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LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
 To move a football which is at rest, someone has to kick it.
MOTION UNDER
FORCE
 To throw a ball upwards, one has to give it an upward push.
 A breeze causes the branches of a tree to swing clearly, so some
external energy is needed to provide force to move a body from rest.
 Likewise, an external force is needed to retard or stop motion.
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 In these examples, the external agency of force is in contact with
the objects.
Force with
contact
MOTION
 A magnet can attract an iron piece from a distance.
Force
 So, it is clear that external agencies can also exert force on a body
even from a distance.
Iron piece
Magnet
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 Force is a pull or a push which generates or tends to generate
motion in a body at rest, stops or tends to stop a body in motion, or
tends to change the shape of the body.
PULL
WHAT IS A FORCE?
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Internal force :
What is Internal Force?
 If the force applying agent is inside a system,
then it is known as internal force.
 It does not provide motion to the system.
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External
push moves
the car
 If you are sitting in a car and you push it, it doesn't move.
 If you come out of the car and apply the same force, then it
moves.
Ex:
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1) What is a force ?
a) It is a pull or a push which tends to move a body.
d) All of the above
c) Tends to change the shape of the body
b) Stops or tends to stop a body in a motion
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d) Both (a) and (c)
2) What is an internal force?
a) Force applying agent is outside a system.
b) Force applying agent is inside a system.
c) It provides motion.
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 This property of the body is called inertia.
What is inertia..?
 Let us first learn about “The first law of motion”. It is also
called as law of inertia.
 To understand Newton's first law we have to learn about inertia at
first.
 If the net external force is zero,
ii) a body in motion continues to move with
a uniform velocity.
i) a body at rest continues to remain at rest and
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Ex:- A ball at rest stays at rest, unless acted upon by a force.
Inertia Mass
MOTION
FORCE
 Inertia means “resistance to change”.
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What is inertia?
Oh! he is not able to move it!!
The man in the
picture is trying to
move a sofa.
Let him go ahead
with his trails.
He can’t move because
it is large in size
Still it is stable
Things like to stay as they are.
It is not willing
to move.
All things in the
universe are lazy.
If things are lazy and they don’t
want to move , how does the
movement happen? No object in this
world starts moving
on its own.
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He is able to move it
as the ball is small.
GOOD!
Even for a small thing
to move, it needs a lot of
effort .
Even a small thing
needs effort to
change its position.
Push and pull are the things
which make the objects move
and change their positions.
If we want to change the
position of an object, an external
force is needed.
If force is not acting on the ball, the
ball will not stop and it keeps on
moving and moving.
When a ball is kicked in
space, it keeps on moving
without stopping. But in
our atmosphere it stops
because of….
air resistance, roughness of the
ground and certain obstacles
coming in its path.
If there is no air resistance or
roughness of the ground or any
obstacle the ball keeps on moving
and it can never stop.
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 Physicists have special name for this laziness.
Inertia means laziness
 If things are at rest, then they want to stay in rest.
 If things are in motion, then they want to move on
and on.
 This is the Newton’s first law of motion. It states…
 “Every body continues its state of rest or its state of
motion until and unless an external force is applied”.
INERTIA
So, how can we say things are
lazy?
Things want to keep
doing what they are
already doing.
Things want to stay
in the same position
as they are.
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1) What is inertia?
d) none of the above
a) It is a property of a body to oppose change in
its state of rest or motion and vice versa
b) It is the property of a body to change
from rest
c) It is the property of a body to change
from motion
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2) Which of the following is also called law of inertia?
d) none of the above
a) Newton’s first law of motion
b) Newton’s second law of motion
c) Newton’s third law of motion
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 According to Newton's first law ,
NEWTON’S FIRST LAW OF MOTION
“Every body continues to be in its state of
rest or of uniform motion in a straight line
unless and until it is compelled by some
external force ”
 This law is also called as
“Law of inertia”
Inertia means that there is a natural
tendency of objects to keep on doing what
they are doing.
All objects resist change in their state
of rest or of uniform motion.
What is the unbalanced force in this figure?
Let us study the skater to understand this a little better.
What happened to the skater in this
figure.
The skater changed his state from
motion to rest .
An external unbalanced force acted on
him in the form of an obstacle (stone).
What does inertia mean ?
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INERTIA CAN BE CLASSIFIED AS
INERTIA OF MOTION
INERTIA OF REST
INERTIA OF DIRECTION
The inability of a
body to change its
state of rest until
and unless an
external force acts
on it.
The inability of a
body to change its
state of motion until
and unless an
external force acts
on it.
The inability of a
body to change its
direction by its own
until and unless an
external force acts
on it.
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Being aware of Newton’s first law of motion
we should wear a seat belt in the car.
First law talks about how to overcome inertia when an
unbalanced force is applied.
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1) Which law of Newton is called the law of equilibrium?
d) Newton’s law of gravitation
a) Newton’s first law of motion
b) Newton’s second law of motion
c) Newton’s third law of motion
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2) You are thrown outwards when your car suddenly
takes a turn. Which law of Newton is involved in this?
d) law of gravitation
a) third law
b) second law
c) first law
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Thank you…
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The motion of a body is represented by two
physical quantities in dynamics.
Momentum (p)
Since the larger stone with
greater mass applies greater
force even though both stones
have same velocity, when they
touch the ground.
If two stones of same mass from different
heights dropped on a vehicle which one
causes greater damage?
Obviously the
larger stone causes
greater damage
(p∝m)
The stone falling from
greater height produces
greater damage.
If two stones of different masses are
dropped from different heights on a
vehicle which causes more damage?
When two stones of different masses
dropped from the same height on a
vehicle, which will cause greater
damage?
Because the stone falling from a greater
height will have larger velocity and it
applies greater force. (p∝v)
The stone of larger mass dropped from
greater height causes greater damage,
since its product of mass and velocity is
larger.
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Momentum
 It is denoted by ‘p’.
 If mass of a body is ‘m’ and velocity is ‘v’ then its momentum is,
p = mv
 Momentum of a body is defined as the quantity of motion contained
in it .
It is measured as the product of mass and velocity.
p = mv
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Units of momentum
 Momentum is a vector quantity whose direction is in the direction
of velocity.
 In CGS system: g–cm s-1
 In SI system: kg–ms-1
Dimensional formula of momentum
 Dimensional formula is [MLT-1].
Note:
Momentum 𝐩= m𝐯
g–cm s-1
kg–ms-1
[MLT-1].
vector
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Newton's second law of motion:
 According to first law of motion, if no external force acts on a
moving body, then it moves with uniform velocity.
 If no net external force acts on the body,
its momentum remains constant.
 Hence, when the momentum of a body
changes, the body must be under the
action of a net external force.
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 NEWTON’S SECOND LAW OF MOTION states that……
“ The rate of change of momentum of a body is directly
proportional to the net external force acting on the body and takes
place in the direction in which the force acts”.
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1) Momentum is …..
d) The product of mass and speed
a) The product of mass and acceleration
b) The product of mass and velocity
c) The product of mass and displacement
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2) If two stones of same mass from different heights
fall on a vehicle which one causes greater damage?
d) none of these
a) stone from a greater height
b) stone from less height
c) (a) and (b) are correct
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3) What is the dimensional formula of momentum?
d) [ML-1T-1]
a) [ML2T-2]
b) [MLT-2]
c) [MLT-1]
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 The rate of change of momentum of a body is directly proportional
to the applied force and takes place in the direction of force.
 The second law of motion tells us that there is a net
SECOND LAW OF MOTION
on a body. It relates to the net external force and the acceleration of a
external force acting
body.
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 Acceleration is produced when a force acts on a mass. The
greater the mass of an object is, the greater is the amount of force
required to accelerate the object.
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 Acceleration is produced when a force acts on a mass. The
greater the mass of an object is, the greater is the amount of force
required to accelerate the object.
 Every one unconsciously knows the second law.
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What does this mean?
Newton's second law gives us
an exact relationship between
force, mass and acceleration.
It talks about how fast an object
moves.
 Every one knows that heavier objects require more force to move the
same distance than lighter objects.
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Mathematically second law can be
expressed as
Newton’s second law of
motion is the most
significant law among
the laws of motion.
F = ma
Or
Force = mass × acceleration
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Units of force
Note: 1N = 105dyne
1) In CGS: gram-cm/sec2 or dyne
2) In MKS or SI: kilogram-meter/sec2 or newton (N)
Dimensional formula of force: MLT-2
Define one
newton? We know that F = ma
If m = 1kg, a = 1m/s2 then
F=1kg ×1m/s2
= 1kg-m/s2 = 1newton
Note: 1N = 105dyne
If a force produces an acceleration
of 1 ms-2 on an object of mass 1 kg,
then the force is called 1 newton.
Second law measures quantity
of force whereas first law of
motion describes concept of
force.
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 Ramu’s car, which weighs 1000kg is out of gas. Ramu is trying to
push the car to a gas station, and he makes the car to move with an
acceleration of 0.05 m/sec2.
Force = 50 N
Example:
a =0.05m/s2
F = ? N
m = 1,000 kg
Using Newton’s second law, you
can compute how much force
Ramu is applying on the car.
F = ma
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1) Which of the following is the most significant law of
motion given by Newton?
d) Newton's law of gravitation
a) Third law of motion
b) Second law of motion
c) First law of motion
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2) A force produces an acceleration of 1 ms-2 on an object
of mass 1 kg. This force is defined as….
d) 1 poundal
a) 1 dyne
b) 1 newton
c) 1 telsa
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“The rate of change of momentum of a body is directly proportional to
the applied external force and takes place in the direction of force”.
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If under the action of external force 𝑭 for time interval 𝒅𝒕 , the
velocity of a body of mass m changes from 𝒗 to 𝒗 + 𝒅𝒗.
Let us have an expression for Newton’s 2nd Law
= 𝒌
𝐦(𝒅𝒗)
𝒅𝒕
𝑭 
𝒅𝒑
𝒅𝒕
𝑭 = 𝒌
𝒅𝒑
𝒅𝒕
If k = 1,
𝑭 = km
𝒅𝒗
𝒅𝒕
𝑭 = 𝒌m𝒂
𝑭 = 𝒎𝒂
According to Newton’s second law
Then its initial momentum 𝒑 = m𝒗 changes by 𝒅𝒑 =𝐦(𝒅𝒗).
𝒌 𝒊𝒔 𝒑𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒏𝒂𝒍𝒊𝒕𝒚
𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝒅𝒑 = 𝐦(𝒅𝒗).
m is constant
𝒂 =
𝒅𝒗
𝒅𝒕
“The rate of change of momentum of a body is
directly proportional to the applied external
force and takes place in the direction of force”.
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Units of force
 In CGS system: g–cm s-2 or dyne
 In SI system: kg–ms-2 or newton(N)
Dimensional formula of force:
 Dimensional formula is [ML𝐓−].
Force F = ma
g–cm s-2 or dyne
kg–ms-2 or newton(N)
[ML𝐓−].
Note: 1N = 105 dyne
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Gravitational units of force :
1Kg.wt or 1kg.f = 9.8 N = g N
1gm.wt or gm.f = 980 dyne = g dyne
 kilogram weight (kg.wt) and gram weight (gm.wt) are called the
gravitational units of force.
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1) What is the relation between force and acceleration?
d) none of these
a) directly proportional
b) inversely proportional
c) equal
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2) What is the SI unit of force?
d) newton/sec
a) dyne
b) newton
c) newton/sec2
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Applications of Newton’s 2nd law:
1) If 𝒖 and 𝒗 are the intial and final velocities of a particle of mass
‘m’ and average force acting on the particle in a time interval ‘dt’ is
𝑭 =
𝒎𝒗 − 𝒎𝒖
𝒅𝒕
=
𝒅𝒑
𝒅𝒕
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If ‘t’ is time of contact of the ball with
the floor, then force on the ball is….
-mv mv
=
𝟐𝒎𝐯
𝒕
F
𝐝𝐩
𝒅𝒕
=
Then the change in linear momentum of the particle is
dp = mv – (-mv) = 2mv. along the normal, away from the
floor
2) If a ball of mass ‘m’ strikes a rigid
floor normally and rebounds with
the same speed v.
floor
-mv
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If a ball of mass m, strikes a rigid wall with a
speed v, making an angle  , with the normal
to the surface and rebounds with the same
speed at the same angle….
Then the magnitude of the change in momentum of
the ball is dp = mvcos - (-mvcos ) = 2mvcos
perpendicular and away from the wall
Normal


m
V
vcos
vsin

wall
m
V
V
vsin
vcos
vcos vcos
-
The change in momentum
along the wall is zero
dp
𝟐𝒎𝒗 𝒄𝒐𝒔
𝒕
F=
If ‘t’ is time of contact of
ball with the wall then
force on the ball is ….. =
𝒅𝒑
𝒅𝒕


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If a ball of mass m, strikes a rigid wall with a
speed v, making an angle  , with the wall and
rebounds with the same speed and same
angle….
Then the magnitude of the change in momentum of
the ball is dp = mvs𝐢𝐧𝜽 - (-mvs𝐢𝐧𝜽) = 2mvs𝐢𝐧𝜽
perpendicular and away from the wall.
Normal
m
V
wall
m
V
V
The change in momentum
along the wall is zero
Since dp = vcos𝜽 - vcos𝜽
= 0
dp
𝟐𝒎𝒗𝐬𝐢𝐧
𝒕
F=
If ‘t’ is the time of contact
of the ball with the wall
then force on the ball is ….. =
𝒅𝒑
𝒅𝒕
If  is the angle between the direction of motion of the ball and the
wall, then the magnitude of change in momentum of the ball
perpendicular to the wall is 2mvsin.


vsin𝜽
vc𝐨𝐬𝜽
vsin𝜽
vc𝐨𝐬𝜽

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 From a nozzle of area of cross-section A, liquid of density  coming
out with velocity V horizontally and strikes a vertical wall, then
=
F Av2
=
F 2Av2
A V

If water does not bounce back i.e., if
it drops down along the wall, then
force exerted on the wall is
If water bounces back with the
same speed then force exerted
on the wall is
v
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1) What is the relation between change in momentum
and the average force with respect to time t seconds?
d) none of these
a) F =
𝐝𝐩
𝐝𝐭
b) F = 𝐩
c) F = m
𝐝𝐩
𝐝𝐭
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2) The motion of a particle of mass m is described by
y = ut + 1/2gt2 . Find the force acting on the
particle.
d) -ma
Given y = ut +
𝟏
𝟐
gt2
We know v =
𝒅𝒚
𝒅𝒕
=
𝒅
𝒅𝒕
(𝒖𝒕 +
𝟏
𝟐
𝒈𝒕2)
= u
𝒅𝒕
𝒅𝒕
+
𝟏
𝟐
g2t
= u + gt
Acceleration a =
𝒅𝒗
𝒅𝒕
=
𝒅
𝒅𝒕
(u + gt) = g
Then force = F = ma = mg
a) mg
b) ma
c) -mg
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Impulse :
Ex:
The product of force and time that produces a finite change in
momentum of the body is called impulse.
Collision between two
billiard balls.
Nail driven into a wall by
striking it with a hammer.
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Units of impulse
 Impulse is a vector quantity whose direction is in the direction of
force.
 In CGS system: dyne - s
 In SI system: N - s
Dimensional formula of impulse
 Dimensional formula is [MLT-1].
Note:
Impulse 𝐉 =𝐅 t
dyne - s
N - s
[MLT-1].
vector
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If the force is constant then impulse
According to Newton’s second law
𝐅 =
𝐝𝐩
𝐝𝐭
𝐅 dt = 𝐝𝐩
= 𝐩𝟐 - 𝐩𝟏 = m 𝐕 − 𝐦 𝐔
 Impulse 𝐉 = 𝐅 dt = m 𝐕 − 𝐦 𝐔
𝑱 = 𝑭𝒕
= m 𝑽 − 𝒎 𝑼
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 If the force is variable then impulse
 The area under F – t graph gives impulse.
 If force 𝑭 acts on a body for time t1, 𝑭𝟐 for t2, 𝑭𝟑for t3,.. then total
change in momentum is equal to 𝑭𝟏𝒕𝟏 + 𝑭𝟐𝒕𝟐 + 𝑭𝟑 𝒕𝟑+…..
𝑱 = 𝒕𝟏
𝒕𝟐
𝑭 𝒅𝒕
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Applications of Impulse:
For a given impulse, the average force is inversely proportional to the
time for which force acts,
a) A person falling on a concrete floor from a certain height receives
injuries more than when he falls from same height on loose soil.
b) Breakable articles like china ware, television sets etc. are
transported from one place to another by packing them with felt,
syfo foam etc.
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c) Shock absorbers are used in vehicles to reduce the
magnitude of impulsive force.
d) A cricketer lowers his hands, while catching the ball to
reduce the impulsive force.
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1) Formula for impulse
b) Force × acceleration
a) Force × time
c) Force × speed
d) None of these
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2) What is the dimensional formula for impulse?
a) MLT-2
b) ML2T-2
c) MLT-1
d) M2L2T-1
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3) A cricketer lowers his hand, while catching the
ball is an example for?
a) impulse
c) force
b) momentum
d) both b and c
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 To every action, there is always an equal and opposite reaction.
NEWTON’S THIRD LAW OF MOTION
ACTION
REACTION
REACTION
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 To every action, there is always an equal and opposite reaction.
NEWTON’S THIRD LAW OF MOTION
What does
this mean?
Action
Reaction
Reaction
Action
This means that for every action there is a reaction which is equal in
magnitude but in opposite direction.
When you push an object, it pushes
you back.
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Characteristics of action and reaction
Action and reaction are
equal in magnitude and
opposite in direction.
Action and reaction do
not act on the same body,
therefore they do not
cancel each other.
Action and reaction
are mutual and act
simultaneously.
Newton’s third law
is not applicable to
pseudo forces.
Newton’s third law defines
the nature of a force and
gives the law of conservation
of linear momentum.
Newton's third law is not
strictly applicable when the
interaction between two
bodies separated by a large
distance.
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Force acting on an object relative to an
observer in a non-inertial frame, without
any interaction with any other object of the
universe is called as pseudo force.
EX:
Centrifugal force,
What is a pseudo force?
deflection of pendulum
gain or loss of weight
relative to accelerating car,
experienced in an accelerating elevator
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Ex:
 A bird flies forward by exerting a force on the air but the air pushes
the wings of the bird back that makes the bird move forward.
 Walking .
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1) Centrifugal force is an example for ?
d) external force
a) gravitational force
b) tangential force
c) pseudo force
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2) Which of the following statements is incorrect for
action and reaction forces?
b) these are equal in magnetite but opposite in direction
a) these act on two different bodies
d) action and reaction never balance each other
c) these act on a single body
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3) The behaviour of a body under zero resultant force is
given by ………
d) Newton’s law of gravitation
a) Newton’s third law of motion
b) Newton’s second law of motion
c) Newton’s first law of motion
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Thank you…
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Motion of cart pulled by a horse
When a horse pulls a cart, which
force helps the horse move forward?
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Cart
R1
W1
F 
The horse pulls the cart by pushing the ground backward with
the feet with a force F making an angle  with horizontal.
 The ground inturn exerts a force R on the horse is equal and
opposite to F . The reaction R can be resolved into two rectangular
components, i.e. R cos  and R sin , as shown in the figure.
R

R cos
R sin
T
W2
R2
The component R cos (=T) tends
to move the cart to the right.
If R cos  > f, the cart will
move forward i.e., to the
right.
Whether the cart moves or not depends upon the
relative magnitudes of Rcos and frictional force
‘f’ between the cart wheel and the ground.
f f
R cos
Note that W1 weight of the cart is
balanced by the reaction R1 of the
ground on the cart.
Similarly, weight W2 of the horse is
balanced by (R2 + R sin ) where R2 is the
reaction of the ground on the horse.
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1) When a horse pulls a cart which force helps the
horse to move forward?
d) None of these
a) The reaction force exerted by the
ground on the horse
b) The action force exerted by the horse
on the ground
c) The resultant force on the ground
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APPLICATIONS OF NEWTON’S LAWS
To apply Newton’s laws to a particle or a body or a system we have to
follow certain steps-
 The system may be a single particle, a block, a combination of two
or more blocks kept one over the other, two or more blocks
connected by strings and a piece of string etc.
 To decide the system on which the laws of motion are to be applied.
 While selecting the system we have to see all its parts should have
same acceleration.
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 To identify the forces acting on the system due to all the objects
other than the system.
The list of forces depends on the physical quantity to be determined.
Internal forces of the system are not to be taken in the list of the forces.
LAWS OF MOTION
 Choose the axes and write equations of motion.
 Draw a free body diagram representing the system by a point in
a separate diagram.
 Different forces acting on the system are represented by vectors
with the point as the common origin.
LAWS OF MOTION
1) What is the common factor in which to describe a
system’s condition, where Newton’s laws are
applicable?
a) acceleration
b) velocity
c) displacement
d) speed
LAWS OF MOTION
2) Which of the steps are correct in applying
Newton's laws to a particle or a body or a system?
d) all the above
a) decide a system on which these laws
are applied
b) identify the forces acting on a
particle or body or a system
c) choose the axes and write the
equations of motion
LAWS OF MOTION
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Inertial frames of reference:
Ex:
Normal force, tension, weight,
spring force, muscular force etc.
 Basic laws of physics are identical in all inertial frames of
reference.
 Inertial frames of reference are called Newtonian or Galilean
frames of reference.
Frames of reference in which
Newton’s Laws of Motion are
applicable and if they move
with the uniform velocity
relative to each other.
LAWS OF MOTION
Non - Inertial frames:
Ex:
Frames of reference in
which Newton’s Laws
are not applicable, such
frames accelerate
relative to inertial
frames.
LAWS OF MOTION
 All rotating frames of reference are non-inertial. The earth is a non-
inertial frame.
 However, in many applications the earth can be approximated as an
inertial frame because acceleration of the earth due to its rotation
and revolution is negligible.
LAWS OF MOTION
a) Inertial frames of reference
1) Frames of reference in which Newton’s laws of
motion are applied are called as?
d) Both (a) and (c)
b) Non-inertial frames of reference
c) Galilean frames of reference
LAWS OF MOTION
2) Which is an example for non-inertial frame of
reference?
d) muscular force
a) normal force
b) tension
c) artificial satellite
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LAWS OF MOTION
Pseudo Force
 It is observed that, when we are traveling in a car and if the car
hits a wall then we feel a push in forward direction.
 Hence we cannot explain this phenomenon using Newton’s laws of
motion, because according to Newton‘s law only external force can
change the motion of a body.
This force cannot be
attributed to any
external source.
However the push that we
experience when the car stops
suddenly is due to our own
inertia.
In order to apply Newton’s laws
to explain this situation we need
to introduce a term called
external force in the equation of
motion.
This added term which does not
have real origin is called pseudo
force.
Force acting on an object relative to an observer in a non-inertial
frame without any interaction with any other object of the universe is
called pseudo force.
LAWS OF MOTION
Applications of pseudo force:
 A block kept on a smooth inclined plane can be kept stationary
relative to the incline by giving a horizontal acceleration of g tan as
shown in the figure .
Fk
a
mg

 mg sin
mg cos
ma 
ma sin
ma cos N
mg sin
mg cos
Pseudo force
Here, ma cos = mg sin
a = g tan
Applied force
F = (M+m)a
= (M+m) g tan 
M is mass of wedge
F
M(Wedge)
m(block)
LAWS OF MOTION

ma
mg
a
 ma
mg
a
 A pendulum is suspended from the roof of a moving car.
a) If the car is moving with uniform velocity, the position of the bob does
not change.
b) If the car moves with acceleration or retardation, the bob moves in
opposite direction to that of acceleration.
LAWS OF MOTION
T sin = ma and T cos = mg
 tan = a/g
For the equilibrium
of the bob,
T sin
ma
T cos
mg
T
 = tan-1 (
𝒂
𝒈
)

The forces acting on the bob are, weight mg, tension in the string T
and pseudo force ma.
LAWS OF MOTION
or a = g tanand T = m( 𝒂𝟐 + 𝒈𝟐)
 In the above case if the car is moving along a horizontal circular
track, pseudo force is mr2.
Here mr2 is centrifugal force.
Then, tan =
𝐦𝐚
𝐦𝐠
=
𝐦𝐫𝟐
𝐦𝐠
or r2 = g tan
LAWS OF MOTION
a

M
m
Here a = g cot.
 An object of mass m is kept on a wedge of mass M. The wedge is
moved with an acceleration ‘a’ such that the object moves down like
a freely falling body (no friction).
LAWS OF MOTION
 A trolley is moving on a horizontal road with uniform acceleration ‘a’.
The pseudo forces acting on the two blocks kept at different positions
are as shown
m1a and m2 a are pseudo forces.
a
m2 a
m1 a
LAWS OF MOTION
1) A ball is suspended by thread from the ceiling of a tram car.
Breaks are applied and the speed of the car changes uniformly
from 36kmph to zero in 5s. The angle by which the ball deviates
from the vertical (g = 10ms-2) is…….
a) Tan-1 𝟏
𝟑
b) Sin-1 𝟏
𝟓
c) Tan-1 𝟏
𝟓
d) Cot-1 𝟏
𝟑
Change in speed of a car is =
36kmph to 0
t = 5s
g = 10ms-2
acceleration a =
𝒗−𝒖
𝒕
a=
𝟏𝟎𝒎/𝒔
𝟓𝒔
= 2m/s2
tan =
𝒂
𝒈
=
𝟐
𝟏𝟎
=
𝟏
𝟓
 = tan-1 𝟏
𝟓
1km = 1000m
36km = 36000m
1hr = 3600sec
v-u =
𝟑𝟔𝟎𝟎𝟎
𝟑𝟔𝟎𝟎
=10m/s
LAWS OF MOTION
Thank you…
LAWS OF MOTION
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FREE BODY DIAGRAM:
 When number of forces act on a body, the resultant force on it
produces acceleration in its direction.
 An object of mass ‘M’ is suspended vertically by a string of negligible
mass from a rigid support as shown in the figure.
M
Mg
Mg
T
M
a) The tension in the stretched string T = Mg.
LAWS OF MOTION
T
M
Mg
a
T
M
Mg
a
b) If the system is pulled upwards with an acceleration ‘a’ then tension in
the string is T = M(g+a)
c) If the system is moved downwards with an acceleration ‘a’ then the
tension in the string is T = M(g-a)
FREE BODY DIAGRAM:
LAWS OF MOTION
d) If the string has mass ‘m’ then the tension in the string at the upper
end is.
T = (M+m)g, if the system is at rest
= (M+m) (g+a) if the system is
moving up with acceleration ‘a’
= (M+m) (g-a) if the system is moving
down with acceleration ‘a’
FREE BODY DIAGRAM:
LAWS OF MOTION
 Masses connected by a string and suspended from a rigid support
then tension in the string when force F is applied down as shown in
the figure.
M1
M2
F
T1
T2
T1 = F + M2g
T2 = F + (M1+ M2)g
LAWS OF MOTION
1) Three blocks of equal masses (each 3kg) are
suspended by weightless strings as shown. If applied
force is 100N, then T1 is equal to (g = 10m/s2)
d) 160N
T3
T2
T1
100N
Given: M1=M2=M3=3kg
F=100N
g=10m/s2
T1=?
Since T1=F+M1g
T1= 3×10+100
T1 = 130N
a) 130N
b) 190N
c) 100N
LAWS OF MOTION
LAWS OF MOTION
MOTION OF THE CONNECTED BODIES:
Contact Force:
i) Masses in contact on a smooth horizontal surface:
M1 M2
F
Acceleration of the system
a =
𝐅
(𝐌𝟏
+𝐌𝟐
)
, f12 = M2a
The force acting between two bodies in contact with each other.
f12
f12  contact force is the force
due to M2 on M1
LAWS OF MOTION
a) Acceleration of the system
a =
𝑭
(𝑴𝟏
+𝑴𝟐
+𝑴𝟑
)
ii) Contact forces as shown in the figure
M1
M2
F
M3
b) Contact force between M1 and M2 is
F12 = (M2 +M3)a
c) Contact force between M2 and M3 is F23 = M3 a
a =
𝐅
(𝐌𝟏
+𝐌𝟐
+𝐌𝟑
)
LAWS OF MOTION
1) Three masses of 16kg, 8kg and 4kg are placed in contact
as shown in the figure. If force of 140N is applied on 4kg
mass, then the force on 16kg will be….
16kg
8kg
140N
4kg
Given m1 = 16kg ,m2 = 8kg, m3=4kg
Force F = 140N
F1 = ?
since a =
𝐅
𝐦𝟏
+𝐦𝟐
+𝐦𝟑
F1=m1a =
𝐦𝟏
𝐅
𝐦𝟏
+𝐦𝟐
+𝐦𝟑
F1 =
𝟏𝟔×𝟏𝟒𝟎𝐍
(𝟔++)
=
𝟏𝟔×𝟏𝟒𝟎
𝟐𝟖
= 𝟖𝟎𝑵
a) 140N
b) 120N
c) 100N
d) 80N
LAWS OF MOTION
LAWS OF MOTION
MOTION OF BODIES CONNECTED BY STRINGS:
i) When two bodies of masses m1 and m2 are tied to the ends of a string
and are kept on smooth horizontal surface are pulled by a force F.
F
a
R1 R2
m1 m2
T T
m1g m2g
a =
𝐅
𝐦𝟏
+𝐦𝟐
and T = m1a =
𝐦𝟏
𝐅
𝐦𝟏
+𝐦𝟐
LAWS OF MOTION
Acceleration produced in the system
ii) If in the above example three bodies are
pulled with a force F.
Tension produced in the first string
a =
𝐅
𝐦𝟏
+𝐦𝟐
+ 𝐦𝟑
T1 =m1a =
𝐦𝟏
𝐅
𝐦𝟏
+𝐦𝟐
+ 𝐦𝟑
m1 m2 m3
T1 T2 T3
F
LAWS OF MOTION
Tension produced in the second string
Tension produced in the third string
T2 =
𝐦𝟏
+𝐦𝟐
𝐅
𝐦𝟏
+𝐦𝟐
+ 𝐦𝟑
(since F – T2 = m3 a)
T3 = F
LAWS OF MOTION
1) Three blocks of masses m1, m2, and m3 are connected
by a mass less string as shown in the figure on a
frictionless table. They are pulled with a force
T3= 40N. If m1 = 10kg, m2 = 6kg and m3 = 4kg, then
tension T2 will be ?
m1 m2 m3
T1 T2 T3
Given, T3 = 40N, m1 = 10kg
m2 = 6kg and m3 = 4kg
m1+m2+m3 = 20kg
Since a =
𝐅
𝐦𝟏
+𝐦𝟐
+𝐦𝟑
T3 = F
a =
𝟒𝟎
𝟏𝟎+𝟔+𝟒
= 𝟐𝒎/𝒔𝒆𝒄𝟐
Now T2 – T1 = m2a
T1 = m1a = 10×2 = 20N
T2 – 20 = 6×2
T2 = 32N
a) 10N
b) 20N
c) 32N
d) 40N
LAWS OF MOTION
LAWS OF MOTION
MOTION OF BODIES CONNECTED BY STRING PASSING
OVER A SMOOTH PULLEY(Atwood’s Machine):
1) Let two bodies of masses m1 and m2 respectively are connected by
a light inextensible string passing over a massless smooth pulley. If
m1> m2 , then mass m1 moves downwards and mass m2 moves
upwards.
Equation of motion
For body A, m1 g – T = m1a
For body B, T – m2 g = m2a
LAWS OF MOTION
a = (
𝐦𝟏
−𝐦𝟐
𝐦𝟏
+𝐦𝟐
) g; and T = (
𝟐𝐦𝟏
𝐦𝟐
𝐦𝟏
+𝐦𝟐
)g
m2
m1
m2g
m1g
a
A
B
a
T
T
The reaction at the pulley R=2T = (
𝟒𝐦𝟏
𝐦𝟐
𝐦𝟏
+𝐦𝟐
)g
Solving, we get
LAWS OF MOTION
2) In the above case, if the pulley is not massless rather it has finite
mass M and radius R, then tensions in two segment strings are
different, say T1 and T2 . Then equations of motion
m2
m1
m2g
m1g
a
A
B
a
T1
T2
R
M
LAWS OF MOTION
For body A, m1g – T1 = m1 a……(1)
For body B, T2 – m2g = m2 a……(2)
 = ( T1 – T2)R = I
 ( T1 – T2)R = 1/2MR2 . (a/R)
For a pulley, Torque  = I
 T1 – T2 =
𝑴𝒂
𝟐
… . (𝟑)
LAWS OF MOTION
Solving equations (1), (2) and (3)
a =
𝐦𝟏
−𝐦𝟐
𝐦𝟏
+𝐦𝟐
+
𝐌
𝟐
g
T1 =
𝐦𝟏
𝟐𝐦𝟐
+
𝐌
𝟐
𝐦𝟏
+𝐦𝟐
+
𝐌
𝟐
g
T2 =
𝐦𝟐
𝟐𝐦𝟏
+
𝐌
𝟐
𝐦𝟏
+𝐦𝟐
+
𝐌
𝟐
g
LAWS OF MOTION
3) Suppose a body B with mass m2 rests on a smooth table. Mass m1 is
connected to m2 by a light string over a pulley fixed at the edge of a
table and it hangs freely, if m1 > m2, then equation of motion
For body A, m1g – T = m1 a
For body B, T= m2 a
Solving , we get
a =
𝐦𝟏
𝐠
𝐦𝟏
+𝐦𝟐
; 𝒂𝒏𝒅 𝑻 =
𝐦𝟏
𝐦𝟐
𝐦𝟏
+𝐦𝟐
g
LAWS OF MOTION
m2
m1
m1g
m2g
R
T
B
If the coefficient of kinetic friction between
the table and mass is , then
𝒂 =
𝐦𝟏
−𝐦𝟐
𝐦𝟏
+𝐦𝟐
𝐠; T =
𝐦𝟏
𝐦𝟐
𝟏+ 𝐠
𝐦𝟏
+𝐦𝟐
LAWS OF MOTION
1) In the figure m1=4m2. The pulleys are smooth and
light. At time t =0, the system is at rest. If the system
is released, the acceleration of mass m1 is
a) 𝒈
𝟐
b) 𝒈
𝟖
c) 𝒈
𝟒
d) 𝒈 m2
m1
T
T
T
T1
Given that m1 = 4m2
t = 0 , a = ?
Since m1 𝒈 – 2T = m1 a….(1)
T-m2 𝒈 = m2(2a)
T = m2(𝒈 +2a)
substituting T in eq (1) we get
m1g - 2× m2(𝒈 +2a) = m1a
Since m2 =
𝒎𝟏
𝟒
m1 𝒈 - 2×
𝒎𝟏
𝟒
𝒈 + 𝟐𝒂 = m1a
2 𝒈 –𝒈 – 2a = 2a
𝒈 = 4a
a =
𝒈
𝟒
LAWS OF MOTION
LAWS OF MOTION
4) Suppose a mass M rests on a frictionless horizontal table. Let the
mass M is connected to two masses m1 and m2 by two light strings
passing over two light pulleys. If m1 > m2 and m1 is moving
downwards with acceleration ‘a’, then the equations of motion .
M
m1
m1g
T1
A
T1
m2
m2g
B
T2
T2
a
LAWS OF MOTION
𝐒𝐨𝐥𝐯𝐢𝐧𝐠 𝐰𝐞 𝐠𝐞𝐭 𝑻𝟏 =
𝐦𝟏
𝟐𝐦𝟐
+𝐌
𝐦𝟏
+𝐦𝟐
+𝐌
g
𝐓𝟐 =
𝐦𝟐
𝟐𝐦𝟐
+𝐌
𝐦𝟏
+𝐦𝟐
+𝐌
g
For body A, m1g – T1 = m1 a
For body B, T2 – m2g = m2 a
For body C, T1 – T2 = Ma
LAWS OF MOTION
5) When the two masses m1 and m2 are hanging as shown in the figure
and m2 > m1.
For body m1, T - m1g sin = m1 a
For body m2, m2g - T = m2 a
m1
m2
m1g sin m2g

T
a
LAWS OF MOTION
Solving these equations, we get
𝐚 =
𝐦𝟐
−𝐦𝟏
𝐬𝐢𝐧
𝐦𝟏
+𝐦𝟐
g and T = m2(g-a);
𝑻 =
𝐦𝟏
𝐦𝟐
𝐠 𝟏+𝐬𝐢𝐧
𝐦𝟏
+𝐦𝟐
g
𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧 𝐟𝐨𝐫𝐜𝐞 𝐨𝐧 𝐩𝐮𝐥𝐥𝐞𝐲
𝑭 = 𝟐𝑻𝒄𝒐𝒔
𝟗𝟎𝟎 − 
𝟐
LAWS OF MOTION
6) If in the system shown in figure, m1 > m2 , the acceleration ‘a’ of the
system is given by
m2 m1
T
T
m2g sinβ m1g sin
A
B
𝐦𝟏𝐬𝐢𝐧 − 𝐦𝟐𝐬𝐢𝐧
𝐦𝟏 + 𝐦𝟐
𝒂 = 𝒈
a
LAWS OF MOTION
And the tension in the string is given by
𝐦𝟏
𝐦𝟐
𝐬𝐢𝐧+𝐬𝐢𝐧
𝐦𝟏
+𝐦𝟐
g
𝑻 = m2(g-a) =
𝒔𝒊𝒏
 + 
𝟐
𝑭 = 2T
Reaction force on pulley
LAWS OF MOTION
a) If a body A of mass M is hanging from one end of a string passing
over a pulley and a person B of same mass starts climbing up along the
same string, the body and the person rises up with the same speed
conserving the momentum.
M
M
Mg sin Mg

T
a
LAWS OF MOTION
b) If in the above example, two persons of equal mass start climbing
up along the string and if one of them climbs faster than the
other, then the two persons reach the pulley simultaneously
conserving the momentum.
LAWS OF MOTION
1) Two masses of 8kg and 4kg are connected to a
frictionless pulley. The acceleration of the system
is.........
8kg
4kg
m2g
300
a) 4m/s2
b) 2m/s2
c) zero
d) 9.8m/s2
LAWS OF MOTION
Thank you…
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
MASSES CONNECTED BY SPRINGS/STRINGS:
i) When two bodies of masses m1 and m2 connected by a spring are
pressed and released m1a1 = m2a2 , where a1 and a2 are their initial
accelerations after release.
ii) If a force F is applied on a body of mass m1 which connected to
another body of mass m2 and the two bodies move with accelerations
a1 and a2 respectively, then
F = m1a1 + m2a2
a2
m2
a1
m1
F
LAWS OF MOTION
iii) A force F is applied on the massless pulley as shown in the figure,
string connected to the block on smooth horizontal surface. Then
m
T
T
F
F = 2T and T = mablock
LAWS OF MOTION
If the block moves a distance ‘x’ the pulley moves x/2 (total length of
the string remains constant)
Therefore acceleration of the pulley = ablock /2
=
𝐓
𝟐𝐦
=
𝐅/
𝟐𝐦
= 𝐅
𝟒𝐦
LAWS OF MOTION
1)In the following figure, the pulley is massless and
frictionless. There is no friction between the body and
the floor. The acceleration produced in the body when it
is displaced through a certain distance with force ‘P’ will
be
a) 𝑷
𝑴
b) 𝑷
𝟐𝑴
c) 𝑷
𝟑𝑴
d) 𝑷
𝟒𝑴
M T
T
p
A
LAWS OF MOTION
LAWS OF MOTION
Apparent weight in a lift or elevator:
A man of mass ‘m’ is inside an elevator:
a) Elevator accelerates up:
Relative to earth N – mg = ma
(M = mass of man)
Apparent weight = N = m(g+a)
Tension in the cable
T = (Melevator +Mman)(g+a).
LAWS OF MOTION
b) Elevator accelerates downward:
Relative to earth mg - N = ma
Apparent weight = N = m(g – a)
Tension in the cable
T = (Melevator +Mman)(g - a).
Same effect is felt when the elevator goes
up with retardation (a = -g)
if an elevator falls freely (cable breaks) a = g hence N =0
i.e apparent weight of a body in a free fall = 0
LAWS OF MOTION
c) Elevator moves up or down with uniform velocity:
Relative to earth N – mg = 0
Apparent weight = N = mg
Tension in the cable T = (Melevator + Mman)g
LAWS OF MOTION
1) A lift is going up with uniform velocity. When brakes
are applied, it slows down. A person in that lift,
experiences……..
a) more weight
b)less weight
c) normal weight
d) zero weight
LAWS OF MOTION
2) A man drops an apple in the lift. He finds that the
apple remains stationary and does not fall. The lift
is….
a) going down with constant speed
b) going up with constant speed
c) going down with constant acceleration
d) going up with constant acceleration
LAWS OF MOTION
LAWS OF MOTION
Man inside an artificial satellite
a) An artificial satellite orbiting the earth in a circular orbit is a freely
falling body because its centripetal acceleration is equal to the
acceleration due to gravity in that orbit.
b) In the above case, apparent weight of the man inside the satellite is
zero.
(V0 = orbital velocity)
𝒎𝑽𝟎
𝟐
𝒓
mg - N =
LAWS OF MOTION
N = 0
𝒈𝒓
v0 =
But
𝒎𝒈𝒓
𝒓
𝒎𝒈 - N=
Hence
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1) The apparent weight of a man is zero in an artificial
satellite orbiting the earth in a circular orbit is a freely
falling body because ?
a) Its centripetal acceleration is equal to g
b) Its centripetal acceleration is not equal to g
c) Its centripetal acceleration is greater than gequal to g
d) Its centripetal acceleration is less than g
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TENSION IN STRING SUPPORTING BOB:
a) A bob of mass ‘m’ is suspended by a string.
b) When the bob is stationary, the tension ‘T’ in the string is given by
T = mg
c) When the bob is pulled up with an acceleration ‘a’ the tension ‘T’
in the string is given by T=m(g +a).
d) When the bob is allowed to move down with an acceleration ‘a’,
the tension ‘T’ in the string is given by T = m(g – a).
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e) When the bob is pulled up or moving down with uniform velocity
‘v’, the tension ‘T’ in the string is given by T = mg.
f) When the bob is falling freely, the tension ‘T’ in the string becomes
zero.
g) When the suspended bob is moving horizontally with an acceleration
‘a’, the tension ‘T’ in the string is given by
T = m 𝒈𝟐 + 𝒂𝟐
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h) A bird is in a wire cage hanging from a spring balance, when the
bird starts flying in the cage, the reading of the balance decreases.
i) In the above case, if the bird is in a closed cage or air – tight cage
and it hovers in the cage, the reading of the spring balance does not
change.
j) In the above case for a closed cage if the
bird accelerates upward reading of the
balance is
R = Wbird + ma, where m is the mass of
the bird and a its acceleration.
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1)A bird is sitting in an air tight cage suspended from a
spring balance. If the bird starts flying, then the
reading of the balance will………
a) becomes zero
b) increases
c) decreases
d) remains unchanged
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Thank you…
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LINEAR MOMENTUM
Linear momentum is a measure of quantity of motion possessed by a
moving body.
 Linear momentum is the product of the mass of a body and its
velocity. 𝑷 = 𝒎𝑽, where ‘m’ is the mass of the body and 𝐕 is its
velocity. This is a product involving a scalar (m) and a vector (𝑽 )
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 Linear momentum is a vector quantity. It has the same direction as
the direction of velocity of a body.
 Units of linear momentum : In SI it is expressed in kg m s-1 or
Newton-second and in C.G.S it is expressed in gm cm s-1 or
dyne second .
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 Dimensional formula of linear momentum is [MLT-1]
 Change in momentum is always calculated as (final momentum–
initial momentum) 𝑷 = 𝑽𝒇 − 𝑽𝒊 . As momentum is a vector, vector
subtraction must be used in finding the change in momentum.
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1) Linear momentum is the product of ?
b) Mass and acceleration
c) Mass and speed
d) Mass and displacement
a) Mass and velocity
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2) What is the dimensional formula for linear
momentum?
a) MLT-2
b) ML-1T-2
c) M1L1T-2
d) MLT-1
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 In the absence of an external force the linear momentum of a
particle or a body remains constant
If 𝑭 = 𝟎 then 𝑷 =constant
The linear momenta of individual particles can change but the total
linear momentum of the whole system remains constant.
LAW OF CONSERVATION OF LINEAR MOMENTUM
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Examples of law of conservation of linear momentum
a) Recoil of gun :If a stationary gun fires a bullet horizontally, the total
momentum of the gun and the bullet is zero before
and after firing.
 According to conservation of linear momentum
m1v1 + m2𝐯𝟐 = 𝟎
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Explosion : In an explosion, linear momentum is conserved. If a
stationary shell breaks into two fragments, they will move
in opposite directions, with velocities in the inverse ratio
of their masses.
 If a shell at rest breaks into three fragments, the total momentum
of any two of the fragments, must be equal and opposite to the
momentum of the third fragment
𝐦𝟐
𝐦𝟏
=
−𝐯𝟏
𝐯𝟐
𝒑𝟏 + 𝒑𝟐 = −𝒑𝟑
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Proof of law of conservation of linear momentum
 Let m1, m2 be the masses of two bodies A and B moving in the same
direction along a straight line with velocities u1 and u2 respectively.
 When they collide with each other, they are in contact for a small
interval of time t and their velocities change to v1 and v2
respectively and they get apart.
A B
A
𝑢1
B
𝑢2
B
𝑣2
A
𝑣1
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 Initial momentum of ‘A’ is 𝑷A = m1 𝒖𝟏;
 Initial momentum of ‘B’ is 𝑷𝐁 = m2 𝒖𝟐;
 Final momentum of ‘A’ is 𝑷𝟏
𝑨 = m1 𝒗𝟏;
 Final momentum of ‘B’ is 𝑷𝟏
𝐁 = m2 𝒗𝟐;
 The force applied on ‘A’ by ‘B’ in t
time =change in momentum of ‘A’
 From Newton’s second law 𝑭𝑨𝑩 =
𝑷𝟏
𝑨
−𝑷𝑨
𝐭
or 𝑭𝑨𝑩 𝐭 = 𝑷𝟏
𝑨 − 𝑷𝑨
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 The force applied on ‘B’ by A in t time = change in momentum
of B
 𝑭𝑩𝑨=
𝑷𝟏
𝑩
−𝑷𝑩
𝐭
or 𝑭𝑩𝑨 𝐭 =𝑷𝟏
𝑩 − 𝑷𝑩
 From Newton’s third law 𝑭𝑨𝑩= - 𝑭𝑩𝑨
𝑷𝟏
𝑨
_ 𝑷𝑨=- 𝑷𝟏
𝑩 − 𝑷𝑩
𝑷𝟏
𝑨+ 𝑷𝟏
𝑩 = 𝑷𝑨+ 𝑷𝑩
or
 The total final momentum of the isolated system is equal to the
total initial momentum.
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 The above equation can also be expressed as
𝐦𝟏𝐮𝟏 + 𝐦𝟐𝐮𝟐 = 𝐦𝟐𝐯𝟐+
𝐦𝟏𝐯𝟏
 The law is applicable in elastic and inelastic collisions.
 In elastic collisions law of conservation of K.E is also applicable.
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1) A ball falls towards the earth, which of the following is
correct?
a) If the system contains ball, the
momentum is conserved
b) If the system contains earth, the
momentum is conserved
c) If the system contains the ball and the earth, the momentum
is conserved
d) If the system contains the ball and the earth and the sun, the
momentum is conserved
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Rocket Propulsion :
a) Rocket propulsion is based on the law of conservation of linear
momentum or Newton’s third law of motion.
b) Rocket is a system in which mass varies with time.
c) Let ‘u0’ be the initial velocity of the rocket of initial mass
‘M0’with its fuel at any time.
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i) Rate of mass of gas ejected
ii) Mass at any time t,
iii) Acceleration of the rocket
r=
𝒅𝑴
𝒅𝒕
M = M0 - rt
a =
𝒖
𝑴
𝒅𝑴
𝒅𝒕
M
𝐌𝟎
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iv) u0 is the initial velocity of the rocket and u is the velocity of ejected
gas then the velocity of the rocket at any time
M0  mass of the rocket at any instant of time ‘t’
d) Thrust on the rocket
V = u0+uloge
𝑴𝟎
𝑴
F = Ma = M
𝒅𝒗
𝒅𝒕
= u
𝒅𝑴
𝒅𝒕 = ru
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a =
𝑭
𝑴
=
𝒖
𝑴
𝒅𝑴
𝒅𝒕
=
𝒓𝒖
𝑴
Acceleration
Velocity
v= uloge
𝑴𝟎
𝑴
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f) If the effect of gravity is considered; then
e) Minimum rate of consumption of fuel.

𝒅𝑴
𝒅𝒕 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 =
𝑴𝒈
𝒖
velocity 𝐯 = 𝐮𝐥𝐨𝐠𝒆(
𝑴𝟎
𝑴
)- gt
Acceleration 𝐚 =
𝐮
𝐌
𝐝𝐌
𝐝𝐭
- g
Force or thrust F = Ma = u
𝐝𝐌
𝐝𝐭
- Mg
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1) The propulsion of a rocket is based on the principle
of law conservation of………….
a) linear momentum
b) velocity
c) angular momentum
d) mass
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2) A 5000 kg rocket is set for vertical firing. The exhaust
speed is 800 ms-1. To give an upward acceleration of
20ms-2, the amount of gas ejected per second to
supply the needed thrust is (g=10ms-2)
a) 127.5 kg s-1
b) 137.5 kg s-1
c) 187.5 kg s-1
d) 185.5 kg s-1
Given m = 5000kg ,
u = 800 m/sec,
a = 20 m/sec2
g = 10 m/sec2
Since a =
𝒖
𝒎
𝒅𝒎
𝒅𝒕
- g
20 =
𝟖𝟎𝟎
𝟓𝟎𝟎𝟎
𝒅𝒎
𝒅𝒕
- 10
30 =
𝟖
𝟓𝟎
𝒅𝒎
𝒅𝒕
𝒅𝒎
𝒅𝒕
=
𝟓𝟎×𝟑𝟎
𝟖
= 187.5kg s-1
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FINDING THE TRUE WEIGHT FROM A FAULTY BALANCE:
A) If the pans are unequal but the arms are of equal length.
W= true wt. of the body, a= length of each arm
S, S' are the wts. of the left and right pans
Place the body in the left pan and let a wt. W1 be placed in the right pan
be required to balance it.
Then (W+S)a=(W1+S')a
W+S=W1+S'  (1)
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Now place the body in the right pan and let W2 be the wt. required to the
placed in the left pan to balance the body. Then
(W2+S)a=(W+S')a  W2+S=W+S'  (2)
Subtracting (2) from (1) W - W2 = W1 - W
 W =
𝐖𝟏
+ 𝐖𝟐
𝟐
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B) If the arms are unequal but the beam is horizontal when the pans
are empty.
Let a and b be the lengths of the left and right arms,
as the beam is horizontal Sa=S'b  (1)
for first weighing (as above) (W+S)a=(W1+S')b  (2)
for second weighing (as above)
(W2+S)a=(W+S')b  (3)
Subtracting (1) from (2)
Wa=W1b  (3)
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subtracting (1) from (3)
Wb = W2b  (5)
Multiplying (4) by (5)
W2ab = W1W2ab
W = 𝐖𝟏𝐖𝟐
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 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, the above statement means that, the
particle is either at rest or in uniform motion.
If two forces F1 and F2, act on a particle, the condition for
equilibrium is
F1 = -F2 F1+F2= 0
or
The two forces on the particle must be equal and opposite.
F2
F1 .
particle
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Equilibrium under three concurrent forces F1, F2 and F3 requires that
the vector sum of the three forces is zero.
𝐅𝟏 + 𝐅𝟐+𝐅𝟑 =
0
Equilibrium under three concurrent forces:
If two or more forces act such that their lines of action pass through a
common point, they are called concurrent forces.
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F1
F2
F3
F1
F2
F3
F2
F1
In other words, the resultant of any two forces say F1 and F2, obtained
by the parallelogram law of forces must be equal and opposite to the
third force F3.
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 The three forces are in equilibrium can be represented by the three
sides of a triangle taken in the same order.
 The result can be generalized to any number of forces.
 A particle is in equilibrium under the action of number of forces
F1,F2,…Fn, if they represent the n sides of a closed polygon taken in
the same order.
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F1x + F2x + F3x = 0
F1y + F2y + F3y = 0
F1z + F2z + F3z = 0
Where F1x, F1y and F1z are the components of F1 along x, y and z
directions respectively.
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1) Equilibrium of a particle in mechanics refers to the
situation where the net external force on the particle
is……
a) zero
b) positive
c) negative
d) none of the above
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Thank you…
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LEONARDO DA VINCI
April 15th 1452 – May2nd 1519
LEONARDO DA VINCI was an Italian renaissance polymath,
painter, sculptor, architect, musician, mathematician, engineer,
inventor, anatomist, geologist, cartographer, botanist and writer.
Leonardo’s great reputation in science and invention is posthumous, based
on the translation and publication of his coded notebooks in the late 19th
and early 20th centuries.
In his lifetime, in addition to his famous paintings, he was known
for his engineering of canal locks, cathedrals and engines of war.
His notable works are – The world famous paintings of Monalisa,
The Last Supper, The Vitruvian Man, Lady with an Ermine.
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What is friction?
Friction is a force by virtue of which it opposes the relative
motion, when a body moves or tends to move over the
surface of another body.
Friction is the property which opposes the relative motion
between the bodies in contact and acting tangentially to the
surfaces in contact.
 A body in motion comes to rest due to opposing forces acting
between the body and the surface in contact .
 The component of the contact force parallel to the surfaces in contact
is friction.
Friction may be defined as the
opposing force which comes
into play tangentially between
two surfaces in contact so as to
destroy the relative motion
between them.
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Bicycle:
When you want to stop a bicycle you press the break and it slows
down because of the friction between the breaks and the wheels.
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Different solid objects
experience different
amounts of friction.
An eraser on a glass top table will experience more friction than a
coin, which will experience more friction than an ice cube on the
same table.
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Why do you slip and fall when the floor
has just been mopped?
This is because liquid
creates a thin layer
between the floor and your
shoes and it reduces the
friction a lot.
We observe many accidents
during monsoon because of
less friction between the
road & the tyres.
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1) We observe many accidents during rainy season
because of…
a) more friction
b) less friction
c) no friction
d) none of these
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2) Friction is always….
a) opposite to the motion of the body
b) perpendicular to the surface of contact
c) tangential to the surface of contact
d) inclined to the surface of contact
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Causes of friction
 The surface of any body appears as a combination of ups (elevation)
and downs (depressions). On placing one body over another body,
few of the elevations on the first body set into few of the depressions
on the second body.
 It looks as the two surfaces interlock with each other.
 Due to this interlocking motion of one body over another body is
restricted.
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On applying sufficient external
force the depressions and
elevations are removed so that
the body slides over the surface
of another body.
Hence, sliding or pulling
force encounters frictional
force.
The greater the
pressure between
the surfaces the
greater will be the
interlocking.
Friction is generally
less, if the surfaces are
smooth.
But if the smoothness is
increased beyond a
limit, relative motion
again becomes difficult.
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When the irregularities on the surfaces are completely removed, the
molecules on either surface will come close and the adhesive or cohesive
forces between them oppose the relative motion and hence the friction
may increase.
If the surfaces are identical in nature (glass over glass) and are very
smooth, cohesive forces increase and that increases the friction.
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The work done in over coming friction by the applied force may result
in heat and cause a sort of thermal welding between the surfaces.
Hence, it is difficult to predict friction exactly.
It depends essentially on the nature of the surfaces in contact and
normal pressure between the surfaces.
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1)
a) less
b) more
c) does not exist
d) none of these
Friction is generally if the surfaces
are smooth.
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2) Frictional force between two surfaces in contact is
due to…..
a) adhesive or cohesive forces between the
molecules
b) only cohesive forces between the molecules
c) repulsive forces between the molecules
d) both (b) and (c)
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Advantages of Friction:
The surfaces in contact will slip on each
other without opposition in the absence
of friction.
Friction is very much
necessary to produce a
controlled motion.
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 Friction between the floor and the feet enables us to walk safely.
 Nails and screws can be driven into the walls due to friction.
 Friction supports the fingers to hold the objects.
 Moving wheels can be stopped due to friction.
Invention of fire was a result
of friction between two stones.
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Friction is greater on rubber surfaces. The
grooves in the rubber tire are made to reduce
the smoothness of the tire and increase
friction which helps avoiding skidding of the
vehicles.
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 Our planet the earth whirls into outer space, if friction was present
in outer space. Friction is absent in outer space. Due to this planets
continue whirling in their space.
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1)
a) friction helps to walk safely
b) friction supports the fingers to hold
the object
c) moving wheels can be stopped due to
friction
d) wear and tear of machines increases
due to friction
Which of the following is not an advantage of
friction?
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Disadvantages of Friction:
 Wear and tear between the machinery parts increases and reduces
their life due to friction.
Friction is disadvantageous. So it produces many undesirable
effects such as;
 Some energy gets converted into heat which goes as waste due to
friction.
 Large amount of power loss in machines & engines, hence brings
down their efficiency.
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1)
a) it opposes motion
b) it increases motion
c) it decreases motion
d) none of these
Friction is disadvantageous because….
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Methods to reduce friction:-
 To reduce friction between the surfaces in contact a thin layer of oil
or a fluid is used. In general, the lubricants such as compressed air,
organic oil are used.
Using lubricants
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Polishing:
 By polishing the surfaces, irregularities on the surfaces are removed,
so friction decreases. Friction increases when bodies are heavily
polished. This is due to increase in adhesive or cohesive force.
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 The freewheels of vehicles are provided with ball bearings to
reduce friction as they convert sliding friction into rolling
friction.
Using Ball Bearings:
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By Streamlining:
 Automobiles and aeroplanes have special design i.e. they are
stream lined to reduce the friction due to air.
 When they are in motion, friction is reduced due to air .
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1)
a) using lubricants
b) polishing the surface
c) using ball bearings
d) all the above
are the methods of reducing friction.
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2)
a) using ball bearings, sliding friction changes to
rolling friction
b) lubricants decrease friction since inter
molecular forces are weak in liquids
c) over polishing increases friction since
molecules come closer
d) none of these
Which of the following is wrong?
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Thank you…
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Static Friction
No actual movement of a body is observed in static friction.
Force of static friction = Applied force
Types of friction
Static
Friction
Sliding
Friction
Rolling
Friction
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Ex:
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Examples for
types of friction
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1)
a) equal to
b) greater than
c) less than
d) none of these
Sliding friction is static friction.
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2)
a) sliding friction
b) static friction
c) rolling friction
d) slanting friction
is not a type of friction.
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Static Friction
 Frictional forces act between two bodies which are in contact but
are not sliding with respect to each other. The friction in such
cases is called static friction.
Example :-
Suppose a person is trying to push a heavy almirah on the floor, he has
to push strongly to overcome the static friction between almirah and
the floor.
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 It is the resistive force that acts, when there is no relative motion
between the surfaces of the bodies in contact under the action of
external force.
 It is a self-adjusting force. It is represented with fs.
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 Consider a block B which is placed on a horizontal table, which is
connected to a small pan by means of a string with the help of a
frictionless pulley as shown in the figure.
 Initially, when there is no weight in the pan, the block does not slide
because the frictional force between the block and table is balanced
by the applied force.
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 When the weight in the pan is increased, the body may still be
static.
 The body does not move because the resultant force on the body is
zero.
 The frictional force is equal in magnitude and opposite in direction
to the applied force ‘P’ and is tangential between the two surfaces.
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 The maximum value of static friction is called limiting friction
The resistance encountered by a body
in static condition while tending to
move under the action of an external
force is called static friction.
Static friction is equal in
magnitude of applied
force but it opposes the
applied force.
Fs(max) = fL (limiting friction)
The coefficient of static friction is usually less than 1
s ⩽ 1
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1)
a) dynamic friction
b) limiting friction
c) rolling friction
d) none of these
The maximum value of static friction is called as …
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2)
a) always negative
b) always greater than one
c) is always less than one
d) usually less than one
The coefficient of static friction is…..
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 At a particular value of weight in the pan, the applied force
overcomes the limiting frictional force and the body is set into
motion. At this condition the body encounters another friction
called Kinetic or Dynamic Friction .
DYNAMIC OR KINETIC FRICTION
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 The resistance encountered by a sliding body on a surface is known
as kinetic friction or dynamic friction (fk) or sliding friction.
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1)
a) static friction
b) rolling friction
c) kinetic friction
d) none of these
Dynamic friction is also called as…….
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2)
a) static friction
b) rolling friction
c) sliding friction
d) none of these
The resistance encountered by a sliding body on a
surface is called as ….
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Difference between Static & Kinetic Friction
 A graph is plotted on taking applied force on x-axis and frictional
force on y-axis. Consider a block on the surface of another body
which is under static friction.
Applied
force (W)
f
(friction) Kinetic
Static
fl
W1
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 When the applied force increases, the frictional force increases and
reaches a maximum value which is represented by a straight line in
the graph.
 In this case the applied force is equal to the magnitude of fl in the
opposite direction.
 Hence, the resultant force on the block is equal to zero and body is
in equilibrium. When applied force crosses W1, body starts sliding
on the surface.
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 This kinetic friction is slightly less than the static friction.
 But the force required to keep the body in motion with uniform
velocity is little less than fl . The friction at this condition is called
kinetic friction.
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Static Friction Kinetic Friction
1) It is equal to the applied
force in magnitude
1) It is equal to the applied force in
magnitude only when it is sliding
with constant velocity.
2) It is opposite to the applied
force in direction.
2) It is always opposite to the
direction of velocity of sliding
Sliding friction < Static friction
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1)
a) always opposite to the direction of velocity of sliding
b) it is opposite to the applied force in direction
c) it is equal to the applied force in magnitude
d) none of these
Kinetic friction is ……
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2)
a) static friction
b) rolling friction
c) dynamic friction
d) none of these
Sliding friction is always less than….
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Even a child can
pull a suitcase, if
the suitcase has
wheels.
Rolling Friction
 The force of friction comes into play, when a body rolls over the
surface of another body.
 It is always easier to roll than to slide a body over another body.
Rollers reduce
friction
It is convenient
to pull a luggage
on rollers than to
carry it.
The force of
friction between
the rolling body
and the surface is
called rolling
friction.
LAWS OF MOTION
Examples
Sliders are replaced in
most machines by rollers
using these ball bearings.
LAWS OF MOTION
 When a spherical body rolls on the horizontal surface then the
speed of the body goes on decreasing and finally it stops.
 The opposing force acting between the surface of the spherical body
and the horizontal surface is named as rolling friction.
LAWS OF MOTION
 The necessary condition to roll the center of mass of a roller at
constant speed on a rough surface is that there should be a
component of the applied force parallel to the surface.
 This force component is numerically equal to the rolling friction
(fR ). This rolling friction is slightly less than the kinetic friction
and which is less than the static friction.
fs>fk>fr
LAWS OF MOTION
1)
a) fs > fk > fr
b) fs < fk < fr
c) fs > fk < fr
d) fs < fk = fr
The order of friction is always as…
LAWS OF MOTION
2)
a) equal to static friction
b) greater than static friction
c) less than static friction
d) none of these
Rolling friction is
LAWS OF MOTION
3)
a) increases
b) add
c) reduce
d) none of these
Rollers friction
LAWS OF MOTION
LAWS OF MOTION
Normal Reaction (N):
 The forces, which are exerted only when two bodies are in contact,
are called contact forces.
 It depends only on the nature of contact but not on the area of
contact.
 The component of the contact force normal to the surfaces in
contact is called normal reaction.
 These forces satisfy Newton’s third law of motion.
R
W
LAWS OF MOTION
 In general all the surfaces are rough at microscopic level, the
actual points of contact are difficult to be ascertained.
 Hence, it is difficult to know the magnitudes and directions
of all these contact forces.
 The resultant of these forces can be estimated.
LAWS OF MOTION
 Let us consider a block of mass ‘m’
resting on a surface as shown in the fig.
 The weight of the block ‘mg’ acts
vertically downwards on it.
 Due to the contacts at various points on the surface a resultant
contact force acts on the body.
 Let the vertical component of this resultant contact force be ‘N’.
N
O
mg
LAWS OF MOTION
 As the body has no vertical motion, this component of the
Normal reaction is the resultant contact force acting on a body placed
on a rigid surface perpendicular to the plane of contact.
N = mg.
contact force (N) must be balanced by the weight of the body
Since it is perpendicular to the plane of contact, it is known as normal
reaction.
LAWS OF MOTION
 Normal reaction depends only on the nature of the contact.
 It does not depend on the area of contact. Even in the event of the
motion of the body on the surface this force remains unaltered.
LAWS OF MOTION
 The weight (mg) of the body acts vertically
downwards and a component mgcos
perpendicular to the plane is exerting contact
pressure.
 Hence, the normal reaction here is given by N = mg cos.
 In the case of the body resting on a plane
surface inclined at an angle  with the
horizontal as shown in fig.
LAWS OF MOTION
1)
a) frictional force
b) normal force
c) force causes motion on the body
d) surface of the table
A body lies on the table. Its weight is balanced by the…
LAWS OF MOTION
2)
a) parallel
b) perpendicular
c) equal
d) none of these
Normal reaction force is to the tangent
drawn to the surface at the point of contact.
LAWS OF MOTION
Thank you…
LAWS OF FRICTION
LAWS OF MOTION
LAWS OF FRICTION
LAWS OF FRICTION
Laws of friction
 Leonardo Da Vinci conducted a number of experiments and studied
the motion of bodies on horizontal planes and inclined planes.
 He proposed the rules governing the motion under friction as laws
of friction.
LAWS OF FRICTION
1. The frictional force between the body and the surface is
independent of the area of contact
2. Frictional force is always directly proportional to the normal
reaction
Both these laws are applicable to Static as well as Kinetic friction.
LAWS OF FRICTION
If ‘N’ is the normal reaction, then the frictional force is given by
f  N or f = N
 = Co-efficient of friction
 It is defined as the ratio of frictional force and normal reaction.
 =
𝐟
𝐍
Co-efficient of friction
LAWS OF FRICTION
 It has no units and dimensions. It can be determined
experimentally.
 It depends upon molecular nature of the surfaces in contact and
their smoothness.
 Dust and impurities on the surface drastically affect the value of
frictional force and hence the co-efficient of friction.
LAWS OF FRICTION
1)
a) nature of surfaces in contact
b) normal reaction between the surfaces
c) area of surfaces in contact
d) all the above
The friction between two surfaces in contact is
independent of ….
LAWS OF FRICTION
2)
a) nature of contact
b) area of contact
c) both (a) & (b)
d) none of these
Normal reaction depends only on the …..
LAWS OF FRICTION
3)
a) inversely proportional
b) directly proportional
c) equal
d) none of these
Frictional force is always to the normal reaction
LAWS OF FRICTION
4)
a) has only units
b) has only dimensions
c) has units and dimensions
d) does not have units and dimensions
The co-efficient of friction .
LAWS OF FRICTION
LAWS OF FRICTION
Laws of static friction
Where s = co-efficient of static frictional force
fL = limiting frictional force
N = normal reaction
 The limiting friction is directly proportional to the magnitude of
normal reaction between the two surfaces.
s =
𝐟𝐋
𝐍
 Limiting friction is independent of the area of contact, so long as the
normal reaction remains the same.
 Limiting friction depends on the materials of the surfaces in contact
and their state of polish.
𝐟𝐋 = s𝐍
LAWS OF FRICTION
 The static frictional force fs ⩽sN when the bodies do not slide
over each other.
LAWS OF FRICTION
1)
a) The ratio of normal force to the
limiting friction
b) The ratio of tangential force to the
limiting friction
c) The ratio of limiting friction to
the normal force
d) none of these
The coefficient of static friction is…..
LAWS OF FRICTION
LAWS OF FRICTION
LAWS OF KINETIC FRICTION:
 Kinetic frictional force is directly
proportional to normal reaction.
 It is independent of area of contact as long as magnitude of normal
reaction remains the same.
𝒇𝒌 ∝ N
𝒇𝒌 = 𝝁𝒌 N
 Kinetic friction is almost independent of velocity, provided the
velocity is not too large or not too small.
LAWS OF FRICTION
Laws of rolling friction
 The rolling friction is directly proportional to the normal reaction.
 If the area of contact is small, the rolling friction is less.
 If the radius of the rolling body is more, then it possesses less
rolling friction.
 The co-efficient of rolling friction is the ratio of rolling frictional
force to the normal reaction.
𝒇𝒓 ∝ 𝑵
 It depends on the area of contact.
LAWS OF FRICTION
Note :-
 For similar surfaces the co-efficient of rolling friction (r) is
smaller than the co-efficient of kinetic friction (k) .
 Let fr be the rolling frictional force and N be the normal reaction
then the co-efficient of rolling friction .
r =
𝐟𝐫
𝐍
=
𝐫𝐨𝐥𝐥𝐢𝐧𝐠 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧
𝐧𝐨𝐫𝐦𝐚𝐥 𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧
 r < k
LAWS OF FRICTION
1)
a) more
b) less
c) equal
d) greater than or equal to
When a body is moving with velocity then
we can apply laws of kinetic friction .
LAWS OF FRICTION
2)
a) normal reaction
b) tangential force
c) parallel force
d) none of these
The co-efficient of kinetic friction is the ratio of kinetic
frictional force to the of a body.
LAWS OF FRICTION
3)
a) inversely proportional
b) directly proportional
c) equal
d) greater
Rolling friction is to the normal force
LAWS OF FRICTION
a) smaller
b) greater
c) equal
d) greater than equal to
4) For same surfaces the co-efficient of rolling friction r is
much _____ than co-efficient of kinetic friction k .
LAWS OF FRICTION
Thank you…
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
Angle of friction()
 The angle made by the resultant of the normal
reaction and the limiting friction with normal
reaction is called angle of friction () .
 Suppose a block of mass ‘m’ is on a rough horizontal surface. The
body is experiencing the limiting frictional force (fL) under the
action of applied force (F).
C A
N
O
Mg
fL
B
R

LAWS OF MOTION
 Let N be the normal reaction acting on the block and R be the
resultant of normal reaction and frictional force.
 In the above figure OA represents N, OB represents fL. Completing
the parallelogram OACB, OC gives the resultant of N and fL.
C A
N
O
Mg
fL
B
R

In OAC
tan =
𝐀𝐂
𝐎𝐀
=
𝐎𝐁
𝐎𝐀
=
𝐟𝐋
𝐍
= 𝐬
tan = 𝐬  = tan-1(𝐬)
LAWS OF MOTION
C A
N
P
O
Mg
fL
B
R

 The greater the angle of friction the greater is the value of
co-efficient of friction.
LAWS OF MOTION
a) angle of repose
b) critical angle
c) angle of friction
d) none of these
1) The greater the ______greater is the value of
co-efficient of friction.
LAWS OF MOTION
2)
a) tan-1k
b) tan-1s
c) tan-1r
d) sin-1s
The angle of friction  = _______.
LAWS OF MOTION
LAWS OF MOTION
I. Motion of a body on a rough horizontal surface
 Consider a rough horizontal surface on
which a block of mass ‘m’ is placed.
 Let a force P which is greater than the limiting frictional force
applied on the body to move it towards right against friction.
 Let a be the acceleration of the body then the net force acting on the
body .
 Here kinetic friction fK will act on the body.
P
a
fK
LAWS OF MOTION
 If ‘N’ is the normal reaction , from the laws of friction
 If the acceleration produced is ‘a’, the resultant force
N
P
a
fK
mg
fk = kN = kmg
LAWS OF MOTION
 Using eq-1, the distance (S) moved in a given time ‘t’ and the velocity
acquired can be calculated with the help of kinematic equations of
motion.
F= P - fk
ma = P - fk
or =
𝐏
𝐦
− 𝐤 g
a (1)
LAWS OF MOTION
LAWS OF MOTION
Motion of a body on rough inclined plane
 The plane which makes an angle  with the horizontal is called
inclined plane.
 Consider a body of mass ‘m’ placed on a rough inclined plane
which makes an angle  with the horizontal.
 When an applied force acts on the body, frictional force comes
into play, if it exceeds the limiting friction, then the body is ready
to slide.
LAWS OF MOTION
 The forces acting on the body are
a) weight(mg) acting vertically downwards. It is resolved into two
components.
N f
mg cos
mg


The component mgcos normal to the inclined plane and the
component mgsin parallel to the inclined plane.
LAWS OF MOTION
b) Normal reaction N perpendicular to the plane.
c) Frictional force (f) in the upward direction parallel to the inclined
plane.
 If the body is at rest on the inclined plane, mgsin balances the
static friction, the normal reaction of the plane N on the block is
balanced by mgcos.
f = mg sin, N = mg cos 
LAWS OF MOTION
LAWS OF MOTION
 The angle of repose is defined as the
angle of inclined plane at which a body
placed on it just begins to slide.
Angle of repose()
N
P
fL
mg cos
mg


Equilibrium on an inclined plane
 Angle of repose doesn’t depend on the mass of the object as well as its
area of contact .
 But it depends on relative smoothness of the surfaces in contact. For
a given pair of surfaces the angle of repose remains constant.
LAWS OF MOTION
 Let us consider a block of mass m, at rest on an inclined plane,
which makes an angle in limiting equilibrium  with the horizontal
as shown in figure.
 The forces acting on the body are
1. Weight of the body ‘mg’ acting vertically
downwards,
2. Normal reaction ‘N’ acting perpendicular to the inclined plane.
3. Force of limiting friction fL acting tangentially to the inclined
plane.
N
P
fL
mg cos
mg


LAWS OF MOTION
 The weight of the body is resolved into two rectangular
components.
i. mg cos perpendicular to the inclined plane,
ii. mg sin parallel to the inclined plane downwards,
where ‘’ is the angle of repose.
LAWS OF MOTION
Thank you…
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
I. Body just ready to slide:
 Since the body is in equilibrium, the net force
on it should be equal to zero.
 Therefore mgsin becomes equal to the
limiting friction (fL).
fL = mg sin and N = mg cos 
From the laws of friction
fL = sN or mgsin = s mgcos
N
mg sinα
mg cosα
LAWS OF MOTION
 Thus, the co-efficient of static friction is numerically equal to the
tangent of the angle of repose.
 In case of inclined plane, the angle of repose () is equal to the
angle of friction (∅).
s =
𝐦𝐠𝐬𝐢𝐧
𝐦𝐠𝐜𝐨𝐬
or s = tan
LAWS OF MOTION
 When the angle of friction is more than the angle of repose, then
the body slides down with uniform acceleration.
II. Body sliding down:
 Consider a body of mass ‘m’ which is sliding down on a rough
inclined plane at an angle  which is greater than the angle of
repose  .
N
fK
mg cos
mg


Body sliding down
on an inclined plane.
LAWS OF MOTION
 The forces acting on the body are
a) Weight acts vertically downwards, one of its component mgcos  acts
normal to the inclined plane and the component mgsin  acts parallel
to the inclined plane.
b) Normal reaction N which is perpendicular to the plane
c) Kinetic frictional force fK acts up the inclined plane.
LAWS OF MOTION
As there is no motion of the body in the perpendicular direction
The body slides down with an acceleration due to net force is
N = mg cos
F = mg sin - fk
 ma = mg sin - kN
 ma = mg sin - kN mgcos
 ma = mg (sin - k cos)
 a = g (sin - k cos)
LAWS OF MOTION
LAWS OF MOTION
Velocity and acceleration of a sliding body:
Acceleration of the body :
 Consider a body of mass ‘m’
placed on a rough inclined plane of
co-efficient of kinetic friction k.
 The body slides down the plane when angle of inclination () is
greater than angle of repose (). It gains an acceleration of
N
fK
mg cos
mg


B C
A
a = g (sin - k cos)
a) When a body slides down
LAWS OF MOTION
Velocity at the bottom of the inclined plane:
 Initially the body starts from rest and attains the velocity v on
reaching the bottom of the inclined plane.
v2 – u2 = 2as
u = 0, s = l , a = g (sin - kcos )l.
v = 𝟐𝐠𝒍(𝐬𝐢𝐧 − 𝐤𝐜𝐨𝐬)
LAWS OF MOTION
 Let ‘t’ be the time taken by the body to
travel a distance ‘l’ ,using
 Time to reach the bottom of the inclined plane
s = ut +
𝟏
𝟐
𝒂𝒕𝟐. 𝑯𝒆𝒓𝒆, 𝒔 =l, u = 0
 l = 0(t) +
𝟏
𝟐
𝒈(𝒔𝒊𝒏 - k cos )t2
 t =
𝟐𝒍
𝐠 𝐬𝐢𝐧 − 𝐤
𝐜𝐨𝐬
LAWS OF MOTION
b) Body sliding up the plane:
If a body is projected with an initial velocity ‘u’ to slide up the plane,
the kinetic frictional force acts down the plane and the body suffers
retardation due to a net force.
Fnet = mgsin + fk
LAWS OF MOTION
N
fK
mg cos α
mg
α
α
B C
A
(u)
 ma = mgsin + kN
 ma = mgsin + kmgcos 
ma = mg (sin + kcos )
 a = g (sin + kcos )
The retardation of the body :
LAWS OF MOTION
Let ‘t’ be the time taken by the body to reach the highest point
v = u + at
v = 0 , u = 𝟐𝒈𝒍(𝒔𝒊𝒏 + 𝒌𝒄𝒐𝒔
v = u + at  0 = u - at
 t =
𝐮
𝐚
=
𝟐𝐠𝒍(𝐬𝐢𝐧+ 𝐊
𝐜𝐨𝐬)
𝐠(𝐬𝐢𝐧+𝐊
𝐜𝐨𝐬)
Here, time of ascent is not equal to time of descent
THE TIME TAKEN BY THE BODY TO REACH THE HIGHEST POINT :
 t =
𝟐𝒍
𝐠(𝐬𝐢𝐧+𝐊𝐜𝐨𝐬)
LAWS OF MOTION
LAWS OF MOTION
Motion of the body on smooth inclined plane
 Consider a body of mass m placed on smooth inclined plane which
makes an angle  as shown in the figure
1. Body sliding down

O
LAWS OF MOTION
 Here the weight ‘mg’ of the body is resolved into two components (i)
mgcos, the vertical component, (ii) mgsin the horizontal
component
 The normal reaction, is balanced by mgcos but mgsin is not
balanced by the applied force, which makes the body slide down
with an acceleration ‘a’ .

O
LAWS OF MOTION
 We have
 But F = mgsin
 Initially the body starts from rest and attains velocity v when it
reaches the bottom of the inclined plane of length l .
F = ma  a =
𝐅
𝐦
 a =
𝐦𝐠𝐬𝐢𝐧
𝐦
ACCELERATION OF THE BODY :
 a =𝒈𝒔𝒊𝒏
VELOCITY OF THE BODY :
LAWS OF MOTION
 We have, v2 – u2 = 2as
 v2 – 02 = 2gsin.l
 Let ‘t’ be the time taken by the body to travel a distance ‘l’
 We have v = u + at
Here u = 0, v = 𝟐𝐠𝐬𝐢𝐧. 𝒍 𝒂 = 𝒈𝒔𝒊𝒏
 𝟐𝒈𝒍𝒔𝒊𝒏 = 𝒈𝒔𝒊𝒏 . t
v = 𝟐𝐠𝐬𝐢𝐧. 𝒍
TIME TO REACH THE BOTTOM :
LAWS OF MOTION
 t =
𝟐𝐠𝐬𝐢𝐧
𝐠𝐬𝐢𝐧
 t =
𝟐𝐥
𝐠𝐬𝐢𝐧
(Time of descent) ….…(1)
LAWS OF MOTION
LAWS OF MOTION
2. BODY SLIDING UPA SMOOTH PLANE -
 Acceleration :As the body is thrown up the plane of length 𝒍 with an
initial velocity u such that it’s final velocity v becomes zero. When
the body moves up the plane it’s acceleration is opposite to the
motion
 Velocity : The initial velocity ‘u’ of the body can be calculated as
 a = -g sin
v2 – u2 = 2as
v = 0 , a = -gsin, s = 𝒍
 u = 𝟐𝒈𝒍 𝒔𝒊𝒏
LAWS OF MOTION
 Time of ascent: Let ‘t’ be the time taken by the body to reach highest
point
v = u + at
v = 0 , u = 𝟐𝐠𝐥𝐬𝐢𝐧, a = -g sin
 0 = 𝟐𝐠𝐥𝐬𝐢𝐧 - gsin  . 𝐭
Hence, time of ascent is equal to time of descent.
 t =
𝟐𝐠𝐥𝐬𝐢𝐧
𝐠𝐬𝐢𝐧
 t =
𝟐𝐥
𝐠𝐬𝐢𝐧
(Time of ascent) ….…(2)
LAWS OF MOTION
LAWS OF MOTION
Pulling the lawn roller
 Consider a lawn roller of mass m on a rough horizontal surface is
pulled by an inclined force F at an angle  as shown in the figure.
N
F
F cos

mg
F sin
Pulling a lawn roller
The force is resolved into two
components, horizontal component F
cos, vertical component F sin .
f
LAWS OF MOTION
In the vertical direction
N + F sin = mg
 N = mg - F sin

𝐟
𝐑
= mg - F sin
 f = 𝐑 (mg - F sin)
LAWS OF MOTION
Pushing the Lawn Roller :-
N
F
F cos

F sin
f
F
Lawn roller
In the vertical direction N = mg + F sin
mg
LAWS OF MOTION
From equations 1 and 2
The frictional force while pulling is less than while pushing.
Hence, it is easier to pull a lawn roller than push it.

𝒇
𝑹
= mg + F sin
 f = 𝑹 (mg + F sin)……(2)
LAWS OF MOTION
1) Pulling of lawn roller is easier than pushing because…
a) Weight of the roller gets decreased when it is
pulled
b) Vertical component of the pulling force acts downwards and
thereby decreases the frictional force during pulling
c) The vertical component of the pulling force acts upwards and
thereby decreases the frictional force during pulling
d) The frictional force acts in the direction of motion of the roller and
thereby making the pulling easier
LAWS OF MOTION
2)
a) always negative
b) always greater than one
c) is always less than one
d) usually less than one
The coefficient of static friction is…..
LAWS OF MOTION
Thank you…
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
CIRCULAR MOTION
 The acceleration of a body moving in a circle of radius R with
uniform speed v is
𝒗𝟐
𝑹
directed towards the centre. According to
the second law, the force 𝒇𝒄 providing this acceleration is : 𝒇𝒄 =
𝒎𝒗𝟐
𝑹
→ (1)
 Where m is the mass of the body. This force directed towards the
centre is called the centripetal force
LAWS OF MOTION
 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 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 a friction
 The circular motion of a car on a flat and banked road given an
interesting application of the laws of motion.
LAWS OF MOTION
 Three forces act on the car
Motion of a car on level road:
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
 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
LAWS OF MOTION
 Note that it is the static friction that provides the centripetal
acceleration. Static friction opposes the impeding motion of the car
moving away from the circle. Using equation (1), we get the result
𝒇 ≤ 𝝁𝒔𝐍
𝒎𝒗𝟐
𝑹
𝒗𝟐 ≥
𝝁𝒔𝑹𝑵
𝒎
= 𝝁𝒔𝑹𝒈 ∴ 𝑵 = 𝒎𝒈
 Which is independent of the mass of the car?
 This shows that for a given value of 𝝁𝒔 and R, there is a maximum
speed of circular motion of the car possible, namely
𝒗𝒎𝒂𝒙 = 𝝁𝒔𝑹𝒈 → (𝟐)
LAWS OF MOTION
 Motion of a car on a banked road
N cosθ = mg + f sinθ → (3)
 We can reduce the contribution of friction to the circular motion of
the car if the road is banked. Since there is no acceleration along the
vertical direction, the net force along this direction must be zero.
Hence,
 The centripetal force is provided by the horizontal components of N
and f.
𝐍𝐬𝐢𝐧𝜽 + 𝒇𝒄𝒐𝒔𝜽 =
𝒎𝒗𝟐
𝑹
𝐁𝐮𝐭 𝐟 ≤ 𝝁𝒔𝐍
LAWS OF MOTION
 Thus to obtain 𝒗𝒎𝒂𝒙 ,we put, f = 𝝁𝒔N
 Then (3) & (4) become Ncosθ = mg + 𝝁𝒔Nsinθ → (5)
 Nsinθ + 𝝁𝒔Ncosθ =
𝒎𝑽𝟐
𝑹
→(6) From equation (5) we obtain
𝑵 =
𝒎𝒈
𝒄𝒐𝒔𝜽 − 𝝁𝒔𝒔𝒊𝒏𝜽
 𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐍 𝐢𝐧 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 (), 𝐰𝐞 𝐠𝐞𝐭
LAWS OF MOTION
𝒎𝒈(𝒔𝒊𝒏𝜽 + 𝝁𝒔𝒄𝒐𝒔𝜽)
𝒄𝒐𝒔𝜽 − 𝝁𝒔𝒔𝒊𝒏𝜽
=
𝒎𝒗𝒎𝒂𝒙
𝟐
𝑹
𝒐𝒓
 Comparing this with Eq (2) we see that maximum possible speed of
a car on a banked road is greater than that on a flat road.
 For 𝝁𝒔 = 0 in Eq (7) 𝒗𝟎 = 𝑹𝒈𝒕𝒂𝒏𝜽 𝟏/𝟐 → (8)
𝒗𝒎𝒂𝒙 = 𝑹𝒈
𝝁𝒔 + 𝒕𝒂𝒏𝜽
𝟏 − 𝝁𝒔𝒕𝒂𝒏𝜽
𝟏/𝟐
→ (𝟕)
LAWS OF MOTION
At this speed, frictional force is not needed at all , when the necessary
centripetal force is provided. 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 < 𝒗𝟎, frictional force will be
up the slope and that of a car can be parked only if tanθ ≤ 𝝁𝒔
LAWS OF MOTION
Thank you…
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
LAWS OF MOTION
1. A force produces an acceleration of 1 ms² on an object of mass 1 kg.
This force is defined as …..
1) 1 dyne 2) 1 newton
3) 1 tesla 4) 1 poundal
LAWS OF MOTION
2. A mass m falls freely from rest. The linear momentum, after it has
fallen through a height h is (g = acceleration due to gravity) ......
1) 𝐦𝐠𝐡 2) 𝐦 𝟐𝐠𝐡
3) 𝐦 𝐠𝐡 4) zero
LAWS OF MOTION
3. Which of the following statements is incorrect for action and reaction
forces ?
1) These act on two different bodies
2) These are equal in magnitude but opposite in direction
3) These act on a single body
4) Action and reaction never balance each other
LAWS OF MOTION
4. A body is said to be in equilibrium if all the force acting on it......
1) Are in the same direction
2) Are equal in magnitude
3) Have zero resultant
4) Can be arranged in
pair
LAWS OF MOTION
5. The quantity of motion of a body is best represented by......
1) Its mass 2) Its speed 3) Its velocity 4) Its linear momentum
LAWS OF MOTION
6. The behaviour of a body under zero resultant force is given by......
1) Newton’s third law of motion
2) Newton’s second law of motion
3) Newton’s first law of motion
4) Newton’s law of gravitation
LAWS OF MOTION
7. Which law of Newton is called the law of equilibrium?
1) Newton’s first law of motion
2) Newton’s second law of motion
3) Newton’s third law of motion
4) Newton’s law of gravitation
LAWS OF MOTION
8. You are thrown outer side when your car suddenly takes a turn.
Which law of Newton is involved in this?
1) Third Law 2) Second Law
3) First Law 4) Law Of Gravitation
LAWS OF MOTION
9. Which of the following is the most significant law of motion given by
Newton?
1) First law of motion
2) Second law of motion
3) Third law of motion
4) Zeroth law of motion
LAWS OF MOTION
10. In translatory motion, Newton's third law can be used to obtain......
1) Newton's first law
2) Law of conservation of momentum
3) law of conservation of mass
4) Law of conservation of kinetic energy
LAWS OF MOTION
11. A frame of reference is said to be inertial if......
1) Newton's laws of motion hold good
2) Newton's laws of motion do not hold good
3) The frame of reference accelerates
4) The frame of reference decelerates
LAWS OF MOTION
12. There are three Newton's laws of motion namely first, second and
third laws. We can derive......
1) Second and third laws from the first law
3) First and second laws from the third law
4) All the laws are independent of each
other
2) Third and first laws from the second law
LAWS OF MOTION
13. A bird is sitting in an air tight cage suspended from a spring balance.
If the bird starts flying, then the reading of the balance will......
1) Becomes zero 3) Decrease 4) Remain unchanged
2) Increase
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LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx
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LAWS OF MOTION class 11th physics theory.pptx
LAWS OF MOTION class 11th physics theory.pptx

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LAWS OF MOTION class 11th physics theory.pptx

  • 3. LAWS OF MOTION  To move a football which is at rest, someone has to kick it. MOTION UNDER FORCE  To throw a ball upwards, one has to give it an upward push.  A breeze causes the branches of a tree to swing clearly, so some external energy is needed to provide force to move a body from rest.  Likewise, an external force is needed to retard or stop motion.
  • 4. LAWS OF MOTION  In these examples, the external agency of force is in contact with the objects. Force with contact MOTION  A magnet can attract an iron piece from a distance. Force  So, it is clear that external agencies can also exert force on a body even from a distance. Iron piece Magnet
  • 5. LAWS OF MOTION  Force is a pull or a push which generates or tends to generate motion in a body at rest, stops or tends to stop a body in motion, or tends to change the shape of the body. PULL WHAT IS A FORCE?
  • 6. LAWS OF MOTION Internal force : What is Internal Force?  If the force applying agent is inside a system, then it is known as internal force.  It does not provide motion to the system.
  • 7. LAWS OF MOTION External push moves the car  If you are sitting in a car and you push it, it doesn't move.  If you come out of the car and apply the same force, then it moves. Ex:
  • 8. LAWS OF MOTION 1) What is a force ? a) It is a pull or a push which tends to move a body. d) All of the above c) Tends to change the shape of the body b) Stops or tends to stop a body in a motion
  • 9. LAWS OF MOTION d) Both (a) and (c) 2) What is an internal force? a) Force applying agent is outside a system. b) Force applying agent is inside a system. c) It provides motion.
  • 11. LAWS OF MOTION  This property of the body is called inertia. What is inertia..?  Let us first learn about “The first law of motion”. It is also called as law of inertia.  To understand Newton's first law we have to learn about inertia at first.  If the net external force is zero, ii) a body in motion continues to move with a uniform velocity. i) a body at rest continues to remain at rest and
  • 12. LAWS OF MOTION Ex:- A ball at rest stays at rest, unless acted upon by a force. Inertia Mass MOTION FORCE  Inertia