Friction intro.pptx

Course: Elements of Civil Engineering and mechanics
Study Material prepared by Mrs Archana K (archanak@bmsit.in)
https://www.youtube.com/channel/UCeJdH2Kc0IjqW2eluEPpp3w
Department of Civil Engineering, BMS Institute of Technology and Management, Bengaluru 560064
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Welcome to Online Learning
@
Department of Civil Engineering

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Vision Statement: To be an Exemplary Centre, disseminating quality education and developing technically competent civil engineers with
professional integrity for the betterment of society
Mission Statements
1. Impart technical proficiency through quality education.
2. Motivate entrepreneurship through enhanced industry -interaction and skill based training.
3. Inculcate human resource values through outreach activities.
Program Educational Objectives (PEOs)
1. Lead a successful career by analyzing, designing and solving various problems in the field of Civil Engineering.
2. Execute projects through team building, communication and professionalism
3. Excel through higher education and research for endured learning.
4. Provide effective solution for sustainable environmental development.
Program Specific Outcomes (PSO)
1. Identify & address the challenges in transportation, sanitation, waste management, and urban flooding in metropolitan cities.
2. Provide solutions related to Civil Engineering built-environment through multidisciplinary approach.
DEPARTMENT OF CIVIL ENGINEERING

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Disclaimer
All the materials mentioned herewith is incepted from online study materials & textbooks, with due gratitude and acknowledgement.
This study material is a conceptual compilation, with several edited intellectual inputs from varying relevant databases.
References have been mentioned, wherever retrievable.
The Author or Institute claims no proprietary rights on this ”academia-use” study material, except for outcomes from the original research
contributions of the author.

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Course & Course Coordinator Details
Elements of Civil Engineering and Mechanics
Mrs Archana K
Department of Civil Engineering
BMS INSTITUTE OF TECHNOLOGY AND MANAGEMENT
(Affiliated to the Visvesvaraya Technological University, Belagavi)
Avalahalli, Bengaluru 560064, Karnataka, INDIA

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Recap to Previous Lecture
Module 1:
Scope of Civil Engineering
Role of Civil Engineers in Infrastructural Development and growth of the nation
Engineering Mechanics
Basic concepts of idealization
System of Forces
Basic Principles:
Physical independence of forces
Superposition Law
Law of Transmissibility
Newton’s Law of Motion
Resolution and composition of forces
Law of parallelogram of forces
Polygonal law

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Recap to Previous Lecture
Module 1:
Resultant of concurrent co-planar force system
Resultant of non-concurrent co-planar force system
Moment of a force
Couple system
Varignon’s Theorem
Module 2:
Equilibrium of forces
- FBD
- Lami’s Theorem
- Equations of equilibrium of concurrent and non-concurrent co-planar force system

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Module 2: FRICTION
FRICTION
- Types of friction
- Laws of dry friction
- Limiting Friction
- Concept of Static and Dynamic Friction
- Numerical on
- Motion of single and connected bodies on plane
- Wedge friction
- Ladder Friction
- Rope and Pulley systems

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FRICTION

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REASONS FOR FRICTIONAL RESISTANCE
1. The Major Cause of the friction is the
interlocking of minute projections on
the surfaces which opposes the
relative motion
2. The amount of friction depends on
the materials from which the two
surfaces are made. The rougher the
surface, the more friction is
produced
3. Position of bodies in contact (
static/kinetic/inclined etc)

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Engineering Applications of friction
1. Moving/Lifting objects from one place to another
2. Tyre friction in automobiles
3. Walking on wet floor ( Slipping)
4. Drag in automobiles
5. Sports like skiing, skating, swimming, slide, running etc
6. Automobile brakes
7. Ice boring
8. Masonry construction
9. Lubricants in vehicles
10. Steel manufacturing
11. Ball bearings, Gears and turbines etc

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FRICTION

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FRICTION

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FRICTION
Friction
•Maximum friction
as motion starts
to impend
Dry/ Coulomb
Friction
•Static Friction
•Dynamic/Kinetic
Friction>>
Fluid Friction
•Skin/Greasy
Friction
•Viscous Friction
Limiting Friction
•Sliding friction
•Rolling Friction
1. Dry friction is a force that opposes the relative
lateral motion of two solid surfaces in contact. Dry
friction is subdivided into static friction between
non-moving surfaces, and kinetic friction between
moving surfaces.
2. Fluid friction is the force that resists motion
either within the fluid itself or of another medium
moving through the fluid. This id further classified
as Skin and Viscous friction.
3. Limiting Friction: The maximum frictional force at
which body starts to impend after overcoming
static friction. ( Fmax = µs * N)

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STATIC FRICTION

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DYNAMIC/KINETIC FRICTION

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FRICTION

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FRICTION

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STATIC & DYNAMIC FRICTION

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FBD& Some Definitions
1. External Force / Applied Force ( P)
2. Frictional Force ( f)
3. Self weight ( W=mg)
4. Normal Reaction force (N)
5. Total Reaction ( R) :
Resultant of Normal and frictional force in
the direction of normal
6. Angle of friction ( Φ) : Angle made by resultant
(R) with normal axis
7. Co-efficient of friction (µ): Ration of frictional
force to normal force

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FBD& Some Definitions
7. Angle of repose ( α) :
When a body is placed on an inclined plane, the angle at which the body is just about to slide
down or impend is called angle of repose
8. Cone of friction: the cone formed by the resultant with angle of friction

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Angle of repose and angle of friction
Hence at limiting friction, when
motion is about impend, after
overcoming static friction,
Angle of repose = angle of limiting
friction, i.e α=Φ

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LAWS OF DRY FRICTION

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LAWS OF DRY FRICTION
• The frictional force always acts in the direction opposite to that of motion trend and
tangential to contact surface
• The value of frictional force (f) is always proportional to normal reaction force between
the surfaces.
F α N
F= µ N
(µ is the co-efficient of friction)
• Frictional force is independent of the area of contact between two surfaces
• Frictional force depends on the roughness of the contact surfaces
• For small velocities, the magnitude of frictional force is independent of the velocity.

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Numerical Problems
Numerical on
1. Single blocks on plane
2. Connected bodies on plane
3. Wedge friction
4. Ladder friction
5. Rope and pulley system

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Numerical Problems
Numerical on
Single blocks/connected bodies on horizontal plane

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Numerical Problems- single blocks on horizontal plane
1. A body of weight 1000 N is placed on a rough horizontal plane. Determine the co-efficient
of friction due to force of 600 N in horizontal direction, which just causes the body to slide.
Given:
W= 1000 N
P=600 N ( horizontal)
µ=?
W=1000N
P= 600 N
N
f = µN
Solution:
1. Resolving forces vertically
Σ V= Σ Fy = 0
N-1000=0
N= 1000 N
2. Resolving forces horizontally
Σ H= Σ Fx = 0
600- f =0
f= 600 N
3. co-efficient of friction
µ= f/N=0.6

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2. A body of weight 100 N is placed on a rough horizontal plane is pulled by a force of 30N
inclined at 15 degrees with horizontal. Determine the co-efficient of friction.
Given:
W= 100 N
P= 30 N ( 15 deg horizontal)
µ=?
W=1000N
P=30 N
N
f = µN
Solution:
Σ V= Σ Fy = 0
N + 30 Sin15-100=0
N= 92.24 N
Σ H= Σ Fx = 0
-f+30cos15=0
f= 28.98 N
3. co-efficient of friction
µ= f/N=0.314
15 deg

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3. Determine the force P required to slide the block B in the arrangement shown in the figure. Also calculate the tension
in the string. Weight of blocks A and B are 500 N and 1000N respectively. Co-efficient of friction is 0.2 for all contact
surfaces.
Given:
WA= 500 N
WB= 1000 N
µ=0.2
P= ?
T=?
String (T)
30 deg
A
B
P
A
30 deg
T
WA= 500 N
N1
f=0.2N1
Motion trend
FBD of Block A:
Solution:
Σ H= Σ Fx = 0
T cos30 -0.2N=0
N1= 4.33T___(1)
Σ V= Σ Fy = 0
-500+Tsin30+N=0____(2)
But N 1=4.33T
Solution:
substituting and solving for T
T= 103.52N_____(3)
Hence by eqn 1, N1 =448.24N

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3. Determine the force P required to slide the block B in the arrangement shown in the figure. Also calculate the tension
in the string. Weight of blocks A and B are 500 N and 1000N respectively. Co-efficient of friction is 0.2 for all contact
surfaces.
Given:
WA= 500 N
WB= 1000 N
µ=0.2
P= ?
T=?
String (T)
30 deg
A
B
P
B
WB= 1000 N
N2
f=0.2N2
Motion trend
FBD of Block B:
Solution:
Σ H= Σ Fx = 0
-N1 -1000+ N2=0
N2= 1448.24 N___(4)
N1
f= 0.2N1
P
Solution:
Σ V= Σ Fy = 0
-P+ 0.2N1+ 0.2N2 = 0____(5)
P=379.30N

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4. A block weighing 2000N is attached to one end of chord which passes round a frictionless pulley as shown, carries a
load of 800 N at the other end and µ=0.35 . Determine value of P, if
1) the motion is impending towards left 2) the motion is impending towards right.
Given:
W= 2000 N
Wp= 800 N
µ=0.35
P= ?
T=?
String (T)
30 deg
800N
2000N
P
FBD of 800 N load: Solution:
Tension in string:
Σ V= Σ Fy = 0
T-800=0____(1)
T=800N
Since the string passes through the
pulley, tension remains same in string
on the other side of the pulley
Frictionless
pulley
T
T
800 N

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Given:
W= 2000 N
Wp= 800 N
µ=0.35
P= ?
T=?
String (T)
30 deg
800N
2000N
P
Σ H= Σ Fx = 0
-P+800cos30+0.35*1600=0
P=1252.82N___(3)
P
Solution:
Σ V= Σ Fy = 0
-2000+N1+800 sin30=0
N1 =1600 N ____(2)
Frictionless
pulley
T
2000 N
Motion trend
N1
f= 0.35N1
2000N
T=800N
30 deg
Case 1: Motion is impending towards left

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Given:
W= 2000 N
Wp= 800 N
µ=0.35
P= ?
T=?
String (T)
30 deg
800N
2000N
P
Σ H= Σ Fx = 0
-P+800cos30-0.35*1600=0
P=132.82N___(5)
P
Solution:
Σ V= Σ Fy = 0
-2000+N2+800 sin30=0
N2 =1600 N ____(4)
Frictionless
pulley
T
2000 N
Motion trend
N2
f= 0.35N1
2000N
T=800N
30 deg
Case 2: Motion is impending towards right

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Numerical Problems
Numerical on
Single blocks/connected bodies on inclined plane

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Numerical Problems- blocks on inclined plane
1. Determine the value of P so that the body will not impend down the plane. Also find the
value of P for the body to impend up the plane. µ=0.3
Given:
W= 100 N
µ=0.3
30 deg
P
Solution:
Σ V= Σ Fy = 0
N 1-100cos30=0
N1 = 86.6 N
W=100N
30 deg
P
W=100N
30 deg
Σ H= Σ Fx = 0
P+0.3N1-100Sin30=0
P=24.2N
Case 1: motion is impending down the plane

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Numerical Problems- blocks on inclined plane
1. Determine the value of P so that the body will not impend down the plane. Also find the
value of P for the body to impend up the plane. µ=0.3
Given:
W= 100 N
µ=0.3
30 deg
P
Solution:
Σ V= Σ Fy = 0
N 2 -100cos30=0
N2 = 86.6 N
W=100N
30 deg
P
W=100N
30 deg
Σ H= Σ Fx = 0
P-0.3N2-100Sin30=0
P=75.98N
Case 2: motion is impending up the plane

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Thank you for your Kind Attention
DEPARTMENT OF CIVIL ENGINEERING

Friction intro.pptx

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