Have you ever tried to analyze motion in outer space ?
Lets see a simple example.
11P05
Laws of Motion
Learning Objectives
Newton’s Laws of Motion and Forces
Applications of Newton’s Laws of Motion
Friction
Circular Motion
5
11P05 Laws of Motion
When an object moves along a straight line with uniform acceleration :
Where u = initial velocity
v = final velocity
s = displacement
a = acceleration
t = time
6
11P05.0 Revision
Q. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find the distance travelled by the train for attaining this velocity.
7
Sol. u =0 m/s , v = 72 km/h =20 m/s and t = 5 minutes = 300 seconds
11P05.0 Revision
Where v= speed
⍵ = angular speed
ac= centripetal acceleration
r = radius of circle
9
Q. If a particle is moving in a circle of radius 0.5 𝑚 then find out its magnitude of radial acceleration if linear speed of the particle is 1 𝑚/𝑠.
Sol.
11P05.1
Newton’s Laws of Motion and Forces
11P05.1 Newton’s Laws of Motion and Forces
Learning Objectives
Force
Newton’s First Law of Motion
Inertia
Newton’s Second Law of Motion
Impulse
Newton’s Third Law of Motion
Common Forces in Mechanics
13
11P05.1
CV 1
Force
11P05.1 Force
Force : It is a physical quantity which either causes change in configuration, change in motion or it tends to cause change in motion.
15
11P05.1 Force
16
Force
Change in state of motion
rest to motion
11P05.1 Force
17
Force
Change in state of motion
motion to rest
rest to motion
11P05.1 Force
18
Force
Change in state of motion
motion to rest
Tendency to cause change in motion
rest to motion
11P05.1 Force
19
Force
Change in state of motion
motion to rest
Tendency to cause change in motion
rest to motion
Change in configuration
11P05.1
CV 2
Newton’s First Law of Motion
11P05.1 Newton’s First Law of Motion
A body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity until unless an external force is applied on it.
22
ConcepTest 01
Ready for a challenge
Q. Calculate the net force acting on a car moving with a constant velocity of 30 km/h on a rough road.
Pause the Video
(Time Duration : 1 Minutes)
Sol.
constant velocity
Sol.
Uniform motion
constant velocity
Sol.
Therefore by Newton’s first law of motion, the net force acting on car is zero.
constant velocity
Uniform motion
No change in state of motion
11P05.1
CV 3
Inertia
11P05.1 Inertia
Inertia : The property of opposition to change in state of motion.
31
Inertia of rest
Inertia
11P05.1 Inertia
Inertia : The property of opposition to change in state of motion.
32
Inertia of motion
Inertia of rest
Inertia
11P05.1 Inertia
Inertia : The property of opposition to change in state of motion.
33
Inertia of motion
Inertia of rest
Inertia of direction
Inertia
11P05.1 Inertia
Mass is the measure of inertia.
34
11P05.1 Inertia
Mass is the measure of inertia.
35
Time in both cases is same
ConcepTest 02
Ready for a challenge
11P05.1 Inertia
Q. When a carpet is beaten with a stick, dust comes out of it. Explain.
37
Pause the Video
(Time Duration : 2 Minutes)
Sol.
Both Carpet and dust particles are at rest
Both Carpet and dust particles are at rest
Case of Inertia of rest
Sol.
Sol.
Both Carpet and dust particles are at rest
Case of Inertia of rest
Carpet is beaten and it comes to motion
Sol.
Both Carpet and dust particles are at rest
Case of Inertia of rest
Dust particles try to remain at rest
Carpet is beaten and it comes to motion
ConcepTest 03
Ready for a challenge
11P05.1 Inertia
Q. Why are the passengers thrown backwards when a stationary bus suddenly speeds
up ?
44
Pause the Video
(Time Duration : 2 Minutes)
Sol.
Bus and passengers are at rest
Sol.
Bus and passengers are at rest
Case of Inertia of rest
Sol.
Bus and passengers are at rest
Case of Inertia of rest
Bus suddenly speeds up in left direction
Sol.
Bus and passengers are at rest
Case of Inertia of rest
Lower part of body comes to motion
Bus suddenly speeds up in left direction
Sol.
As a result, passengers are thrown backwards.
Bus and passengers are at rest
Case of Inertia of rest
Lower part of body comes to motion
Bus suddenly speeds up in left direction
Upper part of body tends to remain in rest
11P05.1
CV 4
Newton’s Second Law of Motion
11P05.1 Newton’s Second Law of Motion
53
If a net force is acting on an object
11P05.1 Newton’s Second Law of Motion
54
If a net force is acting on an object
The object will accelerate.
11P05.1 Newton’s Second Law of Motion
55
11P05.1 Newton’s Second Law of Motion
56
If a net force is acting on an object
11P05.1 Newton’s Second Law of Motion
57
The object will accelerate.The acceleration depends on
If a net force is acting on an object
11P05.1 Newton’s Second Law of Motion
58
The object will accelerate.The acceleration depends on
directly proportional to net force
If a net force is acting on an object
11P05.1 Newton’s Second Law of Motion
59
11P05.1 Newton’s Second Law of Motion
60
Inversely proportional to mass
The object will accelerate.The acceleration depends on
directly proportional to net force
If a net force is acting on an object
11P05.1 Newton’s Second Law of Motion
61
11P05.1 Newton’s Second Law of Motion
62
11P05.1 Newton’s Second Law of Motion
63
11P05.1 Newton’s Second Law of Motion
64
11P05.1 Newton’s Second Law of Motion
The rate of change of momentum of a body is directly proportional to the net applied force and takes place in the direction in which the force acts.
65
ConcepTest 04
Ready for a challenge
11P05.1 Newton’s Second Law of Motion
Q. A stone of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the stone,
67
Pause the Video
(Time Duration : 3 Minutes)
Sol. (a) F = ma=mg
Sol. (a) F = ma=mg
⇒F = 0.05 x 10
⇒F = 0.5 N (downwards)
Sol. (a) F = ma=mg
⇒F = 0.05 x 10
⇒F = 0.5 N (downwards)
(b) F = ma
⇒F= 0.05 x 10
⇒F = 0.5 N (downwards)
Sol. (c)
The direction and magnitude of the net force on the stone will not alter even if it is thrown at 45°, because no other acceleration except g is acting on pebble.
ConcepTest 05
Ready for a challenge
Q. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Pause the Video
(Time Duration : 3 Minutes)
Sol. F1= 8 N and F2= 6 N
8 N
6 N
Sol. F1= 8 N and F2= 6 N
resultant force
8 N
6 N
F
𝛉
Sol. F1= 8 N and F2= 6 N
resultant force
By second law of motion
8 N
6 N
F
𝛉
Sol. F1= 8 N and F2= 6 N
resultant force
By second law of motion
8 N
6 N
F
𝛉
Sol. F1= 8 N and F2= 6 N
resultant force
By second law of motion
8 N
6 N
F
𝛉
Sol. F1= 8 N and F2= 6 N
resultant force
By second law of motion
8 N
6 N
F
𝛉
PSV 01
Q. The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Pause the Video
(Time Duration : 3 Minutes)
Sol. Given,
Total mass (m) = 400+65=465 kg
u = 36 km/h = 10 m/s and v = 0
t = 4 s
Sol. Given,
Total mass (m) = 400+65=465 kg
u = 36 km/h = 10 m/s2 and v = 0
t = 4 s
We know,
acceleration
Sol. Given,
Total mass (m) = 400+65=465 kg
u = 36 km/h = 10 m/s2 and v = 0
t = 4 s
We know,
acceleration
Sol. Given,
Total mass (m) = 400+65=465 kg
u = 36 km/h = 10 m/s2 and v = 0
t = 4 s
We know,
acceleration
Thus, retarding force F = ma = 465 x(-2.5)
Sol. Given,
Total mass (m) = 400+65=465 kg
u = 36 km/h = 10 m/s2 and v = 0
t = 4 s
We know,
acceleration
Thus, retarding force F = ma = 465 x(-2.5)
F= -1162.5 N ≈ -1.2 x 103 N
11P05.1
CV 5
Impulse
11P05.1 Impulse
Impulsive force : It is a large force acting for a short time to produce a finite change in momentum.
92
11P05.1 Impulse- Mathematical formula
Impulse = (force) x (time duration)
= change in momentum
93
ConcepTest 06
Ready for a challenge
Q. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 10m/s. If the mass of the ball is 0.12 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Q. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 10m/s. If the mass of the ball is 0.12 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Pause the Video
(Time Duration : 2 Minutes)
Sol. Given,
initial velocity = -10 m/s
final velocity = 10 m/s
Sol. Given,
initial velocity = -10 m/s
final velocity = 10 m/s
Change in momentum =
= 0.12 x 10 - 0.12 x (-10)
= 2.4 N s
Sol. Given,
initial velocity = -10 m/s
final velocity = 10 m/s
Change in momentum =
= 0.12 x 10 - 0.12 x (-10)
= 2.4 N s
Impulse = 2.4 Ns, in the direction from the batsman to the bowler.
PSV 02
Sol. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Sol. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Pause the Video
(Time Duration : 5 Minutes)
Sol. AO = incident path of the ball
Sol. AO = incident path of the ball
OB = path followed by the ball after deflection
∠AOB = angle between the incident and
deflected paths of the ball
Sol. AO = incident path of the ball
OB = path followed by the ball after deflection
∠AOB = angle between the incident and
deflected paths of the ball
∠AOP= ∠BOP = 22.5°=𝚹
Sol. AO = incident path of the ball
OB = path followed by the ball after deflection
∠AOB = angle between the incident and
deflected paths of the ball
∠AOP= ∠BOP = 22.5°=𝚹
initial and final speed of ball
v=54 km/h =15 m/s
mass of body =m= 0.15 kg
Sol. AO = incident path of the ball
OB = path followed by the ball after deflection
∠AOB = angle between the incident and
deflected paths of the ball
∠AOP= ∠BOP = 22.5°=𝚹
initial and final speed of ball
v=54 km/h =15 m/s
mass of body =m= 0.15 kg
Sol. AO = incident path of the ball
OB = path followed by the ball after deflection
∠AOB = angle between the incident and
deflected paths of the ball
∠AOP= ∠BOP = 22.5°=𝚹
initial and final speed of ball
v=54 km/h =15 m/s
mass of body =m= 0.15 kg
Change in momentum in x-direction = -2mu cos𝚹
Change in momentum in x-direction = -2mu cos𝚹
Change in momentum in y-direction = 0
Change in momentum in x-direction = -2mu cos𝚹
Change in momentum in y-direction = 0
Impulse imparted to ball = change in momentum
Change in momentum in x-direction = -2mu cos𝚹
Change in momentum in y-direction = 0
= -2mu cos𝚹
= -2 x 0.15 x 15 x 0.92
= 4.14 kg-m/s
Impulse imparted to ball = change in momentum
11P05.1
CV 6
Newton’s Third Law of Motion
11P05.1 Newton’s Third Law of Motion
For every action, there is always an equal and opposite reaction.
11P05.1 Newton’s Third Law of Motion
For every action,there is always an equal and opposite reaction.
116
Objects interact
11P05.1 Newton’s Third Law of Motion
For every action,there is always an equal and opposite reaction.
117
Objects interact
Mutual forces of same nature arises on both bodies
11P05.1 Newton’s Third Law of Motion
For every action,there is always an equal and opposite reaction.
118
Objects interact
Mutual forces of same nature arises on both bodies
Forces are equal in magnitude
11P05.1 Newton’s Third Law of Motion
For every action,there is always an equal and opposite reaction.
119
Objects interact
Forces act in opposite direction
Mutual forces of same nature arises on both bodies
Forces are equal in magnitude
11P05.1 Newton’s Third Law of Motion
For every action,there is always an equal and opposite reaction.
120
Objects interact
Forces act in opposite direction
Mutual forces of same nature arises on both bodies
Forces are equal in magnitude
Forces don’t cancel each other because they act on different bodies
ConcepTest 07
Ready for a challenge
Q. Why the boat moves back, when the sailor jumps in forward direction ?
Q. Why the boat moves back, when the sailor jumps in forward direction ?
Pause the Video
(Time Duration : 2 Minutes)
Sol.
Sailor jumps out from boat
Sailor jumps out from boat
Presses the boat with feet(action)
Sailor jumps out from boat
Presses the boat with feet(action)
By third law , there will be an equal and opposite reaction
Sailor jumps out from boat
Presses the boat with feet(action)
By third law , there will be an equal and opposite reaction
Boat exerts a force
ConcepTest 08
Ready for a challenge
Q. The earth pulls a stone downwards due to gravity. Does the stone exert a force on the earth ?
Q. The earth pulls a stone downwards due to gravity. Does the stone exert a force on the earth ?
Pause the Video
(Time Duration : 2 Minutes)
Sol. Yes, The according to Newton’s third law of motion,the stone does exert an equal and opposite force on the earth.
We do not notice it since the earth is very massive and the effect of a small force on its motion is negligible.
F
F
11P05.1
CV 7
Common Forces in Mechanics
11P05.1 Common Forces in Mechanics
134
m
11P05.1 Common Forces in Mechanics
135
m
W=mg
11P05.1 Common Forces in Mechanics
2. Normal reaction(N)
136
11P05.1 Common Forces in Mechanics
2. Normal reaction(N)
137
N
11P05.1 Common Forces in Mechanics
2. Normal reaction(N)
138
N
N
11P05.1 Common Forces in Mechanics
2. Normal reaction(N)
139
N
N
11P05.1 Common Forces in Mechanics
3. Friction (f)
140
F
11P05.1 Common Forces in Mechanics
3. Friction (f)
141
F
F
f
11P05.1 Common Forces in Mechanics
3. Friction (f)
142
F
F
f
f
11P05.1 Common Forces in Mechanics
4. Tension (T)
143
11P05.1 Common Forces in Mechanics
4. Tension (T)
144
T
11P05.1 Common Forces in Mechanics
5. Spring force
145
11P05.1 Common Forces in Mechanics
5. Spring force
146
F
Compressive force
F= -kx
where k = spring constant
x = compression
11P05.1 Common Forces in Mechanics
5. Spring force
147
F
Tensile force
F= -kx
where k = spring constant
x = compression
Note : The negative sign denotes that the force is opposite to the displacement from the unstretched state.
11P05.1 Common Forces in Mechanics
Reference Questions
NCERT : 5.1, 5.2, 5.3, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, 5.11, 5.13, 5.14, 5.20, 5.23
Work Book : 1, 4, 5, 6, 7, 8, 10, 12, 13
148
11P05.2
Applications of Newton’s Laws of Motion
11P05.2 Applications of Newton’s Laws of Motion
Learning Objectives
Free Body Diagram
Equilibrium of a Particle
Conservation of Momentum
151
11P05.2
CV 1
Free Body Diagram
11P05.2 Free Body Diagram
System : A system is that part of a body in which we are interested.
153
11P05.2 Free Body Diagram
System : A system is that part of a body in which we are interested.
154
11P05.2 Free Body Diagram
System : A system is that part of a body in which we are interested.
155
Interested in motion of 15kg
11P05.2 Free Body Diagram
System : A system is that part of a body in which we are interested.
156
Interested in motion of 15kg
11P05.2 Free Body Diagram
Steps of drawing FBD :
157
11P05.2 Free Body Diagram
Steps of drawing FBD :
158
11P05.2 Free Body Diagram
Steps of drawing FBD :
159
11P05.2 Free Body Diagram
Steps :
160
m
11P05.2 Free Body Diagram
Steps :
on the system
161
m
11P05.2 Free Body Diagram
Steps :
on the system
(with direction) on the system
11P05.2 Free Body Diagram
Steps :
on the system
(with direction) on the system
163
ConcepTest 09
Ready for a challenge
Q. A block of mass 𝑚 is suspended with two strings as shown in figure. Draw the free body diagram of the block.
Q. A block of mass 𝑚 is suspended with two strings as shown in figure. Draw the free body diagram of the block.
θ
Pause the Video
(Time Duration : 2 Minutes)
Sol.
T2
T1
mg
θ
PSV 03
Q. Draw the free body diagram of both the masses and pulley for the given sketch.
Pause the Video
(Time Duration : 3 Minutes)
Sol.
m
m
Sol.
Sol.
Sol.
If string and pulley are massless and pulley is frictionless then tension
in string will be same throughout the string.
11P05.2
CV 2
Equilibrium of a Particle
11P05.2 Equilibrium of a Particle
Equilibrium of a particle in mechanics refers to the situation when the net external force on the particle is zero.
179
11P05.2 Equilibrium of a Particle
Equilibrium of a particle in mechanics refers to the situation when the net external force on the particle is zero.
180
11P05.2 Equilibrium of a Particle
If a particle is in equilibrium under the action of forces
181
11P05.2 Equilibrium of a Particle
If a particle is in equilibrium under the action of forces
then,
182
11P05.2 Equilibrium of a Particle
If a particle is in equilibrium under the action of forces
then,
183
11P05.2 Equilibrium of a Particle
If a particle is in equilibrium under the action of forces
then,
184
11P05.2 Equilibrium of a Particle
If a particle is in equilibrium under the action of forces
then,
185
ConcepTest 10
Ready for a challenge
Q. A block of weight 100 N is suspended from a rope tied to two other ropes at point O. One rope is horizontally attached to a wall and the other is fastened to the ceiling. The angle between ceiling and the ropes is 60°. What are the tensions in each of the ropes ? Assume the weights of the ropes and the knot are negligible.
Pause the Video
(Time Duration : 5 Minutes)
Q. A block of weight 100 N is suspended from a rope tied to two other ropes at point O. One rope is horizontally attached to a wall and the other is fastened to the ceiling. The angle between ceiling and the ropes is 60°. What are the tensions in each of the ropes ? Assume the weights of the ropes and the knot are negligible.
Sol. First draw the free body diagram of knot O
In the x-direction
ΣFx = 0
In the x-direction
ΣFx = 0
In the y-direction
ΣFy = 0
From equation (ii) and (iii)
From equation (i) and (iv)
PSV 04
Q. Two blocks of equal mass m are tied to each other using a light inextensible string. One of the blocks is pulled along the line joining them with a constant force 𝐹. Find the tension in the string joining the blocks.
Sol.
Sol.
A
B
Sol.
Sol.
Sol.
Applying Newton’s second law of motion
On block A
Sol.
Applying Newton’s second law of motion
On block A
On block B
Sol.
Applying Newton’s second law of motion
On block A
On block B
From eq. (i) and (ii)
T = F/2
PSV 05
Q. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.(Take g = 10 m/s2)
Q. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.(Take g = 10 m/s2)
Pause the Video
(Time Duration : 5 Minutes)
Sol.
Sol.
Sol.
Sol.
Sol.
Sol.
11P05.2
CV 3
Conservation of Momentum
11P05.2 Conservation of Momentum
214
11P05.2 Conservation of Momentum
215
(No Net External force)
11P05.2 Conservation of Momentum
216
Balls Collide!
11P05.2 Conservation of Momentum
217
Balls Collide!
Collision lasts for a small time “t”.
11P05.2 Conservation of Momentum
218
Balls Collide!
Collision lasts for a small time “t”.
Each ball exerts some force on the other.
11P05.2 Conservation of Momentum
219
Balls Collide!
Collision lasts for a small time “t”.
Each ball exerts some force on the other.
Balls move with final velocity vA and vB after collision.
11P05.2 Conservation of Momentum
220
Initial momentum | mA x uA | mB x uB |
11P05.2 Conservation of Momentum
221
Initial momentum | mA x uA | mA x vA |
Final momentum | mA x vA | mB x vB |
11P05.2 Conservation of Momentum
222
Initial momentum | mA x uA | mA x vA |
Final momentum | mB x uB | mB x vB |
Rate of change of momentum | | |
11P05.2 Conservation of Momentum
223
Rate of change of momentum | | |
11P05.2 Conservation of Momentum
224
Rate of change of momentum | | |
11P05.2 Conservation of Momentum
On simplifying, we get
225
Rate of change of momentum | | |
11P05.2 Conservation of Momentum
The total momentum of an isolated system(i.e. a system with no external force) of interacting particles is conserved.
226
ConcepTest 11
Ready for a challenge
Q. Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 3 ms–1 and 1 ms–1, respectively.They collide and after the collision, the first object moves at a velocity of 1.2 ms–1. Determine the velocity of the second object.
Q. Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 3 ms–1 and 1 ms–1, respectively.They collide and after the collision, the first object moves at a velocity of 1.2 ms–1. Determine the velocity of the second object.
Pause the Video
(Time Duration : 2 Minutes)
Sol. mA = 100 g and mB=200 g
Sol. mA = 100 g and mB=200 g
ConcepTest 12
Ready for a challenge
Q. A bullet of mass 5 g is shot from a gun of mass 5 kg. The muzzle velocity of the bullet is 500 ms-1. What is the recoil velocity of the gun?
Q. A bullet of mass 5 g is shot from a gun of mass 5 kg. The muzzle velocity of the bullet is 500 ms-1. What is the recoil velocity of the gun?
Pause the Video
(Time Duration : 2 Minutes)
Sol.
Sol.
Negative sign indicates that the direction in which the gun would recoil is opposite to that of bullet.
Reference Questions
NCERT : 5.15, 5.16, 5.17, 5.18, 5.19
Work Book : 14, 18, 19
11P05.3
Friction
11P05.3 Friction
Learning Objectives
Friction Basics
Types of Friction
Properties of Friction
240
11P05.3
CV 1
Friction Basics
11P05.3 Friction Basics
11P05.3 Friction Basics
F
11P05.3 Friction Basics
From Newton’s second law,
F
11P05.3 Friction Basics
From Newton’s second law,
F
11P05.3 Friction Basics
From Newton’s second law,
Here m = mass of car
F
11P05.3 Friction Basics
From Newton’s second law,
Here m = mass of car
F
Force applied
11P05.3 Friction Basics
From Newton’s second law,
Here m = mass of car
F
Force applied
acceleration
11P05.3 Friction Basics
From Newton’s second law,
Here m = mass of car
motion
F
Force applied
acceleration
11P05.3 Friction Basics
But why car is not moving ?
11P05.3 Friction Basics
So, there is some force opposing the F known as frictional force.
But why car is not moving ?
11P05.3 Friction Basics
Friction is the resistance to motion or tendency of motion of one object relative to another.
11P05.3 Friction Basics
Friction is the resistance to motion or tendency of motion of one object relative to another.
11P05.3 Friction Basics
Friction is the resistance to motion or tendency of motion of one object relative to another.
Ff ∝ N
11P05.3 Friction Basics
Friction is the resistance to motion or tendency of motion of one object relative to another.
Ff ∝ N
⇒Ff = μN
11P05.3 Friction Basics
Friction is the resistance to motion or tendency of motion of one object relative to another.
Ff ∝ N
⇒Ff = μN
Where μ = coefficient of friction
11P05.3
CV 2
Types of Friction
11P05.3 Types of Friction
Types of Friction
11P05.3 Types of Friction
Static Friction
Types of Friction
11P05.3 Types of Friction
Static Friction
Kinetic Friction
Types of Friction
11P05.3 Types of Friction
Static Friction
Kinetic Friction
Types of Friction
Body is not Moving
11P05.3 Types of Friction
Static Friction
Kinetic Friction
Types of Friction
Body is not Moving
Body is moving
11P05.3 Types of Friction
Static Friction
Kinetic Friction
Types of Friction
Body is not Moving
Body is moving
Ff > Fa
11P05.3 Types of Friction
Static Friction
Kinetic Friction
Types of Friction
Body is not Moving
Body is moving
Ff > Fa
Ff < Fa
11P05.3 Types of Friction
Impending motion
11P05.3 Types of Friction
Impending motion
Moving in frictionless condition but not in actual situation.
11P05.3 Types of Friction
Impending motion
Moving in frictionless condition but not in actual situation.
11P05.3 Types of Friction
Static friction opposes impending motion.
11P05.3 Types of Friction
Static friction opposes impending motion.
F
fs
11P05.3 Types of Friction
Static friction opposes impending motion.
fs = F
F
fs
11P05.3 Types of Friction
Kinetic Friction
11P05.3 Types of Friction
Kinetic Friction
11P05.3 Types of Friction
Kinetic Friction occurs when the two surfaces are moving relative to each other.
11P05.3 Types of Friction
Kinetic Friction occurs when the two surfaces are moving relative to each other.
11P05.3 Types of Friction
11P05.3 Types of Friction
11P05.3 Types of Friction
11P05.3 Types of Friction
11P05.3
CV 3
Properties of Friction
11P05.3 Properties of Friction
F=5N
F=10N
F=20N
f=20N
f=10N
f=5N
Q. Does the value of friction increase continuously or stop at any point ?
11P05.3 Properties of Friction
11P05.3 Properties of Friction
Q. Does the value of friction increase continuously or stop at any point ?
SOL. It is going to stop increasing at the point the applied force become equal to
maximum static friction.
11P05.3 Properties of Friction
Q. Does the value of friction increase continuously or stop at any point ?
SOL. It is going to stop increasing at the point the applied force become equal to
maximum static friction.
Then the kinetic friction will continue.
Applied force vs frictional force
11P05.3 Properties of Friction
11P05.3 Properties of Friction
Static frictional force is a
Self adjusting force
11P05.3 Properties of Friction
Static frictional force is a
Changes according to Applied Force
Self adjusting force
11P05.3 Properties of Friction
Static frictional force is a
Changes according to Applied Force
Self adjusting force
11P05.3 Properties of Friction
Friction force does not depend on contact area.
11P05.3 Properties of Friction
Friction force does not depend on contact area.
11P05.3 Properties of Friction
Friction force does not depend on contact area.
Same Normal Force
11P05.3 Properties of Friction
Friction force does not depend on contact area.
Same Normal Force
+
Same Pair of Interfaces
11P05.3 Properties of Friction
Friction force does not depend on contact area.
Same Frictional Force
Same Normal Force
+
Same Pair of Interfaces
ROLLING FRICTION
The force resisting the motion of rolling body on a surface is known as Rolling Friction.
ROLLING FRICTION
The force resisting the motion of rolling body on a surface is known as Rolling Friction.
ROLLING FRICTION
Rolling friction is considerably weaker than Sliding friction.
ROLLING FRICTION
Rolling friction is considerably weaker than Sliding friction.
ConcepTest 13
Ready for a challenge
Q. Why do we sprinkle powder on the Carrom board ?
Q. Why do we sprinkle powder on the Carrom board ?
Pause the Video
(Time Duration : 1 Minute)
Sol. To reduce the Friction.
PSV 06
Q. An object of mass 1 kg slides down an inclined plane inclined at 30o with constant velocity. Determine coefficient of kinetic friction (μk).
Q. An object of mass 1 kg slides down an inclined plane inclined at 30o with constant velocity. Determine coefficient of kinetic friction (μk).
Pause the Video
(Time Duration : 3 Minutes)
Sol.
w = mg = 9.8N
Sol.
w = mg = 9.8 N
wx = w sin 30o = 4.9 N
wy = w cos 30o = 4.9√3 N
Sol.
w = mg = 9.8N
wx = w sin 30o = 4.9N
Wy w cos 30o = 4.9√3 N
N = wy = 4.9√3 N
Sol.
w = mg = 9.8N
wx = w sin 30o = 4.9N
Wy w cos 30o = 4.9√3 N
N = wy = 4.9√3 N
According to second law of motion,
Sol.
w = mg = 9.8N
wx = w sin 30o = 4.9N
Wy w cos 30o = 4.9√3 N
N = wy = 4.9√3 N
According to second law of motion,
Along the incline
∑F = 0 (constant velocity, so a = 0)
Sol.
w = mg = 9.8N
wx = w sin 30o = 4.9N
Wy w cos 30o = 4.9√3 N
N = wy = 4.9√3 N
According to second law of motion,
∑F = 0 (constant velocity, so a = 0)
wx = μk N
Sol.
w = mg = 9.8N
wx = w sin 30o = 4.9N
Wy w cos 30o = 4.9√3 N
N = wy = 4.9√3 N
According to second law of motion,
∑F = 0 (constant velocity, so a = 0)
wx = μk N
⇒ 4.9= μk (4.9√3)
⇒ μk = 0.58
PSV 07
Q. An object of mass 1 kg is resting on the table. A horizontal force of 1N is applied. Find the value of friction acting on the block from the table. Coefficient of friction between block and surface is 0.2. Take g=10m/s2 .
Q. An object of mass 1 kg is resting on the table. A horizontal force of 1N is applied. Find the value of friction acting on the block from the table. Coefficient of friction between block and surface is 0.2. Take g=10m/s2.
Pause the Video
(Time Duration : 3 Minutes)
Sol.
Sol.
FN = mg
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
F = 1N
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
F = 1N
As F < f ,
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
F = 1N
As F < f ,
a=0m/s2
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
F = 1N
As F < f ,
a=0m/s2
We know friction is a self adjusting force in static condition.
Sol.
FN = mg
FN = 1kg * 10m/s2
FN = 10N
f = μFN
f = 0.2*10
f = 2N
F = 1N
As F < f ,
a=0m/s2
We know friction is a self adjusting force in static condition.
So,
f = 1N
Reference Questions
NCERT : 5.34, 5.35, 5.36, 5.37
Work Book : 2,3,8,15,16,17
11P05.4
Circular Motion
11P05.4 Circular Motion
Learning Objectives
Centripetal Force
Turning of a vehicle on a level road
Turning of a vehicle on a banked road
327
11P05.4
CV 1
Centripetal Force
11P05.4 Centripetal Force
It is a force that makes a body to follow a circular path.
11P05.4 Centripetal Force
It is a force that makes a body to follow a circular path.
11P05.4 Centripetal Force
It is a force that makes a body to follow a circular path.
Net force, Fnet = mac
ac = v2 /r (ac= centripetal acceleration)
11P05.4 Centripetal Force
It is a force that makes a body to follow a circular path.
Net force, Fnet = mac
ac = v2 /r (ac= centripetal acceleration)
so,
11P05.4 Centripetal Force
It is a force that makes a body to follow a circular path.
Net force, Fnet = mac
ac = v2 /r (ac= centripetal acceleration)
so,
WWWWWithout centripetal force no circular motion.t
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Non accelerated frames
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Non accelerated frames
Accelerated frames
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Newton’s law
holds true
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Newton’s law
holds true
Newton’s law does
not holds true
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Newton’s law
holds true
Newton’s law does
not holds true
v= 3m/s
11P05.4 Frame of Reference
Inertial frame of reference
Non-inertial frame of reference
Newton’s law
holds true
Newton’s law does
not holds true
v= 3m/s
11P05.4 Pseudo Force
Pseudo force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference.
11P05.4 Pseudo Force
Pseudo force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference.
11P05.4 Pseudo Force
Pseudo force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference.
Pseudo force
11P05.4 Centrifugal Force
Centrifugal force is a pseudo force.
11P05.4 Centrifugal Force
It occurs in circular motion.
Centrifugal force is a pseudo force.
11P05.4 Centrifugal Force
It occurs in circular motion.
It acts away from the centre.
Centrifugal force is a pseudo force.
11P05.4 Centrifugal Force
It occurs in circular motion.
It acts away from the centre.
Centrifugal force is a pseudo force.
11P05.4 Centrifugal Force
Example of centrifugal force -
11P05.4 Centrifugal Force
Example of centrifugal force -
ConcepTest 14
Ready for a challenge
Q. Which force enables the earth to revolve around the sun.
Q. Which force enables the earth to revolve around the sun.
Pause the Video
(Time Duration : 1 Minute)
Sol. Gravitational force
11P05.4
CV 2
Turning of a Vehicle on a level road
11P05.4 Turning of a Vehicle on a level road
Q. What is the centripetal force acting for a car moving along the curved path ?
11P05.4 Turning of a Vehicle on a level road
Sol. The static friction force between the road and the tires.
Q. What is the centripetal force acting for a car moving along the curved path ?
11P05.4 Turning of a Vehicle on a level road
11P05.4 Turning of a Vehicle on a level road
The centripetal force required for circular motion is provided by the frictional force between road and the car tyres along the surface.
Maximum velcity for which a vehicle not to slip
centripetal force
force of friction
Maximum velcity for which a vehicle not to slip
centripetal force
force of friction
Maximum velcity for which a vehicle not to slip
centripetal force
force of friction
ConcepTest 15
Ready for a challenge
Q. A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius
3 m without reducing the speed. The coefficient of static friction between the
tyres and the road is 0.1. Will the cyclist slip while taking the turn ?
Pause the Video
(Time Duration : 2 Minutes)
Sol : Given , v = 18 km/h = 5 m/s , radius r = 3 m, μs= 0.1
we know the condition for the cyclist not to slip is given by
The condition is not obeyed. The cyclist will slip while taking the circular turn.
11P05.4
CV 3
Turning of a Vehicle on a banked road
11P05.4 Turning of a Vehicle on a banked road
Have you ever wondered why the curved roads on highways are banked?
There are three cases of a moving car on banked road:
11P05.4 Turning of a Vehicle on a banked road
1. when friction is acting towards the centre
2. when friction is zero.
3. when friction is acting away from the centre.
11P05.4 Turning of a Vehicle on a banked road
When friction is acting towards the centre
11P05.4 Turning of a Vehicle on a banked road
Forces acting vertically
11P05.4 Turning of a Vehicle on a banked road
Forces acting horizontally
11P05.4 Turning of a Vehicle on a banked road
Here, forces equations along horizontal and vertical directions are
��
11P05.4 Turning of a Vehicle on a banked road
Here, forces equations along horizontal and vertical directions are
Along the vertical direction
��
11P05.4 Turning of a Vehicle on a banked road
Here, forces equations along horizontal and vertical directions are
Along the vertical direction
��
After solving equations (i) and (ii) ,we will get
Along the horizontal direction
ᐩ
ConcepTest 16
Ready for a challenge
Q. Suppose you want to negotiate a curve with a radius of 50 meters and a bank angle of 15o. If the coefficient of friction between your tires and the pavement is 0.50, what is the maximum speed that you can safely use?
Pause the Video
(Time Duration : 3 Minute)
Sol : We know
So,
When friction is zero
11P05.4 Turning of a Vehicle on a banked road
11P05.4 Turning of a Vehicle on a banked road
At this speed, frictional force is not needed at all to provide the necessary centripetal force.
ConcepTest 17
Ready for a challenge
Q. A curve has a radius of 50 meters and a banking angle of 15o. What is the ideal, or
critical, speed (the speed for which no friction is required between the car's tires
and the surface) for a car on this curve ?
Pause the Video
(Time Duration : 2 Minutes)
Sol :
If the car has a speed of about 11 m/s , it can negotiate the curve without any friction.
When friction is acting away from the centre
11P05.4 Turning of a Vehicle on a banked road
11P05.4 Turning of a Vehicle on a banked road