FRICTION
Contents
Particular's Page No.
Theory 001 – 012
Exercise - 1 013 – 017
Part - I : Subjective Question
Part - II : Only one option correct type
Part - III : Match the column
Exercise - 2 018 – 025
Part - I : Only one option correct type
Part - II : Single and double value integer type
Part - III : One or More than one option correct type
Part - IV : Comprehension
Exercise - 3 025 – 027
Part - I : JEE(Advanced) / IIT-JEE Problems (Previous Years)
Part - II : JEE(Main) / AIEEE Problems (Previous Years)
Answer Key 028
High Level Problems (HLP) 029 – 031
Subjective Question
HLP Answers 031
JEE (ADVANCED) SYLLABUS
Friction : Static and dynamic friction.
JEE (MAIN) SYLLABUS
Friction : Static and Kinetic friction, laws of friction, rolling friction.
©Copyright reserved.
All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the
enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.
, Friction
FRICTION
———————————————————————————————————
1. FRICTION
When two bodies are kept in contact, electromagnetic forces act between the charged particles
(molecules) at the surfaces of the bodies. Thus, each body exerts a contact force on the other. The
magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite
and therefore the contact forces obey Newton’s third law.
The direction of the contact force acting on a particular body is not necessarily perpendicular to the
contact surface. We can resolve this contact force into two components, one perpendicular to the
contact surface and the other parallel to it (figure. The perpendicular component is called the normal
contact force or normal force (generally written as N) and the parallel component is called friction
(generally written as f).
Therefore if R is contact force then R = f 2 N2
2. REASONS FOR FRICTION
(i) nter-locking of extended parts of one object into the extended parts of the other object.
(ii) Bonding between the molecules of the two surfaces or objects in contact.
3. FRICTION FORCE IS OF TWO TYPES.
a. Kinetic b. Static
(a) Kinetic Friction Force
Kinetic friction exists between two contact surfaces only when there is relative motion between the
two contact surfaces. It stops acting when relative motion between two surfaces ceases.
DIRECTION OF KINECTIC FRICTION ON AN OBJECT
t is opposite to the relative velocity of the object with respect to the other object in contact
considered.
Note that its direction is not opposite to the force applied it is opposite to the relative motion
of the body considered which is in contact with the other surface.
MAGNITUDE OF KINETIC FRICTION
The magnitude of the kinetic friction is proportional to the normal force acting between the two
bodies. We can write
fk = k N
where N is the normal force. The proportionality constant k is called the coefficient of kinetic
friction and its value depends on the nature of the two surfaces in contact.
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,Friction
Example 1. Find the direction of kinetic friction force
(a) on the block, exerted by the ground. (b) on the ground, exerted by the block.
Solution : (a)
(b)
where f1 and f2 are the friction forces on the block and ground respectively.
Example 2. In above example correct relation between magnitude of f1 and f2 is
(A) f1 > f2 (B) f2 > f1
(C) f1 = f2 (D) not possible to decide due to insufficient data.
Solution : By Newton‘s third law the above friction forces are action-reaction pair and equal but opposite
to each other in direction. Hence (C).
Also note that the direction of kinetic friction has nothing to do with applied force F.
Example 3. All surfaces as shown in the figure are rough. Draw the friction force on A & B
A 10m/s
B 20m/s
////////////////////////////////////////////////////////
Solution :
Kinetic friction acts in such a way so as to reduce relative motion.
Example 4. Find out the distance travelled by the blocks shown in the figure before it stops.
10 m/s
10 kg
/////////////////////////////////////////
µ k=0.5
Solution : N – 10 g = 0
N = 100 N N
fx = µ kN
µ = µs = µk when not mentioned
fx = 0.5 × 100 = 50 N fk
Fx = ma
50 = 10 a a = 5 10g
v2 = u2 + 2as
02 = 102 + 2 (– 5) (S) S = 10 m
Example 5. Find out the distance travelled by the block on incline before it stops. Initial velocity of the block
is 10 m/s and coefficient of friction between the block and incline is = 0.5.
s
m/
10
37° fixed
µ
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, Friction
Solution : N = mg cos37°
mg sin 37° + µN = ma
a = 10 m/s2 down the incline
Now v2 = u2 + 2as
0 = 102 + 2(–10) S
S=5m
Example 6. Find the time taken in the above example by the block to reach the initial position.
Solution : a = g sin 37° – µg cos 37°
a = 2 m/s2 down the incline
1 2 1 5m
S = ut + at S = × 2 × t2
2 2
t = sec.
Example 7. A block is given a velocity of 10 m/s and a force of
100 N in addition to friction force is also acting on the
block. Find the retardation of the block?
Solution : As there is relative motion
Kinetic friction will act to reduce this relative motion.
fk = µN = 0.1 × 10 × 10 = 10 N
100 + 10 = 10a
a = 5 = 11 m/s2
———————————————————————————————————
(b) STATIC FRICTION
t exists between the two surfaces when there is tendency of relative motion but no relative motion
along the two contact surface.
For example consider a bed inside a room ; when we gently push the bed with a finger, the bed
does not move. This means that the bed has a tendency to move in the direction of applied force
but does not move as there exists static friction force acting in the opposite direction of the applied
force.
Example 8. What is value of static friction force on the block?
Solution : In horizontal direction as acceleration is zero.
Therefore F = 0.
= 0
Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005
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Contents
Particular's Page No.
Theory 001 – 012
Exercise - 1 013 – 017
Part - I : Subjective Question
Part - II : Only one option correct type
Part - III : Match the column
Exercise - 2 018 – 025
Part - I : Only one option correct type
Part - II : Single and double value integer type
Part - III : One or More than one option correct type
Part - IV : Comprehension
Exercise - 3 025 – 027
Part - I : JEE(Advanced) / IIT-JEE Problems (Previous Years)
Part - II : JEE(Main) / AIEEE Problems (Previous Years)
Answer Key 028
High Level Problems (HLP) 029 – 031
Subjective Question
HLP Answers 031
JEE (ADVANCED) SYLLABUS
Friction : Static and dynamic friction.
JEE (MAIN) SYLLABUS
Friction : Static and Kinetic friction, laws of friction, rolling friction.
©Copyright reserved.
All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the
enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.
, Friction
FRICTION
———————————————————————————————————
1. FRICTION
When two bodies are kept in contact, electromagnetic forces act between the charged particles
(molecules) at the surfaces of the bodies. Thus, each body exerts a contact force on the other. The
magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite
and therefore the contact forces obey Newton’s third law.
The direction of the contact force acting on a particular body is not necessarily perpendicular to the
contact surface. We can resolve this contact force into two components, one perpendicular to the
contact surface and the other parallel to it (figure. The perpendicular component is called the normal
contact force or normal force (generally written as N) and the parallel component is called friction
(generally written as f).
Therefore if R is contact force then R = f 2 N2
2. REASONS FOR FRICTION
(i) nter-locking of extended parts of one object into the extended parts of the other object.
(ii) Bonding between the molecules of the two surfaces or objects in contact.
3. FRICTION FORCE IS OF TWO TYPES.
a. Kinetic b. Static
(a) Kinetic Friction Force
Kinetic friction exists between two contact surfaces only when there is relative motion between the
two contact surfaces. It stops acting when relative motion between two surfaces ceases.
DIRECTION OF KINECTIC FRICTION ON AN OBJECT
t is opposite to the relative velocity of the object with respect to the other object in contact
considered.
Note that its direction is not opposite to the force applied it is opposite to the relative motion
of the body considered which is in contact with the other surface.
MAGNITUDE OF KINETIC FRICTION
The magnitude of the kinetic friction is proportional to the normal force acting between the two
bodies. We can write
fk = k N
where N is the normal force. The proportionality constant k is called the coefficient of kinetic
friction and its value depends on the nature of the two surfaces in contact.
Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005
Website : www.resonance.ac.in | E-mail :
ADVFR - 1
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
,Friction
Example 1. Find the direction of kinetic friction force
(a) on the block, exerted by the ground. (b) on the ground, exerted by the block.
Solution : (a)
(b)
where f1 and f2 are the friction forces on the block and ground respectively.
Example 2. In above example correct relation between magnitude of f1 and f2 is
(A) f1 > f2 (B) f2 > f1
(C) f1 = f2 (D) not possible to decide due to insufficient data.
Solution : By Newton‘s third law the above friction forces are action-reaction pair and equal but opposite
to each other in direction. Hence (C).
Also note that the direction of kinetic friction has nothing to do with applied force F.
Example 3. All surfaces as shown in the figure are rough. Draw the friction force on A & B
A 10m/s
B 20m/s
////////////////////////////////////////////////////////
Solution :
Kinetic friction acts in such a way so as to reduce relative motion.
Example 4. Find out the distance travelled by the blocks shown in the figure before it stops.
10 m/s
10 kg
/////////////////////////////////////////
µ k=0.5
Solution : N – 10 g = 0
N = 100 N N
fx = µ kN
µ = µs = µk when not mentioned
fx = 0.5 × 100 = 50 N fk
Fx = ma
50 = 10 a a = 5 10g
v2 = u2 + 2as
02 = 102 + 2 (– 5) (S) S = 10 m
Example 5. Find out the distance travelled by the block on incline before it stops. Initial velocity of the block
is 10 m/s and coefficient of friction between the block and incline is = 0.5.
s
m/
10
37° fixed
µ
Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005
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ADVFR - 2
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, Friction
Solution : N = mg cos37°
mg sin 37° + µN = ma
a = 10 m/s2 down the incline
Now v2 = u2 + 2as
0 = 102 + 2(–10) S
S=5m
Example 6. Find the time taken in the above example by the block to reach the initial position.
Solution : a = g sin 37° – µg cos 37°
a = 2 m/s2 down the incline
1 2 1 5m
S = ut + at S = × 2 × t2
2 2
t = sec.
Example 7. A block is given a velocity of 10 m/s and a force of
100 N in addition to friction force is also acting on the
block. Find the retardation of the block?
Solution : As there is relative motion
Kinetic friction will act to reduce this relative motion.
fk = µN = 0.1 × 10 × 10 = 10 N
100 + 10 = 10a
a = 5 = 11 m/s2
———————————————————————————————————
(b) STATIC FRICTION
t exists between the two surfaces when there is tendency of relative motion but no relative motion
along the two contact surface.
For example consider a bed inside a room ; when we gently push the bed with a finger, the bed
does not move. This means that the bed has a tendency to move in the direction of applied force
but does not move as there exists static friction force acting in the opposite direction of the applied
force.
Example 8. What is value of static friction force on the block?
Solution : In horizontal direction as acceleration is zero.
Therefore F = 0.
= 0
Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005
Website : www.resonance.ac.in | E-mail :
ADVFR - 3
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