CHAPTER FIVE

LAWS OF MOTION

5.1 INTRODUCTION

In the preceding Chapter, our concern was to describe the

motion of a particle in space quantitatively. We saw that

uniform motion needs the concept of velocity alone whereas

non-uniform motion requires the concept of acceleration in

addition. So far, we have not asked the question as to what

governs the motion of bodies. In this chapter, we turn to this

basic question.

Let us first guess the answer based on our common

experience. To move a football at rest, someone must kick it.

To throw a stone upwards, one has to give it an upward

push. A breeze causes the branches of a tree to swing; a

strong wind can even move heavy objects. A boat moves in a

flowing river without anyone rowing it. Clearly, some external

agency is needed to provide force to move a body from rest.

Likewise, an external force is needed also to retard or stop

motion. You can stop a ball rolling down an inclined plane by

applying a force against the direction of its motion.

In these examples, the external agency of force (hands,

wind, stream, etc) is in contact with the object. This is not

always necessary. A stone released from the top of a building

accelerates downward due to the gravitational pull of the

earth. A bar magnet can attract an iron nail from a distance.

This shows that external agencies (e.g. gravitational and

magnetic forces ) can exert force on a body even from a

distance.

In short, a force is required to put a stationary body in

motion or stop a moving body, and some external agency is

needed to provide this force. The external agency may or may

not be in contact with the body.

So far so good. But what if a body is moving uniformly (e.g.

a skater moving straight with constant speed on a horizontal

ice slab) ? Is an external force required to keep a body in

uniform motion?

5.1 Introduction

5.2 Aristotle’s fallacy

5.3 The law of inertia

5.4 Newton’s first law of motion

5.5 Newton’s second law of

motion

5.6 Newton’s third law of motion

5.7 Conservation of momentum

5.8 Equilibrium of a particle

5.9 Common forces in mechanics

5.10 Circular motion

5.11 Solving problems in

mechanics

Summary

Points to ponder

Exercises

Additional exercises

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5.2 ARISTOTLE’S FALLACY

The question posed above appears to be simple.

However, it took ages to answer it. Indeed, the

correct answer to this question given by Galileo

in the seventeenth century was the foundation

of Newtonian mechanics, which signalled the

birth of modern science.

The Greek thinker, Aristotle (384 B.C– 322

B.C.), held the view that if a body is moving,

something external is required to keep it moving.

According to this view, for example, an arrow

shot from a bow keeps flying since the air behind

the arrow keeps pushing it. The view was part of

an elaborate framework of ideas developed by

Aristotle on the motion of bodies in the universe.

Most of the Aristotelian ideas on motion are now

known to be wrong and need not concern us.

For our purpose here, the Aristotelian law of

motion may be phrased thus: An external force

is required to keep a body in motion.

Aristotelian law of motion is flawed, as we shall

see. However, it is a natural view that anyone

would hold from common experience. Even a

small child playing with a simple (non-electric)

toy-car on a floor knows intuitively that it needs

to constantly drag the string attached to the toy-

car with some force to keep it going. If it releases

the string, it comes to rest. This experience is

common to most terrestrial motion. External

forces seem to be needed to keep bodies in

motion. Left to themselves, all bodies eventually

come to rest.

What is the flaw in Aristotle’s argument? The

answer is: a moving toy car comes to rest because

the external force of friction on the car by the floor

opposes its motion. To counter this force, the child

has to apply an external force on the car in the

direction of motion. When the car is in uniform

motion, there is no net external force acting on it:

the force by the child cancels the force ( friction)

by the floor. The corollary is: if there were no friction,

the child would not be required to apply any force

to keep the toy car in uniform motion.

The opposing forces such as friction (solids)

and viscous forces (for fluids) are always present

in the natural world. This explains why forces

by external agencies are necessary to overcome

the frictional forces to keep bodies in uniform

motion. Now we understand where Aristotle

went wrong. He coded this practical experience

in the form of a basic argument. To get at the

true law of nature for forces and motion, one has

to imagine a world in which uniform motion is

possible with no frictional forces opposing. This

is what Galileo did.

5.3 THE LAW OF INERTIA

Galileo studied motion of objects on an inclined

plane. Objects (i) moving down an inclined plane

accelerate, while those (ii) moving up retard.

(iii) Motion on a horizontal plane is an

intermediate situation. Galileo concluded that

an object moving on a frictionless horizontal

plane must neither have acceleration nor

retardation, i.e. it should move with constant

velocity (Fig. 5.1(a)).

(i) (ii) (iii)

Fig. 5.1(a)

Another experiment by Galileo leading to the

same conclusion involves a double inclined plane.

A ball released from rest on one of the planes rolls

down and climbs up the other. If the planes are

smooth, the final height of the ball is nearly the

same as the initial height (a little less but never

greater). In the ideal situation, when friction is

absent, the final height of the ball is the same

as its initial height.

If the slope of the second plane is decreased

and the experiment repeated, the ball will still

reach the same height, but in doing so, it will

travel a longer distance. In the limiting case, when

the slope of the second plane is zero (i.e. is a

horizontal) the ball travels an infinite distance.

In other words, its motion never ceases. This is,

of course, an idealised situation (Fig. 5.1(b)).

Fig. 5.1(b) The law of inertia was inferred by Galileo

from observations of motion of a ball on a

double inclined plane.

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In practice, the ball does come to a stop after

moving a finite distance on the horizontal plane,

because of the opposing force of friction which

can never be totally eliminated. However, if there

were no friction, the ball would continue to move

with a constant velocity on the horizontal plane.

Galileo thus, arrived at a new insight on

motion that had eluded Aristotle and those who

followed him. The state of rest and the state of

uniform linear motion (motion with constant

velocity) are equivalent. In both cases, there is

no net force acting on the body. It is incorrect to

assume that a net force is needed to keep a body

in uniform motion. To maintain a body in

uniform motion, we need to apply an external

force to ecounter the frictional force, so that

the two forces sum up to zero net external

force.

To summarise, if the net external force is zero,

a body at rest continues to remain at rest and a

body in motion continues to move with a uniform

velocity. This property of the body is called

inertia. Inertia means ‘resistance to change’.

A body does not change its state of rest or

uniform motion, unless an external force

compels it to change that state.

5.4 NEWTON’S FIRST LAW OF MOTION

Galileo’s simple, but revolutionary ideas

dethroned Aristotelian mechanics. A new

mechanics had to be developed. This task was

Ideas on Motion in Ancient Indian Science

Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of

motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force

of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent

tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body;

transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps

comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to

be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air

resistance. It was correctly summarised that the different kinds of motion (translational, rotational

and vibrational) of an extended body arise from only the translational motion of its constituent

particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational

and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a

definite (small) displacement. There was considerable focus in Indian thought on measurement of

motion and units of length and time. It was known that the position of a particle in space can be

indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept

of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous

velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly

understood; a current is a motion of particles of water under gravity and fluidity while a wave results

from the transmission of vibrations of water particles.

accomplished almost single-handedly by Isaac

Newton, one of the greatest scientists of all times.

Newton built on Galileo’s ideas and laid the

foundation of mechanics in terms of three laws

of motion that go by his name. Galileo’s law of

inertia was his starting point which he

formulated as the first law of motion:

Every body continues to be in its state

of rest or of uniform motion in a straight

line unless compelled by some external

force to act otherwise.

The state of rest or uniform linear motion both

imply zero acceleration. The first law of motion can,

therefore, be simply expressed as:

If the net external force on a body is zero, its

acceleration is zero. Acceleration can be non

zero only if there is a net external force on

the body.

Two kinds of situations are encountered in the

application of this law in practice. In some

examples, we know that the net external force

on the object is zero. In that case we can

conclude that the acceleration of the object is

zero. For example, a spaceship out in

interstellar space, far from all other objects and

with all its rockets turned off, has no net

external force acting on it. Its acceleration,

according to the first law, must be zero. If it is

in motion, it must continue to move with a

uniform velocity.

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More often, however, we do not know all the

forces to begin with. In that case, if we know

that an object is unaccelerated (i.e. it is either

at rest or in uniform linear motion), we can infer

from the first law that the net external force on

the object must be zero. Gravity is everywhere.

For terrestrial phenomena, in particular, every

object experiences gravitational force due to the

earth. Also objects in motion generally experience

friction, viscous drag, etc. If then, on earth, an

object is at rest or in uniform linear motion, it is

not because there are no forces acting on it, but

because the various external forces cancel out

i.e. add up to zero net external force.

Consider a book at rest on a horizontal surface

Fig. (5.2(a)). It is subject to two external forces :

the force due to gravity (i.e. its weight W) acting

downward and the upward force on the book by

the table, the normal force R . R is a self-adjusting

force. This is an example of the kind of situation

mentioned above. The forces are not quite known

fully but the state of motion is known. We observe

the book to be at rest. Therefore, we conclude

from the first law that the magnitude of R equals

that of W. A statement often encountered is :

“Since W = R, forces cancel and, therefore, the book

is at rest”. This is incorrect reasoning. The correct

statement is

: “Since the book is observed to be at

rest, the net external force on it must be zero,

according to the first law. This implies that the

Galileo Galilei, born in Pisa, Italy in 1564 was a key figure in the scientific revolution

in Europe about four centuries ago. Galileo proposed the concept of acceleration.

From experiments on motion of bodies on inclined planes or falling freely, he

contradicted the Aristotelian notion that a force was required to keep a body in

motion, and that heavier bodies fall faster than lighter bodies under gravity. He

thus arrived at the law of inertia that was the starting point of the subsequent

epochal work of Isaac Newton.

Galileo’s discoveries in astronomy were equally revolutionary. In 1609, he designed

his own telescope (invented earlier in Holland) and used it to make a number of

startling observations : mountains and depressions on the surface of the moon;

dark spots on the sun; the moons of Jupiter and the phases of Venus. He concluded

that the Milky Way derived its luminosity because of a large number of stars not visible to the naked eye.

In his masterpiece of scientific reasoning : Dialogue on the Two Chief World Systems, Galileo advocated

the heliocentric theory of the solar system proposed by Copernicus, which eventually got universal

acceptance.

With Galileo came a turning point in the very method of scientific inquiry. Science was no longer

merely observations of nature and inferences from them. Science meant devising and doing experiments

to verify or refute theories. Science meant measurement of quantities and a search for mathematical

relations between them. Not undeservedly, many regard Galileo as the father of modern science.

normal force R must be equal and opposite to the

weight W ”.

Fig. 5.2 (a) a book at rest on the table, and (b) a car

moving with uniform velocity. The net force

is zero in each case.

Consider the motion of a car starting from

rest, picking up speed and then moving on a

smooth straight road with uniform speed (Fig.

(5.2(b)). When the car is stationary, there is no

net force acting on it. During pick-up, it

accelerates. This must happen due to a net

external force. Note, it has to be an external force.

The acceleration of the car cannot be accounted

for by any internal force. This might sound

surprising, but it is true. The only conceivable

external force along the road is the force of

friction. It is the frictional force that accelerates

the car as a whole. (You will learn about friction

in section 5.9). When the car moves with

constant velocity, there is no net external force.

Galileo Galilei (1564 - 1642)

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It relates the net external force to the

acceleration of the body.

Momentum

Momentum of a body is defined to be the product

of its mass m and velocity v, and is denoted

by p:

p = m v (5.1)

Momentum is clearly a vector quantity. The

following common experiences indicate the

importance of this quantity for considering the

effect of force on motion.

• Suppose a light-weight vehicle (say a small

car) and a heavy weight vehicle (say a loaded

truck) are parked on a horizontal road. We all

know that a much greater force is needed to

push the truck than the car to bring them to

the same speed in same time. Similarly, a

greater opposing force is needed to stop a

heavy body than a light body in the same time,

if they are moving with the same speed.

• If two stones, one light and the other heavy,

are dropped from the top of a building, a

person on the ground will find it easier to catch

the light stone than the heavy stone. The

mass of a body is thus an important

parameter that determines the effect of force

on its motion.

• Speed is another important parameter to

consider. A bullet fired by a gun can easily

pierce human tissue before it stops, resulting

in casualty. The same bullet fired with

moderate speed will not cause much damage.

Thus for a given mass, the greater the speed,

the greater is the opposing force needed to stop

the body in a certain time. Taken together,

the product of mass and velocity, that is

momentum, is evidently a relevant variable

of motion. The greater the change in the

momentum in a given time, the greater is the

force that needs to be applied.

• A seasoned cricketer catches a cricket ball

coming in with great speed far more easily

than a novice, who can hurt his hands in the

act. One reason is that the cricketer allows a

longer time for his hands to stop the ball. As

you may have noticed, he draws in the hands

backward in the act of catching the ball (Fig.

5.3). The novice, on the other hand, keeps

his hands fixed and tries to catch the ball

almost instantly. He needs to provide a much

greater force to stop the ball instantly, and

The property of inertia contained in the First

law is evident in many situations. Suppose we

are standing in a stationary bus and the driver

starts the bus suddenly. We get thrown

backward with a jerk. Why ? Our feet are in touch

with the floor. If there were no friction, we would

remain where we were, while the floor of the bus

would simply slip forward under our feet and the

back of the bus would hit us. However,

fortunately, there is some friction between the

feet and the floor. If the start is not too sudden,

i.e. if the acceleration is moderate, the frictional

force would be enough to accelerate our feet

along with the bus. But our body is not strictly

a rigid body. It is deformable, i.e. it allows some

relative displacement between different parts.

What this means is that while our feet go with

the bus, the rest of the body remains where it is

due to inertia. Relative to the bus, therefore, we

are thrown backward. As soon as that happens,

however, the muscular forces on the rest of the

body (by the feet) come into play to move the body

along with the bus. A similar thing happens

when the bus suddenly stops. Our feet stop due

to the friction which does not allow relative

motion between the feet and the floor of the bus.

But the rest of the body continues to move

forward due to inertia. We are thrown forward.

The restoring muscular forces again come into

play and bring the body to rest.

Example 5.1 An astronaut accidentally

gets separated out of his small spaceship

accelerating in inter stellar space at a

constant rate of 100 m s

–2

. What is the

acceleration of the astronaut the instant after

he is outside the spaceship ? (Assume that

there are no nearby stars to exert

gravitational force on him.)

Answer Since there are no nearby stars to exert

gravitational force on him and the small

spaceship exerts negligible gravitational

attraction on him, the net force acting on the

astronaut, once he is out of the spaceship, is

zero. By the first law of motion the acceleration

of the astronaut is zero. t

5.5 NEWTON’S SECOND LAW OF MOTION

The first law refers to the simple case when the

net external force on a body is zero. The second

law of motion refers to the general situation when

there is a net external force acting on the body.

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this hurts. The conclusion is clear: force not

only depends on the change in momentum,

but also on how fast the change is brought

about. The same change in momentum

brought about in a shorter time needs a

greater applied force. In short, the greater the

rate of change of momentum, the greater is

the force.

Fig. 5.3 Force not only depends on the change in

momentum but also on how fast the change

is brought about. A seasoned cricketer draws

in his hands during a catch, allowing greater

time for the ball to stop and hence requires a

smaller force.

• Observations confirm that the product of

mass and velocity (i.e. momentum) is basic to

the effect of force on motion. Suppose a fixed

force is applied for a certain interval of time

on two bodies of different masses, initially at

rest, the lighter body picks up a greater speed

than the heavier body. However, at the end of

the time interval, observations show that each

body acquires the same momentum. Thus

the same force for the same time causes

the same change in momentum for

different bodies. This is a crucial clue to the

second law of motion.

• In the preceding observations, the vector

character of momentum has not been evident.

In the examples so far, momentum and change

in momentum both have the same direction.

But this is not always the case. Suppose a

stone is rotated with uniform speed in a

horizontal plane by means of a string, the

magnitude of momentum is fixed, but its

direction changes (Fig. 5.4). A force is needed

to cause this change in momentum vector.

This force is provided by our hand through

the string. Experience suggests that our hand

needs to exert a greater force if the stone is

rotated at greater speed or in a circle of

smaller radius, or both. This corresponds to

greater acceleration or equivalently a greater

rate of change in momentum vector. This

suggests that the greater the rate of change

in momentum vector the greater is the force

applied.

Fig. 5.4 Force is necessary for changing the direction

of momentum, even if its magnitude is

constant. We can feel this while rotating a

stone in a horizontal circle with uniform speed

by means of a string.

These qualitative observations lead to the

second law of motion expressed by Newton as

follows :

The rate of change of momentum of a body is

directly proportional to the applied force and

takes place in the direction in which the force

acts.

Thus, if under the action of a force F for time

interval ∆t, the velocity of a body of mass m

changes from v to v + ∆v i.e. its initial momentum

p = m v changes by

∆ = ∆

p v

. According to the

Second Law,

or k

t t

∆ ∆

∝ =

∆ ∆

p p

F F

where k is a constant of proportionality. Taking

the limit ∆t → 0, the term

t∆

∆

becomes the

derivative or differential co-efficient of p with

respect to t, denoted by

. Thus

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(5.2)

For a body of fixed mass m,

(

)

d d d

m m m

t t t

= = =

p v

v a

(5.3)

i.e the Second Law can also be written as

F = k m a (5.4)

which shows that force is proportional to the

product of mass m and acceleration a.

The unit of force has not been defined so far.

In fact, we use Eq. (5.4) to define the unit of force.

We, therefore, have the liberty to choose any

constant value for k. For simplicity, we choose

k = 1. The second law then is

F m

(5.5)

In SI unit force is one that causes an acceleration

of 1 m s

-2

to a mass of 1 kg. This unit is known as

newton : 1 N = 1 kg m s

-2

Let us note at this stage some important points

about the second law :

1. In the second law, F = 0 implies a = 0. The

second law is obviously consistent with the

first law.

2. The second law of motion is a vector law. It is

equivalent to three equations, one for each

component of the vectors :

= =

a m

F ==

(5.6)

This means that if a force is not parallel to

the velocity of the body, but makes some angle

with it, it changes only the component of

velocity along the direction of force. The

component of velocity normal to the force

remains unchanged. For example, in the

motion of a projectile under the vertical

gravitational force, the horizontal component

of velocity remains unchanged (Fig. 5.5).

3. The second law of motion given by Eq. (5.5) is

applicable to a single point particle. The force

F in the law stands for the net external force

on the particle and a stands for acceleration

of the particle. It turns out, however, that the

law in the same form applies to a rigid body or,

even more generally, to a system of particles.

In that case, F refers to the total external force

on the system and a refers to the acceleration

of the system as a whole. More precisely, a is

the acceleration of the centre of mass of the

system about which we shall study in detail in

chapter 7. Any internal forces in the system

are not to be included in F.

Fig. 5.5 Acceleration at an instant is determined by

the force at that instant. The moment after a

stone is dropped out of an accelerated train,

it has no horizontal acceleration or force, if

air resistance is neglected. The stone carries

no memory of its acceleration with the train

a moment ago.

4. The second law of motion is a local relation

which means that force F at a point in space

(location of the particle) at a certain instant

of time is related to a at that point at that

instant. Acceleration here and now is

determined by the force here and now, not by

any history of the motion of the particle

(See Fig. 5.5).

Example 5.2 A bullet of mass 0.04 kg

moving with a speed of 90 m s

–1

enters a

heavy wooden block and is stopped after a

distance of 60 cm. What is the average

resistive force exerted by the block on the

bullet?

Answer The retardation ‘a’ of the bullet

(assumed constant) is given by

–

2 2

– 90 90

m s – 6750 m s

2 0.6

− −

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The retarding force, by the second law of

motion, is

= 0.04 kg × 6750 m s

-2

= 270 N

The actual resistive force, and therefore,

retardation of the bullet may not be uniform. The

answer therefore, only indicates the average

resistive force. t

Example 5.3 The motion of a particle of

mass m is described by y =

ut gt

. Find

the force acting on the particle.

Answer We know

y ut gt

= +

Now,

v u gt

= = +

acceleration,

a g

= =

Then the force is given by Eq. (5.5)

F = ma = mg

Thus the given equation describes the motion

of a particle under acceleration due to gravity

and y is the position coordinate in the direction

of g. t

Impulse

We sometimes encounter examples where a large

force acts for a very short duration producing a

finite change in momentum of the body. For

example, when a ball hits a wall and bounces

back, the force on the ball by the wall acts for a

very short time when the two are in contact, yet

the force is large enough to reverse the momentum

of the ball. Often, in these situations, the force

and the time duration are difficult to ascertain

separately. However, the product of force and time,

which is the change in momentum of the body

remains a measurable quantity. This product is

called impulse:

Impulse = Force × time duration

= Change in momentum (5.7)

A large force acting for a short time to produce a

finite change in momentum is called an impulsive

force. In the history of science, impulsive forces

were put in a conceptually different category from

ordinary forces. Newtonian mechanics has no

such distinction. Impulsive force is like any other

force – except that it is large and acts for a short

time.

Example 5.4 A batsman hits back a ball

straight in the direction of the bowler without

changing its initial speed of 12 m s

–1

If the mass of the ball is 0.15 kg, determine

the impulse imparted to the ball. (Assume

linear motion of the ball)

Answer Change in momentum

= 0.15 × 12–(–0.15×12)

= 3.6 N s,

Impulse = 3.6 N s,

in the direction from the batsman to the bowler.

This is an example where the force on the ball

by the batsman and the time of contact of the

ball and the bat are difficult to know, but the

impulse is readily calculated. t

5.6 NEWTON’S THIRD LAW OF MOTION

The second law relates the external force on a

body to its acceleration. What is the origin of the

external force on the body ? What agency

provides the external force ? The simple answer

in Newtonian mechanics is that the external

force on a body always arises due to some other

body. Consider a pair of bodies A and B. B gives

rise to an external force on A. A natural question

is: Does A in turn give rise to an external force

on B ? In some examples, the answer seems

clear. If you press a coiled spring, the spring is

compressed by the force of your hand. The

compressed spring in turn exerts a force on your

hand and you can feel it. But what if the bodies

are not in contact ? The earth pulls a stone

downwards due to gravity. Does the stone exert

a force on the earth ? The answer is not obvious

since we hardly see the effect of the stone on the

earth. The answer according to Newton is: Yes,

the stone does exert an equal and opposite force

on the earth. We do not notice it since the earth

is very massive and the effect of a small force on

its motion is negligible.

Thus, according to Newtonian mechanics,

force never occurs singly in nature. Force is the

mutual interaction between two bodies. Forces

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always occur in pairs. Further, the mutual forces

between two bodies are always equal and

opposite. This idea was expressed by Newton in

the form of the third law of motion.

To every action, there is always an equal and

opposite reaction.

Newton’s wording of the third law is so crisp and

beautiful that it has become a part of common

language. For the same reason perhaps,

misconceptions about the third law abound. Let

us note some important points about the third

law, particularly in regard to the usage of the

terms : action and reaction.

1. The terms action and reaction in the third law

mean nothing else but ‘force’. Using different

terms for the same physical concept

can sometimes be confusing. A simple

and clear way of stating the third law is as

follows:

Forces always occur in pairs. Force on a

body A by B is equal and opposite to the

force on the body B by A.

2. The terms action and reaction in the third law

may give a wrong impression that action

comes before reaction i.e action is the cause

and reaction the effect. There is no cause-

effect relation implied in the third law. The

force on A by B

and the force on B by A act

at the same instant. By the same reasoning,

any one of them may be called action and the

other reaction.

3. Action and reaction forces act on different

bodies, not on the same body. Consider a pair

of bodies A and B. According to the third law,

= – F

(5.8)

(force on A by B) = – (force on B by A)

Thus if we are considering the motion of any

one body (A or B), only one of the two forces is

relevant. It is an error to add up the two forces

and claim that the net force is zero.

However, if you are considering the system

of two bodies as a whole, F

and F

are

internal forces of the system (A + B). They add

up to give a null force. Internal forces in a

body or a system of particles thus cancel away

in pairs. This is an important fact that

enables the second law to be applicable to a

body or a system of particles (See Chapter 7).

Isaac Newton (1642 – 1727)

Isaac Newton was born in Woolsthorpe, England in 1642, the year Galileo died.

His extraordinary mathematical ability and mechanical aptitude remained hidden

from others in his school life. In 1662, he went to Cambridge for undergraduate

studies. A plague epidemic in 1665 forced the university town to close and Newton

had to return to his mother’s farm. There in two years of solitude, his dormant

creativity blossomed in a deluge of fundamental discoveries in mathematics and

physics : binomial theorem for negative and fractional exponents, the beginning of

calculus, the inverse square law of gravitation, the spectrum of white light, and so

on. Returning to Cambridge, he pursued his investigations in optics and devised a

reflecting telescope.

In 1684, encouraged by his friend Edmund Halley, Newton embarked on writing what was to be one of

the greatest scientific works ever published : The Principia Mathematica. In it, he enunciated the three

laws of motion and the universal law of gravitation, which explained all the three Kepler’s laws of

planetary motion. The book was packed with a host of path-breaking achievements : basic principles of

fluid mechanics, mathematics of wave motion, calculation of masses of the earth, the sun and other

planets, explanation of the precession of equinoxes, theory of tides, etc. In 1704, Newton brought out

another masterpiece Opticks that summarized his work on light and colour.

The scientific revolution triggered by Copernicus and steered vigorously ahead by Kepler and Galileo

was brought to a grand completion by Newton. Newtonian mechanics unified terrestrial and celestial

phenomena. The same mathematical equation governed the fall of an apple to the ground and the

motion of the moon around the earth. The age of reason had dawned.

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Example 5.5 Two identical billiard balls

strike a rigid wall with the same speed but

at different angles, and get reflected without

any change in speed, as shown in Fig. 5.6.

What is (i) the direction of the force on the

wall due to each ball? (ii) the ratio of the

magnitudes of impulses imparted to the

balls by the wall ?

Fig. 5.6

Answer An instinctive answer to (i) might be

that the force on the wall in case (a) is normal to

the wall, while that in case (b) is inclined at 30°

to the normal. This answer is wrong. The force

on the wall is normal to the wall in both cases.

How to find the force on the wall? The trick is

to consider the force (or impulse) on the ball

due to the wall using the second law, and then

use the third law to answer (i). Let u be the speed

of each ball before and after collision with the

wall, and m the mass of each ball. Choose the x

and y axes as shown in the figure, and consider

the change in momentum of the ball in each

case :

Case (a)

(

)

(

)

initial

x y

p mu p

= =

(

)

(

)

final

x y

p mu p

= − =

Impulse is the change in momentum vector.

Therefore,

x-component of impulse = – 2 m u

y-component of impulse = 0

Impulse and force are in the same direction.

Clearly, from above, the force on the ball due to

the wall is normal to the wall, along the negative

x-direction. Using Newton’s third law of motion,

the force on the wall due to the ball is normal to

the wall along the positive x-direction. The

magnitude of force cannot be ascertained since

the small time taken for the collision has not

been specified in the problem.

Case (b)

(

)

cos 30

initial

p m u =



(

)

sin 30

initialy

p m u = −



(

)

– cos 30

final

p m u =



(

)

sin 30

final

p m u = −



Note, while p

changes sign after collision, p

does not. Therefore,

x-component of impulse = –2 m u cos 30°

y-component of impulse = 0

The direction of impulse (and force) is the same

as in (a) and is normal to the wall along the

negative x direction. As before, using Newton’s

third law, the force on the wall due to the ball is

normal to the wall along the positive x direction.

The ratio of the magnitudes of the impulses

imparted to the balls in (a) and (b) is

( )

2 / 2 cos30 1.2

m u m u = ≈



5.7 CONSERVATION OF MOMENTUM

The second and third laws of motion lead to

an important consequence: the law of

conservation of momentum. Take a familiar

example. A bullet is fired from a gun. If the force

on the bullet by the gun is F, the force on the gun

by the bullet is – F, according to the third law.

The two forces act for a common interval of time

∆t. According to the second law, F ∆t is the

change in momentum of the bullet and – F ∆t is

the change in momentum of the gun. Since

initially, both are at rest, the change in

momentum equals the final momentum for each.

Thus if p

is the momentum of the bullet after

firing and p

is the recoil momentum of the gun,

= – p

i.e. p

+ p

= 0. That is, the total

momentum of the (bullet + gun) system is

conserved.

Thus in an isolated system (i.e. a system with

no external force), mutual forces between pairs

of particles in the system can cause momentum

change in individual particles, but since the

mutual forces for each pair are equal and

opposite, the momentum changes cancel in pairs

and the total momentum remains unchanged.

This fact is known as the law of conservation

of momentum :

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LAWS OF MOTION 99

The total momentum of an isolated system

of interacting particles is conserved.

An important example of the application of the

law of conservation of momentum is the collision

of two bodies. Consider two bodies A and B, with

initial momenta p

and p

. The bodies collide,

get apart, with final momenta p′

and p′

respectively. By the Second Law

F p p

AB A A

t∆ =

′

−

and

F p p

BA B B

t∆ =

′

−

(where we have taken a common interval of time

for both forces i.e. the time for which the two

bodies are in contact.)

Since

AB BA

−

by the third law,

(

)

′

− = −

′

−p p p p

A A B B

i.e.

′

A B A B

(5.9)

which shows that the total final momentum of

the isolated system equals its initial momentum.

Notice that this is true whether the collision is

elastic or inelastic. In elastic collisions, there is

a second condition that the total initial kinetic

energy of the system equals the total final kinetic

energy (See Chapter 6).

5.8 EQUILIBRIUM OF A PARTICLE

Equilibrium of a particle in mechanics refers to

the situation when the net external force on the

particle is zero.* According to the first law, this

means that, the particle is either at rest or in

uniform motion.

If two forces F

and F

, act on a particle,

equilibrium requires

= −

(5.10)

i.e. the two forces on the particle must be equal

and opposite. Equilibrium under three

concurrent forces F

, F

and F

requires that

the vector sum of the three forces is zero.

+ F

= 0 (5.11)

* Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational

equilibrium (zero net external torque), as we shall see in Chapter 7.

Fig. 5.7 Equilibrium under concurrent forces.

In other words, the resultant of any two forces

say F

and F

, obtained by the parallelogram

law of forces must be equal and opposite to the

third force, F

. As seen in Fig. 5.7, the three

forces in equilibrium can be represented by the

sides of a triangle with the vector arrows taken

in the same sense. The result can be

generalised to any number of forces. A particle

is in equilibrium under the action of forces F

,... F

if they can be represented by the sides

of a closed n-sided polygon with arrows directed

in the same sense.

Equation (5.11) implies that

+ F

= 0

+ F

= 0

+ F

= 0 (5.12)

where F

, F

and F

are the components of F

along x, y and z directions respectively.

Example 5.6 See Fig. 5.8. A mass of 6 kg

is suspended by a rope of length 2 m

from the ceiling. A force of 50 N in the

horizontal direction is applied at the mid-

point P of the rope, as shown. What is the

angle the rope makes with the vertical in

equilibrium ? (Take g = 10 m s

-2

). Neglect

the mass of the rope.

(a) (b) (c)

Fig. 5.8

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PHYSICS100

Answer Figures 5.8(b) and 5.8(c) are known as

free-body diagrams. Figure 5.8(b) is the free-body

diagram of W and Fig. 5.8(c) is the free-body

diagram of point P.

Consider the equilibrium of the weight W.

Clearly,T

= 6 × 10 = 60 N.

Consider the equilibrium of the point P under

the action of three forces - the tensions T

and

, and the horizontal force 50 N. The horizontal

and vertical components of the resultant force

must vanish separately :

cos

= T

= 60 N

sin

= 50 N

which gives that

Note the answer does not depend on the length

of the rope (assumed massless) nor on the point

at which the horizontal force is applied. t

5.9 COMMON FORCES IN MECHANICS

In mechanics, we encounter several kinds of

forces. The gravitational force is, of course,

pervasive. Every object on the earth experiences

the force of gravity due to the earth. Gravity also

governs the motion of celestial bodies. The

gravitational force can act at a distance without

the need of any intervening medium.

All the other forces common in mechanics are

contact forces.* As the name suggests, a contact

force on an object arises due to contact with some

other object: solid or fluid. When bodies are in

contact (e.g. a book resting on a table, a system

of rigid bodies connected by rods, hinges and

other types of supports), there are mutual

contact forces (for each pair of bodies) satisfying

the third law. The component of contact force

normal to the surfaces in contact is called

normal reaction. The component parallel to the

surfaces in contact is called friction. Contact

forces arise also when solids are in contact with

fluids. For example, for a solid immersed in a

fluid, there is an upward bouyant force equal to

the weight of the fluid displaced. The viscous

force, air resistance, etc are also examples of

contact forces (Fig. 5.9).

Two other common forces are tension in a

string and the force due to spring. When a spring

is compressed or extended by an external force,

a restoring force is generated. This force is

usually proportional to the compression or

elongation (for small displacements). The spring

force F is written as F = – k x where x is the

displacement and k is the force constant. The

negative sign denotes that the force is opposite

to the displacement from the unstretched state.

For an inextensible string, the force constant is

very high. The restoring force in a string is called

tension. It is customary to use a constant tension

T throughout the string. This assumption is true

for a string of negligible mass.

In Chapter 1, we learnt that there are four

fundamental forces in nature. Of these, the weak

and strong forces appear in domains that do not

concern us here. Only the gravitational and

electrical forces are relevant in the context of

mechanics. The different contact forces of

mechanics mentioned above fundamentally arise

from electrical forces. This may seem surprising

* We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are

electrical and magnetic non-contact forces.

Fig. 5.9 Some examples of contact forces in mechanics.

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LAWS OF MOTION 101

since we are talking of uncharged and non-

magnetic bodies in mechanics. At the microscopic

level, all bodies are made of charged constituents

(nuclei and electrons) and the various contact

forces arising due to elasticity of bodies, molecular

collisions and impacts, etc. can ultimately be

traced to the electrical forces between the charged

constituents of different bodies. The detailed

microscopic origin of these forces is, however,

complex and not useful for handling problems in

mechanics at the macroscopic scale. This is why

they are treated as different types of forces with

their characteristic properties determined

empirically.

5.9.1 Friction

Let us return to the example of a body of mass m

at rest on a horizontal table. The force of gravity

(mg) is cancelled by the normal reaction force

(N) of the table. Now suppose a force F is applied

horizontally to the body. We know from

experience that a small applied force may not

be enough to move the body. But if the applied

force F were the only external force on the body,

it must move with acceleration F/m, however

small. Clearly, the body remains at rest because

some other force comes into play in the

horizontal direction and opposes the applied

force F, resulting in zero net force on the body.

This force f

parallel to the surface of the body in

contact with the table is known as frictional

force, or simply friction (Fig. 5.10(a)). The

subscript stands for static friction to distinguish

it from kinetic friction f

that we consider later

(Fig. 5.10(b)). Note that static friction does not

Fig. 5.10 Static and sliding friction: (a) Impending

motion of the body is opposed by static

friction. When external force exceeds the

maximum limit of static friction, the body

begins to move. (b) Once the body is in

motion, it is subject to sliding or kinetic friction

which opposes relative motion between the

two surfaces in contact. Kinetic friction is

usually less than the maximum value of static

exist by itself. When there is no applied force,

there is no static friction. It comes into play the

moment there is an applied force. As the applied

force F increases, f

also increases, remaining

equal and opposite to the applied force (up to a

certain limit), keeping the body at rest. Hence, it

is called static friction. Static friction opposes

impending motion. The term impending motion

means motion that would take place (but does

not actually take place) under the applied force,

if friction were absent.

We know from experience that as the applied

force exceeds a certain limit, the body begins to

move. It is found experimentally that the limiting

value of static friction

(

)

max

is independent of

the area of contact and varies with the normal

force(N) approximately as :

(

)

max

s s

f N

(5.13)

where

is a constant of proportionality

depending only on the nature of the surfaces in

contact. The constant

is called the coefficient

of static friction. The law of static friction may

thus be written as

≤

N (5.14)

If the applied force F exceeds

(

)

max

the body

begins to slide on the surface. It is found

experimentally that when relative motion has

started, the frictional force decreases from the

static maximum value

(

)

max

. Frictional force

that opposes relative motion between surfaces

in contact is called kinetic or sliding friction and

is denoted by f

Kinetic friction, like static

friction, is found to be independent of the area

of contact. Further, it is nearly independent of

the velocity. It satisfies a law similar to that for

static friction:

k k

f N

(5.15)

where

k′

the coefficient of kinetic friction,

depends only on the surfaces in contact. As

mentioned above, experiments show that

less than

. When relative motion has begun,

the acceleration of the body according to the

second law is ( F – f

)/m. For a body moving with

constant velocity, F = f

. If the applied force on

the body is removed, its acceleration is – f

and it eventually comes to a stop.

The laws of friction given above do not have

the status of fundamental laws like those for

gravitational, electric and magnetic forces. They

are empirical relations that are only

friction.

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PHYSICS102

approximately true. Yet they are very useful in

practical calculations in mechanics.

Thus, when two bodies are in contact, each

experiences a contact force by the other. Friction,

by definition, is the component of the contact force

parallel to the surfaces in contact, which opposes

impending or actual relative motion between the

two surfaces. Note that it is not motion, but

relative motion that the frictional force opposes.

Consider a box lying in the compartment of a train

that is accelerating. If the box is stationary

relative to the train, it is in fact accelerating along

with the train. What forces cause the acceleration

of the box? Clearly, the only conceivable force in

the horizontal direction is the force of friction. If

there were no friction, the floor of the train would

slip by and the box would remain at its initial

position due to inertia (and hit the back side of

the train). This impending relative motion is

opposed by the static friction f

. Static friction

provides the same acceleration to the box as that

of the train, keeping it stationary relative to the

train.

Example 5.7 Determine the maximum

acceleration of the train in which a box

lying on its floor will remain stationary,

given that the co-efficient of static friction

between the box and the train’s floor is

0.15.

Answer Since the acceleration of the box is due

to the static friction,

ma = f

≤

N =

m g

i.e. a ≤

∴ a

max

g = 0.15 x 10 m s

–2

= 1.5 m s

–2

Example 5.8 See Fig. 5.11. A mass of 4 kg

rests on a horizontal plane. The plane is

gradually inclined until at an angle

= 15°

with the horizontal, the mass just begins to

slide. What is the coefficient of static friction

between the block and the surface ?

Fig. 5.11

Answer The forces acting on a block of mass m

at rest on an inclined plane are (i) the weight

mg acting vertically downwards (ii) the normal

force N of the plane on the block, and (iii) the

static frictional force f

opposing the impending

motion. In equilibrium, the resultant of these

forces must be zero. Resolving the weight mg

along the two directions shown, we have

m g sin

= f

, m g cos

= N

increases, the self-adjusting frictional force

increases until at

max

, f

achieves its

maximum value,

(

)

max

Therefore,

tan

max

= tan

–1

When

becomes just a little more than

max

there is a small net force on the block and it

begins to slide. Note that

max

depends only on

and is independent of the mass of the block.

For

max

= 15°,

= tan 15°

= 0.27 t

Example 5.9 What is the acceleration of

the block and trolley system shown in a

Fig. 5.12(a), if the coefficient of kinetic friction

between the trolley and the surface is 0.04?

What is the tension in the string? (Take g =

10 m s

-2

). Neglect the mass of the string.

(a)

(b) (c)

Fig. 5.12

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LAWS OF MOTION 103

is the reason why discovery of the wheel has

been a major milestone in human history.

Rolling friction again has a complex origin,

though somewhat different from that of static

and sliding friction. During rolling, the surfaces

in contact get momentarily deformed a little, and

this results in a finite area (not a point) of the

body being in contact with the surface. The net

effect is that the component of the contact force

parallel to the surface opposes motion.

We often regard friction as something

undesirable. In many situations, like in a

machine with different moving parts, friction

does have a negative role. It opposes relative

motion and thereby dissipates power in the form

of heat, etc. Lubricants are a way of reducing

kinetic friction in a machine. Another way is to

use ball bearings between two moving parts of a

machine [Fig. 5.13(a)]. Since the rolling friction

between ball bearings and the surfaces in

contact is very small, power dissipation is

reduced. A thin cushion of air maintained

between solid surfaces in relative motion is

another effective way of reducing friction (Fig.

5.13(a)).

In many practical situations, however, friction

is critically needed. Kinetic friction that

dissipates power is nevertheless important for

quickly stopping relative motion. It is made use

of by brakes in machines and automobiles.

Similarly, static friction is important in daily

life. We are able to walk because of friction. It

is impossible for a car to move on a very slippery

road. On an ordinary road, the friction between

the tyres and the road provides the necessary

external force to accelerate the car.

Answer As the string is inextensible, and the

pully is smooth, the 3 kg block and the 20 kg

trolley both have same magnitude of

acceleration. Applying second law to motion of

the block (Fig. 5.12(b)),

30 – T = 3a

Apply the second law to motion of the trolley (Fig.

5.12(c)),

– f

= 20 a.

Now f

= µ

Here

= 0.04,

N = 20 x 10

= 200 N.

Thus the equation for the motion of the trolley is

T – 0.04 x 200 = 20 a Or T – 8 = 20a.

These equations give a =

2 2

2 3

m s

–2

= 0.96 m s

-2

and T = 27.1 N. t

Rolling friction

A body like a ring or a sphere rolling without

slipping over a horizontal plane will suffer no

friction, in principle. At every instant, there is

just one point of contact between the body and

the plane and this point has no motion relative

to the plane. In this ideal situation, kinetic or

static friction is zero and the body should

continue to roll with constant velocity. We know,

in practice, this will not happen and some

resistance to motion (rolling friction) does occur,

i.e. to keep the body rolling, some applied force

is needed. For the same weight, rolling friction

is much smaller (even by 2 or 3 orders of

magnitude) than static or sliding friction. This

Fig. 5.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine.

(b) Compressed cushion of air between surfaces in relative motion.

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5.10 CIRCULAR MOTION

We have seen in Chapter 4 that acceleration of

a body moving in a circle of radius R with uniform

speed v is v

/R directed towards the centre.

According to the second law, the force f

providing

this acceleration is :

f =

(5.16)

where m is the mass of the body. This force

directed forwards the centre is called the

centripetal force. For a stone rotated in a circle

by a string, the centripetal force is provided by

the tension in the string. The centripetal force

for motion of a planet around the sun is the

is the static friction that provides the centripetal

acceleration. Static friction opposes the

impending motion of the car moving away from

the circle. Using equation (5.14) & (5.16) we get

the result

= ≤

f N

v Rg

≤ =

[

∵

N = mg]

which is independent of the mass of the car.

This shows that for a given value of

and R,

there is a maximum speed of circular motion of

the car possible, namely

max s

v Rg

(5.18)

(a) (b)

Fig. 5.14 Circular motion of a car on (a) a level road, (b) a banked road.

gravitational force on the planet due to the sun.

For a car taking a circular turn on a horizontal

road, the centripetal force is the force of friction.

The circular motion of a car on a flat and

banked road give interesting application of the

laws of motion.

Motion of a car on a level road

Three forces act on the car (Fig. 5.14(a):

(i) The weight of the car, mg

(ii) Normal reaction, N

(iii) Frictional force, f

As there is no acceleration in the vertical

direction

N – mg = 0

N = mg (5.17)

The centripetal force required for circular motion

is along the surface of the road, and is provided

by the component of the contact force between

road and the car tyres along the surface. This

by definition is the frictional force. Note that it

Motion of a car on a banked road

We can reduce the contribution of friction to the

circular motion of the car if the road is banked

(Fig. 5.14(b)). Since there is no acceleration along

the vertical direction, the net force along this

direction must be zero. Hence,

N cos

= mg + f sin

(5.19a)

The centripetal force is provided by the horizontal

components of N and f.

N sin

+ f cos

(5.19b)

But f

≤

Thus to obtain v

max

we put

f N

Then Eqs. (5.19a) and (5.19b) become

N cos

= mg +

sin

(5.20a)

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LAWS OF MOTION 105

N sin

cos

= mv

/R (5.20b)

From Eq. (5.20a), we obtain

–

cos sin

θ µ θ

Substituting value of N in Eq. (5.20b), we get

(

)

max

–

mg sin cos

cos sin R

θ µ θ

max

1 –

tan

v Rg

tan

µ θ

 

 

 

(5.21)

Comparing this with Eq. (5.18) we see that

maximum possible speed of a car on a banked

road is greater than that on a flat road.

For

= 0 in Eq. (5.21 ),

( R g tan θ )

(5.22)

At this speed, frictional force is not needed at all

to provide the necessary centripetal force.

Driving at this speed on a banked road will cause

little wear and tear of the tyres. The same

equation also tells you that for v < v

, frictional

force will be up the slope and that a car can be

parked only if tan

≤

Example 5.10 A cyclist speeding at

18 km/h on a level road takes a sharp

circular turn of radius 3 m without reducing

the speed. The co-efficient of static friction

between the tyres and the road is 0.1. Will

the cyclist slip while taking the turn?

Answer On an unbanked road, frictional force

alone can provide the centripetal force needed

to keep the cyclist moving on a circular turn

without slipping. If the speed is too large, or if

the turn is too sharp (i.e. of too small a radius)

or both, the frictional force is not sufficient to

provide the necessary centripetal force, and the

cyclist slips. The condition for the cyclist not to

slip is given by Eq. (5.18) :

≤ µ

R g

Now, R = 3 m, g = 9.8 m s

-2

, µ

= 0.1. That is,

R g = 2.94 m

-2

. v = 18 km/h = 5 m s

-1

; i.e.,

= 25 m

-2

. The condition is not obeyed.

The cyclist will slip while taking the circular

turn. t

Example 5.11 A circular racetrack of

radius 300 m is banked at an angle of 15°.

If the coefficient of friction between the

wheels of a race-car and the road is 0.2,

what is the (a) optimum speed of the race-

car to avoid wear and tear on its tyres, and

(b) maximum permissible speed to avoid

slipping ?

Answer On a banked road, the horizontal

component of the normal force and the frictional

force contribute to provide centripetal force to

keep the car moving on a circular turn without

slipping. At the optimum speed, the normal

reaction’s component is enough to provide the

needed centripetal force, and the frictional force

is not needed. The optimum speed v

is given by

Eq. (5.22):

= (R g tan

)

1/2

Here R = 300 m,

= 15°, g = 9.8 m s

-2

; we

have

= 28.1 m s

-1

The maximum permissible speed v

max

is given by

Eq. (5.21):

5.11 SOLVING PROBLEMS IN MECHANICS

The three laws of motion that you have learnt in

this chapter are the foundation of mechanics.

You should now be able to handle a large variety

of problems in mechanics. A typical problem in

mechanics usually does not merely involve a

single body under the action of given forces.

More often, we will need to consider an assembly

of different bodies exerting forces on each other.

Besides, each body in the assembly experiences

the force of gravity. When trying to solve a

problem of this type, it is useful to remember

the fact that we can choose any part of the

assembly and apply the laws of motion to that

part provided we include all forces on the chosen

part due to the remaining parts of the assembly.

We may call the chosen part of the assembly as

the system and the remaining part of the

assembly (plus any other agencies of forces) as

the environment. We have followed the same

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PHYSICS106

method in solved examples. To handle a typical

problem in mechanics systematically, one

should use the following steps :

(i) Draw a diagram showing schematically the

various parts of the assembly of bodies, the

links, supports, etc.

(ii) Choose a convenient part of the assembly

as one system.

(iii) Draw a separate diagram which shows this

system and all the forces on the system by

the remaining part of the assembly. Include

also the forces on the system by other

agencies. Do not include the forces on the

environment by the system. A diagram of

this type is known as ‘a free-body diagram’.

(Note this does not imply that the system

under consideration is without a net force).

(iv) In a free-body diagram, include information

about forces (their magnitudes and

directions) that are either given or you are

sure of (e.g., the direction of tension in a

string along its length). The rest should be

treated as unknowns to be determined using

laws of motion.

(v) If necessary, follow the same procedure for

another choice of the system. In doing so,

employ Newton’s third law. That is, if in the

free-body diagram of A, the force on A due to

B is shown as F, then in the free-body

diagram of B, the force on B due to A should

be shown as –F.

The following example illustrates the above

procedure :

Example 5.12 See Fig. 5.15. A wooden

block of mass 2 kg rests on a soft horizontal

floor. When an iron cylinder of mass 25 kg

is placed on top of the block, the floor yields

steadily and the block and the cylinder

together go down with an acceleration of

0.1 m s

–2

. What is the action of the block

on the floor (a) before and (b) after the floor

yields ? Take g = 10 m s

–2

. Identify the

action-reaction pairs in the problem.

Answer

(a) The block is at rest on the floor. Its free-body

diagram shows two forces on the block, the

force of gravitational attraction by the earth

equal to 2 × 10 = 20 N; and the normal force

R of the floor on the block. By the First Law,

the net force on the block must be zero i.e.,

R = 20 N. Using third law the action of the

block (i.e. the force exerted on the floor by

the block) is equal to 20 N and directed

vertically downwards.

(b) The system (block + cylinder) accelerates

downwards with 0.1 m s

-2

. The free-body

diagram of the system shows two forces on

the system : the force of gravity due to the

earth (270 N); and the normal force R′ by the

floor. Note, the free-body diagram of the

system does not show the internal forces

between the block and the cylinder. Applying

the second law to the system,

270 – R′

= 27 × 0.1N

ie. R′

= 267.3 N

Fig. 5.15

By the third law, the action of the system on

the floor is equal to 267.3 N vertically downward.

Action-reaction pairs

For (a): (i) the force of gravity (20 N) on the block

by the earth (say, action); the force of

gravity on the earth by the block

(reaction) equal to 20 N directed

upwards (not shown in the figure).

(ii) the force on the floor by the block

(action); the force on the block by the

floor (reaction).

For (b): (i) the force of gravity (270 N) on the

system by the earth (say, action); the

force of gravity on the earth by the

system (reaction), equal to 270 N,

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LAWS OF MOTION 107

directed upwards (not shown in the

figure).

(ii) the force on the floor by the system

(action); the force on the system by the

floor (reaction). In addition, for (b), the

force on the block by the cylinder and

the force on the cylinder by the block

also constitute an action-reaction pair.

The important thing to remember is that an

action-reaction pair consists of mutual forces

which are always equal and opposite between

two bodies. Two forces on the same body which

happen to be equal and opposite can never

constitute an action-reaction pair. The force of

gravity on the mass in (a) or (b) and the normal

force on the mass by the floor are not action-

reaction pairs. These forces happen to be equal

and opposite for (a) since the mass is at rest.

They are not so for case (b), as seen already.

The weight of the system is 270 N, while the

normal force R′ is 267.3 N. t

The practice of drawing free-body diagrams is

of great help in solving problems in mechanics.

It allows you to clearly define your system and

consider all forces on the system due to objects

that are not part of the system itself. A number

of exercises in this and subsequent chapters will

help you cultivate this practice.

SUMMARY

1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A

force is necessary in practice to counter the opposing force of friction.

2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and

arrived at the law of inertia. Newton’s first law of motion is the same law rephrased

thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line,

unless compelled by some external force to act otherwise”. In simple terms, the First Law

is “If external force on a body is zero, its acceleration is zero”.

3. Momentum (p

) of a body is the product of its mass (m) and velocity (v) :

p = m v

4. Newton’s second law of motion :

The rate of change of momentum of a body is proportional to the applied force and takes

place in the direction in which the force acts. Thus

k k m

= =

F a

where F is the net external force on the body and a its acceleration. We set the constant

of proportionality k = 1 in SI units. Then

= =

F a

The SI unit of force is newton : 1 N = 1 kg m s

-2

(a) The second law is consistent with the First Law (F = 0 implies a = 0)

(b) It is a vector equation

is the total external force on the system and a is the acceleration of the system as

a whole.

(d) F at a point at a certain instant determines a at the same point at that instant.

That is the Second Law is a local law; a at an instant does not depend on the

history of motion.

5. Impulse is the product of force and time which equals change in momentum.

The notion of impulse is useful when a large force acts for a short time to produce a

measurable change in momentum. Since the time of action of the force is very short,

one can assume that there is no appreciable change in the position of the body during

the action of the impulsive force.

6. Newton’s third law of motion:

To every action, there is always an equal and opposite reaction

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PHYSICS108

In simple terms, the law can be stated thus :

Forces in nature always occur between pairs of bodies. Force on a body A by body

B is equal and opposite to the force on the body B by A.

Action and reaction forces are simultaneous forces. There is no cause-effect

relation between action and reaction. Any of the two mutual forces can be

called action and the other reaction. Action and reaction act on different

bodies and so they cannot be cancelled out. The internal action and reaction

forces between different parts of a body do, however, sum to zero.

7. Law of Conservation of Momentum

The total momentum of an isolated system of particles is conserved. The law

follows from the second and third law of motion.

8. Friction

Frictional force opposes (impending or actual) relative motion between two

surfaces in contact. It is the component of the contact force along the common

tangent to the surface in contact. Static friction f

opposes impending relative

motion; kinetic friction f

opposes actual relative motion. They are independent

of the area of contact and satisfy the following approximate laws :

(

)

max

f f R

s s s

≤ =

f R

(co-efficient of static friction) and µ

(co-efficient of kinetic friction) are

constants characteristic of the pair of surfaces in contact. It is found

experimentally that µ

is less than µ

POINTS TO PONDER

1. Force is not always in the direction of motion. Depending on the situation, F

may be along v, opposite to v, normal to v or may make some other angle with

v. In every case, it is parallel to acceleration.

2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that

force or acceleration are necessarily zero at that instant. For example, when a

ball thrown upward reaches its maximum height, v = 0 but the force continues

to be its weight mg and the acceleration is not zero but g.

3. Force on a body at a given time is determined by the situation at the location of

the body at that time. Force is not ‘carried’ by the body from its earlier history of

motion. The moment after a stone is released out of an accelerated train, there is

no horizontal force (or acceleration) on the stone, if the effects of the surrounding

air are neglected. The stone then has only the vertical force of gravity.

4. In the second law of motion F = m a, F stands for the net force due to all

material agencies external to the body. a is the effect of the force. ma should

not be regarded as yet another force, besides F.

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LAWS OF MOTION 109

5. The centripetal force should not be regarded as yet another kind of force. It is

simply a name given to the force that provides inward radial acceleration to a

body in circular motion. We should always look for some material force like

tension, gravitational force, electrical force, friction, etc as the centripetal force

in any circular motion.

6. Static friction is a self-adjusting force up to its limit

N (f

≤ µ

N). Do not put

= µ

without being sure that the maximum value of static friction is coming

into play.

7. The familiar equation mg = R for a body on a table is true only if the body is in

equilibrium. The two forces mg and R can be different (e.g. a body in an

accelerated lift). The equality of mg and R has no connection with the third

law.

8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for

simultaneous mutual forces between a pair of bodies. Unlike their meaning in

ordinary language, action does not precede or cause reaction. Action and reaction

act on different bodies.

9. The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’,

‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’

in different contexts. For clarity, every force and its equivalent terms

encountered in mechanics should be reduced to the phrase ‘force on A by B’.

10. For applying the second law of motion, there is no conceptual distinction between

inanimate and animate objects. An animate object such as a human also

requires an external force to accelerate. For example, without the external

force of friction, we cannot walk on the ground.

11. The objective concept of force in physics should not be confused with the

subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of

our body are subject to an inward force, but we have a feeling of being pushed

outward – the direction of impending motion.

EXERCISES

(For simplicity in numerical calculations, take g = 10 m s

-2

)

5.1 Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g floating on water,

(d) a car moving with a constant velocity of 30 km/h on a rough road,

(e) a high-speed electron in space far from all material objects, and free of

electric and magnetic fields.

5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction

and magnitude of the net force on the pebble,

(a) during its upward motion,

(b) during its downward motion,

change if the pebble was thrown at an angle of 45° with the horizontal

direction?

Ignore air resistance.

5.3 Give the magnitude and direction of the net force acting on a stone of mass

0.1 kg,

(a) just after it is dropped from the window of a stationary train,

(b) just after it is dropped from the window of a train running at a constant

velocity of 36 km/h,

(c ) just after it is dropped from the window of a train accelerating with 1 m s

-2

(d) lying on the floor of a train which is accelerating with 1 m s

-2

, the stone

being at rest relative to the train.

Neglect air resistance throughout.

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PHYSICS110

5.4 One end of a string of length l is connected to a particle of mass m and the

other to a small peg on a smooth horizontal table. If the particle moves in a

circle with speed v the net force on the particle (directed towards the centre)

is :

(i) T, (ii)

−

, (iii)

, (iv) 0

T is the tension in the string. [Choose the correct alternative].

5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving

initially with a speed of 15 m s

-1

. How long does the body take to stop ?

5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s

-1

to 3.5 m s

-1

in 25 s. The direction of the motion of the body remains

unchanged. What is the magnitude and direction of the force ?

5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N.

Give the magnitude and direction of the acceleration of the body.

5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child

standing in the middle of the road and brings his vehicle to rest in 4.0 s just

in time to save the child. What is the average retarding force on the vehicle ?

The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial

acceleration of 5.0 m s

-2

. Calculate the initial thrust (force) of the blast.

5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s

-1

the north is subject to a constant force of 8.0 N directed towards the south

for 30 s. Take the instant the force is applied to be t = 0, the position of the

body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

5.11 A truck starts from rest and accelerates uniformly at 2.0 m s

-2

. At t = 10 s, a

stone is dropped by a person standing on the top of the truck (6 m high from

the ground). What are the (a) velocity, and (b) acceleration of the stone at t =

11s ? (Neglect air resistance.)

5.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is

set into oscillation. The speed of the bob at its mean position is 1 m s

-1

What is the trajectory of the bob if the string is cut when the bob is (a) at one

of its extreme positions, (b) at its mean position.

5.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of 10 m s

-1

(b) downwards with a uniform acceleration of 5 m s

-2

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled

down freely under gravity ?

5.14 Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is

the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and

t = 4 s ? (Consider one-dimensional motion only).

Fig. 5.16

5.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal

surface are tied to the ends of a light string. A horizontal force F = 600 N is

applied to (i) A, (ii) B along the direction of string. What is the tension in the

string in each case?

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LAWS OF MOTION 111

5.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible

string that goes over a frictionless pulley. Find the acceleration of the masses, and

the tension in the string when the masses are released.

5.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates

into two smaller nuclei the products must move in opposite directions.

5.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s

-1

collide and rebound with the same speed. What is the impulse imparted to each ball due

to the other ?

5.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the

shell is 80 m s

-1

, what is the recoil speed of the gun ?

5.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is

equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.)

5.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius

1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the

string ? What is the maximum speed with which the stone can be whirled around if

the string can withstand a maximum tension of 200 N ?

5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible

value, and the string breaks suddenly, which of the following correctly describes the

trajectory of the stone after the string breaks :

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

speed of the particle ?

5.23 Explain why

(a) a horse cannot pull a cart and run in empty space,

(b) passengers are thrown forward from their seats when a speeding bus stops

suddenly,

(d) a cricketer moves his hands backwards while holding a catch.

Additional Exercises

5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a

suitable physical context for this motion. What is the time between two consecutive

impulses received by the body ? What is the magnitude of each impulse ?

Fig. 5.17

5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor

belt that is accelerating with 1 m s

-2

. What is the net force on the man? If the

coefficient of static friction between the man’s shoes and the belt is 0.2, up to what

acceleration of the belt can the man continue to be stationary relative to the belt ?

(Mass of the man = 65 kg.)

Fig. 5.18

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PHYSICS112

Fig. 5.19

5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R.

The net forces at the lowest and highest points of the circle directed vertically

downwards are : [Choose the correct alternative]

Lowest Point Highest Point

(a) mg – T

mg + T

(b) mg + T

mg – T

– (m v

) / R mg – T

+ (m v

) / R

(d) mg – T

– (m v

) / R mg + T

+ (m v

) / R

and v

denote the tension and speed at the lowest point. T

and v

denote

corresponding values at the highest point.

5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s

-2

. The crew

and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

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

5.28 A stream of water flowing horizontally with a speed of 15 m s

-1

gushes out of a tube of

cross-sectional area 10

-2

, and hits a vertical wall nearby. What is the force exerted

on the wall by the impact of water, assuming it does not rebound ?

5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m.

Give the magnitude and direction of

(a) the force on the 7

coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7

coin by the eighth coin,

coin on the 7

coin.

5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked

at 15°. What is the radius of the loop ?

5.31 A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h.

The mass of the train is 10

kg. What provides the centripetal force required for this

purpose — The engine or the rails ? What is the angle of banking required to prevent

wearing out of the rail ?

5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in

Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor

yields to a normal force of 700 N, which mode should the man adopt to lift the block

without the floor yielding ?

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LAWS OF MOTION 113

5.33 A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which

can stand a maximum tension of 600 N. In which of the

following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s

-2

(b) climbs down with an acceleration of 4 m s

-2

-1

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with

each other rest on a table against a rigid wall (Fig. 5.21). The

coefficient of friction between the bodies and the table is

0.15. A force of 200 N is applied horizontally to A. What are

(a) the reaction of the partition (b) the action-reaction forces

between A and B ? What happens when the wall is removed?

Does the answer to (b) change, when the bodies are in motion?

Ignore the difference between µ

and µ

5.35 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction

between the block and the trolley is 0.18. The trolley accelerates from rest with

0.5 m s

-2

for 20 s and then moves with uniform velocity. Discuss the motion of the

block as viewed by (a) a stationary observer on the ground, (b) an observer moving with

the trolley.

5.36 The rear side of a truck is open and a box of 40 kg

mass is placed 5 m away from the open end as shown

in Fig. 5.22. The coefficient of friction between the

box and the surface below it is 0.15. On a straight

road, the truck starts from rest and accelerates with

2 m s

-2

. At what distance from the starting point

does the box fall off the truck? (Ignore the size of

the box).

5.37 A disc revolves with a speed of

rev/min, and has a radius of 15 cm. Two coins are

placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction

between the coins and the record is 0.15, which of the coins will revolve with the record ?

5.38 You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-

well’ (a hollow spherical chamber with holes, so the spectators can watch from outside).

Explain clearly why the motorcyclist does not drop down when he is at the uppermost

point, with no support from below. What is the minimum speed required at the

uppermost position to perform a vertical loop if the radius of the chamber is 25 m ?

5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of

radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of

friction between the wall and his clothing is 0.15. What is the minimum rotational

speed of the cylinder to enable the man to remain stuck to the wall (without falling)

when the floor is suddenly removed ?

5.40 A thin circular loop of radius R rotates about its vertical diameter with an angular

frequency

. Show that a small bead on the wire loop remains at its lowermost point

for

≤ g / R

. What is the angle made by the radius vector joining the centre to

the bead with the vertical downward direction for

= 2g / R

? Neglect friction.

Fig. 5.20

Fig. 5.21

Fig. 5.22

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