# What are the laws of motion?

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We all know that stationary objects remain wherever they are placed. These objects cannot move on their own from one place to another place unless forced to change their state of rest.
1. MODULE - 1 Laws of Motion
Motion, Force and Energy
3
Notes
LAWS OF MOTION
In the previous lesson you learnt to describe the motion of an object in terms of
its displacement, velocity and acceleration. But an important question is : what
makes an object to move? Or what causes a ball rolling along the ground to come
to a stop? From our everyday experience we know that we need to push or pull
an object if we wish to change its position in a room. Similarly, a football has to
be kicked in order to send it over a large distance. A cricket ball has to be hit hard
by a batter to send it across the boundary for a six. You will agree that muscular
activity is involved in all these actions and its effect is quite visible.
There are, however, many situations where the cause behind an action is not
visible. For example, what makes rain drops to fall to the ground? What makes
the earth to go around the sun? In this lesson you will learn the basic laws of
motion and discover that force causes motion. The concept of force developed in
this lesson will be useful in different branches of physics. Newton showed that force
and motion are intimately connected. The laws of motion are fundamental and enable
us to understand everyday phenomena.
OBJECTIVES
After studying this lesson, you should be able to :
z explain the significance of inertia;
z state Newton’s laws of motion and illustrate them with examples;
z explain the law of conservation of momentum and illustrate it with examples;
z understand the concept of equilibrium of concurrent forces;
z define coefficient of friction and distinguish between static friction, kinetic
friction and rolling friction;
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z suggest different methods of reducing friction and highlight the role of friction
in every-day life; and
z analyse a given situation and apply Newton’s laws of motion using free body
diagrams.
3.1 CONCEPTS OF FORCE AND INERTIA
Notes
We all know that stationary objects remain wherever they are placed. These objects
cannot move on their own from one place to another place unless forced to
change their state of rest. Similarly, an object moving with constant velocity has
to be forced to change its state of motion. The property of an object by which
it resists a change in its state of rest or of uniform motion in a straight line is
called inertia. Mass of a body is a measure of its inertia.
In a way, inertia is a fantastic property. If it were not present, your books or
classnotes could mingle with those of your younger brother or sister. Your
wardrobe could move to your friend’s house creating chaos in life. You must
however recall that the state of rest or of uniform motion of an object are not
absolute. In the previous lesson you have learnt that an object at rest with respect
to one observer may appear to be in motion with respect to some other observer.
Observations show that the change in velocity of an object can only be brought,
if a net force acts on it.
You are very familiar with the term force. We use it in so many situations in our
everyday life. We are exerting force when we are pulling, pushing, kicking, hitting
etc. Though a force is not visible, its effect can be seen or experienced. Forces are
known to have different kinds of effects :
(a) They may change the shape and the size of an object. A balloon changes
shape depending on the magnitude of force acting on it.
(b) Forces also influence the motion of an object. A force can set an object
into motion or it can bring a moving object to rest. A force can also change
the direction or speed of motion.
(c) Forces can rotate a body about an axis. You will learn about it in lesson
seven.
3.1.1 Force and Motion
Force is a vector quantity. For this reason, when several forces act on a body
simultaneously, a net equivalent force can be calculated by vector addition, as
discussed in lesson 1.
Motion of a body is characterised by its displacement, velocity etc. We come
across many situations where the velocity of an object is either continuously
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Motion, Force and Energy
increasing or decreasing. For example, in the case of a body falling freely, the
velocity of the body increases continuously, till it hits the ground. Similarly, in the
case of a ball rolling on a horizontal surface, the velocity of the ball decreases
continuously and ultimately becomes zero.
From experience we know that a net non-zero force is required to change the
state of a body. For a body in motion, the velocity will change depending on
Notes the direction of the force acting on it. If a net force acts on a body in motion, its
velocity will increase in magnitude, if the direction of the force and velocity are
same. If the direction of net force acting on the body is opposite to the direction
of motion, the magnitude of velocity will decrease. However, if a net force acts
on a body in a direction perpendicular to its velocity, the magnitude of velocity of
the body remains constant (see Sec 4.3). Such a force changes only the direction
of velocity of the body. We may therfore conclude that velocity of a body changes
as long as a net force is acting on it.
3.1.2 First Law of Motion
When we roll a marble on a smooth floor, it stops after some time. It is obvious
that its velocity decreases and ultimately it becomes zero. However, if we want it
to move continuously with the same velocity, a force will have to be constantly
applied on it.
We also see that in order to move a trolley at constant velocity, it has to be
continuously pushed or pulled. Is there any net force acting on the marble or
trolley in the situations mentioned here?
Motion and Inertia
Galileo carried out experiments to prove that in the absence of any external
force, a body would continue to be in its state of rest or of uniform motion in
a straight line. He observed that a body is accelerated while moving down an
inclined plane (Fig. 3.1 a) and is retarded while moving up an inclined plane
(Fig. 3.1 b). He argued that if the plane is neither inclined upwards nor
downwards (i.e. if it is a horizontal plane surface), the motion of the body will
neither be accelerated not retarded. That is, on a horizontal plane surface, a
body will move with a uniform speed/velocity (if there is no external force).
Fig. 3.1 : Motion of a body on inclined and horizontal planes
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In another thought experiment, he considered two inclined planes facing each
other, as shown in Fig. 3.2. The inclination of the plane PQ is same in all the
three cases, whereas the inclination of the plane RS in Fig. 3.2 (a) is more than
that in (b) and (c). The plane PQRS is very smooth and the ball is of marble.
When the ball is allowed to roll down the plane PQ, it rises to nearly the same
height on the face RS. As the inclination of the plane RS decreases, the balls
moves a longer distance to rise to the same height on the inclined plane (Fig. Notes
3.2b). When the plane RS becomes horizontal, the ball keeps moving to attain
the same height as on the plane PQ, i.e. on a horizontal plane, the ball will
keep moving if there is no friction between the plane and the ball.
Initial Position Final Position Final Position
P A B S P A B S
h h h
h Q R Q R
(a) (b)
P
Where is the final position?
h
Q R S
(c)
Fig. 3.2 : Motion of a ball along planes inclined to each other
Sir Issac Newton
(1642–1727)
Newton was born at Wollsthorpe in England in 1642. He
studied at Trinity College, Cambridge and became the most
profound scientist. The observation of an apple falling towards
the ground helped him to formulate the basic law of gravitation.
He enunciated the laws of motion and the law of gravitation.
Newton was a genius and contributed significantly in all fields
of science, including mathematics. His contributions are of a
classical nature and form the basis of the modern science. He wrote his book
“Principia” in Latin and his book on optics was written in English.
You may logically ask : Why is it necessary to apply a force continuously to the
trolley to keep it moving uniformly? We know that a forward force on the cart is
needed for balancing out the force of friction on the cart. That is, the force of
friction on the trolley can be overcome by continuously pushing or pulling it.
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Isaac Newton generalised Galileo’s conclusions in the form of a law known as
Newton’s first law of motion, which states that a body continues to be in a state
of rest or of uniform motion in a straight line unless it is acted upon by a net
external force.
As you know, the state of rest or motion of a body depends on its relative position
with respect to an observer. A person in a running car is at rest with respect to
Notes another person in the same car. But the same person is in motion with respect to
a person standing on the road. For this reason, it is necessary to record
measurements of changes in position, velocity, acceleration and force with respect
to a chosen frame of reference.
A reference frame relative to which a body in translatory motion has constant
velocity, if no net external force acts on it, is known as an inertial frame of
reference. This nomenclature follows from the property of inertia of bodies due
to which they tend to preserve their state (of rest or of uniform linear motion). A
reference frame fixed to the earth (for all practical purposes) is considered an
inertial frame of reference.
Now you may like to take a break and answer the following questions.
INTEXT QUESTIONS 3.1
1. Is it correct to state that a body always moves in the direction of the net
external force acting on it?
2. What physical quantity is a measure of the inertia of a body?
3. Can a force change only the direction of velocity of an object keeping its
magnitude constant?
4. State the different types of changes which a force can bring in a body when
applied on it.
3.2 CONCEPT OF MOMENTUM
You must have seen that a fielder finds it difficult to stop a cricket ball moving
with a large velocity although its mass is small. Similarly, it is difficult to stop a
truck moving with a small velocity because its mass is large. These examples
suggest that both, mass and velocity of a body, are important, when we study the
effect of force on the motion of the body.
The product of mass m of a body and its velocity v is called its linear momentum
p. Mathmatically, we write
p = mv
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In SI units, momentum is measured in kg ms–1. Momentum is a vector quantity.
The direction of momentum vector is the same as the direction of velocity vector.
Momentum of an object, therefore, can change on account of change in its
magnitude or direction or both. The following examples illustrate this point.
Example 3.1: Aman weights 60 kg and travels with velocity 1.0 m s–1 towards
Manoj who weights 40 kg, and is moving with 1.5 m s–1 towards Aman. Calculate
their momenta. Notes
Solution : For Aman
momentum = mass × velocity
= (60 kg) × (1.0 m s–1)
= 60 kg ms–1
For Manoj
momentum = 40 kg × (– 1.5 ms–1)
= – 60 kg ms–1
Note that the momenta of Aman and Manoj have the same magnitude but they
are in opposite directions.
Example 3.2: A 2 kg object is allowed to fall freely at t = 0 s. Callculate its
momentum at (a) t = 0, (b) t = 1 s and (c) t = 2 s during its free-fall.
Solution : (a) As velocity of the object at t = 0 s is zero, the initial momentum of
the object will also be zero.
(b) At t = 1s, the velocity of the object will be 9.8 ms–1 [use v = v0 + at] pointing
downward. So the momentum of the object will be
p1 = (2 kg) × (9.8 ms–1) = 19.6 kg ms–1 pointing downward.
(c) At t = 2 s, the velocity of the object will be 19.6 m s–1 pointing downward. So
the momentum of the object will now be
p2 = (2 kg) × (19.6 ms–1) = 39.2 kg ms–1 pointing downward.
Thus, we see that the momentum of a freely-falling body increases continuously
in magnitude and points in the same direction. Now think what causes the
momentum of a freely-falling body to change in magnitude?
Example 3.3: A rubber ball of mass 0.2 kg strikes a rigid wall with a speed of
10 ms–1 and rebounds along the original path with the same speed. Calculate the
change in momentum of the ball.
Solution : Here the momentum of the ball has the same magnitude before and
after the impact but there is a reversal in its direction. In each case the magnitude
of momentum is (0.2 kg)×(10 ms–1) i.e. 2 kg ms–1.
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If we choose initial momentum vector to be along + x axis, the final momentum
vector will be along –x axis. So pi = 2 kg ms–1, pf = –2 kg ms–1. Therefore, the
change in momentum of the ball, pf – pi = (–2 kgms–1) – (2 kgms–1) = – 4 kgms–1.
Here negative sign shows that the momentum of the ball changes by 4 kg ms–1 in
the direction of –x axis. What causes this change in momentum of the ball?
In actual practice, a rubber ball rebounds from a rigid wall with a speed less than
Notes
its speed before the impact. In such a case also, the magnitude of the momentum
will change.
3.3 SECOND LAW OF MOTION
You now know that a body moving at constant velocity will have constant
momentum. Newton’s first law of motion suggests that no net external force
acts on such a body.
In Example 3.2 we have seen that the momentum of a ball falling freely under
gravity increases with time. Since such a body falls under the action of gravitational
force acting on it, there appears to be a connection between change in momentum
of an object, net force acting on it and the time for which it is acting. Newton’s
second law of motion gives a quantitative relation between these three physical
quantities. It states that the rate of change of momentum of a body is directly
proportional to the net force acting on the body. Change in momentum of the
body takes place in the direction of net external force acting on the body.
This means that if Δp is the change in momentum of a body in time Δt due to a net
external force F, we can write
Δp
F ∝
Δt
Δp
or F =k
Δt
where k is constant of proportionality.
By expressing momentum as a product of mass and velocity, we can rewrite this
result as
⎛ Δv ⎞
F = k m⎜ ⎟
⎝ Δt ⎠
Δv
F =kma (as = a) (3.1)
Δt
The value of the constant k depends upon the units of m and a. If these units are
chosen such that when the magnitude of m = 1 unit and a = 1 unit, the magnitude
of F is also be 1 unit. Then, we can write
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1 = k . 1. 1
i.e., k =1
Using this result in Eqn. (3.1), we get
F =ma (3.2)
In SI units, m = 1 kg, a = 1 m s–2. Then magnitude of external force
Notes
F = 1 kg × 1 ms–2 = 1 kg ms–2
= 1 unit of force (3.3)
This unit of force (i.e., 1 kg m s–2) is called one newton.
Note that the second law of motion gives us a unit for measuring force. The SI
unit of force i.e., a newton may thus, be defined as the force which will produce
an acceleration of 1 ms–2 in a mass of 1 kg.
Example 3.3: A ball of mass 0.4 kg starts rolling on the ground at 20 ms–1 and
comes to a stop after 10s. Calculate the force which stops the ball, assuming it to
be constant in magnitude throughout.
Solution : Given m= 0.4 kg, initial velocity u = 20 ms–1, final velocity v = 0
m s–1 and t = 10s. So
m(v – u ) 0.4 kg ( − 20 ms –1 )
|F| = m|a| = =
t 10 s
= – 0.8 kg m s–2 = – 0.8 N
Here negative sign shows that force on the ball is in a direction opposite to that of
its motion.
Example 3.4: A constant force of magnitude 50 N is applied to a body of 10 kg
moving initially with a speed of 10 m s–1. How long will it take the body to stop
if the force acts in a direction opposite to its motion.
Solution : Given m = 10 kg, F = –50 N, v0 = 10 ms–1 and v = 0. We have to
calculate t. Since
F = ma
we can write
⎛ v − v0 ⎞
F = m⎜ ⎟
⎝ t ⎠
⎛ 0 − 10 m s –1 ⎞
∴ –50 N = 10 kg ⎜ ⎟
⎝ t ⎠
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–100 kgms –1 100 kg ms –1
or t = = = 2 s.
–50 N 50 kg m s –2
It is important to note here that Newton’s second law of motion, as stated here is
applicable to bodies having constant mass. Will this law hold for bodies whose
mass changes with time, as in a rocket?
Notes
INTEXT QUESTIONS 3.2
1. Two objects of different masses have the same momentum. Which of them is
moving faster?
2. A boy throws up a ball with a velocity v0. If the ball returns to the thrower
with the same velocity, will there be any change in
(a) momentum of the ball?
(b) magnitude of the momentum of the ball?
3. When a ball falls from a height, its momentum increases. What causes increase
in its momentum?
4. In which case will there be larger change in momentum of the object?
(a) A 150 N force acts for 0.1 s on a 2 kg object initially at rest.
(b) A 150 N force acts for 0.2 s on a 2 kg. object initially at rest.
5. An object is moving at a constant speed in a circular path. Does the object
3.4 FORCES IN PAIRS
It is the gravitational pull of the earth, which allows an object to accelerate towards
the earth. Does the object also pull the earth? Similarly when we push an almirah,
does the almirah also push us? If so, why don’t we move in the direction of that
force? These situations compel us to ask whether a single force such as a push or
a pull exists? It has been observed that actions of two bodies on each other are
always mutual. Here, by action and reaction we mean ‘forces of interaction’. So,
whenever two bodies interact, they exert force on each other. One of them is
called ‘action’ and the other is called ‘reaction’. Thus, we can say that forces
always exist in pairs.
3.4.1 Third Law of Motion
On the basis of his study of interactions between bodies, Newton formulated
third law of motion: To every action, there is an equal and opposite reaction.
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Here by ‘action’ and ‘reaction’ we mean force. F2
Thus, when a book placed on a table exerts
some force on the table, the latter, also exerts
a force of equal magnitude on the book in the
F1
upward direction, as shown in Fig. 3.3. Do
the forces F1 and F2 shown here cancel out?
It is important to note that F1 and F2 are acting
Notes
on different bodies and therefore, they do not Fig 3.3 : A book placed on a table
cancel out. exerts a force F1 (equal to its
weight mg) on the table,
The action and reaction in a given situation while the table exerts a
appear as a pair of forces. Any one of them force F2 on the book.
cannot exist without the other.
If one goes by the literal meaning of words, reaction always follows an action,
whereas action and reaction introduced in Newton’s third law exist simultaneously.
For this reason, it is better to state Newton’s third law as when two objects
interact, the force exerted by one object on the other is equal in magnitude
and opposite in direction to the force exerted by the latter object on the former.
Vectorially, if F12 is the force which object 1 experiences due to object 2 and F21
is the force which object 2 experiences due to object 1, then according to Newton’s
third law of motion, we can write
F12 = –F21 (3.4)
3.4.2 Impulse
The effect of force applied for a short duration is called impulse. Impulse is defined
as the product of force (F) and the time duration (Δt) for which the force is
i.e., Impulse = F.Δt
If the initial and final velocities of body acted upon by a force F are u and v
respectively then we can write
mv −mu
Impulse = . Δt
Δt
= mv – m u
= pf – pi
= Δp
That is, impulse is equal to change in linear momentum.
Impulse in a vector quantity and its SI unit is kgms–1 (or N s).
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INTEXT QUESTIONS 3.3
1. When a high jumper leaves the ground, where does the force which throws
the jumper upwards come from?
2. Identify the action - reaction forces in each of the following situations:
Notes
(a) A man kicks a football
(b) Earth pulls the moon
(c) A ball hits a wall
3. “A person exerts a large force on an almirah to push it forward but he is not
pushed backward because the almirah exerts a small force on him”. Is the
argument given here correct? Explain.
3.5 CONSERVATION OF MOMENTUM
It has been experimentally shown that if two bodies interact, the vector sum of
their momenta remains unchanged, provided the force of mutual interaction is
the only force acting on them. The same has been found to be true for more than
two bodies interacting with each other. Generally, a number of bodies interacting
with each other are said to be forming a system. If the bodies in a system do not
interact with bodies outside the system, the system is said to be a closed system
or an isolated system. In an isolated system, the vector sum of the momenta of
bodies remains constant. This is called the law of conservation of momentum.
Here, it follows that it is the total momentum of the bodies in an isolated system
remains unchanged but the momentum of individual bodies may change, in
magnitude alone or direction alone or both. You may now logically ask : What
causes the momentum of individual bodies in an isolated system to change? It is
due to mutual interactions and their strengths.
Conservation of linear momentum is applicable in a wide range of phenomena
such as collisions, explosions, nuclear reactions, radioactive decay etc.
3.5.1 Conservation of Momentum as a Consequence of Newton’s Laws
According to Newton’s second law of motion, Eqn. (3.1), the change in momentum
Δp of a body, when a force F acts on it for time Δt, is
Δp = F Δt
This result implies that if no force acts on the body, the change in momentum of
the body will be zero. That is, the momentum of the body will remain unchanged.
This agrument can be extended to a system of bodies as well.
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Newton’s third law can also be used to arrive at the same result. Consider an
isolated system of two bodies A and B which interact with each other for time Δt.
If FAB and FBA are the forces which they exert on each other, then in accordance
with Newton’s third law
FAB = – FBA
Δp A Δp B
or =– Notes
Δt Δt
or ΔpA + ΔpB = 0 or
or Δptotal = 0
or ptotal = constant
That is, there is no change in the momentum of the system. In other words, the
momentum of the system is conserved.
3.5.2 A Few Illustrations of Conservation of Momentum
a) Recoil of a gun : When a bullet is fired from a gun, the gun recoils. The
velocity v2 of the recoil of the gun can be found by using the law of conservation
of momentum. Let m be the mass of the bullet being fired from a gun of mass M.
If v1 is the velocity of the bullet, then momentum will be said to be conserved if
the velocity v2 of the gun is given by
mv1 + Mv2 = 0
or mv1 = – Mv2
m
or v2 = − v1 (3.5)
M
Here, negative sign shows that v2 is in a direction opposite to v1. Since m << M,
the recoil velocity of the gun will be considerably smaller than the velocity of the
b) Collision : In a collision, we may regard the colliding bodies as forming a
system. In the absence of any external force on the colliding bodies, such as the
force of friction, the system can be considered to be an isolated system. The
forces of interaction between the colliding bodies will not change the total
momentum of the colliding bodies.
Collision of the striker with a coin of carrom or collision between the billiared
balls may be quite instructive for the study of collision between elastic bodies.
Example 3.5 : Two trolleys, each of mass m, coupled together are moving with
initial velocity v. They collide with three identical stationary trolleys coupled
together and continue moving in the same direction. What will be the velocity of
the trolleys after the impact?
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Solution : Let v′ be the velocity of the trolleys after the impact.
Momentum before collision = 2 mv
Momentum after collision = 5 mv′
In accordance with the law of conservation of momentum, we can write
2mv = 5 mv′
Notes 2
or v′ = v
5
c) Explosion of a bomb : A bomb explodes into fragments with the release of
huge energy. Consider a bomb at rest initially which explodes into two fragments
A and B. As the momentum of the bomb was zero before the explosion, the total
momentum of the two fragments formed will also be zero after the explosion.
For this reason, the two fragments will fly off in opposite directions. If the masses
of the two fragments are equal, the velocities of the two fragments will also be
equal in magnitude.
d) Rocket propulsion : Flight of a rocket is an important practical application
of conservation of momentum. A rocket consists of a shell with a fuel tank, which
can be considered as one body. The shell is provided with a nozzle through which
high pressure gases are made to escape. On firing the rocket, the combustion of
the fuel produces gases at very high pressure and temperature. Due to their high
pressure, these gases escape from the nozzle at a high velocity and provide thrust
to the rocket to go upward due to the conservation of momentum of the system.
If M is the mass of the rocket and m is the mass of gas escaping per second with
a velocity v, the change in momentum of the gas in t second = m vt.
If the increase in velocity of the rocket in t second is V, the increase in its momentum
= MV. According to the principle of conservation of momentum,
mvt + MV = 0
V mv
or =a=–
t M
i.e., the rocket moves with an acceleration
mv
a =–
M
3.5.3 Equilibrium of Concurrent Forces
A number of forces acting simultaneously at a point are called Concurrent
Forces. Such forces are said to be in equilibrium, if their resultant is zero.
Let F1, F2 and F3 be three concurrent forces acting at a point P, as shown
in Fig. 3.4.
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F1
F
F2 F (F1 + F2 )
F2
Fig. 3.4 Notes
The resultant of F1 and F2 , obtained by the parallelogram law, is shown by PA
(i.e. PA = F1 + F2 )
For equilibrium, the sum (F1 + F2 ) must be equal and opposite to F3 i.e.
F3 = −(F1 + F2 ) or F1 + F2 + F3 = 0
Or, the sum or resultant of two forces must be equal and opposite to the third
force or for equilibrium, their vector sum must be zero.
3.6 FRICTION
You may have noticed that when a batsman hits a ball to make it roll along the
ground, the ball does not continue to move forever. It comes to rest after travelling
some distance. Thus, the momentum of the ball, which was imparted to it during
initial push, tends to be zero. We know that some force acting on the ball is
responsible for this change in its momentum. Such a force, called the frictional
force, exists whenever bodies in contact tend to move with respect to each other.
It is the force of friction which has to be overcome when we push or pull a body
horizontally along the floor to change its position.
Force of friction is a contact force and always acts along the surfaces in a
direction opposite to that of the motion of the body. It is commonly known that
friction is caused by roughness of the surfaces in contact. For this reason deliberate
attempts are made to make the surfaces rough or smooth depending upon the
Friction opposes the motion of objects, causes wear and tear and is responsible
for loss of mechanical energy. But then, it is only due to friction that we are able
to walk, drive vehicles and stop moving vehicles. Friction thus plays a dual role in
our lives. It is therefore said that friction is a necessary evil.
3.6.1 Static and Kinetic Friction
We all know that certain minimum force is required to move an object over a
surface. To illustrate this point, let us consider a block resting on some horizontal
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surface, as shown in Fig.3.5. Let some external force Fext be applied on the block.
Initially the block does not move. This is possible only if some other force is
acting on the block. The force is called the force of static friction and is represented
by symbol fs. As Fext is increased, fs also increases and remains equal to Fext in
magnitude until it reaches a critical value fs(max). When Fext is increased further, the
block starts to slide and is then subject to kinetic friction. It is common experience
that the force needed to set an object in motion is larger than the force needed to
Notes
keep it moving at constant velocity. For this reason, the maximum value of
static friction fs between a pair of surfaces in contact will be larger than the
force of kinetic friction fk between them. Fig. 3.6 shows the variation of the
force of friction with the external force.
For a given pair of surfaces in contact, you may like to know the factors on
which fs(max) and fk depend? It is an experimental fact that fs(max) is directly
proportional to the normal force FN. i.e.
fs(max) α FN or fs(max) = μs FN (3.6)
where μs is called the coefficient of static friction. The normal force FN of the
surface on the block can be found by knowing the force with which the block
presses the surface. Refer to Fig. 3.5. The normal force FN on the block will be
mg, where m is mass of the block.
Since fs = Fext for fs < fs max, we can write
fs < μs FN.
It has also been experimentally found that maximum force of static friction
between a pair of surfaces is independent of the area of contact.
fs
FN
fs (max)
Fext fk
smooth sliding
fs
Fext
mg (at rest)
Fig. 3.5 : Forces acting on the block Fig. 3.6 : Variation of force of friction
with external force
Similarly, we can write
fk = μk FN
where μk is the coefficient of kinetic friction. In general, μs > μk. Moreover,
coefficients μs and μk are not really constants for any pair of surfaces such as
74 PHYSICS
16. Laws of Motion MODULE - 1
Motion, Force and Energy
wood on wood or rubber on concrete, etc. Values of μs and μk for a given pair
of materials depend on the roughness of surfaces, there cleanliness,
temperature, humidity etc.
FN
Notes
mg
Fig. 3.7 : Normal force on the block
Example 3.6: A 2 kg block is resting on a horizontal surface. The coefficient of
static friction between the surfaces in contact is 0.25. Calculate the maximum
magnitude of force of static friction between the surfaces in contact.
Solution :
Here m = 2 kg and μs= 0.25. From Eqn. (3.6), we recall that
fs(max) = μsFN
= μs mg
= (0.25) (2 kg) (9.8 ms–2)
= 4.9 N.
Example 3.7: A 5 kg block is resting on a horizontal surface for which μk = 0.1.
What will be the acceleration of the block if it is pulled by a 10 N force acting on
it in the horizontal direction?
Solution :
As fk= μk FN and FN = mg, we can write
fk = μk mg
= (0.1) (5 kg) (9.8 ms–2)
= 4.9 kg ms–2 = 4.9 N
Net force on the block = Fext – fk
= 10 N – 4.9 N
= 5.1 N
Fnet 51
. N
acceleration = a = = 5kg = 1.02 ms–2
m
So the block will have an acceleration of 1.02 ms–2 in the direction of externally applied
PHYSICS 75
17. MODULE - 1 Laws of Motion
Motion, Force and Energy
3.6.2 Rolling Friction
It is a common experience that pushing or pulling objects such as carts on wheels
is much easier. The motion of a wheel is different from sliding motion. It is a
rolling motion. The friction in the case of rolling motion is known as rolling
friction. For the same normal force, rolling friction is much smaller than sliding
friction. For example, when steel wheels roll over steel rails, rolling friction is
Notes about 1/100th of the sliding friction between steel and steel. Typical values for
coefficient of rolling friction μr are 0.006 for steel on steel and 0.02 – 0.04 for
rubber on concrete.
We would now like you to do a simple activity :
ACTIVITY 3.1
Place a heavy book or a pile of books on a table and try to push them with your
fingers. Next put three or more pencils below the books and now push them
again. In which case do you need less force? What do you conclude from your
experience?
3.6.3 Methods of Reducing Friction
Wheel is considered to be greatest invention of mankind for the simple reason
that rolling is much easier than sliding.
Because of this, ball bearings are used in
machines to reduce friction. In a ball-
bearing, steel balls are placed between two
co-axial cylinders, as shown in Fig.3.8.
Generally one of the two cylinders is
allowed to turn with respect to the other.
Here the rotation of the balls is almost
frictionless. Ball-bearings find application
in almost all types of vehicles and in electric
motors such as electric fans etc.
Use of lubricants such as grease or oil Fig. 3.8 : Balls in the ball-bearing
between the surfaces in contact reduces
friction considerably. In heavy machines, oil is made to flow over moving parts.
It reduces frictional force between moving parts and also prevents them from
getting overheated. In fact, the presence of lubricants changes the nature of friction
from dry friction to fluid friction, which is considerably smaller than the former.
Flow of compressed and purified air between the surfaces in contact also reduces
friction. It also prevents dust and dirt from getting collected on the moving parts.
76 PHYSICS
18. Laws of Motion MODULE - 1
Motion, Force and Energy
Fluid Friction
Bodies moving on or through a liquid or gas also face friction. Shooting stars
(meteors) shine because of the heat generated by air-friction. Contrary to solid
friction, fluid friction depends upon the shape of the bodies. This is why fishes
have a special shape and fast moving aeroplanes and vehicles are also given a
fish-like shape, called a stream-line shape. Fluid friction increases rapidly with
increase in speed. If a car is run at a high speed, more fuel will have to be burnt Notes
to overcome the increased fluid (air) friction. Car manufactures advise us to
drive at a speed of 40-45 km h–1 for maximum efficiency.
3.7 THE FREE BODY DIAGRAM TECHNIQUE
Application of Newton’s laws to solve problems in mechanics becomes easier by
use of the free body diagram technique. A diagram which shows all the forces
acting on a body in a given situation is called a free body diagram (FBD). The
procedure to draw a free body diagram, is described below :
1. Draw a simple, neat diagram of the system as per the given description.
2. Isolate the object of interest. This object will be called the Free Body now.
3. Consider all external forces acting on the free body and mark them by arrows
touching the free body with their line of action clearly represented.
4. Now apply Newton’s second law ΣF = m a
(or ΣFx = m ax and ΣFy = m ay )
Remember : (i) A net force must be acting on the object along the direction of
motion. (ii) For obtaining a complete solution, you must have as many independent
equations as the number of unknowns.
Example 3.8 : Two blocks of masses m1 and m2 are connected by a string and placed
on a smooth horizontal surface. The block of mass m2 is pulled by a force F acting
parallel to the horizontal surface. What will be the acceleration of the blocks and the
tension in the string connecting the two blocks (assuming it to be horizontal)?
Solution : Refer to Fig. 3.9. Let a be the acceleration of the blocks in the direction
of F and let the tension in the string be T. On applying ΣF = ma in the component
form to the free body diagram of system of two bodies of masses m1 and m2, we
N – (m1 + m2) g = 0
and F = (m1 + m2)a
F
⇒ a = m +m
1 2
PHYSICS 77
19. MODULE - 1 Laws of Motion
Motion, Force and Energy
N
m1
T T m2 F m1 + m2 F
(m 1 + m 2 ) g
Notes
Fig 3.9: Free body diagram for two blocks connected by a string
On applying ΣF = ma in the component form to the free body diagram of m1 we
get
N1 – m1g = 0 and T = m1a
N1 N2
a a

F F I
T = m GH m + m JK
T T F
1 1 2
m1g m2g
or
F m I
T = GH m + m JK .F
1 Fig 3.10
1 2
Apply ΣF = ma once again to the free body diagram of m2 and see whether you
get the same expressions for a and T.
Example 3.9 : Two masses m1 and m2 (m1 > m2) are connected
at the two ends of a light inextensible string that passes over
a light frictionless fixed pulley. Find the acceleration of the
masses and the tension in the string connecting them when
the masses are released. T
T
m1
Solution : Let a be acceleration of mass m1 downward. The
acceleration of mass m2 will also be a only but upward. m2
(Why?). Let T be the tension in the string connecting the two
T
masses.
a
On applying ΣF = ma to m1 and m2 we get m1g
m1g – T = m1a
T
T – m2g = m2a a
On solving equations (1) and (2) for a and T we get m2g
FG m - m IJ
1 2 FG 2m m IJ
1 2
Fig 3.11
1H
a = m + m .g
2 K T = m +m a
1 2 H K
At this stage you can check the prediction of the results thus obtained for the
extreme values of the variables (i.e. m1 and m2). Either take m1 = m2 or m1 >> m2
and see whether a and T take values as expected.
78 PHYSICS
20. Laws of Motion MODULE - 1
Motion, Force and Energy
Example 3.10 : A trolley of mass M = 10 kg is connected to a block of mass m =
2 kg with the help of massless inextensible string passing over a light frictionless
pulley as shown in Fig. 3.12 (a). The coefficient of
kinetic friction between the trolley and the surface (μk)
M
= 0.02. Find,
a) acceleration of the trolley, and (a) m
b) tension in the string. Notes
FN
Solution : Fig (b) and (c) shows the free body diagrams a
T
of the trolley and the block respectively. Let a be the T a
acceleration of the block and the trolley. FK
Mg mg
For the trolley, FN = Mg and (b) (c)
T – fk = Ma where fk = μk FN Fig. 3.12
= μk Mg
So T – μk Mg = Ma ...(1)
For the block mg – T = ma ...(2)
On adding equations (1) and (2) we get mg – μk Mg = (M + m) a
mg − μ k Mg (2 kg) (9.8ms −2 ) − (0.02) (10 kg) (9.8 ms −2 )
or a = M +m =
(10 kg + 2 kg)
19.6 kg ms −2 − 1.96 kg ms −2
= 12 kg
= 1.47 ms–2
So a = 1.47 ms–2
From equation (2) T = mg – ma = m (g – a)
= 2 kg (9.8 ms–2 – 1.47 ms–2)
= 2 kg (8.33 ms–2)
So T = 16.66 N
INTEXT QUESTIONS 3.4
1. A block of mass m is held on a rough inclined surface of inclination θ. Show
in a diagram, various forces acting on the block.
2. A force of 100 N acts on two blocks A and B of masses
F
2 kg and 3 kg respectively, placed in contact on a smooth A B
horizontal surface as shown. What is the magnitude of Fig. 3.13
force which block A exerts on block B?
PHYSICS 79
21. MODULE - 1 Laws of Motion
Motion, Force and Energy
3. What will be the tension in the string when a 5kg object suspended from it is
pulled up with
(a) a velocity of 2ms–1?
(b) an acceleration of 2ms–2?
3.8 ELEMENTARY IDEAS OF INERTIAL AND NON
Notes
INERTIAL FRAMES
To study motion in one dimension (i.e. in a straight line) a reference point (origin)
is enough. But, when it comes to motions in two and three dimensions, we have
to use a set of reference lines to specify the position of a point in space. This set
of lines is called frame of reference.
Every motion is described by an observer. The description of motion will change
with the change in the state of motion of the observer. For example, let us consider
a box lying on a railway platform. A person standing on the platform will say that
the box is at rest. A person in a train moving with a uniform velocity v will say
that the box is moving with velocity –v. But, what will be the description of the
box by a person in a train having acceleration (a). He/she will find that the box is
moving with an acceleration (– a). Obviously, the first law of motion is failing for
this observer.
Thus a frame of reference is fixed with the observer to describe motion. If the
frame is stationary or moving with a constant velocity with respect to the object
under study (another frame of reference), then in this frame law of inertia holds
good. Therefore, such frames are called inertial frames. On the other hand, if the
observer’s frame is accelerating, then we call it non-inertial frame.
For the motion of a body of mass m in a non-inertial frame, having acceleration
(a), we may apply second law of motion by involving a psuedo force m a. In a
rotating body, this force is called centrifugal force.
INTEXT QUESTIONS 3.5
1. A glass half filled with water is kept on a horizontal table in a train. Will the
free surface of water remain horizontal as the train starts?
2. When a car is driven too fast around a curve it skids outwards. How would
a passenger sitting inside explain the car’s motion? How would an observer
standing on a road explain the event?
3. A tiny particle of mass 6 × 10–10 kg is in a water suspension in a centrifuge
which is being rotated at an angular speed of 2π × 103 rad s–1. The particle is
80 PHYSICS
22. Laws of Motion MODULE - 1
Motion, Force and Energy
at a distance of 4 cm from the axis of rotation. Calculate the net centrifugal
force acting on the particle.
4. What must the angular speed of the rotation of earth so that the centrifugal
force makes objects fly off its surface? Take g = 10 m s–2.
5. In the reference frame attached to a freely falling body of mass 2 kg, what is
the magnitude and direction of inertial force on the body?
Notes
WHAT YOU HAVE LEARNT
z The inertia of a body is its tendency to resist any change in its state of rest or
uniform motion.
z Newton’s first law states that a body remains in a state of rest or of uniform
motion in a straight line as long as net external force acting on it is zero.
z For a single particle of mass m moving with velocity v we define a vector
quantity p called the linear momentum as p = m v.
z Newton’s second law states that the time rate of change of momentum of a
body is proportional to the resultant force acting on the body.
z According to Newton’s second law, acceleration produced in a body of
constant mass is directly proportional to net external force acting on the
body : F = m a.
z Newton’s third law states that if two bodies A and B interact with each
other, then the force which body A exerts on body B will be equal and opposite
to the force which body B exerts on body A.
z According to the law of conservation of momentum, if no net external force
acts on a system of particles, the total momentum of the system will remain
constant regardless of the nature of forces between them.
z A number of forces acting simultaneously at a point are called concurrent
force. Such forces are said to be in equilibrium if their resultant is zero.
z Frictional force is the force which acts on a body when it attempts to slide, or
roll along a surface. The force of friction is always parallel to the surfaces in
contact and opposite to the direction of motion of the object.
z The maximum force of static friction fs(max) between a body and a surface is
proportional to the normal force FN acting on the body. This maximum force
occurs when the body is on the verge of sliding.
z For a body sliding on some surface, the magnitude of the force of kinetic
friction fk is given by fk = μk FN where μk is the coefficient of kinetic friction
for the surfaces in contact.
PHYSICS 81
23. MODULE - 1 Laws of Motion
Motion, Force and Energy
z Use of rollers and ball-bearings reduces friction and associated energy losses
considerably as rolling friction is much smaller than kinetic friction.
z Newton’s laws of motion are applicable only in an inertial frame of reference.
An inertial frame is one in which an isolated object has zero acceleration.
z For an object to be in static equilibrium, the vector sum of all the forces
acting on it must be zero. This is a necessary and sufficient conditions for
Notes
point objects only.
TERMINAL EXERCISE
1. Which of the following will always be in the direction of net external force
acting on the body?
(a) displacement (b) velocity
(c) acceleration (d) Change is momentum.
2. When a constant net external force acts on an object, which of the following
may not change?
(a) position (b) speed
(c) velocity (d) acceleration
3. A 0.5 kg ball is dropped from such a height that it takes 4s to reach the
ground. Calculate the change in momentum of the ball.
4. In which case will there be larger change in momentum of a 2 kg object:
(a) When 10 N force acts on it for 1s ?
(b) When 10 N force acts on it for 1m ?
Calculate change in momentum in each case.
5. A ball of mass 0.2 kg falls through air with an acceleration of 6 ms–2. Calculate
the air drag on the ball.
6. A load of mass 20 kg is lifted with the help of a rope at a constant acceleration.
The load covers a height of 5 m in 2 seconds. Calculate the tension in the
rope. In a rocket m changes with time. Write down the mathmatical form of
Newton’s law in this case and interpret it physically.
7. A ball of mass 0.1 kg moving at 10 m s–1 is deflected by a wall at the same
speed in the direction shown. What is the magnitude of the change in
momentum of the ball?
82 PHYSICS
24. Laws of Motion MODULE - 1
Motion, Force and Energy
45º
Wall
45º
Notes
Fig. 3.14
8. Find the average recoil force on a machine gun that is firing 150 bullets per
minute, each with a speed of 900 m s–1. Mass of each bullet is 12 g.
9. Explain why, when catching a fast moving ball, the hands are drawn back
while the ball is being brought to rest.
10. A constant force of magnitude 20 N acts on a body of mass 2 kg, initially at
rest, for 2 seconds. What will be the velocity of the body after
(a) 1 second from start? (b) 3 seconds from start?
11. How does a force acting on a block in the direction shown here keep the
block from sliding down the vertical wall?
30°
Wall
Fig 3.15
12. A 1.2 kg block is resting on a horizontal surface. The coefficient of static
friction between the block and the surface is 0.5. What will be the magnitude
and direction of the force of friction on the block when the magnitude of the
external force acting on the block in the horizontal direction is
(a) 0 N ? (b) 4.9 N ? (c) 9.8 N ?
13. For a block on a surface the maximum force of static friction is 10N. What
will be the force of friction on the block when a 5 N external force is applied
to it parallel to the surface on which it is resting?
14. What minimum force F is required to keep a 5 kg block at rest on an inclined
plane of inclination 300. The coefficient of static friction between the block
and the inclined plane is 0.25.
PHYSICS 83
25. MODULE - 1 Laws of Motion
Motion, Force and Energy
15. Two blocks P and Q of masses m1 = 2 kg and m2 = 3 kg respectively are
placed in contact with each other on horizontal frictionless surface. Some
external force F = 10 N is applied to the block P in the direction parallel to
the surface. Find the following
(a) acceleration of the blocks
(b) force which the block P exerts on block Q.
Notes 16. Two blocks P and Q of masses m1 = 2 kg and m2
m1 m2
= 4 kg are connected to a third block R of mass
M as shown in Fig. 3.16 For what maximum
value of M will the system be in equilibrium? P Q R
M
The frictional force acting on each block is half
Fig. 3.16
the force of normal reaction on it.
17. Explain the role of friction in the case of bicycle brakes. What will happen if
a few drops of oil are put on the rim?
18. A 2 kg block is pushed up an incline plane of inclination θ = 370 imparting it
a speed of 20 m s–1. How much distance will the block travel before coming
to rest? The coefficient of kinetic friction between the block and the incline
plane is μk = 0.5.
Take g = 10 m s–2 and use sin 370 = 0.6, cos 370 = 0.8.
3.1
1. No. The statement is true only for a body which was at rest before the
application of force.
2. Inertial mass
3. Yes, as in uniform circular motion.
4. A force can change motion. It can also deform bodies.
3.2
1. Object of smaller mass
2. (a) Yes (b) No.
3. Momentum of the falling ball increases because gravitational force acts on it
in the direction of its motion and hence velocity increases.
84 PHYSICS
26. Laws of Motion MODULE - 1
Motion, Force and Energy
4. In case (b) the change in momentum will be larger. It is the F Δt product that
⎛ ⎞ Δp
gives the change in momentum. ⎜⎝ as F ∝ Δt ⎟⎠
5. No. Though the speed is constant, the velocity of the object changes due to
change in direction. Hence its momentum will not be constant.
3.3 Notes
1. The jumper is thrown upwards by the force which the ground exerts on the
jumper. This force is the reaction to the force which the jumper exerts on the
ground.
2. (a) The force with which a man kicks a football is action and the force
which the football exerts on the man will be its reaction.
(b) The force with which earth pulls the moon is action and the force which
the moon exerts on the earth will be its reaction.
(c) If the force which the ball exerts on the wall is the action then the force
which the wall exerts on the ball will be its reaction.
3. No. The arguement is not correct. The almirah moves when the push by the
person exceeds the frictional force between the almirah and the floor. He
does not get pushed backward due to a large force of friction that he
experiences due to the floor. On a slippery surface, he will not be able to
push the almirah foward.
FN
fs
sin q
mg q q
mg mg cos q
Fig. 3.17
2. 40 N
3. (a) (5 × 9.8) N
(b) F = (5 × 2) N + (5 × 9.8) N = 59 N
PHYSICS 85
27. MODULE - 1 Laws of Motion
Motion, Force and Energy
3.5
(1) When the train starts it has an acceleration, say a. Thus the total force acting
on water in the frame of reference attached to the train is
F=mg–ma
where m is the mass of the water and the glass. (Fig. 3.16). The surface of
the water takes up a position normal to F as shown.
Notes
–ma ma
F mg
Fig. 3.18
(2) To the passenger sitting inside, a centrifugal force (–mv2/r) acts on the car.
The greater v is the larger r would be. To an observer standing on the road,
the car moving in a curve has a centripetal acceleration given by v2/r. Once
again, the greater is v, the larger will be r.
(3) The net centrifugal force on the particle is F = mω2r = (6 × 10–10 kg) × (2π ×
103 rad s–1)2 × (0.04 m) = 9.6 × 10–4 N.
(4) For an object to fly off centrifugal force (= centripetal force) should be just
mv 2
more than the weight of a body. If r is the radius of the earth then = mg
r
as v = rω
r 2ω 2
=g
r
or, angular speed ω = g/r
∴ Any angular speed more than g / r will make objects fly off.
5. Zero (as it is a case of free fall of a body).
1. (d)
2. (a) if internal forces developed within the material counter bank the external
force. A it happens in case of force applied on a wall.
(b) It force is applied at right angles to the direction of motion of the body,
the force changes the direction of motion of body and not to speed.
3. v = 0 + (–g) × 4
86 PHYSICS
28. Laws of Motion MODULE - 1
Motion, Force and Energy
|v| = 40 m s–1
∴ ΔP = m (v – u) = (0.5 × 40) = 20 kg m s–2
4. When 10 N force acts for 1s.
5. 0.76 N
7. 250 N.
Notes
8. 27 N
10. (a) 10 m s–1 (b) 20 m s–1
12. (a) 0 N (b) 4.9 N (c) ~7.5 N
13. 5 N
14. 14.2 N
15. (a) 2 m s–2 (b) 6 N
16. 3 kg
18. 20 m
PHYSICS 87