's pull on the Earth. The table's push is equal and opposite too, but it acts on the same object, the book: that pair is balanced, not a third-law pair.

6.7 Forces Acting on a System of Objects

At an airport, a porter hitches two luggage trolleys together with a short rope and pulls the front one. How quickly will the pair speed up? Call the mass of the front trolley m1 and of the rear one m2, and the porter's horizontal pull F. The floor is smooth enough for friction to be ignored (Fig. 6.15).

The rope is stretched, and a stretched rope pulls on whatever is tied to its ends. It tugs the rear trolley forwards and the front trolley backwards with equal forces. This pull is called tension, T; the two tugs are a third-law pair passed along the rope.

2/3Draw a boundary round both: the two tensions are internal and cancel, so only F accelerates the pair.

Fig. 6.15Inside one boundary, the tension forces cancel; only external forces accelerate the system.NCERT Fig. 6.34 · p. 111

Two trolleys stand on a floor, joined by a rope, with trolley 1 on the right and trolley 2 on the left; a blue arrow labelled F pulls trolley 1 to the right. Step 1: two short arrows labelled T show the rope pulling trolley 1 backwards and trolley 2 forwards. Step 2: a dashed boundary is drawn round both trolleys and the rope; the tension arrows fade, because inside the system they cancel, and the acceleration is F divided by the total mass. Step 3: the vertical external forces are added, the total weight acting down and the floor's total normal force acting up, and they balance.

Now step back and look at both trolleys and the rope as one object, a system. Ask which forces come from outside it. Along the floor, only the porter's pull does: it is an external force. The two tension forces are exchanged between parts of the system, so they are internal; being equal and opposite, they add up to zero for the system. The whole pair, of total mass m1+m2, is therefore accelerated by F alone:

a=Fm1+m2
(6.4)
where
a
acceleration of both trolleys together
F
the external pull
m1+m2
the total mass of the system

Working with each trolley separately gives the same answer, only with more steps; the example below checks this. Up and down there are external forces too, but they cancel: the total weight (m1+m2)g acting down is balanced by the floor's normal forces N1+N2 acting up.

Definition
Tension

The pulling force that a stretched string, rope or chain exerts on the objects tied to its ends.

Definition
External force

A force on a chosen system that comes from something outside it. Forces between parts of the system are internal forces, and they cancel in pairs.

Example

Pulling two trolleys

Trolley 1 (20 kg) and trolley 2 (30 kg) are tied together on a smooth floor. A pull of 100 N acts on trolley 1. Find the acceleration of the pair, then check it by looking at each trolley on its own.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A 5 kg block and a 3 kg block are tied together on a frictionless table and pulled with 24 N. What is their acceleration?
m s⁻²
Show answer

Answer 3 m s⁻²

a=F/(m1+m2)=24/8=3 m s−2. Using only one block's mass ignores the fact that both must be accelerated.
Recap

How Forces Affect Motion: key points and summary

  1. A force is a push or a pull with magnitude and direction, measured in newtons, for example with a spring balance (Section 6.1).
  2. Forces along one line combine into a net force: add those in the same direction, subtract opposite ones (Section 6.2).
  3. Balanced forces give zero net force and cannot change motion; unbalanced forces can (Section 6.2).
  4. Friction acts where surfaces touch and opposes sliding; it is weaker between smoother surfaces (Section 6.3).
  5. On a level floor an object's weight is balanced by the normal force from the floor (Section 6.3).
  6. First law: with no net force, an object stays at rest or keeps a constant velocity; this resistance to change is inertia (Section 6.4).
  7. Second law: a=F/m, with the acceleration along the net force; 1 N=1 kg m s−2 (Section 6.5).
  8. Weight is F=mg, where g=9.8 m s−2 near the Earth's surface for every object ((6.3)).
  9. A longer stopping time means a smaller force: giving hands, airbags and soft landings (Fig. 6.9).
  10. Third law: forces come in equal and opposite pairs acting on two different objects, so a pair never cancels (Section 6.6).
  11. Walking, rowing, rockets and recoil are third-law pairs at work; equal forces on unequal masses give unequal accelerations (Example 6.8).
  12. Joined objects can be treated as one system: a=F/(m1+m2), ignoring internal forces such as tension (Section 6.7).
Practice · 20 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises Q1

    A worker drags a sack of wheat across a godown floor at a steady speed with a horizontal pull of 90 N. How large is the friction on the sack?

    N
    Show answer

    Answer 90 N

    At constant velocity the net force is zero, so friction equals the pull: 90 N, acting opposite to the motion.
  2. Quick answerEasyNCERT Exercises Q2

    An air-hockey puck glides along an almost frictionless table. Complete each statement.

    1. With no net force on it, the puck's velocity will…

      Show answer

      Answer (A) stay the same

      First law: zero net force, no change in velocity.
    2. With a net force along its direction of motion, its speed will…

      Show answer

      Answer (B) increase

      The acceleration points along the motion, so the puck speeds up.
    3. With a net force opposite to its direction of motion, its speed will…

      Show answer

      Answer (C) decrease

      The acceleration points against the motion, so the puck slows down.
  3. Quick answerEasyNCERT Exercises Q3

    On a smooth track, trolley X rolls along at a steady 2 m s−1. Trolley Y has forces of 8 N towards the east and 5 N towards the west acting on it. Which statement is correct?

    Show answer

    Answer (B) Only Y has a net force on it

    X has a constant velocity, so its net force is zero. Y's forces are unbalanced: 8−5=3 N towards the east.
  4. NumericalMediumNCERT Exercises Q4

    At a chariot festival, 40 devotees pull the ropes of a wooden chariot. 36 of them pull forwards, but 4 get turned round in the crowd and pull backwards. Each person pulls horizontally with 150 N. Find the net force of the ropes on the chariot. (Ignore friction and other forces.)

    Show solution

    Answer4800 N forwards

    Ffwd=36×150=5400 NFback=4×150=600 N
    Fnet=5400−600=4800 N

    The net force is 4800 N in the forward direction. (Shortcut: 32 people's worth of pull, 32×150.)

  5. Quick answerEasyNCERT Exercises Q5

    A net force F acts on a body of mass m. The body accelerates…

    Show answer

    Answer (B) along F, more strongly when F is larger

    The second law: the acceleration points along the net force and is proportional to it (a=F/m). A larger mass gives a smaller acceleration, not a larger one.
  6. Quick answerMediumNCERT Exercises Q6

    Each graph below shows how the position of one object changes with time; every object travels along a straight line. Which object has a net force acting on it?

    Fig. 6.16Position–time graphs of four objects.NCERT Fig. 6.37 · p. 113

    Four small position–time graphs in a two-by-two grid. A is a horizontal line. B is a straight line rising from near the origin. C is a straight line falling from high up to low down. D is a curve that rises steeply at first and then flattens out.

    Show answer

    Answer (D) D

    A is at rest; B and C are straight lines, so they move with constant velocity (C back towards the start). Zero net force on all three. D's slope keeps changing (it is slowing down), so its velocity changes and a net force acts on it.
  7. Short answerMediumNCERT Exercises Q7

    Priya stands on a skateboard and throws a heavy school bag forwards to her friend. What happens to Priya and the skateboard? Explain.

    Show solution

    AnswerShe rolls backwards.

    To throw the bag forwards, Priya pushes it forwards. By the third law, the bag pushes Priya backwards with an equal force. The skateboard's wheels let her move easily, so she and the skateboard roll backwards, opposite to the bag.

  8. Short answerMediumNCERT Exercises Q8

    Why should you bend your knees as you land when you jump down from a wall onto hard ground?

    Show solution

    Your downward velocity has to drop to zero when you land. Bending your knees spreads this stop over a longer time. A longer stopping time means a smaller deceleration and so, by F=ma, a smaller force on your legs and joints. Landing with straight legs stops you almost instantly, with a much larger force.

  9. Quick answerEasyNCERT Exercises Q9

    A heavily loaded bullock cart bumps into an empty cycle-rickshaw. During the bump…

    Show answer

    Answer (C) each pushes the other with a force of the same size

    The two forces form a third-law pair, so they are equal in size and opposite in direction. The lighter rickshaw gets the larger acceleration, which is why it seems to be hit harder.
  10. DiagramMediumNCERT Exercises Q10

    A student applies the same force to blocks of different masses on a frictionless track and records the accelerations in the table.

    Table 6.5
    Mass (kg)Acceleration ( m s−2)
    0.512
    16
    23
    41.5
    1. What force did the student apply?

      N
      Show answer

      Answer 6 N

      For every row F=ma: 0.5×12=1×6=2×3=4×1.5=6 N.
    2. Sketch the force–mass graph for these readings.

      Show solution

      Every row gives the same force, 6 N. So the force–mass graph is a horizontal straight line at F=6 N, parallel to the mass axis, from 0.5 kg to 4 kg. (The acceleration–mass graph is a curve, because a=6/m falls as m rises.)

  11. NumericalMediumNCERT Exercises Q11

    The velocity–time graph of a 4 kg cart moving along a straight line is shown. Find the net force on the cart.

    Fig. 6.17Velocity–time graph of the cart.NCERT Fig. 6.41 · p. 114

    A velocity–time graph with time from 0 to 7 seconds and velocity from 0 to 16 metres per second. A straight line starts at 2 metres per second at time zero and rises to 14 metres per second at 6 seconds, where dashed guide lines meet the axes.

    N
    Show solution
    a=14−26−0=2 m s−2
    F=ma=4×2=8 N

    The net force is 8 N along the direction of motion, and it is constant because the graph is a straight line.

  12. NumericalHardNCERT Exercises Q12

    An arrow of mass 30 g flying at 60 m s−1 hits a bale of hay and stops after going 20 cm into it. Assuming a constant deceleration, find the average stopping force on the arrow.

    N
    Show solution

    Here u=60 m s−1, v=0, s=0.20 m and m=0.030 kg.

    0=602+2a(0.20)a=−36000.40=−9000 m s−2
    F=ma=0.030×9000=270 N

    The hay pushes back on the arrow with about 270 N, opposite to its motion.

  13. NumericalHardNCERT Exercises Q13

    A hockey player hits a ball of mass 0.16 kg lying at rest. It leaves the stick at 90 km h−1. If the average force of the stick on the ball is 400 N, for how long were the stick and the ball in contact?

    s
    Show solution
    v=90×10003600=25 m s−1
    a=Fm=4000.16=2500 m s−2
    t=v−ua=25−02500=0.01 s

    The contact lasts only about 0.01 s, a hundredth of a second.

  14. NumericalHardNCERT Exercises Q14

    A 5 kg box sliding at 6 m s−1 moves onto a rough mat, where the friction on it is 12 N. At the same moment a child pushes against its motion with a steady 8 N. How far does the box travel on the mat before it stops?

    m
    Show solution

    Both forces act against the motion, so the net force points backwards:

    F=12+8=20 N
    a=−205=−4 m s−2
    0=62+2(−4)ss=368=4.5 m

    The box slides 4.5 m on the mat before stopping.

  15. Long answerHardNCERT Exercises Q15

    A winch on a riverbank pulls with a steady force F. Pulling a raft alone, it gives the raft an acceleration a1. Pulling a small boat alone, it gives the boat an acceleration a2. Now the raft is tied behind the boat and the same force F pulls both together. Ignore the drag of the water.

    1. Find the new acceleration in terms of a1 and a2.

      Show solution

      Answera=a1a2a1+a2

      From F=ma, the masses are m1=F/a1 (raft) and m2=F/a2 (boat). For the joined pair, Eq. (6.4) gives:

      a=Fm1+m2=FF/a1+F/a2
      a=11/a1+1/a2=a1a2a1+a2
    2. If a1=3 m s−2 and a2=6 m s−2, find the new acceleration.

      m s⁻²
      Show answer

      Answer 2 m s⁻²

      a=(3×6)/(3+6)=18/9=2 m s−2, smaller than either one alone, as expected for a bigger total mass.
  16. Short answerMediumNCERT Exercises Q16

    Hold a strong magnet a few centimetres above a steel paper clip lying on a table. The clip jumps up to the magnet, but you hardly feel the magnet move. The magnet and the clip pull on each other with equal forces, so why does only the clip move noticeably?

    Show solution

    By the third law the forces are equal and opposite. The clip's mass is tiny, so the magnet's pull gives it a large acceleration (a=F/m), and the pull is enough to overcome its weight. The magnet is much heavier, and your hand holds it: your hand supplies a balancing force, so the net force on the magnet stays almost zero and it barely moves.

  17. Quick answerEasy

    An object moving with a constant velocity must have no forces at all acting on it.

    Show answer

    Answer False

    Forces may act, but they must balance so that the net force is zero, like the push and friction on a steadily moving handcart.
  18. Quick answerEasy

    One newton is the net force that gives a mass of kg an acceleration of 1 m s−2.

    Show answer

    Answer 1

    1 N=1 kg×1 m s−2.
  19. Quick answerMedium

    Put these in order of the acceleration produced, smallest first.

    1. 10 N on 20 kg
    2. 40 N on 8 kg
    3. 6 N on 4 kg
    4. 12 N on 4 kg
    Show answer
    Answer
    1. 10 N on 20 kg
    2. 6 N on 4 kg
    3. 12 N on 4 kg
    4. 40 N on 8 kg
    Using a=F/m: 0.5, 1.5, 3 and 5 m s−2.
  20. Quick answerMedium

    A car of mass 1200 kg speeds up from 10 m s−1 to 20 m s−1 in 5 s on a straight road. What is the net force on it?

    N
    Show answer

    Answer 2400 N

    a=(20−10)/5=2 m s−2, so F=1200×2=2400 N.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Pause and Ponder Q1Pause and ponder 1–2 · Q1
NCERT Pause and Ponder Q2Pause and ponder 1–2 · Q2
NCERT Pause and Ponder Q3Pause and ponder 3–5 · Q3
NCERT Pause and Ponder Q4Pause and ponder 3–5 · Q4
NCERT Pause and Ponder Q5Pause and ponder 3–5 · Q5
NCERT Pause and Ponder Q6Pause and ponder 6–8 · Q6
NCERT Pause and Ponder Q7Pause and ponder 6–8 · Q7
NCERT Pause and Ponder Q8Pause and ponder 6–8 · Q8
NCERT Pause and Ponder Q9Pause and ponder 9–10 · Q9
NCERT Pause and Ponder Q10Pause and ponder 9–10 · Q10
NCERT Exercises Q1Exercises · Q1
NCERT Exercises Q2Exercises · Q2
NCERT Exercises Q3Exercises · Q3
NCERT Exercises Q4Exercises · Q4
NCERT Exercises Q5Exercises · Q5
NCERT Exercises Q6Exercises · Q6
NCERT Exercises Q7Exercises · Q7
NCERT Exercises Q8Exercises · Q8
NCERT Exercises Q9Exercises · Q9
NCERT Exercises Q10Exercises · Q10
NCERT Exercises Q11Exercises · Q11
NCERT Exercises Q12Exercises · Q12
NCERT Exercises Q13Exercises · Q13
NCERT Exercises Q14Exercises · Q14
NCERT Exercises Q15Exercises · Q15
NCERT Exercises Q16Exercises · Q16

Quick check: MCQs on How Forces Affect Motion

  1. 6.1.1 Measuring the magnitude of a force3 questions
  2. 6.2 Balanced and Unbalanced Forces3 questions
  3. 6.3 The Force of Friction: Often Overlooked but Always Present3 questions
  4. 6.4 Newton’s First Law of Motion3 questions
  5. 6.5 Newton’s Second Law of Motion4 questions
  6. 6.6 Newton’s Third Law of Motion4 questions
  7. 6.7 Forces Acting on a System of Objects3 questions
Glossary · 12

Key terms (English–Hindi)

action–reaction pairक्रिया-प्रतिक्रिया युग्मkriya-pratikriya yugm
The two equal and opposite forces that two interacting objects exert on each other. They act on different objects.First seen in the chapter →
balanced forcesसंतुलित बलsantulit bal
Forces on an object whose net force is zero, so together they cannot change its motion. If the net force is not zero, the forces are unbalanced.First seen in the chapter →
external forceबाह्य बलbaahya bal
A force on a chosen system from something outside it. Forces between parts of the system are internal and cancel in pairs.First seen in the chapter →
forceबलbal
A push or a pull on an object that comes from its interaction with another object. It has magnitude and direction.First seen in the chapter →
frictionघर्षणgharshan
The force between two surfaces in contact that opposes their sliding over each other.First seen in the chapter →
inertiaजड़त्वjadatva
The property of an object by which it resists any change in its velocity, whether it is at rest or moving.First seen in the chapter →
momentumसंवेगsanveg
Mass times velocity, p=mv. It points in the direction of the velocity.First seen in the chapter →
net forceपरिणामी बलparinaami bal
The single force that has the same effect as all the forces acting on an object together.First seen in the chapter →
newtonन्यूटनnyootan
The SI unit of force (symbol N): the net force that gives a 1 kg mass an acceleration of 1 m s−2.First seen in the chapter →
normal forceअभिलंब बलabhilamb bal
The push of a surface on an object resting on it, acting at right angles to the surface.First seen in the chapter →
tensionतनावtanaav
The pulling force that a stretched string, rope or chain exerts on the objects tied to its ends.First seen in the chapter →
weightभारbhaar
The pull of gravity that the Earth exerts on an object, equal to mg.First seen in the chapter →