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PhysicsCh 4 Laws of Motion

Laws of Motion: Class 11 Physics · Chapter 4

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 100 min
  • 29 practice questions
  • 30 MCQs

What you will learn in Laws of Motion

  • Use Galileo's reasoning to explain why uniform motion continues without any net force, and state Newton's first law
  • Define momentum and impulse, and apply F = dp/dt = ma in vector and component form
  • State Newton's third law, identify action–reaction pairs correctly, and derive the conservation of momentum
  • Apply the condition for equilibrium of a particle using force triangles and components
  • Describe the common forces of mechanics and use the laws of static, kinetic and rolling friction
  • Analyse circular motion on level and banked roads, finding the optimum and the safe range of speeds
  • Solve problems on connected and accelerating bodies systematically with free-body diagrams

Original companion text aligned to NCERT Physics Class 11, Chapter 4, pp. 49–70. Not affiliated with or endorsed by NCERT.

4.1 Introduction

In Chapters 2 and 3 we described motion: where a body is, how fast it goes and how its velocity changes. We never asked why it moves as it does. That question, about the causes of motion, is the subject of this chapter, and its answer is the foundation of all of mechanics (गति के नियम, laws of motion).

Everyday experience suggests a first answer. A carrom coin moves only when the striker hits it, and a rolling ball stops when a fielder blocks it. Sometimes the push or pull needs contact, as with the striker; sometimes it acts across empty space, as when a mango drops from a branch or a magnet tugs a pin towards itself. So starting and stopping both need a force from outside. But what about a coin that is already gliding steadily across a well-powdered board? Does something have to keep pushing it?

4.2 Aristotle's fallacy

The Greek philosopher Aristotle (384–322 BCE) answered yes. In his scheme, a body moves only for as long as something keeps acting on it; take the mover away and the motion dies. He even claimed that air rushing in behind a flying arrow keeps shoving it forward. Put simply, the Aristotelian rule says: to keep a body moving, an outside force must keep acting on it.

It is easy to see why this felt obvious. A vegetable seller's handcart keeps rolling only while the seller keeps pushing; the moment he lets go, it slows and stops. On Earth, things left alone always seem to come to rest.

The flaw is that the cart is never left alone. The road and the axles exert friction on it, and the air drags on it. While the cart rolls at a steady speed, the seller's push only cancels these opposing forces, so the net force on the cart is zero (Fig. 4.1). When he lets go, friction is the only horizontal force left, and it slows the cart. Remove friction and drag entirely, and the cart would roll on with no push at all. Aristotle's observations were accurate; what his bookkeeping lacked was a force. The breakthrough came from asking what would happen if that missing force could be switched off, and the person who asked it was Galileo.

1/3Pull and friction are equal and opposite: zero net force, constant velocity.

Fig. 4.1Why a pushed cart seems to need a force to keep going: the push only cancels friction.NCERT p. 50

A side view of a handcart carrying two crates on a floor. Step 1: a blue pull arrow to the right and an orange friction arrow to the left have equal length, and a green velocity arrow points right; the label says steady pull, net force zero. Step 2: the pull arrow is gone, the cart has moved on, only friction acts and the velocity arrow is half as long; the label says friction slows it. Step 3: the floor is drawn as an ideal smooth surface, there is no friction arrow and no pull, and the velocity arrow is back to full length; the label says it keeps moving.

4.3 The law of inertia

Early in the seventeenth century, Galileo Galilei timed balls rolling on smooth, gently sloping tracks and spotted a pattern. Downhill, a ball gains speed; uphill, it loses speed; and the steeper the track, the faster the change. A track that slopes neither way should then cause no change at all, provided friction could be removed: the ball would roll on at a constant velocity.

A thought experiment with two facing ramps makes the same point (Fig. 4.2). Let a ball go from some height on the left ramp. It runs down, crosses the bottom and climbs the right ramp until it is almost back at its starting level; the smoother the ramps, the closer it gets, so a frictionless ball would reach that level exactly. Now tilt the right ramp lower. To regain the same level the ball has to travel further along it. Lay the right ramp flat, and the level can never be regained, so the ball should roll on without end. Real friction stops it eventually, but the trend leaves little doubt about the ideal case.

2/3A gentler slope: the same height, reached after a much longer path.

Fig. 4.2Galileo's double slope (frictionless ideal): the ball always tries to regain its starting height.NCERT Fig. 4.1 · p. 50

A fixed slope on the left with a dashed outline of a ball at the release point, and a dashed horizontal line marking the starting height across the whole figure. A second slope on the right is hinged at its foot. Step 1: the second slope is steep and the ball stops on it at the starting height after a short climb. Step 2: the second slope is lowered to a gentle angle and the ball again stops at the starting height, but much further along. Step 3: the second slope lies flat and the ball rolls along it with an arrow pointing onward, labelled never reaches that height.

This overturned two thousand years of thinking. Sitting still and gliding steadily are the same state as far as forces go: either way, the forces on the body add up to zero. What needs a force is a change of velocity. The handcart seller in Section 4.2 pushes only to cancel friction. Every body tends to keep doing what it is already doing, staying put or cruising straight on at a steady speed; this tendency is called inertia (जड़त्व).

Definition
Inertia

The tendency of a body to keep its velocity unchanged: to stay at rest if it is at rest, or to keep moving straight on at a steady speed if it is moving. More mass means more inertia.

ExampleA loaded truck is much harder to start or stop than a scooter.

Quick check
1 of 3
A trolley rolls at a steady 2 m/s along a corridor while a child pushes it with 15 N. What is the net force on the trolley?
Show answer

Answer (D) Zero

Constant velocity means zero acceleration, so the forces balance: the 15 N push is cancelled by 15 N of friction and drag. Thinking the push is the net force is exactly Aristotle's mistake.

4.4 Newton's first law of motion

Galileo had shown what was wrong with the old mechanics; the new one was built largely by Isaac Newton (1642–1727), whose Principia (1687) set out three laws of motion. Newton took Galileo's law of inertia as his first law.

LawNewton's first law of motion

A body remains at rest, or continues to move in a straight line at constant speed, unless a net external force acts on it. Put another way, zero net external force means zero acceleration, and only a net external force can make a body accelerate.

We use the law in two directions. Sometimes we know the forces cancel and conclude that the acceleration is zero: a dust grain drifting in deep space, far from every star, feels essentially no force, so it keeps whatever velocity it has. More often we cannot list every force in advance, but we can see that a body is not accelerating, and then the law tells us the forces on it must add up to zero. On Earth gravity always acts, and moving things meet friction and drag, so a body at rest or in uniform motion is not free of forces; its forces simply cancel.

Take a phone lying on a desk. Two forces act on it: its weight W downwards and the push of the desk, the normal force N, upwards. Nobody measured N beforehand; the desk supplies whatever push is needed, and it adjusts itself to the load. The correct chain of reasoning runs one way only: the phone is observed to be at rest, so by the first law the net force is zero, so N=W. Saying the phone stays at rest because N equals W gets the logic backwards.

Now consider a car pulling away from a traffic light, then cruising at a steady speed on a straight road. While it gains speed it accelerates, so a net external force must act on it. Its engine and gears are internal: on their own they could no more accelerate the car than you can lift yourself by pulling on your shoelaces. The engine turns the wheels, the tyres push backwards on the road, and the road pushes the car forwards through friction (Section 4.9). Once the car cruises steadily, that forward friction exactly balances air drag and rolling resistance, and the net force is zero again.

Inertia explains a familiar jolt in a bus (Fig. 4.3). When a standing passenger's bus starts suddenly, friction from the floor drags the feet forward with the bus, but the upper body tends to stay where it was, so the passenger tips backwards relative to the bus until the muscles bring the body along. When the bus brakes hard, the feet are stopped by friction while the upper body carries on forwards, and the passenger lurches ahead. Holding a handrail helps because the arm can then supply the force the upper body needs.

2/3Friction drags the feet forward; the upper body lags behind and the passenger tips backwards.

Fig. 4.3Friction moves the feet with the bus; inertia keeps the rest of the body doing what it was doing.NCERT p. 52

A side view of the inside of a bus with a standing passenger in the middle. Step 1: the bus is at rest and the passenger stands upright. Step 2: a blue acceleration arrow points forward, to the right; an orange friction arrow at the feet also points forward, and the passenger's upper body leans back, labelled body lags. Step 3: the acceleration arrow points backward, the friction arrow at the feet points backward, and the upper body leans forward, labelled body runs on.

Example 4.1NCERT Example 4.1 · p. 52

A package left behind by a probe

A space probe far from any star or planet fires its thrusters and accelerates at a steady 3.0 m/s2. Its robotic arm lets go of a small instrument package. (a) What is the package's acceleration just after release? (b) How far apart are the probe and the package 4.0 s later, and how fast are they separating?

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

Quick check
1 of 3
A lamp hangs at rest from a ceiling by a cord. Which reasoning is correct?
Show answer

Answer (B) The lamp is at rest, so the tension must equal the weight

We see the lamp at rest; the first law then says the net force is zero, which fixes the tension. The forces are not a third-law pair: both act on the lamp.

4.5 Newton's second law of motion

The first law covers the special case of zero net force. The second law handles every other case: it tells us how much a given net external force changes a body's motion. To state it we need one new quantity.

Momentum

Multiply a body's mass by its velocity and you get its momentum (संवेग):

Momentum
𝐩=m𝐯
(4.1)
where
𝐩
momentum, in kg m s−1
m
mass
𝐯
velocity

Because velocity is a vector, so is momentum, and it points the same way as the velocity. Why is this particular product worth naming? Several everyday observations point to it:

  • Mass matters. Pushing a loaded tempo up to walking pace takes far more effort, or far more time, than pushing an empty scooter to the same pace. Stopping the tempo is just as much harder.
  • Speed matters. A tennis ball lobbed gently is easy to stop with your palm; the same ball arriving at 40 m/s stings. For one mass, more speed means more force to stop it in a given time.
  • Time matters. Jump off a wall onto a sand pit and you hardly feel it; land stiff-legged on concrete and your knees jar. The change in momentum is the same, but sand and bent knees spread it over a longer time, so the force is smaller.
  • The same force for the same time gives the same momentum change. Apply one steady force for one second to a light trolley and to a heavy one, both starting at rest. The light trolley ends up faster, but both end up with exactly the same momentum. This is the key clue to the second law.

Momentum can also change in direction alone. An athlete spinning a hammer (a steel ball on a wire) before the throw keeps its speed steady for a moment, yet its direction turns all the time, and the athlete must keep pulling inwards on the wire to make it turn. Spin it faster, or on a shorter wire, and the pull needed grows. A faster change in momentum, whether in size or in direction, needs a larger force. We return to this pull in Section 4.10.

LawNewton's second law of motion

How fast a body's momentum changes is proportional to the net external force on it, and the momentum changes in the direction in which that force acts.

DerivationFrom the second law to F = ma

Let a net force 𝐅 act for a short time Δt, changing the momentum by Δ𝐩. The law says 𝐅∝Δ𝐩/Δt, so 𝐅=k Δ𝐩/Δt for some constant k. Letting Δt→0, the ratio becomes the derivative of 𝐩 with respect to time:

𝐅=k d𝐩dt
(4.2)

For a body whose mass does not change, m comes outside the derivative:

d𝐩dt=ddt(m𝐯)=m d𝐯dt=m𝐚
(4.3)

so the law reads

𝐅=k m𝐚
(4.4)

We have not yet defined a unit of force, so we are free to choose k. The simplest choice is k=1, which defines the unit of force through this very equation:

Newton's second law
𝐅=d𝐩dt=m𝐚
(4.5)

The SI unit of force, the newton, is the force that gives a 1 kg mass an acceleration of 1 m/s2: 1 N=1 kg m s−2. An average apple weighs roughly 1 N.

Four features of the second law deserve attention before we use it.

  1. It contains the first law. Put 𝐅=0 in (4.5) and you get 𝐚=0.
  2. It is a vector law. It stands for three separate equations, one for each axis ((4.6)). A force changes only the velocity component along its own direction; the component at right angles is untouched. For a projectile, gravity acts vertically, so the horizontal velocity stays constant. The force is always along the acceleration, but it may point along the velocity, against it, at right angles to it or at any angle.
  3. It applies to systems too. For a rigid body, or any group of particles, 𝐅 means the total external force and 𝐚 the acceleration of the centre of mass (Chapter 6). Forces between parts of the system are internal and are left out.
  4. It is local in time. The acceleration at an instant is fixed by the force at that same instant and place. A body keeps no memory of forces that acted on it earlier (Fig. 4.4).
Second law in components
Fx=dpxdt=maxFy=dpydt=mayFz=dpzdt=maz
(4.6)

2/3The instant it is released, the only force is gravity: no horizontal force, no horizontal acceleration.

Fig. 4.4Once released, the packet keeps the horizontal velocity it had at that instant, but not the helicopter's acceleration (air resistance neglected).NCERT Fig. 4.5 · p. 55

A helicopter flies to the right with a blue acceleration arrow. Step 1: a yellow packet hangs from its hook and carries the same blue acceleration arrow, labelled held, accelerates with it. Step 2: the hook is gone; the packet now has only an orange gravity arrow pointing down and a green velocity arrow pointing right, labelled released, only gravity acts. Step 3: the helicopter has moved far ahead, while the packet has fallen along a dashed curved path; it still has the same green horizontal velocity arrow and the orange gravity arrow, labelled packet now lags behind.

Example 4.2NCERT Example 4.2 · p. 55

Driving a nail

A hammer head of mass 1.2 kg hits a nail at 6.0 m/s, and the two stop together after the nail has gone 1.5 cm into a wooden beam. Treat the retarding force as constant and the nail's mass as negligible. Find (a) the average force on the hammer head while it stops and (b) the stopping time.

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

Example 4.3NCERT Example 4.3 · p. 55

Force from a position–time law

A 2.0 kg trolley moves along the x-axis so that its position, in SI units, is x=4.0+3.0t−0.50t2. (a) Find the force on it. (b) When is it momentarily at rest, and what is the force at that instant?

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

Impulse

A bat strikes a ball for only a few milliseconds, and during that time the force rises and falls in a way that is very hard to record. The ball's velocity before and after, however, is easy to measure, and the second law links the two: adding up force × time over the whole contact gives the change in momentum. This product is called the impulse of the force:

Impulse
𝐉=𝐅av Δt=Δ𝐩
(4.7)
where
𝐅av
average force over the contact time Δt
Δ𝐩
change in momentum, final minus initial

When the force varies, the impulse is the area under the force–time graph. The same area can be a tall, narrow spike or a low, wide hump (Fig. 4.5): this is why bending your knees on landing, a cricketer letting the hands give as the ball arrives, a car's crumple zone or a helmet's foam lining all reduce the force by stretching out the time.

3/3The two areas are equal: the same change in momentum in both landings.

Fig. 4.5The area under a force–time graph is the impulse. Spreading the same impulse over a longer time lowers the peak force.NCERT p. 56

A graph with force on the vertical axis and time on the horizontal axis. Step 1: a tall, narrow orange spike labelled stiff legs, short and large. Step 2: the spike fades and a low, wide blue hump appears, labelled bent knees, long and gentle. Step 3: both shapes are shown together with the label equal areas, equal impulse, because the jumper's change in momentum is the same either way.

A force that is very large but lasts only an instant is called an impulsive force. It is not a different kind of force: apart from its size and brevity, the second law treats it exactly like a gentle push. Because the contact time is so short, the body hardly moves while it acts, so we can treat its position as fixed during the impact.

Example 4.4NCERT Example 4.4 · p. 56

A football bouncing off a wall

A 0.43 kg football hits a wall head-on at 8.0 m/s and bounces straight back at 6.0 m/s. The contact lasts 20 ms. Find the impulse on the ball and the average force of the wall on it.

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

Quick check
1 of 4
A 0.50 kg ball speeds up from 4.0 m/s to 6.0 m/s in a straight line in 0.25 s. What average net force acts on it, in newtons?
N
Show answer

Answer 4 N

Δp=0.50×2.0=1.0 N s, so F=1.0/0.25=4.0 N.

4.6 Newton's third law of motion

The second law tells us what a force does; the third tells us where forces come from. Every force on a body is exerted by some other body, and the influence always runs both ways. Step off a small boat onto a jetty and the boat drifts backwards as you move forwards. With bodies that do not touch, the two-way nature is harder to see. The Earth pulls a falling coconut down; Newton's claim is that the coconut pulls the Earth up just as hard. For a 1.2 kg coconut that force is about 12 N, which gives the Earth, of mass 6×1024 kg, an acceleration of only about 2×10−24 m/s2. That is why we never notice it.

So a force is never a one-sided affair; it is an interaction between two bodies, and the two sides of it are always equal in size and opposite in direction. Newton put it this way:

To every Action there is always opposed an equal Reaction.

Isaac Newton, Law IIIPrincipia (1687), in Andrew Motte's English translation (1729)
LawNewton's third law of motion

Forces always occur in pairs. If body B exerts a force on body A, then A exerts on B a force of the same magnitude in the opposite direction, at the same instant.

𝐅AB=−𝐅BA
(4.8)
where
𝐅AB
the force on A by B
𝐅BA
the force on B by A

Because Newton's wording is so short, it is often misread. Three points prevent most mistakes:

  1. The two forces act on different bodies. When you study the motion of A alone, only 𝐅AB matters, so the pair can never cancel each other in A's equation of motion. Only when A and B are treated together as one system do the two become internal forces that add to zero. In the same way, all the internal forces inside a body cancel in pairs, which is what lets the second law work for a whole body.
  2. Neither comes first. The two forces arise together and vanish together; one does not cause the other. Either may be called the action.
  3. Action and reaction are just forces. The words add nothing else. It is clearest to name every force as "the force on A by B".

Fig. 4.6 shows the law at work. Two skaters on smooth ice push palms. Each pushes the other with the same force, so the lighter skater, with less inertia, gets the larger acceleration. They glide apart in opposite directions, and since the forces act for the same time, they end up with equal and opposite momenta: the 40 kg skater moves off 1.5 times as fast as the 60 kg one.

2/3B pushes A exactly as hard as A pushes B, in the opposite direction.

Fig. 4.6The push is one interaction: equal and opposite forces on the two skaters, at the same instant.NCERT p. 57

Two skaters stand face to face on ice: A on the left, labelled 60 kg, and the smaller B on the right, labelled 40 kg. Step 1: their hands touch and both are at rest. Step 2: two orange force arrows of equal length start at the hands, one pointing left on A and one pointing right on B. Step 3: the skaters have moved apart; A has a short green velocity arrow pointing left, labelled slower, and B has a longer one pointing right, labelled faster.

Example 4.5NCERT Example 4.5 · p. 57

A ball bouncing off a smooth floor

A 60 g ball strikes a smooth horizontal floor at 10 m/s, its velocity making 60∘ with the vertical. A smooth floor can push only at right angles to itself. The ball leaves the floor with an upward velocity component of 3.0 m/s. Find (a) the impulse on the ball, (b) the direction of the force on the floor and (c) the speed and direction of the rebound.

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

Quick check
1 of 3
A book rests on a table. By the third law, the partner of the Earth's pull on the book (its weight) is
Show answer

Answer (C) the book's gravitational pull on the Earth

Weight is the force on the book by the Earth, so its partner is the force on the Earth by the book. The table's push also acts on the book, so it cannot be the partner, even though it happens to be equal here.

4.7 Conservation of momentum

Put the second and third laws together and a powerful result drops out. Fire a cannon. While the powder burns, the cannon pushes the ball forward with a force 𝐅, and the ball pushes the cannon back with −𝐅, for the same short time Δt. The ball's momentum changes by 𝐅Δt and the cannon's by −𝐅Δt. Both started at rest, so afterwards the ball's momentum and the cannon's recoil momentum add up to zero, just as before the shot.

LawLaw of conservation of momentum

If no net external force acts on a group of interacting particles (an isolated system), their total momentum never changes.

DerivationMomentum in a collision

Two bodies A and B with momenta 𝐩A and 𝐩B collide and separate with momenta 𝐩 A′ and 𝐩 B′. During the contact time Δt each feels only the other's force. By the second law:

𝐅AB Δt=𝐩 A′−𝐩A𝐅BA Δt=𝐩 B′−𝐩B

Adding, the left side is zero because 𝐅AB=−𝐅BA ((4.8)). Rearranging:

Momentum is conserved
𝐩 A′+𝐩 B′=𝐩A+𝐩B
(4.9)

Nothing in the argument depends on what happens to the energy, so the result holds whether the bodies bounce apart (an elastic collision) or crumple and stick (an inelastic one). In an elastic collision the total kinetic energy is conserved as well (Chapter 5). For many particles, the internal forces still cancel pair by pair, so the total momentum cannot change unless an external force acts.

3/4An elastic bounce: A ends at −13 m/s, B at 143 m/s. Total: 4.

Fig. 4.7A collision between a 2.0 kg trolley at 3.0 m/s and a 1.0 kg trolley at 2.0 m/s coming the other way. Momentum moves from one to the other, but the total stays 4.0 kg m s−1.NCERT p. 58

Trolley A of 2 kg and trolley B of 1 kg run on a track; below them are horizontal bars for the momentum of A, of B and the total, with a dashed line marking the total. Step 1, before: A moves right with a long arrow and B moves left; bar A is 6 units to the right, bar B is 2 units to the left and the total bar reaches 4. Step 2, in contact: equal and opposite orange force arrows; both trolleys share one velocity, bars A and B are 2.7 and 1.3 units and the total is still 4. Step 3, they bounce: A moves slowly left and B fast right; bar A is slightly negative, bar B is 4.7 and the total is 4. Step 4, they stick: both move right together; the bars are 2.7 and 1.3 and the total is again 4.

Example

Coupling railway wagons

In a goods yard, a 20 t wagon rolling at 1.5 m/s runs into a stationary 30 t wagon, and the two couple together. Ignoring friction during the brief impact, find their common speed and the impulse each wagon receives.

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

Quick check
1 of 3
A 0.20 kg ball at 3.0 m/s hits a stationary 0.30 kg ball head-on and stops dead. How fast does the second ball move off, in m/s?
m/s
Show answer

Answer 2 m/s

0.20×3.0=0.30 v, so v=2.0 m/s.

4.8 Equilibrium of a particle

A particle is in equilibrium (साम्यावस्था) when the net external force on it is zero. By the first law it is then either at rest or moving with constant velocity. With two forces this simply means they are equal and opposite:

𝐅1=−𝐅2
(4.10)

With three forces acting at one point, equilibrium needs their vector sum to vanish:

𝐅1+𝐅2+𝐅3=0
(4.11)

Read geometrically, this says that the resultant of any two of the forces (by the parallelogram law) is equal and opposite to the third. Equivalently, if we draw the three forces head to tail, they close into a triangle (Fig. 4.8). The idea extends to any number: a particle is in equilibrium under n forces exactly when their arrows, joined head to tail, form a closed polygon.

3/3The last arrow ends where the first began: the vector sum is zero.

Fig. 4.8Forces in equilibrium, drawn head to tail, form a closed triangle. The tensions here are those of Example 4.6.NCERT Fig. 4.7 · p. 58

On the left, a knot pulled by three forces: a long blue arrow up and to the left, a shorter dashed orange arrow up and to the right, and a vermilion arrow straight down. Step 1 shows only these three forces. Step 2: copies of the blue and orange arrows slide to the right panel and are joined head to tail, labelled head to tail. Step 3: a copy of the downward arrow slides across and its head lands exactly on the starting point, closing a triangle, labelled closed, sum is zero.

For calculations we usually resolve along axes instead. (4.11) then splits into one scalar equation per axis:

F1x+F2x+F3x=0F1y+F2y+F3y=0F1z+F2z+F3z=0
(4.12)

For an extended body this is only half the story. A body can feel zero net force and still start to spin, so its equilibrium also needs zero net torque, which we study in Chapter 6. A particle has no size, so for it zero net force is enough.

Example 4.6NCERT Example 4.6 · p. 58

A lantern on two cords

A 5.0 kg lantern hangs from a knot K. One cord runs from K to the ceiling at 30∘ to the vertical; another runs to a wall at 60∘ to the vertical. Find the tension in each cord. Take g=10 m/s2 and treat the cords as massless.

Fig. 4.9The set-up and the free-body diagram of the knot.

Left: a ceiling and a wall bracket with two cords meeting at a knot K; the cord to the ceiling makes 30 degrees with a dashed vertical line and the cord to the wall makes 60 degrees with it. A 5.0 kilogram lantern hangs below the knot. Right: the free-body diagram of the knot, with tension T1 along the first cord, a shorter dashed tension T2 along the second cord and the weight W pointing down.

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

Quick check
1 of 3
Forces of 3 N and 4 N act at right angles on a particle. What single third force keeps it in equilibrium?
Show answer

Answer (D) 5 N, opposite to the resultant of the two

The resultant is 32+42=5 N. The third force must be equal and opposite to it. Adding magnitudes (7 N) ignores directions.

4.9 Common forces in mechanics

Gravity is special: it acts between bodies that are not touching, across empty space. Every object near the Earth feels it, and it also steers the Moon round the Earth and the planets round the Sun. (Charged and magnetised bodies also feel electric and magnetic forces at a distance; we leave them aside here.)

Every other force we meet in mechanics is a contact force: it exists only while one object touches another, solid or fluid, and it comes in third-law pairs. Where two solid surfaces press together, it is useful to split the contact force into two parts (Fig. 4.10). The part perpendicular to the surfaces is the normal reaction; the part along them is friction. Fluids exert contact forces too: an object in water or air feels an upward buoyant force that matches the weight of the fluid it pushes aside, and a body moving through a fluid feels a viscous drag, which in air we call air resistance.

2/3A gentle push: friction appears, so the contact force leans back against the push.

Fig. 4.10One contact force, two components: normal (perpendicular to the floor) and friction (along it). The crate has not started to slide.NCERT Fig. 4.9 · p. 59

A crate stands on a floor with an orange weight arrow pointing down from its centre and a purple arrow for the contact force from the floor. Step 1: no push; the contact force points straight up. Step 2: a short blue push arrow acts from the left; the purple contact force now leans back, against the push, and dashed lines split it into a vertical normal part and a horizontal friction part along the floor. Step 3: a longer push arrow; the contact force leans back further and its friction part is longer, while the normal part stays the same.

Two other contact forces appear in almost every problem. A spring that is stretched or squashed pushes or pulls back towards its natural length. For small deformations the restoring force is proportional to the displacement x from the natural length:

F=−kx
where
k
the spring constant (force constant), in N/m
x
the extension (positive) or compression (negative)

The minus sign says the force always opposes the displacement. A string behaves like a spring with an enormous k: it stretches almost not at all, and the pull it exerts is called tension. In problems we usually take a string to be massless and inextensible. Then the tension is the same all along it, and a smooth, light pulley changes only the direction of the pull, not its size.

Physicists recognise four fundamental forces: gravitational, electromagnetic, strong and weak. The strong and weak forces act only inside atomic nuclei, so in mechanics only gravity and electromagnetism matter. Remarkably, every contact force, from the normal reaction and friction to tension, spring forces and drag, is electrical at heart: it comes from the forces between the charged electrons and nuclei of the atoms in the two surfaces. Working them out atom by atom would be hopelessly complicated, so mechanics describes each one by a simple law found from experiment.

4.9.1 Friction

Lean on a steel almirah standing on the floor: nothing happens. Lean harder: still nothing. Since the almirah does not accelerate, the first law says the horizontal forces on it must balance, so the floor is pushing back on it with a force exactly equal to yours. That force is static friction fs. With no push there is no static friction at all; as you push harder it grows to match you. It opposes the impending motion, the sliding that would happen if the floor were frictionless.

Definition
Static friction

The friction force between two surfaces that are not sliding over each other. It adjusts itself to oppose any tendency to slide, up to a maximum value μsN.

Static friction cannot grow without limit: push hard enough and the almirah starts to slide. Experiments show that the largest possible static friction does not depend on the area of contact and is, to a good approximation, proportional to the normal force:

Limiting static friction
(fs)max⁡=μsN
(4.13)

The constant μs, the coefficient of static friction, depends only on the pair of surfaces in contact. The rule for static friction is therefore an inequality:

Law of static friction
fs≤μsN
(4.14)

Once sliding starts, the friction force drops below the static maximum. This kinetic friction (or sliding friction) also does not depend on the contact area, is nearly independent of the speed, and follows a similar law:

Kinetic friction
fk=μkN
(4.15)
Definition
Kinetic friction

The friction force between two surfaces sliding over each other, fk=μkN, directed against the relative motion. For a given pair of surfaces, μk is less than μs.

Because μk<μs, it takes more force to start a body sliding than to keep it sliding (Fig. 4.11). Once it is moving, the net force on it is F−μkN, so a block can glide at a steady speed under a push smaller than the one that started it, and it slows to rest at μkg once the push stops. Try both regimes in the friction lab below (Fig. 4.12).

2/3At the peak, static friction reaches its limit μsN.

Fig. 4.11Friction on a block as the push grows. Static friction matches the push up to μsN; after slipping, kinetic friction stays near μkN.NCERT Fig. 4.10 · p. 60

A graph with applied force on the horizontal axis and friction on the vertical axis. A blue line rises at 45 degrees from the origin through a shaded region labelled at rest, up to a peak; a dashed drop leads down to an orange horizontal line in a region labelled sliding. Step 1: a dot sits on the rising line, where friction equals the push. Step 2: the dot is at the peak, the limiting static friction. Step 3: the sliding region brightens and the dot sits on the lower horizontal line, the constant kinetic friction.

InteractiveFriction lab

An interactive block-and-friction lab with two modes. In the push mode a block rests on a level floor; a slider sets the applied force, and arrows show the push, the friction, the normal force and the weight. In the tilt mode a slider tilts the board, and the arrows show the normal force perpendicular to the board, friction up the slope and the vertical weight. Sliders set the mass and the two friction coefficients. A graph plots friction against the push or the tilt angle, with a dot for the present state, and readouts give the normal force, the friction and the acceleration. A status line says whether the block is at rest or sliding, and once it slides it keeps sliding until the push or tilt falls below the kinetic limit.

Fig. 4.12Push a block on a level floor, or tilt the board under it. Watch static friction match the push until it reaches μsN.
Table 4.1Typical coefficients of friction (dry surfaces; real values vary a lot with cleanliness and roughness)
Surfacesμsμk
Rubber on dry concrete1.00.8
Steel on steel0.740.57
Wood on wood0.25–0.50.2
Ice on ice0.10.03
Teflon on Teflon0.040.04

Unlike Newton's law of gravitation, these friction rules are not deep laws of nature. They are rules of thumb distilled from measurements, and real surfaces can depart from them noticeably. For most calculations, though, they are all we need.

Notice what friction actually opposes: relative motion between the surfaces, not motion as such. A crate on the open bed of a delivery truck speeds up with the truck. The only horizontal force on the crate is friction from the truck bed, and it points forwards, in the direction of motion. If the bed were frictionless, the truck would slide out from under the crate. Static friction opposes that impending slip, and in doing so drags the crate forward with the truck.

Example 4.7NCERT Example 4.7 · p. 61

A crate on a truck bed

A crate rests on the flat bed of a truck. The coefficients of friction between them are μs=0.35 and μk=0.28. (a) What is the greatest acceleration the truck can have without the crate slipping? (b) The driver goes from rest to 54 km/h in 4.0 s. Does the crate slip, and if so, what is its acceleration relative to the truck? Take g=9.8 m/s2.

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

Example 4.8NCERT Example 4.8 · p. 61

A tiffin box on a tilted board

A steel tiffin box sits on a wooden board 1.20 m long. One end of the board is raised slowly, and the box just begins to slide when that end is 0.50 m higher than the other. (a) Find μs. (b) If μk=0.30, what is the box's acceleration once it slides, with the board held at that angle? Take g=9.8 m/s2.

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

Example 4.9NCERT Example 4.9 · p. 61

A block and a hanging pan

A 4.0 kg block on a horizontal table is tied by a light string, passing over a smooth pulley at the table's edge, to a 1.5 kg pan hanging freely. Between block and table, μs=0.30 and μk=0.20. The system is released from rest. Does it move? If so, find the acceleration and the tension. Take g=9.8 m/s2.

Fig. 4.13The forces along the direction of motion.

A 4.0 kilogram block sits on a table. A string runs from it horizontally to a small pulley at the right edge of the table and then down to a hanging pan of 1.5 kilograms. A green arrow shows the tension pulling the block to the right, an orange arrow shows friction on the block pointing left, and blue arrows show the block accelerating to the right and the pan accelerating downward.

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

Rolling friction

Drag a brick across the floor and then roll a ball of the same weight: the ball keeps going far longer. In the ideal case a hard wheel rolling on a hard road touches it at a single point that is momentarily at rest on the road, so nothing slides and there is no sliding friction at all. Real wheels still slow down, because the tyre flattens a little and the road dents a little. They now touch over a small patch instead of a point, and the contact force on that patch tilts slightly backwards (Fig. 4.14). Its backward component is rolling friction. For the same load it is typically hundreds to a thousand times smaller than sliding friction, which is why loads move so much more easily on wheels or rollers than when dragged.

2/3Both surfaces deform; the contact force on the patch leans back, opposing the rolling.

Fig. 4.14Rolling friction comes from deformation. Ball bearings replace sliding between a shaft and its housing by rolling.NCERT Fig. 4.13 · p. 62

Step 1: an ideal rigid wheel touches a flat road at a single marked point, labelled point contact. Step 2: the tyre is flattened at the bottom and the road is slightly dented, so they meet over a contact patch; a green arrow shows the wheel rolling to the right, and a purple contact-force arrow from the patch leans backwards, towards the left. Step 3: a ball bearing appears beside the wheel: an outer ring, a central shaft and a ring of eight balls between them, labelled ball bearing.

Friction is often a nuisance. In a machine, sliding parts waste power as heat and wear each other away. Engineers fight it with lubricating oil, with ball bearings that turn sliding into rolling, and with a thin cushion of air between surfaces, as in a hovercraft or an air-hockey table.

Just as often, friction is exactly what we need. You can walk only because the ground's friction pushes your foot forward; people and animals need an external force to accelerate, just like carts. The same friction from the road speeds up a car, steers it round a bend and, through its brakes, brings it to a stop. On a road covered with oil or ice, a car can do none of these.

Compare

Friction: foe and friend

Reduce it

  • Engines and gearboxes: oil films
  • Bicycle hubs and fans: ball bearings
  • Hovercraft: an air cushion
  • Polished, well-fitted moving parts

Need it

  • Walking and running
  • Tyres gripping, steering and braking
  • Nails, screws and knots holding
  • Writing with a pen or a pencil
Quick check
1 of 4
A 10 kg box rests on a floor with μs=0.50. You push it sideways with 30 N and it does not move. How large is the friction force, in newtons? Take g=9.8 m/s2.
N
Show answer

Answer 30 N

The box is at rest, so friction just balances the push: 30 N. The limit μsN=49 N is not reached.

4.10 Circular motion

In Chapter 3 we found that a body moving at constant speed v round a circle of radius R accelerates towards the centre, with magnitude v2/R. By the second law, a net force must point towards the centre too. Its size is

Centripetal force
fc=mv2R
(4.16)

This inward net force is called the centripetal force (अभिकेंद्र बल). The name describes a job, not a new kind of force. In every circular motion some real force, exerted by a real object, does the job: the tension in a string for a whirled stone, the Sun's gravity for a planet, the Earth's gravity for the Moon or a satellite, friction for a car on a flat bend, and part of the normal force for a car on a banked track or a rider on a maut ka kuan (wall of death). Never draw "centripetal force" as an extra arrow beside these; it is their resultant towards the centre.

Definition
Centripetal force

The net force towards the centre needed to keep a body of mass m moving in a circle of radius R at speed v; its magnitude is mv2/R. It is always supplied by real forces such as tension, gravity, friction or a normal force.

Motion of a car on a level road

DerivationMaximum speed on a flat bend

Three forces act on the car: its weight mg, the normal force N and friction f from the road. Nothing accelerates vertically, so N−mg=0:

N=mg
(4.17)

The only horizontal force is the friction of the road on the tyres, which must supply the whole centripetal force. The tyres do not slide sideways, so this is static friction, opposing the car's impending skid outwards. Static friction is limited by (4.14):

mv2R=fs≤μsN=μsmg

so v2≤μsRg. The mass cancels, and the highest safe speed on the bend is

Maximum speed, level road
vmax⁡=μsRg
(4.18)

A loaded truck and an empty one, on the same tyres and road, share the same limit. Rain or sand lowers μs and with it vmax⁡; a sharper bend (smaller R) lowers it too.

Example 4.10NCERT Example 4.10 · p. 64

A scooter on a dry road and a wet road

A scooter rider takes a flat turn of radius 25 m at 36 km/h. Is the turn safe (a) on a dry road with μs=0.45 and (b) after rain, when μs=0.20? (c) What is the safe limit on the wet road? Take g=9.8 m/s2.

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

Motion of a car on a banked road

Relying on friction alone is risky, so highway curves and racetracks are banked: the outer edge of the road is built higher than the inner edge. The normal force, being perpendicular to the road, then tilts towards the centre of the curve, and its horizontal component helps to supply the centripetal force (Fig. 4.15).

DerivationMaximum speed on a banked road

Let the road be banked at angle θ. At high speed the car tends to skid up and outwards, so friction f acts down the slope. Vertically there is no acceleration:

Ncos⁡θ=mg+fsin⁡θ
(4.19a)

Horizontally, the components of N and f towards the centre supply the centripetal force:

Nsin⁡θ+fcos⁡θ=mv2R
(4.19b)

Since f≤μsN, the speed is greatest when f=μsN:

Ncos⁡θ−μsNsin⁡θ=mg
(4.20a)
Nsin⁡θ+μsNcos⁡θ=mvmax⁡2R
(4.20b)

Divide (4.20b) by (4.20a). Both N and m cancel:

vmax⁡2Rg=sin⁡θ+μscos⁡θcos⁡θ−μssin⁡θ

Dividing the top and bottom of the fraction by cos⁡θ gives the result:

Maximum speed, banked road
vmax⁡2Rg=μs+tan⁡θ1−μstan⁡θ
(4.21)

With θ=0 this reduces to (4.18); any banking raises the limit. With no friction at all (μs=0), only one speed works:

Optimum speed
v0=Rgtan⁡θ
(4.22)

At the optimum speed v0 the tilted normal force alone provides mv2/R, so the tyres need no sideways friction at all and wear least. Drive faster than v0 and friction must act down the slope; drive slower and the car tends to slide down the bank, so friction acts up the slope. Repeating the derivation with f reversed gives the lowest safe speed:

Minimum speed, banked road
vmin⁡2=Rg tan⁡θ−μs1+μstan⁡θ

If tan⁡θ≤μs the right-hand side is zero or negative, so there is no lower limit: a car can even be parked on the bank without sliding down.

2/4Banked, at v0: normal force plus weight add to a horizontal net force; no friction is needed.

Fig. 4.15Who supplies mv2/R? Friction on a level road; the tilted normal force on a banked road, with friction helping in one direction or the other.NCERT Fig. 4.14 · p. 63

A car is seen from behind on a road; an arrow at the top left points towards the centre of the turn. Step 1, level road: the blue normal force points straight up, the vermilion weight straight down, and a green friction arrow at the tyres points towards the centre. Step 2, banked road at the ideal speed: the road and car tilt up towards the outside; the normal force tilts towards the centre, there is no friction arrow, and a dashed black arrow shows the net force pointing horizontally towards the centre. Step 3, banked and faster: a green friction arrow points down the slope. Step 4, banked and slower: the friction arrow points up the slope.

InteractiveBanked road lab

A rear view of a car on a road whose bank angle you set with a slider. Sliders also set the speed, the radius of the turn and the coefficient of static friction. Arrows show the normal force perpendicular to the road, the vertical weight and the friction needed, which points down the slope, up the slope or vanishes; it turns orange when more friction is needed than the tyres can give. Readouts show the optimum speed, the lowest and highest safe speeds and the centripetal acceleration, and a status line says whether the car grips, skids outwards or slides inwards. Buttons set the speed to the optimum, flatten the road or make it icy.

Fig. 4.16Set the radius, bank angle, friction and speed. The lab finds the friction the tyres must supply and whether they can.
Example 4.11NCERT Example 4.11 · p. 64

Designing a highway curve

A highway curve of radius 120 m is banked at 10∘, and μs=0.30 between tyres and road. Find (a) the optimum speed, (b) the maximum safe speed and (c) whether a car can stop and wait on the curve without sliding down. Take g=9.8 m/s2.

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

Quick check
1 of 4
A car goes round a level bend at constant speed. Which force supplies its centripetal force?
Show answer

Answer (C) Static friction from the road

The tyres grip without sliding, so the inward force is static friction. On a level road the normal force is vertical and cannot point to the centre; centrifugal force is not a real force on the car.

4.11 Solving problems in mechanics

With three laws you can now tackle a huge range of problems. A typical one involves several bodies, linked by strings, stacked on one another or pressing against supports, each also pulled by gravity. The key freedom is this: you may apply the laws of motion to any part of the arrangement you like, which we call the system, as long as you include every force exerted on it by everything else, the environment. The worked examples so far have quietly used this idea. Here it is as a routine:

  1. Sketch the whole arrangement: the bodies, strings, pulleys and supports.
  2. Choose a convenient part as your system.
  3. Draw it on its own with every force acting on it from outside: from bodies in contact, from strings and from gravity. Leave out the forces the system exerts on other things. This is a free-body diagram; it does not imply that the net force is zero.
  4. Mark the forces you know for sure (weight straight down, tension along the string, normal force perpendicular to the surface) and label the rest as unknowns. Name each force as "on X by Y": friction, tension, thrust, buoyancy and weight are all just forces.
  5. If you need more equations, choose another system. If A's diagram shows a force 𝐅 by B, then B's diagram must show −𝐅 by A.
  6. Write 𝐅=m𝐚 along convenient axes for each system, solve, and check the units and a limiting case.

3/4On the scale: the floor's push N2 up, the student's push N1 down and its own small weight W2.

Fig. 4.17The same lift seen as three different systems. Forces between the parts of a system cancel and drop out of its diagram (the scale's weight is enlarged for visibility).NCERT Fig. 4.15 · p. 65

Left: a lift cabin hanging from a cable, with a blue acceleration arrow pointing up beside it; inside, a student stands on a bathroom scale. Step 1: the right panel says choose a system, draw its forces. Step 2, student alone: a box with a long green arrow up, the normal force from the scale, and a shorter vermilion arrow down, the student's weight. Step 3, scale alone: a flat box with a long purple arrow up from the floor, a green arrow down from the student and a tiny vermilion arrow down for the scale's own weight. Step 4, student plus scale: a long purple arrow up from the floor and a vermilion arrow down for the total weight; the forces between student and scale no longer appear.

Example 4.12NCERT Example 4.12 · p. 65

A bathroom scale in a lift

A 50 kg student stands on a 2.0 kg bathroom scale on the floor of a lift that accelerates upward at 1.5 m/s2. Find (a) the force between the student and the scale, (b) the force between the scale and the floor, and (c) list the action–reaction pairs. Take g=9.8 m/s2.

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

The example shows why the familiar result "normal force equals weight" must be used with care. It holds only when the body is in equilibrium (on a desk, or in a lift moving at constant velocity), and it has nothing to do with the third law. Free-body diagrams make such traps visible; the exercises below will turn them into a habit.

Quick check
1 of 3
A 50 kg person stands on a scale in a lift accelerating downward at 2.0 m/s2. The scale reads (g=10 m/s2)
Show answer

Answer (C) 400 N

N=m(g−a)=50×8.0=400 N. 600 N would be for an upward acceleration; zero only in free fall.
Recap

Laws of Motion: key points and summary

  1. Aristotle thought a moving body needs a continuing force. In fact a force is needed only to balance friction and drag; with zero net force, motion continues unchanged (Section 4.2).
  2. Galileo's slope experiments led to the law of inertia: rest and uniform motion in a straight line are equivalent states of zero net force (Section 4.3).
  3. First law: a body's velocity changes only when a net external force acts on it (Section 4.4).
  4. Momentum is 𝐩=m𝐯. Second law: 𝐅=d𝐩/dt=m𝐚, with 1 N=1 kg m s−2 ((4.5)).
  5. The second law is a vector law (one equation per axis), applies to systems with 𝐅 the total external force, and is local: force now fixes acceleration now (four features).
  6. Impulse =𝐅avΔt=Δ𝐩, the area under a force–time graph; a longer contact time means a smaller force (impulse).
  7. Third law: forces come in simultaneous, equal and opposite pairs that act on different bodies, so they never cancel for one body (Section 4.6).
  8. The total momentum of an isolated system is conserved, in elastic and inelastic collisions alike ((4.9)).
  9. A particle is in equilibrium when the forces on it sum to zero: drawn head to tail they close a polygon (Section 4.8).
  10. Gravity is the only non-contact force of mechanics; normal reaction, friction, tension, spring forces and drag are contact forces of electrical origin (Section 4.9).
  11. Static friction adjusts itself up to μsN; kinetic friction is μkN with μk<μs; rolling friction is far smaller (Section 4.9.1).
  12. Centripetal force mv2/R is supplied by real forces. On a level road vmax⁡=μsRg; on a banked road v0=Rgtan⁡θ and vmax⁡ is higher (Section 4.10).
  13. Solve problems with free-body diagrams: choose a system, draw every force on it from outside, keep third-law pairs consistent, then apply 𝐅=m𝐚 (Section 4.11).
Flash cards · 10

Formula revision

1 / 10
Practice · 29 questions

Practice questions with answers (NCERT “Exercises”)

Take g=10 m/s2 unless a question says otherwise, and neglect air resistance. Questions 24–29 are in the style of JEE Main and NEET.

  1. Short answerEasyNCERT Exercises 4.1

    For each case, state the size and direction of the resultant force.

    1. A skydiver who has reached a steady falling speed.

      Show solution

      AnswerZero.

      Steady speed in a straight line means zero acceleration, so air resistance exactly balances the weight. Net force zero.

    2. A 25 g plastic duck floating at rest in a bucket of water.

      Show solution

      AnswerZero.

      It is at rest, so the buoyant force (0.25 N upwards) balances its weight (0.25 N downwards).

    3. A drone hovering motionless above a cricket ground.

      Show solution

      AnswerZero.

      No acceleration, so the upward push of the air on the rotors balances its weight.

    4. An 8000 kg bus going round a roundabout of radius 20 m at a steady 18 km/h.

      Show solution

      Answer1.0×104 N, towards the centre.

      Its speed is steady but its direction changes, so it accelerates towards the centre. With v=5.0 m/s: F=mv2/R=8000×25/20=1.0×104 N, towards the centre of the roundabout.

  2. Short answerEasyNCERT Exercises 4.2

    A 0.20 kg ball is thrown at 60∘ above the horizontal. What resultant force (size and direction) acts on it

    1. on its way up;

      N
      Show answer

      Answer 2 N

      Only gravity acts: mg=0.20×10=2.0 N, vertically downwards.
    2. at the highest point of its path;

      Show solution

      Answer2.0 N, vertically downwards.

      Still only gravity. The ball's velocity there is horizontal, not zero, but even a ball momentarily at rest would feel mg.

    3. on its way down;

      Show solution

      Answer2.0 N, vertically downwards.

      The direction of motion does not matter: the force is the weight throughout the flight.

    4. Would any answer change if the ball were thrown straight up?

      Show solution

      AnswerNo.

      The net force is mg downwards at every point of any flight, whatever the launch angle. Only the velocity differs.

  3. Short answerMediumNCERT Exercises 4.3

    Find the net force (magnitude and direction) on a 0.25 kg water bottle in each case.

    1. Just after it falls out of the window of a bus moving at a steady 54 km/h.

      Show solution

      Answer2.5 N, vertically down.

      Once free, only gravity acts: mg=0.25×10=2.5 N downwards. Its horizontal velocity of 15 m/s needs no force.

    2. Just after it falls out of the window of a bus that is speeding up at 2.0 m/s2.

      Show solution

      Answer2.5 N, vertically down.

      The same: the bus's acceleration does not travel with the bottle. Force depends only on the situation now, so there is no horizontal force.

    3. Standing on the floor of that speeding-up bus without sliding.

      Show solution

      Answer0.50 N, forwards.

      It accelerates with the bus, so F=ma=0.25×2.0=0.50 N forwards, supplied by static friction. Weight and normal force cancel.

    4. Sliding over that bus's floor, with μk=0.10.

      Show solution

      Answer0.25 N, forwards.

      It slides backwards relative to the floor, so kinetic friction on it points forwards: μkmg=0.10×0.25×10=0.25 N. That is the net force.

  4. Quick answerEasyNCERT Exercises 4.4

    A small ball tied to a pin by a string moves in a horizontal circle on a smooth table. If its speed is doubled while the radius stays the same, the tension in the string becomes

    Show answer

    Answer (D) four times as large

    T=mv2/r, so doubling v multiplies T by 4. The net inward force is just T; there is no extra outward force to subtract.
  5. NumericalEasyNCERT Exercises 4.5

    A 1500 kg car moving at 20 m/s brakes with a steady retarding force of 6.0 kN. How long, in seconds, does it take to stop?

    s
    Show solution

    a=F/m=6000/1500=4.0 m/s2, so t=v/a=20/4.0=5.0 s. It travels v2/2a=50 m while stopping.

  6. NumericalEasyNCERT Exercises 4.6

    A steady force speeds a 0.80 kg trolley up from 1.2 m/s to 3.0 m/s in 6.0 s, without changing its direction. Find the force in newtons.

    N
    Show solution

    F=mΔv/Δt=0.80×1.8/6.0=0.24 N, in the direction of motion.

  7. NumericalMediumNCERT Exercises 4.7

    Two forces act on a 2.0 kg puck on smooth ice: 5.0 N towards the east and 12 N towards the north. Find the magnitude and direction of its acceleration.

    Show solution

    Answer6.5 m/s2, at about 67∘ north of east.

    F=5.02+122=13 N

    So a=13/2.0=6.5 m/s2, along the resultant force, at tan⁡−1(12/5.0)≈67∘ north of east.

  8. NumericalEasyNCERT Exercises 4.8

    An e-rickshaw of mass 350 kg carries a 60 kg driver and two passengers of 55 kg each. Travelling at 27 km/h, it stops in 3.0 s to avoid a cow on the road. Find the average retarding force, in newtons.

    N
    Show solution

    Total mass 350+60+110=520 kg; 27 km/h=7.5 m/s. So a=7.5/3.0=2.5 m/s2 and F=520×2.5=1.3×103 N.

  9. NumericalMediumNCERT Exercises 4.9

    A 1.5 kg model rocket leaves the launch pad with an upward acceleration of 25 m/s2. What is the thrust of its motor, in newtons?

    N
    Show solution

    T−mg=ma, so T=m(g+a)=1.5×(10+25)=52.5 N. Forgetting the weight gives ma=37.5 N; that thrust would leave a net force of only 22.5 N and an acceleration of 15 m/s2.

  10. NumericalHardNCERT Exercises 4.10

    A 0.50 kg puck glides east at a steady 6.0 m/s. At t=0, when it is at x=0 (east positive), a constant 2.0 N force towards the west starts to act on it; the force is removed at t=10 s. Find the puck's position at:

    1. t=−4.0 s;

      Show solution

      Answerx=−24 m.

      Before t=0 no force acts, so x=6.0t=−24 m.

    2. t=5.0 s;

      Show solution

      Answerx=−20 m.

      a=−2.0/0.50=−4.0 m/s2, so x=6.0(5.0)−12(4.0)(5.0)2=30−50=−20 m. The puck has already turned round (at t=1.5 s).

    3. t=20 s.

      Show solution

      Answerx=−480 m.

      At t=10 s: x=60−200=−140 m and v=6.0−40=−34 m/s. Then it moves uniformly: x=−140−34×10=−480 m.

  11. NumericalMediumNCERT Exercises 4.11

    A pickup van starts from rest and accelerates uniformly at 1.5 m/s2 along a straight road. At t=8.0 s a ball is released from rest relative to the van, 5.0 m above the ground. Find the ball's (a) velocity and (b) acceleration at t=8.5 s.

    Show solution

    Answer(a) 13 m/s at about 23∘ below the horizontal. (b) 10 m/s2, vertically down.

    At release, vx=1.5×8.0=12 m/s and vy=0. The ball takes 2×5.0/10=1.0 s to land, so at t=8.5 s it is still in the air, 0.50 s after release.

    (a) vx=12 m/s (no horizontal force), vy=10×0.50=5.0 m/s downwards, so v=122+52=13 m/s at tan⁡−1(5/12)≈23∘ below the horizontal.

    (b) Only gravity acts: 10 m/s2 downwards. The van's 1.5 m/s2 plays no part.

  12. Short answerMediumNCERT Exercises 4.12

    A 0.20 kg toy hangs from a 1.5 m cord and swings to and fro; its speed at the lowest point is 2.0 m/s. Describe its path if the cord snaps:

    1. as the toy passes the lowest point;

      Show solution

      AnswerA parabola, launched horizontally.

      Its velocity is 2.0 m/s horizontally, and afterwards only gravity acts, so it moves like a projectile launched horizontally: a downward-curving parabola.

    2. at one of the extreme positions of the swing.

      Show solution

      AnswerStraight down.

      There it is momentarily at rest, so with only gravity acting it falls vertically from rest.

  13. NumericalMediumNCERT Exercises 4.13

    A 60 kg woman stands on a weighing scale (reading in newtons) in a lift. What does it read when the lift is

    1. moving down at a steady 3.0 m/s?

      N
      Show answer

      Answer 600 N

      Zero acceleration, so N=mg=600 N. The speed itself does not matter.
    2. moving down but slowing at 2.5 m/s2?

      N
      Show answer

      Answer 750 N

      Slowing while moving down means the acceleration points up: N=m(g+a)=60×12.5=750 N.
    3. moving up but slowing at 2.5 m/s2?

      N
      Show answer

      Answer 450 N

      The acceleration points down: N=m(g−a)=60×7.5=450 N.
    4. falling freely after its cable snaps?

      N
      Show answer

      Answer 0 N

      Woman and scale both fall with acceleration g, so N=m(g−g)=0: she feels weightless.
  14. NumericalMediumNCERT Exercises 4.14

    The graph shows the position of a 2.0 kg particle moving along a line.

    Fig. 4.18Position against time for Question 14.

    A position–time graph. The position is 0 metres before time zero, rises in a straight line to 6 metres at 3 seconds, falls in a straight line to 3 metres at 6 seconds, and then stays at 3 metres.

    1. What is the force on the particle for t<0, for 0<t<3 s, for 3<t<6 s and for t>6 s?

      Show solution

      AnswerZero in every interval.

      Each segment is straight, so the velocity is constant within it (0, +2.0, −1.0 and 0 m/s) and the acceleration, and hence the force, is zero.

    2. What impulse does it receive at t=0, at t=3 s and at t=6 s?

      Show solution

      Answer+4.0, −6.0 and +2.0 N s.

      J=mΔv. At t=0: 2.0(2.0−0)=+4.0 N s. At t=3 s: 2.0(−1.0−2.0)=−6.0 N s. At t=6 s: 2.0(0+1.0)=+2.0 N s. Positive means along +x.

  15. NumericalMediumNCERT Exercises 4.15

    Crates P (15 kg) and Q (45 kg) on a smooth floor are joined by a light rope. A horizontal 240 N force pulls along the rope's line. Find the tension when the force pulls

    1. on P (so P leads)?

      N
      Show answer

      Answer 180 N

      a=240/60=4.0 m/s2. The rope alone accelerates Q: T=45×4.0=180 N.
    2. on Q (so Q leads)?

      N
      Show answer

      Answer 60 N

      Same acceleration; now the rope accelerates P: T=15×4.0=60 N. Pulling the heavier crate keeps the rope's tension low.
  16. NumericalMediumNCERT Exercises 4.16

    Masses of 3.0 kg and 5.0 kg hang at the ends of a light string over a smooth, light pulley and are released.

    1. Find the acceleration of the masses.

      m/s²
      Show solution

      50−T=5.0a and T−30=3.0a. Adding: 20=8.0a, so a=2.5 m/s2.

    2. Find the tension in the string.

      N
      Show solution

      T=30+3.0×2.5=37.5 N, between the two weights, as it must be.

    3. What force does the pulley's support carry while they move?

      N
      Show solution

      Both sides of the string pull down on the pulley with T, so the support carries 2T=75 N, less than the total weight of 80 N because the system's centre of mass is accelerating downward.

  17. Short answerMediumNCERT Exercises 4.17

    A firecracker shell at rest on a smooth frozen pond bursts into two fragments that slide along the ice. Show that they must move in exactly opposite directions. If their masses are in the ratio 1:3, find the ratio of their speeds.

    Show solution

    AnswerOpposite directions; speeds in the ratio 3:1 (the lighter fragment is faster).

    The forces of the burst are internal and the horizontal external force is negligible, so the total momentum stays zero: m1𝐯1+m2𝐯2=0, that is, 𝐯2=−(m1/m2)𝐯1. The velocities are antiparallel. With m1:m2=1:3, v1:v2=m2:m1=3:1.

  18. NumericalEasyNCERT Exercises 4.18

    Two 0.50 kg trolleys move towards each other at 3.0 m/s each, collide head-on and rebound at 2.0 m/s each. What impulse, in N s, does each receive from the other?

    N s
    Show solution

    For each trolley J=m(v−u)=0.50 [2.0−(−3.0)]=2.5 N s, directed back the way it came. The two impulses are equal and opposite, as the third law requires.

  19. NumericalEasyNCERT Exercises 4.19

    A 1.2 t cannon on a fort's rampart fires a 6.0 kg ball horizontally at 240 m/s. With what speed, in m/s, does the cannon start to recoil?

    m/s
    Show solution

    1200 V=6.0×240, so V=1.2 m/s, opposite to the ball.

  20. NumericalHardNCERT Exercises 4.20

    A hockey player deflects a 0.16 kg ball through 60∘ without changing its speed of 72 km/h. Find the magnitude (in N s) and the direction of the impulse on the ball.

    N s
    Show solution

    v=20 m/s. The vectors 𝐯i and 𝐯f have equal length and 60∘ between them, so ∣Δ𝐯∣=2vsin⁡30∘=v=20 m/s.

    J=m∣Δ𝐯∣=0.16×20=3.2 N s. It points along 𝐯f−𝐯i, at 120∘ to the original direction of motion, along the bisector of the angle between the final velocity and the reversed initial velocity.

  21. NumericalMediumNCERT Exercises 4.21

    A 0.40 kg ball on a 1.2 m cord is whirled in a horizontal circle at 60 revolutions per minute (ignore the cord's slight downward tilt).

    1. Find the tension in the cord, in newtons.

      N
      Show solution

      ω=2π rad/s, so T=mω2r=0.40×(2π)2×1.2≈19 N.

    2. The cord breaks at 150 N. What is the highest speed, in m/s, the ball can have?

      m/s
      Show solution

      vmax⁡=Tmax⁡ r/m=150×1.2/0.40=450≈21 m/s.

  22. Quick answerEasyNCERT Exercises 4.22

    In Question 21 the ball is whirled too fast and the cord snaps. Seen from above, the ball then moves

    Show answer

    Answer (A) along the tangent to the circle at the point where it was released

    With the tension gone there is no horizontal force, so the ball keeps the velocity it had, which was along the tangent. Flying straight outwards would need an outward force that does not exist.
  23. Short answerMediumNCERT Exercises 4.23

    Explain each of the following using the laws of motion.

    1. A bullock cannot get its cart moving on perfectly smooth ice.

      Show solution

      To accelerate the bullock and cart together, an external force must act on that system. On land it is the ground's friction on the bullock's hooves. The pull between bullock and cart is internal. On frictionless ice the hooves just slip, so there is no external horizontal force and the system cannot accelerate.

    2. Standing passengers lurch forward when a metro train brakes suddenly.

      Show solution

      Friction from the floor slows their feet with the train, but nothing acts at once on the upper body, which keeps moving forward by inertia until the muscles stop it.

    3. A loaded suitcase trolley is easier to pull at an angle than to push at the same angle.

      Show solution

      A pull at an angle above the horizontal has an upward component that reduces the normal force, and so reduces friction. A push at the same angle presses the trolley down, increasing the normal force and the friction.

    4. A long-jumper lands in a pit of soft sand.

      Show solution

      The jumper's change in momentum on landing is fixed. Sand stretches the stopping time, so the average force Δp/Δt on the body is much smaller and causes no injury.

  24. Quick answerMedium

    Assertion (A): A body moving in a circle at constant speed has zero acceleration. Reason (R): Its speed does not change.

    Show answer

    Answer (D) A is false but R is true

    The speed is constant (R is true), but the direction of the velocity changes, so there is a centripetal acceleration v2/R (A is false).
  25. Quick answerMedium

    A block slides down a plane inclined at 30∘ with μk=0.25. What is its acceleration, in m/s²?

    m/s²
    Show answer

    Answer 2.83 m/s²

    a=g(sin⁡30∘−0.25cos⁡30∘)=10(0.500−0.217)≈2.8 m/s2.
  26. Quick answerMedium

    Blocks of 2.0 kg and 3.0 kg touch each other on a smooth floor. A horizontal 20 N force pushes the 2.0 kg block, which pushes the other. The force between the blocks is

    Show answer

    Answer (C) 12 N

    a=20/5.0=4.0 m/s2; the contact force alone accelerates the 3.0 kg block, so it is 3.0×4.0=12 N. 8 N is what the 2.0 kg block needs, not the contact force.
  27. Quick answerMedium

    At what angle, in degrees, should a curve of radius 90 m be banked so that cars at 15 m/s need no friction?

    °
    Show answer

    Answer 14.04 °

    tan⁡θ=225/(90×10)=0.25, so θ≈14∘.
  28. Quick answerEasy

    Match each circular motion with the force that supplies its centripetal force.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • Moon round the Earth : gravity
    • Car on a flat bend : static friction
    • Stone whirled on a string : tension
    • Rider on a wall of death : normal force
    Centripetal force is always one of these real forces, never an extra force of its own.
  29. Quick answerEasy

    Assertion (A): It takes less force to keep a heavy box sliding across a floor than to start it sliding. Reason (R): The coefficient of kinetic friction is smaller than the coefficient of static friction.

    Show answer

    Answer (A) Both A and R are true, and R is the correct explanation of A

    Starting needs more than μsN; keeping it sliding needs only μkN, and μk<μs. So R explains A.

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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.

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NCERT Exercises 4.1Practice Q1
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NCERT Exercises 4.20Practice Q20
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NCERT Exercises 4.23Practice Q23

Quick check: MCQs on Laws of Motion

  1. 4.3 The law of inertia3 questions
  2. 4.4 Newton's first law of motion3 questions
  3. Impulse4 questions
  4. 4.6 Newton's third law of motion3 questions
  5. 4.7 Conservation of momentum3 questions
  6. 4.8 Equilibrium of a particle3 questions
  7. Rolling friction4 questions
  8. Motion of a car on a banked road4 questions
  9. 4.11 Solving problems in mechanics3 questions
Glossary · 15

Key terms (English–Hindi)

centripetal forceअभिकेंद्र बलabhikendra bal
The net force towards the centre, of size mv2/R, that keeps a body moving in a circle.First seen in the chapter →
coefficient of frictionघर्षण गुणांकgharshan gunaank
For a pair of surfaces, the ratio of the limiting friction to the normal force: μs for static friction and μk for sliding.First seen in the chapter →
equilibriumसाम्यावस्थाsaamyaavastha
The state of a particle on which the net external force is zero, so that it is at rest or moves with constant velocity.First seen in the chapter →
forceबलbal
A push or pull exerted on a body by another body, measured in newtons. A net force changes the body's velocity.First seen in the chapter →
free-body diagramमुक्त पिंड आरेखmukt pind aarekh
A sketch of one chosen body or system on its own, showing every external force that acts on it.First seen in the chapter →
frictionघर्षणgharshan
The component of a contact force that lies along the touching surfaces. It opposes actual or impending sliding between them.First seen in the chapter →
impulseआवेगaaveg
The product of a force and the time it acts, equal to the change in momentum it produces; measured in N s.First seen in the chapter →
inertiaजड़त्वjadatva
The tendency of a body to resist any change in its state of rest or uniform straight-line motion; mass measures it.First seen in the chapter →
kinetic frictionगतिज घर्षणgatij gharshan
Friction between sliding surfaces, fk=μkN, directed against the relative motion.First seen in the chapter →
momentumसंवेगsanveg
The product of mass and velocity, 𝐩=m𝐯, a vector measured in kg m s−1.First seen in the chapter →
newtonन्यूटनnyootan
The SI unit of force: the force that gives a mass of 1 kg an acceleration of 1 m/s2.First seen in the chapter →
normal forceअभिलंब बलabhilamb bal
The part of a contact force perpendicular to the surfaces in contact. It adjusts itself to stop one surface moving into the other.First seen in the chapter →
rolling frictionलोटनिक घर्षणlotanik gharshan
Resistance to rolling caused by the slight deformation of a wheel and the surface it rolls on; far smaller than sliding friction.First seen in the chapter →
static frictionस्थैतिक घर्षणsthaitik gharshan
Friction between surfaces that are not sliding; it adjusts itself up to a maximum of μsN.First seen in the chapter →
tensionतनावtanaav
The pull exerted by a stretched string, rope or cable. In a massless string it is the same all along its length.First seen in the chapter →