1
ENGINEERING MECHANICS
[SUBJECT CODE-ME101] FIRST YEAR FIRST SEMESTER
2 SYLLABUS MODULE 1
Importance of Mechanics in engineering; Introduction to Statics; Concept of Particle and Rigid Body; Types of forces: collinear, concurrent, parallel, concentrated, distributed; Vector and scalar quantities; Force is a vector; Transmissibility of a force (sliding vector). Introduction to Vector Algebra; Parallelogram law; Addition and subtraction of vectors; Lami’s theorem; free vector; Bound vector; Representation of forces in terms of i, j, k; Cross product and Dot product and their applications.
Two dimensional force system; Resolution of forces; Moment; Varignon’s theorem; Couple; Resolution of a coplanar force by its equivalent force-couple system; Resultant of forces. MODULE 2
Concept and Equilibrium of forces in two dimensions; Free body concept and diagram; Equations of equilibrium.
Concept of Friction; Laws of Coulomb friction; Angle of Repose; Coefficient of friction MODULE 3
Distributed Force: Centroid and Centre of Gravity; Centroid of a triangle, circular sector, quadrilateral, composite areas consisting of above figures.
Moments of inertia: MI of plane figure with respect to an axis in its plane, MI of plane
figure with respect to an axis perpendicular to the plane of the figure; Parallel axis theorem; Mass moment of inertia of symmetrical bodies’ e.g. cylinder, sphere, cone.
Concept of simple stresses and strains: Normal stress, Shear stress, bearing stress, Normal strain, Shearing strain; Hooke’s law; Poisson’s ratio; Stress-strain diagram of ductile and brittle materials; Elastic limit; Ultimate stress; Yielding; Modulus of elasticity; Factor of safety. MODULE 4
Introduction to Dynamics: Kinematics and Kinetics; Newton’s laws of motion; Law of gravitation & acceleration due to gravity; Rectilinear motion of particles; determination of position, velocity and acceleration under uniform and non-uniformly accelerated rectilinear motion; construction of x-t, v-t and a-t graphs.
Plane curvilinear motion of particles: Rectangular components (Projectile motion); Normal and tangential components (circular motion).
Kinetics of particles: Newton’s second law; Equation of motion; D.Alembert’s principle and free body diagram; Principle of work and energy; Principle of conservation of energy; Power and efficiency.
3 LECTURE CONTENTS
SL Lecture No. Lecture Title Page
1. Lecture 1
Introduction to mechanics, Statics, Concept of Particle and Rigid body, Force, Transmissibility of force
6-7 2. Lecture 2 Classification and type of forces 8-11
3. Lecture 3
Introduction to Vector Algebra, Parallelogram Law, Addition and Subtraction of vector
12-16
5. Lecture 4
Lami’s theorem, Free vector, Bound vector, Representation of forces in term of I,j & k
17-18
5. Lecture 5 Cross product, Dot product, examples 19-20
6. Lecture 6 Examples 21-22
7. Lecture 7
Two dimensional force system, Resolution of forces, Moment, Varignon’s theorem
23-24
8. Lecture 8 Couple, Example 25-29
9. Lecture 9
Resolution of a coplanar force by its equivalent force-couple system,
Resolution of forces, Example 30-33
10. Lecture10 Example 34-38
11. Lecture11 Concept and equilibrium of forces, Free body diagram
39-40
12. Lecture12 Free body diagram , example 41-42
13 Lecture13 Equation of equilibrium, example 43-49
14 Lecture14
Concept of friction, Columb law of
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15 Lecture15 Coefficient of friction, Kinetic friction 55-59
16 Lecture16 Examples 60-68
17 Lecture17 Distributed force, Centroid & Centre of Gravity
69-73
18 Lecture18 Determination of Centroid & C.G. 74-78 19 Lecture19 Centroid of an arc, Centroid of a
triangle, Example
79-82
20 Lecture20 Examples 83-86
21 Lecture21 Moment of Inertia, Parallel axis theorem 87-90
22 Lecture22 Moment of inertia of plain figure, Perpendicular axis theorem, Radius of gyration
91-97
23 Lecture23 Examples 98-101
24 Lecture24 Concept of stress, different types of stresses & strains, Hook’s law
102-110
25 Lecture25 Possoin’ratio,Stress-Strain curve for ductile & brittle material, Example
111-117
26 Lecture26 Projectiles 118-120
27 Lecture27 Projectiles at inclined plane 121-126
28 Lecture28 Examples of Projectiles, Kinematics, Rectilinear motion
127-129
29 Lecture29 Relative motion and example 130-138
30 Lecture30 Kinetics of particle, curvilinear motion, example
5
31 Lecture31 Rolling motion 142-145
32 Lecture32 Solution to kinetic problem, 146-148
33 Lecture33 Solution to problems 149-151
34 Lecture34 Example 152-157
35 Lecture35 Work, Power, Energy 158-163
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Lecture 1 : Introduction to Mechanics
STATICS
Statics is the branch of mechanics concerned with the analysis of loads (force, torque/moment) on physical systems in static equilibrium, that is, in a state where the relative positions of
subsystems do not vary over time, or where components and structures are at a constant velocity. When in static equilibrium, the system is either at rest, or its center of mass moves at constant velocity. The study of moving bodies is known as dynamics.
By Newton's first law, this situation implies that the net force and net torque (also known as moment of force) on every body in the system is zero. From this constraint, such quantities as stress or pressure can be derived. The net forces equaling zero is known as the first condition for equilibrium, and the net torque equaling zero is known as the second condition for
equilibrium. See statically determinate. Solids
Statics is used in the analysis of structures, for instance in architectural and structural engineering. Strength of materials is a related field of mechanics that relies heavily on the application of static equilibrium. A key concept is the centre of gravity of a body at rest: it
represents an imaginary point at which all the mass of a body resides. The position of the point relative to the foundations on which a body lies determines its stability towards small
movements. If the center of gravity falls outside the foundations, then the body is unstable because there is a torque acting: any small disturbance will cause the body to fall or topple. If the centre of gravity falls within the foundations, the body is stable since no net torque acts on the body. If the center of gravity coincides with the foundations, then the body is said to be met stable.
Particle:
A body of infinitely small volume i.e. negligible dimensions but having mass concentrated at a point is called particle. Such a body cannot exist theoretically.
Rigid Body:
A rigid body may be defined as a body in which the relative positions of any two particles do not change under the action of forces. Nobody is perfectly rigid. A body when subjected to external forces, it must undergo some form of deformation, however small it may be.
7 Force
It is the agent which changes or tends to change the state of rest or motion of a body. The force is a vector quantity. The characteristics of force are magnitude, direction and point of application. The unit of force is ‘N’, ‘KN’ ‘Kgf’ etc.
Principle of transmissibility of Forces
The state of rest or of a motion of a rigid body is unaltered; if a force acting on a body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. If deformation of the body is to be considered, the law of transmissibility will not hold good. By transmission of force, only the state of the body is
unaltered, but not the internal stresses which may develop in the body. This force is a sliding vector as this has a unique line of action in space but not a unique point of application.
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Lecture 2 : Classification & Types Of Forces
1.1.1 CLASSIFICATION OF FORCE
The general classification of force is presented below:
Gravitational force Electromagnetic force Strong force Weak force
But engineering mechanics point of view, a system of forces can be classified as:
System of Forces
Coplanar Forces Non-Coplanar Forces (lying on the same plane) (do not lie on the same) plane)
Coplanar forces may be Non-Coplanar forces may be ■ Collinear ■ Concurrent ■ Non-Concurrent ■ Parallel Force ■ Concurrent ■ Non-Concurrent ■ Parallel ■ Non-Parallel
9 ■ Non-Parallel
Table 1.2 TYPES & CHARACTERISTICS OF FORCE SYSTEM
Force System Line Diagram Characteristics
Coplanar forces
Y
x
Lines of action of all forces lie on the same plane
Non-Coplanar forces y F1 x F fF z F2
Lines of action of all forces do not lie on the same plane
Collinear Forces Lines of action of all forces
lie on the same st. line
Concurrent Forces Lines of action of all forces
10 Parallel Forces
F1 F2 F3
Lines of action all forces are parallel to each other
According to the effect produced by the force a) External Force
This force is applied to externally to a body. b) Internal Force
This force is developed into the body to resist the deformation or change of shape of a body. c) Active Force
This force causes a body to move or to change its shape. d) Passive Force
This force represents the motion of a body. According to the nature of force
a) Action or Reaction Force
When two bodies come into contact with each other, each body will exert a force on the other body. Out of these forces one is known as action and other is known as reaction. It is seen that action and reaction are equal.
b) Attraction and Repulsion
The non-contact force exerted by one body on the another body without any visible medium are known as attraction or repulsion force e.g. magnetic force.
c) Tension or Compression
When a pull is applied on a structural member, it tends to elongate/increase in length, pull is known as tensile force. If a structure member is subjected to two equal and opposite pushes and the member tends to shorten/decrease in length, member is said to be in compression.
According to the force applied to a point or over an area
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b) Distributed Force: The force which is distributed over an area, that force is known as distributed force.
Multiple Choice Questions
1. Forces are called concurrent when their lines of action meet in a) One plane, b) One point, c) Different planes , d) Two points 2. Forces are called coplanar when all the forces acting on a body a) One plane, b) One point, c) Different planes, d) Different points 3. A force is completely defined when we specify
a) Its magnitude, b) Its direction, c) Its point of application, d) All of the above 4. The amount of matter contained in a body is called its
a) Mass, b) Volume, c) Weight, d) None of the above
5. Two non collinear parallel equal forces acting in opposite direction a) Balances each other, b) Constitute a couple, c) Constitute a moment d) Constitute a moment of couple
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Lecture 3 : Introduction To Vector Algebra
Introduction to Vector Algebra
We use two kinds of quantities in mechanics-scalars and vectors. Scalar quantities are those with which only a magnitude is associated. Examples of scalar quantities are time, volume, density, speed energy and mass. Vector quantities, on the other hand, possess direction as well as magnitude, and must obey the parallelogram law of addition. Examples of vector
quantities are displacement, velocity, acceleration, force, moment and momentum. Velocity is specified by a direction as well as speed. However, the mass of a body is completely specified by a magnitude, and hence mass is a scalar quantity. Since the weight of a body is the force with which it is attracted by the earth, weight has a downward direction and thus is a vector quantity. Since weight and mass are different physical concepts, they should not be measured in the same units. The gram is a unit of mass. The force with which the earth attracts a one-gram mass at a standard location sometimes is called a “gram-weight” of force. * Since weight is proportional to mass in any given locality, this experiment is not affected by the slight variations consequent to laboratory conditions
Parallelogram Law of forces
If two forces acting at a point be represented in magnitude, direction and sense by the two adjacent sides of a parallelogram, their resultant is represented in magnitude, direction and sense by the diagonal of the parallelogram passing through that point .
In the figure vectors P and Q are represent in B C Magnitude, direction and sense by OA and OB
respectively.
The resultant R is represented by OC in magnitude, Q R and direction.
α
We now find magnitude of resultant R. Ø P α O A D
From C we draw CD perpendicular to OA produced. Let α = angle between two forces P and Q = <AOB Now <DAC = <AOB = α (corresponding angle)
In parallelogram OACB, AC is parallel and equal to OB So, AC = Q
13 In ∆OCD, OC2 = OD2 +DC2
Or, R2 = (OA+AD)2 +DC2
Or, R2 = OA2 +2. OA.AD + AD2 +DC2 Or, R2 = P2+2PQcosα +Q2cos2α+ Q2 sin2α Or, R = (P2 + Q2 +2PQcosα) ½
Direction of resultant vector R
Let Ø = angle made by resultant with OA
Then from ∆OCD, tan Ø = CD/OD = Qsinα/ P+Qcosα So, Ø = tan¯1 [Qsinα/ P+Qcosα]
Composition and Resolution of Concurrent Forces by Vector Methods
In order to add scalar quantities, one has merely to make the algebraic addition. When one wishes to add two vector quantities, the process is more difficult because their directions must be considered. The vector sum of two vector quantities is the single vector quantity that would produce the same result as the original pair.
The addition of vector quantities is greatly simplified by representing the vector quantity graphically. A vector is the line segment whose length represents the magnitude of a vector quantity and whose direction is that of the vector quantity. The sense along the line is indicated by an arrow. For example, a force of 100lb. acting at an angle of 30° above the horizontal may be represented by the line OA. Fig. 1, which is 5 units long and has the correct direction. Each unit of length thus represents 20lb.
When vectors do not have the same line of action, their vector sum is not their algebraic sum but a geometric sum.
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This geometric sum may be determined by either graphical or analytical methods. Graphical methods are simple and direct but are limited in precision to that obtainable by drawing instruments. Analytical methods have no such inherent limitations. In this experiment both graphical and analytical methods will he applied to forces as examples of vector quantities, but the same methods apply to all vector quantities.
The vector sum, or resultant, of a set of forces is the single force that will have the same effect,
Vector Summation by Graphical Methods: As an example of vector addition let us consider
the case of two forces acting on a body in such a direction that the forces are concurrent, that is their lines of action, if projected would intersect at a point. The vectors OA and OB representing two such forces are shown in Fig. 2. Their vector sum or resultant R, is found by constructing a parallelogram having the two vectors as sides and drawing the concurrent diagonal, as shown in Fig. 3. This diagonal vector R represents in magnitude and direction the single force that is equivalent to the origina1 pair that is their vector sum. When the resultant of more than two vectors is to be obtained graphically a polygon method is used. This is illustrated in Fig. 4. The
vector A is first constructed by the use of a chosen scale and reference direction. Then, from the head of A, the vector B is drawn. It is clear that the vector M is the resultant of vectors A and B, since M would be the concurrent diagonal of a
parallelogram if such a parallelogram had been drawn, as was done in Fig. 3. Similarly, it follows that the vector R is the resultant of M and C or of A, B, and C. When the resultant of several forces is required this method is simpler than the
parallelogram method. It should be noted that when the parallelogram method is used, the arrows, with their tails together, all radiate from a common point. But in the polygon method the tail of the second arrow coincides with the head of the first, etc.
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Summation of Vectors by Analytical Methods: The resultant of two vectors may be
determined analytically by the use of the trigonometric laws of sines and cosines. Consider the vectors A and B in Fig. 5. The magnitude of the resultant R can be obtained by the application of the law of cosines:
R2 = A2 + B2 + 2ABcosβ (1)
The direction of R can then be obtained from the law of sines: Sinφ/Sinβ = B/R
Since sinβ = sinθ, Sinφ = (B/R) sinθ (2)
Components of Vectors: Any single force may be replaced by two or more forces whose joint
action will produce the same effect as the single force. These various forces are said to be components of the single force. The most useful set of components is usually a pair at right angles to each other, as shown in Fig. 6.
The force B is the resultant of Forces Bx and By. Therefore conditions are
unchanged by replacing the single force B by forces Bx and By, called their X- and Y- components. It is obvious from Fig. 6 that Bx = B cosβ and By = B sinβ.
Component Method for Addition of Vectors: Fig. 7 illustrates the component method of computing the resultant of A,
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B, and C. The X- axis is so chosen that it coincides with the vector A, and the vectors B and C are resolved into X- and Y-components. The three forces A, B, and C have been replaced by five forces (A has no Y- component). The slim of the component along either axis may be computed by algebraic addition. Calling the sum of the X-components Fx and the sum of the Y-components Fy, it follows that the resultant R is given by the equation
R2 = (FX) 2 + (FY) 2 (3)
and that the angle φ- the angle that R makes with the X-axis may be determined from the equation
tanφ = Fy/Fx (4)
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Lecture 4 : Lami’s Theorem
Lami’s Theorem Q
If a body is in equilibrium under the action of γ
three forces, each force is proportional to the P sine of the angle between other two forces. α β
Suppose the three forces P, Q & R are acting at a point O and they are in equilibrium as shown
in figure. According to the Lami’s theorem, R
P/sinα = Q/sinβ = R/sinγ
α Proof of Lami’s theorem
The three forces acting on a point are in (Π- α)
Equilibrium and hence they can be R Q represented by the three sides of a γ
triangle taken in the same order. In Π - γ figure .applying sine rule, we get Π –β
P/ sin (Π -α) = Q/sin (Π -β) = R/sin (Π - γ) β P Or P/sinα = Q/sinβ = R/sinγ
Free Vector
Situation in which vector may be positioned anywhere in space without loss or change of meaning provided that magnitude and direction are kept intact. It is not constrained to any particular direction and location. It can be moved anywhere in space without rotation. Example 1. The velocity of water particle (having turbulent motion).
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2. Couple (the change in location of application couple in a plane does not change its effect)
Bound Vector
A bound or fixed vector has a definite point of application. It is specified by magnitude, direction and its point of application. Change in the point of application of force will alter its effect.
y Representation of forces in terms of I,j & k
FY
β Many problems in mechanics require analysis in three dimensions and for such problem it is α
often necessary to resolve a force into its three FX x mutually perpendicular components. The force O F acting at point O has the rectangular components FZ
FX, FY, FZ.
γ z
Fx = F cosα, Fy = F cosβ, Fz = F cosγ Also F = √ (Fx2 + FY2 + FZ2)
The cosines of α, β and γ are known as the direction cosine of vector F and denoted by l= cosα, m= cosβ, n= cosγ
The three angles are related by, cos2α +cos2β +cos2γ = 1 or l2 +m2 + n2= 1 Let i= vector of unit length in the +tive x direction
j= vector of unit length in the +tive y direction k= vector of unit length in the +tive z direction The force F is represented by, F = Fxi + FY j+ Fzk Magnitude of unit vector F = l F l = √ (Fx2 + FY2 + FZ2 ) But Fx = F cosα, Fy = F cosβ, Fz = F cosγ
Substituting these values in earlier equation, we get
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Lecture 5 : Cross Product & Dot Product
Cross Product or Vector Product
The cross product of vectors A & B is a vector quantity and is defined as the product of the magnitude of the vector A, the magnitude of the vector B and the sine of the smaller angle between the two vectors. Its direction is perpendicular to the plane containing the vector.
Now if n is the unit vector which gives the n B
direction of the resultant vector R,
θ R = A X B = lA l IB l sinθ. n = AB sinθ. n A
Dot Product or Scalar Product
Dot product of A and B is a scalar quantity and is defined as the product of the magnitude the vectors and cosines of their included angle.
A. B = lA l IB l cosθ = AB cosθ Following points is to be noted
(i) When θ = 00, and vectors A & B are along same direction, A. B = AB cos0 = AB
(ii) When θ = 900, and vectors A & B are perpendicular to each other, A. B = AB cos90 = 0 (iii) A. B = A times projection of B on A projection of A
On B = B times projection of A on B
(iv) The angle between the vector A and B is O Cos θ =A. B / lAl lBl Projection of B on A
= A. B/AB
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The dot product is used to find the component of a vector along an arbitrary direction and to define the term work.
Example.1
Add a 20N force in the positive x direction to a 50N force at an angle 45° to axis in the first quadrant and directed away from origin.
F1 = 20N, F2 = 50N
B C We add vectorally, F1 + F2 = F 50N
To get the sum, we may use the law of cosines 45° F for the triangular portions of the parallelogram
by using the ∆OCA, α A O 20N lFl = √ (202 + 502 + 2.20.50.cos45)
= 65.68267N (Ans)
The direction of the vector may be described by giving angle and sense. The angle is determined by employing the law of sine.
For ∆OBA, 50/sinα = 65, 68267/sin135 Or α = 32.566° (Ans)
Example.2
Force A (given as a horizontal 10N force) and B (vertical) and add up to a force C that has magnitude of 20N. What is the magnitude of force B? C y 20N Let OA represents force A (FA) = 10N B A
Let OB “ “ B (FB) =? 10N Let OC “ “ C (FC) = 20N x B 20N OC2 = OA2 + OB2 B C Or OB= √ (202-102) FB 10N = 17.32N O A Tanθ = 17.32/10 or θ = 60°
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Lecture 6 : Solved Examples
Example3.
A force vector of magnitude 100N has a line of action with direction cosines, l= 0.7, m=0.2 & n=0.59 relative to reference xyz. The vector points away from the origin. What is the component of the force vector along a direction ‘a’ having direction cosines, l=-0.3, m=0.1 & n=0.95 for the xyz reference?
F = 100 (li +mj+nk) = 100 (0.7i +0.2j+0.59k) = 70i+20j+59k
Unit vector along ‘a’ = li+mj+nk = -0.3i+0.1j+0.55k R
Component of F along a is Fcosθ = [F] [a] cosθ = F. a θ = (70i+20J+59k) (-0.3i+0.1j+0.55k) N = 37.05N Fcosθ
Example.4
Making use of the cross product, give the unit vector n normal to the inclined surface ABC. Given OB=8cm, OC=10cm, <ABY=150°
From figure we can write, <OBA=180-150=30°, <BOA= 90° So AO = OB tan30 = 8 x tan30m =4.618m
Coordinate of the points A,B &C are A(4.618,0,0), (0,8,0) , (0,0,10) respectively. z
AB = -4.618i+8j+0k and AC = 4.618i+0j+10k C
Area = ½.AB.AC = ½(-4.618i+8j+0k) (4.618i+0j+10k) O B y
=(40i+23.09j+18k A
x 150° Unit Vector = (40i+23.09j+18k)/ √ (402 + 23.092 +182) = 0.804i+.464j+.371k) (ans)
Assignment 1. Subtract the 20N force in example.1 from50N force. 400N
22 2. What is the sum of forces transmitted by the structural rods to the pin at A?
3. Given the vectors
A= 6i+3j+10k, B=2i-5j+5k, C=5i-2i+7k What vector D gives the following results D.A=20, D.B=5, D.C=10
Multiple Choice Questions 1. Moment of a force is a
a) Scalar quantity, b) Vector quantity, c) Either a or b, d) None of the above 2. A vector having a unit magnitude is known as
a) Null vector, b)Free vector, c)Unit vector,d)None of the above 3. The dot product of two orthogonal vectors is
a) 1, b) o, c) No definite value, c) None of the above 4. Two vectors are said to be equivalent if they produces
a) Same magnitudes, b) Same directions, c) Same effect in a certain respect d) None of these
.
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Lecture 7 : Two Dimensional Forces
Two dimensional force system & Resolution of forces
When several forces of different magnitudes and directions act upon a body, they constitute a system of forces. If all the forces in a system lie in a single plane is known as coplanar force system.
If all the forces in a system lie on the same plane and the lines of action of all forces do not pass through a single point, the system is known as coplanar non-concurrent force system.
The most common two-dimensional y resolution of a force vector is into j rectangular components. It follows
from the parallelogram rule that the F vector F (in figure) may be written as FY
F=FX+Fy θ x FX I
Where Fx & Fy are vector components of F in the x and y directions. Each of the two vector components may be written as a scalar times the appropriate unit vector. In terms of the unit vector I and j, FX=Fxi & FY =Fyj and thus we may write
F = Fxi +Fyj
Where the scalars FX and FY are the x and y scalar components of the vector F.
The scalar components can be positive or negative, depending on the quadrant into which F points. For the force vector shown above the x and y scalar components are both positive are related to the magnitude and direction of F by
FX = Fcosθ & FY = Fsinθ F = √ (FX2 + Fy2), θ = tan¯1 (FX/ FY) Moment of a force
The product of the magnitude of a force and the perpendicular distance of the line .O
of action the force from appoint is known d as moment of a force about that point.
moment of a force about a point is the measure of its rotational effect. The
rotational effect of a force becomes very P important when we deal with non concurrent
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Moment ‘M’ of a force P about point O as shown in figure is given by, M = P x d where d is the perpendicular distance between the line of action of the force and the moment centre. The tendency of this moment is to rotate the body in anticlockwise direction about O.
Varignon's theorem
One of the most useful principles of mechanics is Varignon’s theorem.
The algebraic sum of the moments of a system of coplanar forces about a moment centre in their plane is equal to the moment of their resultant force about the same moment centre. Proof
According to the figure, let F be the Y Resultant of the forces F1 and F2
acting at A. Let us consider any point B lying in the plane of the forces, as a moment centre. Let I, l1 and l2 be the moment arms
of the forces F, F1 and F2 respectively F from the moment centre B. F2
θ2
We join AB and consider it as y-axis F1 And draw x-axis at right angle to it θ θ1
at A as shown in figure. O X FX2
FX1 FX
Now moment of the force F about B = F X l = F (BAcosθ) = BA (Fcosθ) = BAFX
Moment of the force F1 about B = F1 X l1 = F1 (BAcosθ1) = BA (F1 cosθ1) = BAFX1….1 Moment of the force F2 about B = F2 X l2 = F1 (BAcosθ2) = BA (F2 cosθ2) = BAFX2...2
Now adding equation 1 and 2, F1 X l1 + F2 X l2 = BA (FX1 + FX2)
But the sum of the x components of forces F1 and F2 = x components of the resultant F So FX = FX1 + FX2 or BAFx = BA (FX1 + FX2)
So from Fx l = F1 X l1 + F2 X l2 hence proved
If a system of forces consists of more than two forces, the above proof can be extended by taking into consideration all forces.
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Lecture 8 : Couples
Couple
Two parallel forces of equal magnitude 1 P d P but opposite in direction are separated .
by a finite distance to form a couple. d1 The algebraic sum of the forces
forming a couple is zero, which means
the translatory effect is zero. The algebraic d2 sum of the moments of the two forces of a
couple is independent of the position in the
plane of the couple, of the moment centre P d3 2 d4 and is equal to the moment centre and is
always equal to the product of the magnitude of the forces and the perpendicular distance d between the two forces.
Let the magnitude of the forces forming a couple P is P and the perpendicular distance between
the two forces is d. Considering the moment of the two forces forming a couple about point 1as shown. Le t the moment be M1. So M1 = Pd2 –Pd1 =P (d2 –d1) = P d
Now we consider the moment of the forces about point 2 as shown. Let M2 is the moment, then M2 = P (d3+d4) = P. d
The moment of a couple about any point is the same. Since the only effect of a couple is a
moment (rotational effect) and this moment is the same about any point, the action of a couple is unchanged if
(a) The perpendicular distance of the forces i.e. arm is turned in the plane of couple through any angle about one of its end.
(b) The magnitude of the forces and the arm of the couple both are changed in such a way that the moment of the couple remains unchanged.
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Example 1: A person is holding a 100N weight (that is roughly a 10kg mass) by a light weight (negligible mass) rod AB. The rod is 1.5m long and weight is hanging at a distance of 1m from the end A, which is on a table (see figure 6). How much force should the person apply to hold the weight?
Let the normal reaction of the table on the rod be N and the force by the point be F1. Then the two equilibrium conditions give
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(Note: If the brick did not provide friction, the force applied cannot be only in the vertical direction as that would not be sufficient to cancel the horizontal component of N). Let us see what happens if the brick offered no friction and we applied a force in the vertical direction. The fulcrum applies a force N perpendicular to the rod so if we apply only a vertical force, the rod will tend to slip to the left because of the component of N in that direction. Try it out on a smooth corner and see that it does happen. However, if the friction is there then the rod will not slip. Let us apply the equilibrium conditions in such a situation. The balance of forces gives
Let us choose the fulcrum as the point about which we balance the torque. It gives Then
The normal force and the frictional force can now be calculated with the other two equations obtained above by the force balance equation.
Example2: To balance a heavy weight of 5000 N, two persons dig a hole in the ground and put a pole of length l in it so that the hole acts as a socket. The pole makes an angle of 30° from the ground. The weight is tied at the mid point of the pole and the pole is pulled by two horizontal ropes tied at its ends as shown in figure 2. Find the tension in the two ropes and the reaction forces of the ground on the pole.
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To solve this problem, let me first choose a co-ordinate system. I choose it so that the pole is over the y-axis in the (y-z) plane (see figure 2).
The ropes are in (x.y) direction with tension T in each one of them so that tension in each is written as
You may be wondering why I have taken the tension to be the same in the two ropes. Actually it arises from the torque balance equation; if the tensions were not equal; their component in the x-direction will give a nonzero torque.
Let the normal reaction of the ground be (Nx, Ny, Nz). Then the force balance equation gives
Taking torque about point O and equating it to zero, we get
which gives
29
Lecture 9 : Resolution Of Coplanar Forces
Resolution of a coplanar force by its equivalent force-couple system
The effect of force acting on a body is the tendency to push or pull the body in the direction of the force, and to rotate the body about any fixed axis which does not intersect the line of force. We can represent this dual effect more easily by replacing the given force by an equal parallel force and a couple to compensate for the change in the moment of the force.
This replacement is illustrated in figure where the given force F acting at point A is replaced by an equal force F and at some point B and the counterclockwise couple M= F d.
B F . B F F
A. A d F
Resultant of forces
The most common type of force system occurs when the forces all act in a single plane, say, the x-y plane. We obtain the magnitude and direction of the resultant force R by forming the force polygon. Thus for any system of coplanar forces we may write
R = F1 +F2 + F3 + ………. =∑F
Rx = ∑FX RY = ∑FY where R = √ [(∑FX) ²+ (∑FY) ²]
30 Example 3:
The picture below shows the forces acting on a parked car. If the weight of the car acts exactly halfway between the two wheels and the weight is 1000lbs how much force is exerted on the rear wheel? What about the front wheel?
Writing the force equations
There are no forces in the x direction Writing moment equation about front wheel
Subbing RB back into the sum Fy
Please note that Rf and RB are distributed over two wheels. Each front wheel supports half of Rf and each back wheel supports half of RB.
Example 4. A weight of 15kN hangs from a point C, by two strings BC and AC (Fig. to Exmpl.5.1). Determine the tensions in the strings.
B B C B T1 T2 75° 135° 150° A 15KN
31
The load 15kN force acts downwards inducing tensions T1and T2 in the strings CB & CA (Fig to Exmpl.5)
Applying Lami’s theorem:
= sin =
Or, T1 = 15kN (sin150o) / sin75o = 7.76kN Ans. & T2 = 15kN (sin 135o) / sin 75o = 10.98kN Ans.
Example 5. ABCD is a square of 1m side. It is being acted upon by a number of forces as shown (Fig. to Exampl.4.3). The system is at equilibrium. Find the magnitude of
i) P & Q ii)The resultant couple. P D C Q 200kN A B 100KN 1m Fig. to Example. 5
Resolving the system of forces horizontally and computing Σ H = 0 100kN –100kN (cos45o) – P = 0
Or, P = 29.3kN ns.
Resolving the forces vertically and computing Σ V = 0 200kN –100kN (sin45o) –Q = 0 Or, Q = 129.3kN Ans. 15kN sin75o T1 sin150o T2 sin135o
32 ii) Couple
Moment of a couple = Σ M of constituents forces about any point
Σ MA = + 200kN x 1m + P x 1m = 200 + 29.3 kN = 229.3kN Ans. Since the moment is + tive, the couple is counterclockwise.
Example 6. At what point on the beam a weight of 2kN is to be placed so that one of the strings may just snap?
The weight of the beam 4kN acts halfway between A and B.
TA TB 3m x A D C B AC = BC 2kN 4kN Fig to Example 6
33
TA and TB are the tensions in the two strings, which balance the net downward force. When either TA or, TB becomes 3.5kN, the corresponding string snaps. Let string at end A snaps.
Taking moment about B and for balance: ΣMB = 0 3.5kN x 3m = 2kN x (3 – x) + 4kN x 1.5m Or, x = 0.75m Ans.
34
Lecture 10 : Solved Examples
Example7.The screw eye in Fig. is subjected to two forces. Determine the magnitude and direction of the resultant force.
10° Ø
Parallelogram Law. The parallelogram law of addition is shown in Figure Two unknowns are the magnitude of FR and the angle θ(theta).
FR = √ (100 N) 2 + (150 N) 2 – 2x100 Nx150 N cos 115° =213N (Ans)
The angle is determined by applying the law of sines, using the computed value of FR. 150 N 212.6 N --- = --- sin θ sin 115° 150 N sin θ = --- (0.9063) 212.6 N F2=150N F1= 100N 15° 15° 10° θ 90°-25°=65° 65° 125° 150N 100N FR
35 θ = 39.8°
Thus, the direction Ø (phi) of FR, measured from the horizontal, is Ø = 39.8° + 15.0° = 54.8° (ans)
Example8.The force F acting on the frame shown in Figure has a magnitude
of 500 N and is to be resolved into two components acting along members AB and AC. Determine the angle measured below the horizontal, so that the component is directed from A toward C and has a magnitude of 400 N.
By using the parallelogram law, the vector addition of the two components yielding the resultant is shown in Figure A. Note carefully how the resultant force is resolved into the two components FAB and FAC which have specified lines of action. The angle Ø can be determined by using the law of sines. The corresponding vector triangle is shown in Fig.
400 N 500N --- = --- sin Ø sin 60° 400 N sin Ø = --- x sin 60 = 0.6928 or Ø = 43.9° 500 N Hence, θ = 180° - 60° - 43.9° = 76.1° (ans)
Using this value for θ we apply the law of cosines or the law of sines to find the value of FAB which has a magnitude of 561N .
A B C 500N θ 30° FAC=400N FAB 500N θ 60° 30° FAC=400 500N FAB θ 60° Ø FIG A FIG B
36
Example9. The system in Figure shown below is in equilibrium with the string in the center exactly horizontal. Find (a) tension T1, (b) tension T2, (c) tension T3 and (d) angle θ.
T3 T1 T2 T2 (b) 40N (a) 50N
(a) Forces at the left junction of the strings. (b) Forces acting at the right junction of the strings..
We have to solve for four unknowns (T1, T2, T3 and θ).
We consider the points where the strings meet; the left junction is shown in Figure (a). Since a string under tension pulls inward along its length with a force given by the string tension, the forces acting at this point are as shown.
Since this junction in the strings is in static equilibrium, the (vector) sum of the forces acting on it must give zero. Thus the sum of the x components of the forces is zero: −T1 sin 35_ + T2 = 0 ………1
The sum of the y components of the forces is zero: +T1 cos 35- 40N = 0 ………2
Now we look at the right junction of the strings; the forces acting here are shown in Figure (b). Again, the sum of the x components of the forces is zero:
−T2 + T3 sin θ = 0 …………3
The sum of the y components of the forces is zero: +T3 cos θ − 50N = 0……….4
And at this point we are done with the physics because we have four equations for four
40N 50N
T1
T2
T3
37 unknowns. We will do algebra to solve for them. From Eq. 2, we get, T1 = 40N/cos 35 = 48.8N
and then Eq. 1 gives us T2, 2 = T1 sin 35 = (48.8N) sin 35 = 28.0N . We now rewrite Eq. 3. As: T3 sinθ = T2 = 28.0N……..5
And Eq. 4 as: T3 cosθ = 50.0N……….6
Now if we divide the left and right sides of 5 by the left and right sides of 6 we get: tanθ = (28.0N)/(50.0N) = 0.560
And then θ = tan− 1(0.560) = 29.3
Finally, we get T3 from Eq. 3.9: T3 = 50.0Ncos 29.3 = 57.3N Summarizing, we have found:
T1 = 48.8N, T2 = 28.0N, T3 = 57.3N, θ= 29.3
Assignment
1A. The vertical force F acts downward at A on the two-membered frame. Determine the magnitudes of the two
components of F directed along the axes of AB and AC. Set F=500N, <ABC=45°, <ACB=60°
1B. Solve Prob. 1A with F = 350 lb.
2. For the stepladder shown in Figure, sides AC and CE are
each 8.0 ft long and hinged at C. Bar BD is a tie–rod 2.5 ft long, halfway up. A man weighing 192 lb climbs 6.0 ft along
the ladder. Assuming that the floor is frictionless and
neglecting the weight of the ladder, find (a) the tension B in the tie–rod and the forces exerted on the ladder by the
floor at (b) A and (c) E. Hint: It will help to isolate parts A E of the ladder in applying the equilibrium conditions.
Multiple Choice Questions
1. A force can be replaced by
a) A force of same magnitude and couple
b) A force-couple combination so that equilibrium is maintained c) A force of different magnitude and couple
d) None of the above
A
B
C C
38
2. Two non-collinear parallel equal forces acting in opposite direction a) Balance each other
b) Constitute a couple c) Constitute a moment d) None of the above
3. The force polygon for coplanar concurrent forces a) Must Close
b) Must not close c) Anyone of the above d) None of the above 4. Moment of a force
a) Varies directly with its distance from the pivot b) Varies inversely with its distance from the pivot c) Is independent of its distance from the pivot d) None of the above
5. A couple consists of
a) Two like parallel forces of same magnitude b) Two like parallel forces of different magnitude c) Two unlike parallel forces of same magnitude d) Two unlike parallel forces of different magnitude
39
Lecture 11 : Concept Of Equillibrium
Equilibrium: Many problems that concern the physicist and engineer involve several forces acting on a body under circumstances in which they produce no change in the motion of the body. This condition is referred to as equilibrium. The body does not necessarily have to be at rest, but its motion must retain the same velocity; hence both magnitude and direction of motion are unchanged.
First Condition for Equilibrium: Insofar as linear motion is concerned, a body is in equilibrium if there is no resultant force acting upon it that is if the vector sum of all the forces is zero. This statement is called the first condition for equilibrium. This condition is satisfied if the vector polygon representing all the external forces acting on the body is a closed figure. Analytically
this condition is satisfied if each set of rectangular components of the forces separately adds to zero, or
Rx = ΣFx = 0 (5)
Ry = ΣFy = 0 (6)
Mechanical equilibrium
A standard definition of static equilibrium is:
A system of particles is in static equilibrium when all the particles of the system are at rest and the total force on each particle is permanently zero.[1]
This is a strict definition, and often the term "static equilibrium" is used in a more relaxed manner interchangeably with "mechanical equilibrium", as defined next.[2]
A standard definition of mechanical equilibrium for a particle is:
The necessary and sufficient conditions for a particle to be in mechanical equilibrium are that the net force acting upon the particle is zero.[3]
The necessary conditions for mechanical equilibrium for a system of particles are: (i)The vector sum of all external forces is zero;
(ii) The sum of the moments of all external forces about any line is zero.[3] As applied to a rigid body, the necessary and sufficient conditions become:
A rigid body is in mechanical equilibrium when the sum of all forces on all particles of the system is zero, and also the sum of all torques on all particles of the system is zero.
40
A rigid body in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.
However, this definition is of little use in continuum mechanics, for which the idea of a particle is foreign. In addition, this definition gives no information as to one of the most important and interesting aspects of equilibrium states – their stability.
An alternative definition of equilibrium that applies to conservative systems and often proves more useful is:
A system is in mechanical equilibrium if its position in configuration space is a point at which the gradient with respect to the generalized coordinates of the potential energy is zero.
41
Lecture 12 : Free Body Diagram
Free body diagram
Block on a ramp (top) and corresponding free body diagram of just the block (bottom).A free body diagram is a pictorial representation often used by physicists and engineers to analyze the forces acting on a free body. A free body diagram shows all contact and non-contact forces acting on the body. Drawing such a diagram can aid in solving for the unknown forces or the equations of motion of the body. Creating a free body diagram can make it easier to understand the forces, and moments, in relation to one another and suggest the proper concepts to apply in order to find the solution to a problem. The diagrams are also used as a conceptual device to help identify the internal forces—for example, shear forces and bending moments in beams— which are developed within structures.
Construction
A free body diagram consists primarily of a sketch of the body in question and arrows representing the forces applied to it. The selection of the body to sketch may be the first important decision in the problem solving process. For example, to find the forces on the pivot joint of a simple pair of pliers, it is helpful to draw a free body diagram of just one of the two pieces, not the entire system, replacing the second half with the forces it would apply to the first half.
What is included
The sketch of the free body need include only as much detail as necessary. Often a simple outline is sufficient. Depending on the analysis to be performed and the model being employed, just a single point may be the most appropriate.
All external contacts, constraints, and body forces are indicated by vector arrows labeled with appropriate descriptions. The arrows show the direction and magnitude of the various forces. To the extent possible or practical, the arrows should indicate the point of application of the force they represent.
Only the forces acting on the object are included. These may include forces such as friction, gravity, normal force, drag, or simply contact force due to pushing. When in a non-inertial reference frame, fictitious forces, such as centrifugal force may be appropriate.
A coordinate system is usually included, according to convenience. This may make defining the vectors simpler when writing the equations of motion. The x direction might be chosen to point down the ramp in an inclined plane problem, for example. In that case the friction force only has an x component, and the normal force only has a y component. The force of gravity will still have components in both the x and y direction: mgsinθ in the x and mgcosθ in the y, where theta is the angle between the ramp and the horizontal.
42 What is excluded
All external contacts and constraints are left out and replaced with force arrows as described above.
Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the ball applies a force to the table, and the table applies an equal and opposite force to the ball. The FBD of the ball only includes the force that the table causes on the ball.
Internal forces, forces between various parts that make up the system that is being treated as a single body, are omitted. For example, if an entire truss is being analyzed to find the reaction forces at the supports, the forces between the individual truss members are not included.
Any velocity or acceleration is left out. These may be indicated instead on a companion diagram, called "Kinetic diagrams", "Inertial response diagrams", or the equivalent.
Assumptions
The free body diagram reflects the assumption and simplifications made in order to analyze the system. If the body in question is a satellite in orbit for example, and all that is required is to find its velocity, then a single point may be the best representation. On the other hand, the brake dive of a motorcycle cannot be found from a single point, and a sketch with finite dimensions is required.
Force vectors must be carefully located and labeled to avoid assumptions that presuppose a result. For example, in the accompanying diagram of a block on a ramp, the exact location of the resulting normal force of the ramp on the block can only be found after analyzing the motion or by assuming equilibrium.
Other simplifying assumptions that may be considered include two-force members and three-force members.
43
Lecture 13 : Equation Of Equilibrium
Equation of equilibrium
When a body is in equilibrium, resultant of all forces and moments acting on that body is zero. Stated in another way, a body is in equilibrium if all forces and moments applied to it are in balance. These requirements are contained in the vector equations of equilibrium which in two dimensions may be written in scalar from as
∑FX = 0, ∑FY = 0, ∑MO =0
The third one represents the zero sums of the moments of all forces about any point O on or off the body.
Let's look at a truss
P Q S
A B
C D
44 P Q S Ax C D W Ay By y y x x y A A F A F B M => = Σ => = Σ => = Σ 0 0 0
Additional equations could be written. 0
=
ΣMB Does not provide any new info. This is not an independent equation. We can use ΣMB =0to replace one of the above 3.
Example 1.Given Find: Reactions 2 m 2 kN 2 kN 1.5 m 1.5 m A B kN 4 0 2 2 0 − = = + + = Σ Ax Ax x R R F 0 3 0 = + + − = Σ B Ay y R R F
45 FBD
Substituting RB into y-equation: RAy = -1.5 KN
Example2.Given Find: Reactions at A FBD 2 m 2 kN 2 kN 1.5 m 1.5 m A B RB RAy RAx 2 ft 3 ft 2000 lbs B A C 3000 ft lbs 2 ft 3 ft 2000 lbs B A C 3000 ft lbs RAy RAx MA 0 0 = = Σ Ax x R F lbs 2000 0 2000 0 = = − = Σ Ay Ay y R R F lbs ft 000 , 7 0 ) 5 ( 2000 3000 0 = = − + = Σ A A A M M M
46
A simple free body diagram, shown above, of a block on a ramp illustrates this.
• All external supports and structures have been replaced by the forces they generate. These include:
• Mg: the product of the mass of the block and the constant of gravitation acceleration: its weight.
• N: the normal force of the ramp. • Ff: the friction force of the ramp.
• The force vectors show direction and point of application and are labeled with their magnitude.
• It contains a coordinate system that can be used when describing the vectors.
Example 3. Two loads 400N and 500N are suspended in a vertical plane by three springs as shown in Figure. Find the tension in the strings
47 Fig for Example.1
Obviously tension in OB = TOB= 500N Resolving the forces at O vertically POA.sin300 + POB.sin30o = 400N Or, POA = 300N
Resolving the forces at O horizontally POC + POA.cos30o = POB.cos30O
or, POC + 300N(cos30o) = (500N)cos30o POC = 100√3 N
Tension in OA = 300N Tension in OB = 500N Tension in OC = 100√3 N
Assignment
Problem. 1 An electric light fixture weighing 15 Newton hangs from a point C, by two strings AC
and BC. AC is inclined at 60° to the horizontal and BC at 45° to the vertical as shown in Fig. /Prob.1.Using Lami’s theorem or otherwise determine the forces in the strings AC and BC.
400N 500N B C O 30° 30° A
48 15N
Example2 Given: The loading car weight is 5500 lbs and its CG is at point G.
Find: tension in cable and reactions at wheels.
Multiple Choice Questions
1. Free body diagram can be applied only in a) Dynamic equilibrium problem
b) Static equilibrium problem
c) Both static & dynamic equilibrium problems d) None of these
2. If the body is in equilibrium ,we may conclude that a) No force acting in it
b) Moment of all forces about any point is zero c) The resultant of all forces acting on it is zero d) Both b&c
B
A
45O 60OC
30" 25o G 25" 24" 25" T49
3. The algebraic sum of the moments of two forces about any point in their plane is equal to the moment of their resultant about that point is known as
a) Principle of moments b) Varignon’s theorem c) Lamis theorem d) None of these
4. Two coplanar couples having equal and opposite moments a) Produce a couple and an unbalanced force
b) Are equivalent c) Balance each other d) None of these
5. A free body diagram of a body represents
a) With its surroundings and external forces acting on it
b) Isolated from its surroundings and all external forces acting on it c) Isolated from all external actions
d) None of these
6.The force which meet at one point and their lines of action also lie on the same plane are known as…………forces. a) Coplanar concurrent b) Coplanar non-concurrent c) Non-coplanar non-concurrent d) None of these
50
Lecture 14 : Friction
Friction is the force distribution at the surface of contact between two bodies that prevents or impedes sliding motion of one body relative to the other. This force distribution is tangent to the contact surface and has, for the body under consideration, a direction at every point in the contact surface that is in opposition to the possible or existing slipping motion of the body at that point.
Types of friction
• Dry friction resists relative lateral motion of two solid surfaces in contact. Dry friction is also subdivided into static friction between non-moving surfaces, and kinetic friction (sometimes called sliding friction or dynamic friction) between moving surfaces.
• Lubricated friction or fluid friction resists relative lateral motion of two solid surfaces separated by a layer of gas or liquid.
• Fluid friction is also used to describe the friction between layers within a fluid that are moving relative to each other.
• Skin friction is a component of drag, the force resisting the motion of a solid body through a fluid.
• Internal friction is the force resisting motion between the elements making up a solid material while it undergoes deformation.
51
Consider a solid block of mass m resting on a horizontal surface as shown. Assume that the contacting surfaces are rough. As we gradually increase the load, the block remains static till the load reaches a threshold value. Since the force in the x-direction has to be balanced, it is apparent that as the magnitude of F increases from zero, the friction force also increases. Friction force, is thus, self adjusting. However, the friction force cannot increase beyond a limit. Thus there is a limiting value of friction. The maximum value of friction force, which comes into play, when the motion is impending, is known as limiting friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional force is called static friction, which may have any value between zero and the limiting friction. If the value of the applied force exceeds the limiting friction, the body starts moving over the other body and the frictional resistance experienced by the body while moving is known as Dynamic friction. Dynamic friction is found to be less than limiting friction. See the following animation to understand the phenomenon of dry friction.
It is experimentally found that the magnitude of limiting friction bears a constant ratio to the normal relation between the two surfaces and this ratio is called coefficient of Friction.
Coefficient of friction = = µ
Where F is limiting friction and N is the normal reaction between the contact surfaces.
52 Columb Laws of friction:
(1) The force of friction always acts in a direction opposite to that in which the body tends to move.
(2) Till the limiting value is reached, the magnitude of friction is exactly equal to the force which tends to move the body.
(3) The magnitude of the limiting friction bears a constant ratio to the normal reaction between the two surfaces.
(4) The force of friction depends upon the roughness/smoothness of the surfaces. (5) The force of friction is independent of the area of contact between the two surfaces.
Angle of static friction:
Consider the block on the following surface.
The free body diagram is shown. The direction of resultant R measured from the direction of N is specified by tan a=F/N. When the friction force reaches its limiting static value Fmax, the angle a reaches a maximum.
53 Value of fs. Thus
tan fs = ms
The angle fs is called the angle of static friction. Angle of kinetic friction:
When slippage is occurring, the angle a has a value fR corresponding to the kinetic friction force. tan fR = mR
Cone of friction:
When a body is having impending motion in the direction of P the frictional force will be the limiting friction and the resultant reaction R will make limiting friction angle a with the normal as shown in the following figure. If the body is having impending motion in some other direction, again the resultant reaction makes limiting frictional angle a with the normal in that direction. Angle of Repose:
The maximum inclination of the plane on which a body, free from external forces, experiences repose (sleep) is called Angle of Repose.
54
Now consider the equilibrium of the block shown above. Since the surface of contact is not smooth, not only normal reaction, but frictional force also develops. Since the body tends to slide downward, the frictional resistance will be up the plane.
∑forces normal to the plane =0, gives
N=Wcos θ …………1 ∑forces normal to the plane =0, gives
F=Wsin θ ………2
Dividing equ (2) by equ (1), we get
If N is the value of normal force when motion is impending, frictional force will be µN and hence
Hence, to avoid free sliding, the inclination angle should be less than the friction angle.
55
Lecture 15 : Coefficient Of Friction
Coefficient of friction
The coefficient of friction (COF), also known as a frictional coefficient or friction coefficient, symbolized by the Greek letter µ, is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient of friction, while rubber on pavement has a high coefficient of friction.
The coefficient of friction is an empirical measurement – it has to be measured experimentally, and cannot be found through calculations. Rougher surfaces tend to have higher effective values. Most dry materials in combination have friction coefficient values between 0.3 and 0.6. Values outside this range are rarer, but teflon, for example, can have a coefficient as low as 0.04. A value of zero would mean no friction at all, an elusive property – even magnetic levitation vehicles have drag. Rubber in contact with other surfaces can yield friction coefficients from 1 to 2. Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. For example, silicone rubber or acrylic rubber-coated surfaces have a coefficient of friction that can be substantially larger than 1.
Both static and kinetic coefficients of friction depend on the pair of surfaces in contact; their values are usually approximately determined experimentally. For a given pair of surfaces, the coefficient of static friction is usually larger than that of kinetic friction; in some sets the two coefficients are equal, such as teflon-on-teflon.
Kinetic friction
Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). The coefficient of kinetic friction is typically denoted as µk, and is usually less than the coefficient of static friction for the same materials.
Examples of kinetic friction:
• Kinetic friction is when two objects are rubbing against each other. Putting a book flat on a desk and moving it around is an example of kinetic friction.
56
• Fluid friction is the interaction between a solid object and a fluid (liquid or gas), as the object moves through the fluid. The skin friction of air on an airplane or of water on a swimmer are two examples of fluid friction. This kind of friction is not only due to rubbing, which generates a force tangent to the surface of the object (such as sliding friction). It is also due to forces that are orthogonal to the surface of the object. These orthogonal forces significantly (and mainly, if relative velocity is high enough) contribute to fluid friction. Fluid friction is the classic name of this force. This name is no longer used in modern fluid dynamics. Since rubbing is not its only cause, in modern fluid dynamics the same force is typically referred to as drag or fluid resistance, while the force component due to rubbing is called skin friction. Notice that a fluid can in some cases exert, together with drag, a force orthogonal to the direction of the relative motion of the object (lift). The net force exerted by a fluid (drag + lift) is called fluid dynamic force (aerodynamic if the fluid is a gas, or hydrodynamic if the fluid is a liquid).
Application of Friction 1) Wedges
Wedges are small pieces of material with two of its opposite surfaces not parallel. They are used to lift heavy blocks, machinery, precast beam etc., slightly, required for final alignment or to make place for inserting lifting devices. The weight of the wedge is very small compared to the weight lifted. Hence, in all the problems, weight of wedges may be neglected. The following figure is showing a wedge:
57
The body is acted upon by three forces: weight W, R1 (the resultant of normal wall reaction and friction force) and R2 ( the resultant of the normal reaction of the wedge and the friction force). In the free body diagram, the resolved components of R1and R2 are shown by thin lines. If the friction angle is f, the R1and R2 will make angle with the respective normals to the surfaces. Note that coefficient of friction is given by .
As the body is acted by three non-parallel forces, the forces must be coplanar and concurrent. The relation between the forces can be found by Lami's theorem. The following figure shows the three forces meeting at a common point:
58 By Lami's theorem:
Now, let us make the free body diagram of the wedge. Three forces acting on the wedge are shown. R1and R2 are the resultants of normal and surface forces.
59 The Lami's theorem gives us:
This analysis pertains to load being lifted by the wedge. If the load is lowered, the direction of friction forces and P will reverse. The analysis is similar, except that f will be replaced by - . For , P will be positive. That is some force will be needed to lower the load. In other words, without applied P, the load W will not get lowered and the wedge is called self-locking.
