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ISBN : 978-81-909042-4-7 - www.airwalkpublications.com

ANNA UNIVERSITY - R2013 GE6253 ENGINEERING MECHANICS UNIT I: BASICS AND STATICS OF PARTICLES 12 Introduction Units and Dimensions Laws of Mechanics Lami s theorem, Parallelogram and triangular Law of forces Vectorial representation of forces Vector operations of forces -additions, subtraction, dot product, cross product Coplanar Forces rectangular components Equilibrium of a particle Forces in space Equilibrium of a particle in space Equivalent systems of forces Principle of transmissibility. UNIT II EQUILIBRIUM OF RIGID BODIES 12 Free body diagram Types of supports Action and reaction forces stable equilibrium Moments and Couples Moment of a force about a point and about an axis Vectorial representation of moments and couples Scalar components of a moment Varignon s theorem Single equivalent force -Equilibrium of Rigid bodies in two dimensions Equilibrium of Rigid bodies in three dimensions UNIT III PROPERTIES OF SURFACES AND SOLIDS 12 Centroids and centre of mass Centroids of lines and areas - Rectangular, circular, triangular areas by integration T section, I section, - Angle section, Hollow section by using standard formula Theorems of Pappus Area moments of inertia of plane areas Rectangular, circular, triangular areas by integration T section, I section, Angle section, Hollow section by using standard formula Parallel axis theorem and perpendicular axis theorem Principal moments of inertia of plane areas Principal axes of inertia-mass moment of inertia mass moment of inertia for prismatic, cylindrical and spherical solids from first principle Relation to area moments of inertia. UNIT IV DYNAMICS OF PARTICLES 12 Displacements, Velocity and acceleration, their relationship Relative motion Curvilinear motion - Newton s laws of motion Work Energy Equation Impulse and Momentum Impact of elastic bodies. UNIT V FRICTION AND ELEMENTS OF 12 RIGID BODY DYNAMICS Friction force Laws of sliding friction equilibrium analysis of simple systems with sliding friction wedge friction - Rolling resistance -Translation and Rotation of Rigid Bodies Velocity and acceleration General Plane motion of simple rigid bodies such as cylinder, disc/wheel and sphere. (iii)

CONTENTS Chapter 1 - www.airwalkpublications.com Basics and Statics of Particles 1.0 Introduction... 1.1 1.1 Classification of Engineering Mechanics... 1.1 1.2 Fundamental Concepts of Mechanics... 1.3 1.3 Units and Dimensions... 1.4 1.4 Laws of Mechanics or Basic Principles of Mechanics... 1.13 1.4.1 Newton s Three Fundamental Laws... (a) Newton s First Law... 1.13 (b) Newton s Second Law... 1.14 (c) Newton s Third Law... 1.15 1.4.2 Newton s Law of Gravitation... 1.15 1.4.3 Parallelogram Law of Forces... 1.16 1.4.4 Triangular Law of Forces... 1.16 1.4.5 Polygon Law of Forces... 1.18 1.4.6 Law (or) Principle of Transmissibility of Forces... 1.19 1.4.7 Law of conservation of energy... 1.20 1.4.8 Principle of work and energy... 1.21 1.5 Force and Force System... 1.21 (a) External forces... 1.23 (b) Body forces... 1.24 1.5.3 Free body diagram... 1.24 1.5.5 Types of force system... 1.26 (a) Coplanar forces... 1.27 (b) Concurrent forces.... 1.27 (vi)

(c) Coplanar - concurrent force system... 1.27 (d) Non concurrent forces... 1.28 (e) Coplanar - Non concurrent forces... 1.28 (f) Collinear forces... 1.28 (g) Parallel forces... 1.28 (h) Non coplanar force system... 1.29 (i) Non coplanar concurrent forces... 1.29 (j) Non coplanar Non concurrent forces... 1.30 (k) Non coplanar parallel forces... 1.30 (l) Non coplanar Non concurrent and Non parallel forces... 1.30 1.5.6 Statics of particles - Composition of forces 1.31 1.5.7 Resultant of two coplanar concurrent forces... 1.31 1.5.8 Rectangular components of force system (Resolution)... 1.36 1.5.9 Resultant of several concurrent forces... 1.38 1.5.10 Summary... 1.40 Solved Problems... 1.42 to 1.64 1.6 Equilibrium of Particle... 1.65 1.6.1 Equilibrant... 1.65 1.6.2 Equations of Equilibrium of a particle... 1.66 1.6.3 Equilibrium of two force body... 1.67 1.6.4 Equilibrium of a three force body... 1.67 1.6.5 Condition for Three forces in equilibrium 1.68 1.6.6 Lami s Theorem... 1.68 Solved Problems... 1.69 to 1.96 1.6.7 Graphical method... 1.96 Solved Problems... 1.97 to 1.102 1.7 Review of Vector Algebra... 1.103 (vii)

- www.airwalkpublications.com 1.7.1 Scalar and vector quantities... 1.103 1.7.2 Types of vectors... 1.104 1.7.3 Vector operations... 1.105 Addition of vectors... 1.105 Parallelogram law of vector addition... 1.105 Triangular law of vector addition... 1.106 Vector subtraction... 1.106 Polygon law of vector addition... 1.107 1.7.4 Vectorial representation of forces... 1.107 1.7.5 Dot product or Scalar product... 1.109 1.7.6 Cross or vector product... 1.111 Solved Problems... 1.114 to 1.119 Equilibrium of Particles in Space and Rigid bodies in Three dimensions... 1.120 1.7.8 Resolution of Vector... 1.120 Solved Problems... 1.122 to 1.127 1.8 Forces in Space... 1.127 1.8.1 Forces in three dimensional space (Component of force in space)... 1.129 1.8.2 Resultant of several concurrent forces... 1.129 1.8.3 Equilibrium of particle in space... 1.130 1.8.4 Equivalent system of forces... 1.131 Solved Problems... 1.131 to 1.176 Chapter 2 Equilibrium of Rigid Bodies 2.1 Introduction... 2.1 2.2 Free Body Diagram... 2.1 (viii)

2.3 Types of Forces... 2.3 2.4 Types of Supports and their Reactions... 2.4 1. Frictionless Support... 2.5 2. Roller and Knife Edge Supports... 2.5 3. Hinged Support... 2.5 4. Built-in Support or Fixed support... 2.6 2.4.1 Various types of Loadings... 2.9 1. Concentrated or Point Load... 2.9 2. Uniformly Distributed Load (UDL)... 2.9 3. Uniformly Varying Load... 2.9 2.5 Requirements of Stable Equilibrium... 2.19 2.6 Equilibrium of Rigid Bodies in Two Dimensions... 2.21 2.7 Moments and Couples... 2.25 2.7.1 Moment of a Force... 2.25 2.7.2 Principle of Moments (or Varignon s Principle)... 2.26 2.7.3 Couple... 2.27 2.7.4 Moment of a Couple... 2.28 2.7.5 Resolution of a Force into a Force and a Couple... 2.29 2.7.6 Parallel force system... 2.72 2.8 Moment of a Force about a Point [Vectorial Treatment]... 2.93 2.9 Moment of a Force about a Given Axis... 2.97 2.9.1 Varignon s - Theorem... 2.100 2.10 Rectangular Components of the Moment of a Force... 2.100 2.11 Moment of force about origin O in three dimensions (Simple Method)... 2.109 (ix)

- www.airwalkpublications.com 2.11.1 Moment about any other point (other than origin O)... 2.110 2.11.2 Moment of a Force about an axis.... 2.127 2.12 Moment of a Couple... 2.139 2.13 Addition of Couples... 2.140 2.14 Vector Representing Couple... 2.141 2.15 Couple in three Dimensional Space... 2.144 2.16 Resultant of Non Current Non Coplanar System... 2.147 2.17 Equivalent System of Forces... 2.148 2.18 Single Equivalent Force... 2.148 2.19 Equilibrium of Rigid Body in two Dimensions... 2.156 2.20 Equilibrium of Rigid Body in three Dimensions... 2.157 2.21 Resultant Force of Non-concurrent and Non-parallel Forces... 2.157 2.22 Equilibrium of Non-concurrent and Non-parallel Spatial Forces... 2.162 Chapter 3 Analysis of Plane Truss 3.1 Introduction... 3.1 3.2 The Classifications of Structure... 3.1 3.2.1 Frame... 3.1 3.2.2 Truss... 3.2 3.2.3 Classification of Dimensional configuration... 3.3 3.2.4 Classification of Determinacy... 3.4 (x)

Statically determinate truss... 3.4 Statically indeterminate truss... 3.4 3.2.5 Classification of supports... 3.4 Simply supported truss... 3.4 Cantilever truss... 3.5 3.3 Perfect Frame... 3.5 3.3.1 Mathematical equation for perfect frame. 3.6 3.3.2 Imperfect frame... 3.6 3.4 Trusses for Various Applications... 3.7 3.5 Assumptions for Analysing Perfect Frame... 3.8 3.6 Analysis of Forces in a Frame... 3.8 3.6.1 Determination of the reactions at the supports... 3.8 3.6.2 Determination of the internal forces in the members of the frame... 3.9 3.6.3 Analytical method... 3.9 3.6.4 Method of joints... 3.10 3.6.5 Solved problems on Method of joints 3.11 to 3.29 3.6.6 Method of sections... 3.29 3.6.7 Solved problems on Method of sections and Method of joints... 3.31 to 3.62 Chapter 4 Properties of Surfaces and Solids 4.1 Centre of Gravity or Centroid... 4.1 4.1.1 First Moment of Area... 4.2 4.2 Centroid of a Uniform Lamina... 4.2 4.3 Centre of Gravity... 4.3 (xi)

- www.airwalkpublications.com Centroids of Various shapes of Area (or) Centre of gravity... 4.4 4.4 Solved problems on centroids... 4.13 to 4.40 4.5 Moment of Inertia (Second Moment of Area). 4.41 4.6 Polar Moment of Inertia and Perpendicular Axis Theorem... 4.42 4.6.1 Perpendicular Axis theorem... 4.42 4.7 Radius of Gyration of an Area... 4.43 4.8 Parallel Axis Theorem... 4.44 4.9 Solved Problems on Moment of Inertia 4.46 to 4.55 Moments of Inertia of Important Figures... 4.56 4.10 Moment of Inertia of Composite Section... 4.58 Solved Problems... 4.60 to 4.88 4.11 Product of Inertia... 4.89 4.12 Principal Moment of Inertia... 4.90 4.13 Solved Problems on Product of Inertia and Principal Moment of Inertia... 4.91 to 4.103 4.14 Moment of Inertia of a Mass... 4.104 4.14.1 Parallel Axis Theorem... 4.105 Mass Moment of Inertia of Important Bodies... 4.105 4.15 Solved Problems on Mass Moment of Inertia... 4.107 to 4.116 Chapter 5 Friction 5.1 Introduction... 5.1 5.2 Limiting Friction... 5.1 5.3 Coloumb s Law of Dry Friction... 5.2 (xii)

5.4 Angle of Friction... 5.3 5.5 Rolling Resistance... 5.4 Solved Problems... 5.6 to 5.10 5.7 Inclined Plane and Angle of Repose... 5.10 5.8 External force P Applied Parallel to the Plane... 5.12 5.9 P Applied at an Angle up the Plane... 5.13 5.10 Solved Problems... 5.15 to 5.51 5.11 Screw Jack... 5.52 Solved Problems... 5.54 to 5.58 5.12 Types of Belt Drives... 5.59 5.12.1 Open Belt Drive... 5.59 5.12.2 Cross belt drive... 5.60 5.12.3 Ratio of Tensions for Flat Belts... 5.60 Solved Problems and Exercises... 5.61 to 5.80 Chapter 6 Dynamics of Particles - Kinematics 6.1 Dynamics... 6.1 6.2 Rectilinear Motion... 6.1 6.2.2 Displacement... 6.2 6.2.3 Average velocity... 6.3 6.2.4 Average acceleration... 6.4 6.2.5 Instantaneous acceleration (or) Simply acceleration: a... 6.4 6.2.6 Motion Under Variable Acceleration Solved Problems... 6.6 to 6.15 (xiii)

- www.airwalkpublications.com 6.2.7 Equations of motion with uniform acceleration... 6.15 Three equations for uniform accelerated motion... 6.18 Solved Problems... 6.18 to 6.25 6.2.8 Freely falling body... 6.26 6.2.9 Sign convention:... 6.26 Solved Problems... 6.27 to 6.51 6.2.10 Relative Motion... 6.33 6.3 Curvilinear Motion... 6.52 6.3.1 Motion of a projectile... 6.52 6.3.2 Study of the motion of a projectile... 6.54 Solved Problems... 6.56 to 6.72 6.3.3 Motion along a Circular Path... 6.72 Solved Problems... 6.73 to 6.79 6.3.4 Radial and Transverse components of motion... 6.80 Solved Problems and Exercises... 6.81 to 6.94 Chapter 7 Dynamics of Particles - Kinetics 7.1 Newton s Second Law... 7.1 7.1.1 Linear Momentum L... 7.3 7.1.2 Unit of Force... 7.3 7.1.3 Equation of Motion... 7.4 7.2 Dynamic Equilibrium... 7.5 Solved Problems... 7.6 to 7.35 7.3 Lift Motion... 7.36 (xiv)

Solved Problems... 7.36 to 7.40 7.4 Work and Power... 7.41 7.4.1 Graphical representation of Work:... 7.42 7.4.2 Power... 7.42 7.5 Work-energy Equation of Particles: (or) Principle of Work-energy... 7.42 Solved Problems... 7.44 to 7.67 7.6 Impulse - Momentum Principle... 7.68 Solved Problems... 7.71 to 7.90 7.7 Conservation of Momentum... 7.91 Solved Problems... 7.91 to 7.103 7.8 Conservation of Energy... 7.104 7.8.1 Energy... 7.104 7.8.2 Forms of energy... 7.104 7.8.3 Conservation of energy... 7.105 7.8.4 Law of conservation of Energy... 7.105 Solved Problems... 7.105 to 7.112 7.9 Collision of Elastic Bodies... 7.113 7.9.1 Elasticity... 7.113 7.9.2 Restitution... 7.113 7.9.3 Coefficient of restitution... 7.113 Solved Problems... 7.115 to 7.126 7.9.4 Oblique Impact or Indirect Impact... 7.126 Solved Problems... 7.127 to 7.133 7.9.5 Impact of a body on a fixed plane... 7.113 Solved Problems and Exercises... 7.134 to 7.144 (xv)

- www.airwalkpublications.com Chapter 8 Elements of Rigid Body Dynamics 8.1 Rotation about a fixed axis... 8.1 Translation and Rotation Motion... 8.3 Equations of Rotational Motion... 8.4 Angular Displacement... 8.5 Angular Velocity... 8.6 Angular acceleration... 8.7 8.2 Summary... 8.7 Equations of Translation and Rotational Motion with Uniform acceleration... 8.7 Solved Problems... 8.8 to 8.26 8.3 General Plane Motion 8.27 Solved Problems and Exercises... 8.29 to 8.80 Short Questions and Answers University Solved Papers (xvi)

Basics and Statics of Particles 1.1 Chapter 1 Basics and Statics of Particles 1.0 INTRODUCTION Mechanics can be defined as that science which describes and predicts the condition of rest or motion of bodies under the action of forces. Mechanics is divided into three major parts. 1. Mechanics of rigid bodies 2. Mechanics of deformable bodies 3. Mechanics of fluids Engineering mechanics concerns itself mainly with the application of the principles of mechanics to the solution of problems commonly encountered in the field of engineering practice. 1.1 CLASSIFICATION OF ENGINEERING MECHANICS Engineering mechanics may be classified based upon the type or nature of the body involved and is shown in Fig. 1.1. Rigid body is one which can retain its shape, size or one which does not undergo any deformation under the loads. Statics is branch of mechanics which deals with the force and its effects on bodies at rest. Dynamics is branch of mechanics which deals with the force and its effects on bodies in motion.

1.2 Engineering Mechanics - www.airwalkpublications.com Engineering Mechanics Mechanics of solids Mechanics of fluid Rigid bodies Deformable bodies Ideal fluid Viscous fluid Compressible fluid Statics Dynamics Strength of Materials Theory of Elasticity Theory of Plasticity Kinetics Kinematics Fig. 1.1 Classification of Engineering Mechanics Kinetics is the branch of mechanics which deals with the body in motion when the forces which cause the motion are considered. Kinematics is the branch of mechanics which deals with the body in motion, when the forces causing the motion are not considered. Deformable bodies are one which undergoes deformation under application of forces. The branch of mechanics which deals with the study of internal force distribution, stress and strain developed in the bodies is called mechanics of deformable bodies or mechanics of materials. Fluid mechanics is branch of mechanics which deals with study of fluids both liquids and gases at rest or in motion.

Basics and Statics of Particles 1.3 1.2 FUNDAMENTAL CONCEPTS OF MECHANICS The basic concepts used in everyday mechanics are based on newtonian mechanics. The basic concepts used in newtonian mechanics are space, time, mass and force. These are absolute concepts because they are independent of each other. These concepts cannot be truly defined and should be accepted on the basis of our intuition and experience and used as a mental frame of reference for our study of mechanics. Space is a geometric region in which events involving bodies occur. Space is associated with the position of a point P. The position of P can be defined by three lengths measured from a certain reference point or origin in three given directions. These lengths are known as coordinates of P. Time is a measure of the succession of events. The standard unit used for its measurement is the second s, which is based on duration. Mass is used to characterize and compare bodies on the basis of certain fundamental mechanical experiments. Two bodies of the same mass, will be attracted by the earth in the same manner and, they will also offer same resistance to a change in translation motion. Mass is quantitative measure of inertia. Mass of a body is constant regardless of its location. Force is the action of one body on another. Each body tends to move in the direction of external force acting on it. A force is characterized by its point of application, its magnitude and its direction. A force is represented by a vector. Simply force is push or pull,

1.4 Engineering Mechanics - www.airwalkpublications.com which by acting on a body changes or tends to change its state of rest or motion. Weight is the force with which a body is attracted towards the centre of earth by the gravitational pull. The relation between the mass and weight of a body is given in the equation 1.1. Weight (W) mass (m) gravity (g)... (1.1) where W weight in Newton m mass of body in kg g acceleration due to gravity i.e 9.81 m/s 2 1.3 UNITS AND DIMENSIONS Unit is defined as the numerical standard used to express the qualitative measure of a physical quantity. Scalar quantities are those physical quantities that have only magnitude and no direction eg. mass, time, volume, etc. Vector quantities are those physical quantities that have both magnitude and direction eg. Displacement, velocity, acceleration, momentum, force, etc. The four basic concepts in mechanics namely, space, time, mass and force are related to each other by equation 1.2. Force (F) mass (m) Acceleration (a)...(1.2) F ma or F mass Length Time 2

Basics and Statics of Particles 1.5 The different system of units are (i) SI units International System of units (ii) F.P.S units Foot - Pound - Second system (iii) C.G.S units Centimeter - Gram - Second system (iv) M.K.S units Metre - Kilogram - Second system The three units of basic concepts can be defined arbitrarily and are referred as base units. The fourth unit must be selected as per the equation 1.2 and is referred as derived unit. In SI system, the base units are length, mass and time expressed in Metre (m), Kilogram (kg) and Second (s) respectively. The force is derived unit and is expressed in Newton (N). 1.3.1 SI units SI units are absolute system of units which are independent of the location where the measurement are made. SI units in abbreviation International System of Units. SI units have 7 base units, 2 supplementary units and many number of desired units. The various SI units are shown in the Table 1.1.

1.6 Engineering Mechanics - www.airwalkpublications.com Table 1.1 SI units Parameter Symbol Unit Base unit Length l m Mass m kg Time t sec Derived units Force F N (i.e., 1 N = 1 kg m/s 2 Pressure P N/m 2 or Pascal Density kg/m 3 Sp. wt w N/m 3 Energy E Nm or Joule Area A m 2 Velocity v, u m/s Moment or couple M N m Angle rad Angular velocity rad/s Acceleration a m/s 2 Angular acceleration rad/s 2 Torque T N-m Power P Watts or J/S Frequency f Hz Volume V m 3 Work W 1-2 N-m Impulse kg - m/s Moment of force N-m Stress N/m 2

Basics and Statics of Particles 1.7 Multiples and sub multiples of SI units are obtained using prefixes given in the Table 1.2. In the problems of mechanics, magnitude of physical quantities involved may be very large or very small. Hence the prefixes are found to be useful. Table 1.2 Prefixes in SI units Multiplication factor Prefix Symbol 10 12 tera T 10 9 giga G 10 6 mega M 10 3 kilo k 10 2 hecto h 10 1 deca da 10 1 deci d 10 2 centi c 10 3 milli m 10 6 micro 10 9 nano n 10 12 pico p * should be avoided if possible 1.3.2 Dimensional Analysis Any physical variable can be described by using qualitative and quantitative approach. The qualitative description of physical variable is known as dimension.

1.8 Engineering Mechanics - www.airwalkpublications.com The quantitative description of physical variable is known as unit. Dimensional analysis is a branch of mathematics which deals with dimensions of quantities. The fundamental basic dimensions are length, mass, force & time. Combinations of these determine the secondary dimensions in which all other quantities can be expressed. Dimensions are classified as Absolute system (MLT system) and Grativational system (FLT system). Absolute system (MLT system) is one in which system of units are defined on basis of length, time and mass. SI system is an absolute system. Gravitational system (FLT system) is one in which system of units are defined on basis of length, time and force. Force is based on gravitational acceleration which depends upon the location. In MLT system In FLT system F ML T 2 M FT2 L The basic and derived dimensions of various quantities in MLT and FLT system are shown in Table 1.3

Basics and Statics of Particles 1.9 Table 1.3 Dimensions of physical quantities in MLT & FLT system Sl. No. Physical quantity Unit MLT system FLT system 1. Length (l) metre L L 2. Mass (m) kg M FL 1 T 2 3. Time (T) sec T T 4. Force (F) N kg m MLT 2 F s 2 2 T 2 5. Acceleration (a) m s 2 LT 6. Angular rad T 2 acceleration s 2 7. Angular rad T 1 T 1 velocity s 8. Area (A) m 2 L 2 L 2 LT 2 9. Density kg ML 3 FT 3 L 4 m 3 10. Energy (E) N m kgm s 2 m ML 2 T 2 FL 11. Impulse kg m MLT 1 FT s 12. Moment of force (M) 13. Moment of inertia (Area) 14. Moment of inertia (Mass) N-m ML 2 T 2 FL m 4 L 4 L 4 kg m 2 ML 2 FLT 2

1.10 Engineering Mechanics - www.airwalkpublications.com Sl. No. Physical quantity Unit MLT system FLT system 15. Momentum kg m s MLT 1 FT 16. Modulus of Elasticity (E) N m 2 kg m s 2 1 m 2 ML 1 T 2 FL 2 17. Modulus of rigidity (K) 18. Power (P) N m 2 ML 1 T 2 Watts N m s FL 2 ML 2 T 3 FLT 1 19. Pressure (p) N m 2 ML 1 T 2 20. Specific height (w) N ML 2 T 2 m 3 21. Stress N m 2 ML 1 T 2 FL 2 FL 3 FL 2 22. Velocity (v, u) m LT 1 LT 1 s 23. Volume (V) m 3 L 3 L 3 24. Weight (w) N MLT 2 F 1.3.3 Dimensional Homogeneity An equation is said to be dimensionally homogeneous, if the dimensions of various terms on the left and right side of the equation are identical. For example, consider an equation X V W. In the above equation Dimension of X Dimension of V Dimension of W.

Basics and Statics of Particles 1.11 Consider an equation y K X where K is constant and is dimensionless quantity and therefore Dimension of y Dimension of X Problem 1.1: Verify whether the following equations are dimensionally homogeneous (a) v 2 u 2 2as (b) v 2gh (c) s ut 1 10 t2 2 Solution: (a) v 2 u 2 2as Substituting dimensions of the various terms LT 1 2 LT 1 2 [LT 2 ] [L] [L 2 T 2 ] [L 2 T 2 ] [L 2 T 2 ] Equation is dimensionally homogenous. (b) V 2 g h Substituting the dimensions of various terms [LT 1 ] [LT 2 L] 1/2 [LT 1 ] [L 2 T 2 ] 1/2 [LT 1 ] [LT 1 ] Equation is dimensionally homogeneous. (c) S ut 1 10 t2 2 Substituting the dimensions of various terms [L] [LT 1 ] [T] [T 2 ]

1.12 Engineering Mechanics - www.airwalkpublications.com [L] [L] [T 2 ] [L] [L] [T 2 ] The equation is not dimensionally homogeneous. Problem 1.2: A physical phenomenon is given by relation x P c.v 2 y g z where p - pressure, c - constant, v - velocity, - mass density, g - acceleration due to gravity. Find the dimensions of x, y, z using MLT system. Solution: Refer table 1.3 for the dimensions of P ML 1 T 2, v LT 1, ML 3, g LT 2 Dimension of x Dimension of P Dimension of v 2 y Dimension of gz x ML 1 T 2 y LT 1 2 ML 3 LT 2 z Dimension of x ML 1 T 2 Dimension of y ML 1 T 2 L 2 T 2 ML 3 Dimension of ML 1 T 2 z ML 3 LT 2 L Problem 1.3: What is the dimension of Newton s law of gravitational attraction G. Solution: F G M 1 M 2 R 2

Basics and Statics of Particles 1.13 G F R2 M 1 M 2 Substituting the various dimensions of the above equation (Table 1.3) G [MLT 2 ] [L 2 ] [M] [M] [M 1 L 3 T 2 ] 1.4 LAWS OF MECHANICS OR BASIC PRINCIPLES OF MECHANICS The study of mechanics rests on the following fundamental principles based on the experimental evidences. (i) Newton s three fundamental laws (First law, Second law and, Third law) (ii) Newton s Law of Gravitation (iii) The parallelogram law of addition of forces (iv) The principle of transmissibility of forces (v) The triangular law of forces (vi) The law of conservation of Energy (vii) The principle of work and Energy 1.4.1 Newton s Three Fundamental Laws (a) Newton s First Law Newton s first law states that If the net force or the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion).

1.14 Engineering Mechanics - www.airwalkpublications.com Or in other words, Every particle continues in a state of rest or of motion in a straight line unless it is compelled to change that state by a force imposed on the body. So first law uses to define the forces. (b) Newton s Second Law Newton s second law states that If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and move in the direction of this resultant force. F Resultant force or Net force F ma where m Mass of the body a Acceleration of the body This law is used to measure a force. Newton s Second Law in other words: The rate of change of momentum of a body is directly proportional to the imposed force and it takes place in the direction in which the force acts. i.e., Force Rate of change of momentum Now, Momentum Mass Velocity Rate of change of momentum mass rate of change of velocity

Basics and Statics of Particles 1.15 Mass Acceleration Force mass acceleration i.e., F d dt mv m dv dt m a F K ma K Constant of proportionality 1 in S.I. units (c) Newton s Third Law Each and every action (FORCE) has equal and opposite reaction. This means that the forces of action and reaction between two bodies are equal in magnitude but opposite in direction. 1.4.2 Newton s Law of Gravitation This states that two particles of mass M and m are mutually attracted with equal and opposite forces F and F. This force F is defined as follows: F G Mm r 2 where r Distance between the two particles. G Universal constant called as the Constant of Gravitation. Refer Fig 1.2. M -F The Example The attraction of the earth on a particle located r Fig.1.2. F m

1.16 Engineering Mechanics - www.airwalkpublications.com on its surface. The force F exerted by the earth on the particle is known as weight W. W mg where W weight of the body in Newton N m mass of the body in kg g 9.81 m/sec 2 acceleration due to gravity. 1.4.3 The Parallelogram Law of Force Construct a parallelogram using a P force P and a force Q as two sides of the parallelogram. Now the Q diagonal passing through Fig.1.3. A A represents the resultant. This is called as parallelogram law for addition of two forces. Refer Fig 1.3. Or in other words If two forces P and Q acting at a point A are represented by the adjacent sides of a parallelogram, then the diagonal of the parallelogram passing through that point of intersection represents the resultant. Resultant R P 2 Q 2 2PQ cos 1.4.4 Triangular Law of Forces If two forces acting simultaneously on a body are represented by the sides of a triangle taken in order, their resultant is represented by the closing side of the triangle taken in the opposite order.

Basics and Statics of Particles 1.17 Refer the Fig. 1.4 (a). The sum of two forces P and Q may be obtained by arranging P and Q in tip to tail fashion and then connecting the tail of P with tip of Q. Addition of Two Vectors (Forces) P and Q Refer Fig.1.4 (a) and (b) P Q Even if we P R esultant P +Q cha nge the orde r we get same resultant Resulta nt Q +P (a) Fig.1.4. Q (b) Now P Q Q P. So addition of two forces is commutative. The subtraction of a force is obtained by the addition of corresponding negative force. The force P Q is obtained by adding to P with the negative force Q as shown in Fig. 1.5 (a) & (b). P P Q (a) P- Q -Q (b) subtraction Fig.1.5. Resultant P- Q

1.18 Engineering Mechanics - www.airwalkpublications.com So, P Q P Q 1.4.5 Polygon Law of Forces If a number of concurrent forces acting simultaneously on a body, are represented in magnitude and direction by the sides of a polygon taken in order, then the resultant is represented in magnitude and direction by the closing side of the polygon, taken in opposite order. Graphically arrange all the given forces in the tip-to-tail fashion in any order. To obtain the resultant, join the tail of the second force with the tip of the first force and so on. Concurrent Forces acting on a body. Refer Fig. 1.6(a) F 5 F 2 F 5 F 4 Fig.1.6.(a) F 3 By using Polygon Law of Forces Vectorial Addition of F 1, F 2, F 3, F 4, and F 5 Resultant The resultant R does not depend upon the order in which the forces are chosen to draw the polygon.

Basics and Statics of Particles 1.19 F 2 R esultant = F + F +F +F + F 1 2 3 4 5 F 5 F 3 o By using Polygon law of forces F 1 F 4 F 1 F 2 Both Resultant are equal Change the order : o R esultant = F +F +F + F + F F 4 F 5 1 2 3 4 5 Fig.1.6.(b) Fig.1.6.( c ) F 3 This means: We can add the forces as in Fig 1.6(b) We can also add the forces as in Fig. 1.6(c). F 2 F 1 F 4 F 3 F 5 Resultant. Here order is changed. Even then, we get the same resultant. Fig. 1.6 (b) and Fig. 1.6 (c) gives the same resultant. 1.4.6 Law (or) Principle of Transmissibility of Forces The condition of equilibrium or motion of a rigid body will remain unchanged if the point of application of a force acting on the rigid body is transferred to any other point along its line of action. Refer Fig. 1.7 Line of action a F A B b = a A B Same Effect (Push) Fig.1.7. (Pull) F b

1.20 Engineering Mechanics - www.airwalkpublications.com A force F acting on a body at point A is transferred to point B along the same line of action without changing its net effect on the rigid body. Principle of Transmissibility in other words (Refer Fig. 1.8) F A line of action Fig.1.8. = line of action B F The principle of transmissibility states that the conditions of equilibrium or the motion of a rigid body will remain unchanged if a force F acting at a given point of the rigid body is replaced by a force F of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action. The two forces F and F have the same effect on the rigid body and are said to be equivalent. 1.4.7 Law of conservation of energy It states that the energy can neither can be created nor be destroyed, it can only be transformed from one form to another. In other words, the total energy possessed by a body remains constant provided no energy is added to or taken from it.

Basics and Statics of Particles 1.21 At any instant, Total energy potential energy kinetic energy. 1.4.8 Principle of work and energy The work done by a system of forces on a particle is equal to the change in the kinetic energy of the particle. This is called principle of work and energy mathematically. K E 1 W 1 2 K E 2 where K E 1 Kinetic energy of particle before application of force K E 2 Kinetic energy of particle after application of force W 1 2 Work done in movement of particle from position 1 to 2 1.5 FORCE AND FORCE SYSTEM A particle is defined as an idealized model of actual physical body of real world, as an entity having only mass and location in space. 1.5.1 Force Force is an action that changes or tends to change the state of the body on which it acts. Simply, force is a push or pull on a body. Force cannot be seen and only its effect on the object upon which it acts can be seen. A force can also change or tend to change the size and shape of a deformable bodies.

1.22 Engineering Mechanics - www.airwalkpublications.com The characteristics of forces are (i) Magnitude (ii) Direction (slope) (iii) Sense (iv) Point of application. These characteristics distinguish one force from another. Representation of a force is shown in the Fig. 1.9. line of action M agnitude(f) F O = inclination of force with horizontal point of action Fig. 1.9 Force Space diagram is the diagram in which the forces are represented and their magnitudes are written along their lines of action. (Fig. 1.10). C C=20N B B=30N C B B A=40N A (a) Sp ace diagram Fig 1.10 C A A (b) Force diagram

Basics and Statics of Particles 1.23 Force diagram is the diagram in which the forces are drawn to scale, parallel to their respective line of action. 1.5.2 Types of forces Forces may be classified as (i) External forces (contacting or applied forces) (ii) Body forces (non-contacting or non applied forces) (a) External forces Forces which act on a body due to external causes or agency are termed as external forces. The external forces acting on the surface of a body are called surface forces. The equilibrium of the body is affected by all the external forces (body forces and contact forces) which act on the body. A body pressing against another is subjected to both normal and tangential force. Normal forced is perpendicular to the surface of contact. The tangential force of the contact surfaces is called frictional force. The applied forces are further classified as (i) concentrated or point forces (ii) Distributed forces F 1 Point forces are those forces which act at specific points on the body. Body F 2 Example - force exerted by wheels of vehicle on road. Fig. 1.11. F 3 Fig 1.11 Point force

1.24 Engineering Mechanics - www.airwalkpublications.com Distributed forces are those forces which are distributed over a definite areas of surface of the body. Example - a beam of a ceiling of a floor Fig. 1.12. X a W/m Simply supported beam b Fig 1.12 Distributed force (b) Body forces The external forces acting on each and every particle of a body are termed as body force or volume force. Examples are gravitational pull on all physical bodies, magnetic force, etc. If the force acting on an area of specific size and if the area is significant, the resultant force per unit area is termed as pressure or stress and is described as force per unit area N/m 2. Y 1.5.3 Free body diagram A body may consist of more than one element and support. Each element or support can be isolated from the rest of the system by incorporating the net effect of the remaining system through a set of forces. This diagram of isolated element or a portion of the body along with the net effects of forces on it is called a Free Body Diagram (FBD). (a) Body Resting on surface (b) Free body diagram (smooth surface) (c) Free body diagram (Rough surface)

Basics and Statics of Particles 1.25 Surface (smooth or rough) Block P W P W P F (a) Body resting on surface Fig 1.13 Free Body diagram (F.B.D) R N (b) Free body diagram (Smooth surface) R N (c) Free body diagram (Rough surface) Here P pull force W Weight of body R N Normal reaction F Frictional force 1.5.4 Free body diagram in simple words To study the forces clearly, we draw free body diagram. Free body diagram is drawn by removing all the supports i.e., wall, floor, hinge or any other supports and replacing them by reactions. All the external forces acting on the body are shown in this free body diagram. The self weight (acting downward always) is also considered in this diagram. The self weight (W mg) is assumed to act at the centre of gravity of the body.

1.26 Engineering Mechanics - www.airwalkpublications.com Actual diagram Support rope Free body diagram T(Tension force) (Totally two forces are acting on the body) Suspended Body Actual diagram Body (d) w (Weight of body acting downward always) w (wt of body acting downward always) Floor Actual diagram (Totally two forces acting on the body) (e) R (Reaction force) Side Support A 1 2 Floor B Side Support C Freebody diag ram R C (Reaction R w2 A Force at C) w1 Reaction Force at A (Totally 5 forces acting on the body) (g) R B (Reaction Force at B) Fig.1.13. Free Body Diagram 1.5.5 Types of force system When several forces are acting upon a body simultaneously, they constitute a system of forces. Force

Basics and Statics of Particles 1.27 system is classified based upon the two dimensional or three dimensional space of forces and the orientation of line of action of forces. The classification is given below. I. Coplanar force system (a) Concurrent (b) Parallel (c) Non-concurrent (d) Collinear II. Non coplanar force system (a) Concurrent (b) Parallel (c) Non-concurrent (a) Coplanar forces A system of forces that are contained in a single plane or system of forces having their line of actions in the single plane are called coplanar forces (Fig. 1.14) Forces F 1, F 2, F 3 are coplanar forces. F 2 F 1 F 3 Fig 1.14 Coplanar forces Plane Forces (b) Concurrent forces When the lines of action of all the forces of a system intersect at a common point, the system of forces are said to be concurrent. Plane (c) Coplanar - concurrent force system When the system of forces lie in the same plane and the line of action of forces pass through the common point, then the system of forces is called coplanar - concurrent F 3 F 2 2 O F 1 Fig 1.15 Coplanarconcurrent forces

1.28 Engineering Mechanics - www.airwalkpublications.com force system (Fig. 1.15) Forces F 1, F 2, F 3 are in same plane and pass through common point O. (d) Non concurrent forces When the system of forces whose line of action does not pass through a common point are called non concurrent forces. (e) Coplanar - Non concurrent forces When the system of forces is acting in a single plane and does not have their line of action passing through a common point are called coplanar non concurrent forces. Fig. 1.14 is an example of coplanar - non concurrent forces. (f) Collinear forces When the system of forces acting in a single plane with a common line of action are called collinear forces. Fig. 1.16 shows collinear force system. O F 2 F 1 F 3 F 4 F 5 Plane Fig 1.16 Collinear forces Common line of action Forces Plane (g) Parallel forces When the line of action of all the forces of the system are parallel, then the system is called parallel force system. When the line of action of all the F 1 F 2 F 3 F 4 Fig 1.17 Coplanar like parallel forces

Basics and Statics of Particles 1.29 forces are parallel and all of them act in one and same direction, then the force system is called like parallel forces. When the line of action of all the forces are parallel and some forces act in one direction while others in opposite direction are called unlike parallel force system. Fig. 1.17 shows an example of coplanar like parallel force system and Fig. 1.18 shows an example of coplanar unlike parallel force system. (h) Non coplanar force system A system of forces in which the forces lie in different plane or in a three dimensional space are called Non coplanar force system. F 1 F 2 F 3 Plane Fig 1.18 Coplanar unlike parallel forces forces (i) Non coplanar concurrent forces A system of forces lying in different planes with their line of action intersecting at a common point are called Non coplanar concurrent force system. Fig. 1.19 shows such an example. Y O Z F 2 F 1 O Planes Fig 1.19 Non Coplaner Concurrent forces X Forces

1.30 Engineering Mechanics - www.airwalkpublications.com (j) Non coplanar Non concurrent forces Z Planes A system of forces lying in different planes with different lines of actions are called non coplanar non concurrent forces (Fig. 1.20) F 3 O F 1 Forces X F 2 (k) Non coplanar parallel forces Y Fig 1.20 Non Coplanar non concurrent forces A system of forces lying in different plane but their line of action parallel to each other are called non coplanar parallel forces. In Fig. 1.21 forces F 1 and F 2 are F 4 F 3 O Z F 1 F 2 X non coplanar like parallel force and F 3 and F 4 are non coplanar unlike parallel forces. Y Fig 1.21 Non Coplanar non parallel forces Z (l) Non coplanar Non concurrent and Non parallel forces F 1 A system of forces that are not lying in same plane with different line of action and not passing through same common point and are not parallel Y F 3 O F 2 Fig 1.22 Skew force X

Basics and Statics of Particles 1.31 to each other are called Non coplanar Non concurrent and non parallel forces called as skew forces. 1.5.6 Statics of particles - Composition of forces Composition of a force system is a process of finding a single force, known as resultant, that can produce the same effect on the particle as that of the system of forces. For example Fig. 1.23 shows a system of F 3 F 2 three forces acting on a particle with the resultant R which can produce the same effect. A particle means that the size and shape of body does not significantly affect the solution of problems and all the forces are assumed to act at the same point. Particle R (Resultant ) F 1 = Fig 1.23 Resultant force 1.5.7 Resultant of two coplanar concurrent forces Resultant of two coplanar concurrent forces can be obtained by the following methods. 1. Analytical method (a) Parallelogram law of forces (b) Triangular law of force 2. Graphical method (a) Parallelogram law of forces (b) Triangular law of forces

1.32 Engineering Mechanics - www.airwalkpublications.com I. Analytical method - parallelogram law of forces When two forces F 1, F 2 acting on a particle are represented by two adjacent sides of a parallelogram, the diagonal connecting the two sides represents the Resultant force R in magnitude and direction Fig.1.24(a) F 2 F 2 R R A F 1 Hence, the relationship between F 1, F 2 and R can be derived as follows [Fig. 1.24 (b)]. Consider the parallelogram OACB. Let OA and OB represent the forces F 1 and F 2 acting at a point O. The diagonal OC represents the resultant R which can be expressed as, OC 2 OA AD 2 CD 2 OA 2 2 OA AD AD 2 CD 2 OA 2 2 OA AD AC 2 [... AC 2 AD 2 CD 2 R 2 F 1 2 2 F 1 F 2 cos F 2 2 [... AC OB] [... AD F 2 cos ] 2 Hence R F 1 2 F2 2F 1 F 2 cos F 1 Fig 1.24(a) Parallelogram law A F 2 B C R O F 1 A D Fig. 1.24(b) Resultant of two forces

Basics and Statics of Particles 1.33 CD Also tan OA AD F 2 sin F 1 F 2 cos Consider the following special cases. Note (1): If F 1, F 2 are at right angles, then 90, cos 90 0 2, R F 1 2 F2 tan F 2 F 1 Note (2): If F 1, F 2 are collinear and are in the same direction, then 0, cos 1 R 2 2 F 1 2 F2 2F1 F 2 Resultant R F 1 F 2, tan 0 or 0 Case (3): If F 1 F 2 are collinear and are in opposite directions F 1 F 2, then 180 R 2 F 1 2 F2 2 2F1 F 2, R F 1 F 2 tan 0: 0 Problem 1.4: The maximum and minimum resultant of two forces acting on a particle are 50 kn and 10 kn respectively. If 50 kn is the magnitude of the resultant for the given system of forces F 1 and F 2, find the angle between F 1 & F 2. Solution: We know that Resultant of two coplanar concurrent forces is

1.34 Engineering Mechanics - www.airwalkpublications.com R 2 F 1 2 F2 2 2F1 R 2 cos When R is maximum 0 i.e., 50 F 1 F 2 When R is minimum, 180 i.e., 10 F 1 F 2 Solving the above two equations, we get F 1 30 kn and F 2 20 kn If the resultant of forces F 1, F 2 acting at an angle is 50 kn, 50 2 30 2 20 2 2 30 20 cos cos 1 or 0 Thus F 1 and F 2 are at collinear to each other when the resultant is 50 kn. II Analytical Method Triangle Law of Forces If two forces F 1, F 2 acting simultaneously on a particle can be represented by the two sides of a triangle (in magnitude and direction) taken in order, then, the third side (closing side) represents the resultant in the opposite order. (Fig. 1.25) F 2 R F 2 C B F F 1 1 Fig 1.25 Triangle law of forces A Fig 1.26 Representation of forces

Basics and Statics of Particles 1.35 Thus all the trigonometric relations can be applied. From Fig. 1.26. A sin B sin C sin A 2 B 2 C 2 2BC cos B 2 A 2 C 2 2AC cos C 2 A 2 B 2 2AB cos III Graphical method - parallelogram law of forces Consider two forces F 1 & F 2 acting at a point O as shown in Fig. 1.27 (a). Draw two sides of parallelogram representing the forces F 1 & F 2 to some scale. Fig. 1.27 (b) as OA and OB. Complete the parallelogram by drawing AC, BC equal to OB and OA to the same scale. Draw the diagonal OC representing the resultant of the two force system. The length of diagonal measured to the chosen scale represents the magnitude of the resultant acting at an angle from the x axis. F 2 B C F 2 R O F 1 (a) O F 1 x axis A (b) Fig 1.27 Graphical method - parallelogram law of forces

1.36 Engineering Mechanics - www.airwalkpublications.com IV Graphical method - Triangle law of force Consider two forces F 1 and F 2 as shown in Fig. 1.28 (a). Draw OA to some scale as shown in the Fig. 1.28 (b). From the end OA, draw OB to same scale. Join OB as the closing side of triangle. The length OB to the same scale represents the magnitude of the resultant of the two forces F 1 & F 2. The angle represents the angle of resultant taken from x axis. F 2 B O F 1 O R esultant(r) F 1 Fig 1.28 Triangle law of forces A F 2 x axis 1.5.8 Rectangular components of force system (Resolution) The process of replacing a single force F acting on a particle by two or more forces which together have the same effect of a single force is called resolution of Y force into components. Theoretically a force can be resolved into an infinite number of F y F component sets, however in practice a force is resolved into two O X F x components. Fig 1.29 Resolution of forces

Basics and Statics of Particles 1.37 Consider a force F acting at an angle from x axis as shown in Fig. 1.29. The two rectangular components of forces are resolved along x and y axis and are given as In the vector form F x F cos ; F y F sin Force F F x i F y j [Vectors will be explained in detail in the later part of this chapter] Considering resolution of force on an inclined plane as shown in Fig. 1.30. The force F is resolved into two components F n and F t Normal component, F n F cos F F t F n x Fig 1.30 Resolution of forces on inclined plane Tangential component, F t F sin Writing in vector form F F t Fnsub In simple words, Resolution of Forces into components A single force can be resolved into two components which give F Horizontal component of F F cos = = F sin Fig.1.30.(a)

1.38 Engineering Mechanics - www.airwalkpublications.com the same effect on the particle. Refer Fig. 1.30 (a). Now a force F is resolved as - Horizontal component of F i.e. F cos and Vertical component of F i.e. F sin. 1.5.9 Resultant of several concurrent forces Consider a particle shown in Fig. 1.31 (a) subjected to four forces F 1, F 2, F 3, F 4. The Resultant of each four forces is shown in Fig. 1.31 (b). Y Y F 2 F 2 F 1 F 2 y F 1 F 1 y 2 1 X 3 o 4 F 3x F 2x o F 1x F 4x X F 3 F 3 y F 3 F 4 F 4 y F 4 Particle (a) (b) Fig 1.31 Reactant of Several Concurrent forces The Resultant R R 1 R2 R3 R4 R 1 F1x i F 1y j F 1 cos 1 i F 1 sin 1 j R 2 F2x i F 2y j F 2 cos 2 i F 2 sin 2 j ( ve sign is due to x axis) R 3 F3x i F 3y j F 3 cos 3 i F 3 sin 3 j ( ve sign is due to III quadrant x, y axis)

Basics and Statics of Particles 1.39 R 4 F4x i F 4y j F 4 cos 4 i F 4 sin 4 j ( ve sign is due to IV quadrant x, y axes) R R F1x F 2x F 3x F 4x i F 1y F 2y F 3y F 4y j R Rx i R y j F x i F y j [Vectors will be explained in detail in later part of this chapter) The Resultant of the above forces is shown in Fig. 1.31 (c). The Angle made by Resultant R is given by tan R y R x F y F x R y O Y R R x Fig 1.31 (c) Resultant X The magnitude of the Resultant F F x 2 F y 2 For n number of forces acting on the particle, then the Resultant is given as R F F1x F 2x F nx i F 1y F 2y F ny j R Rx i R y j F x i F y j and tan F y F x Note: If angle of the various forces are taken from positive x axis, then the (Fig. 1.32)

1.40 Engineering Mechanics - www.airwalkpublications.com Resultant n R x F x F i cos i i 1 n R y F y F i sin i i 1 R R x i R y j 2 Magnitude R R x e Ry Angle of inclination of Resultant tan 1 R y R x F 2 F 3 Y 2 3 4 F 1 1 F 4 Fig 1.32 Force system X 1.5.10 Summary Resultant Force Two or more forces on a particle may be replaced by a single force called resultant force which gives a same effect. P 2 2 R = P + Q The resultant force, of a given system of forces, is found out by the method of resolution as followed: Fig.1.33. Q 1. Resolve all the forces vertically and add all the vertical components (i.e., Find F y ) 2. Resolve all the forces horizontally and add all the horizontal components (i.e., Find F x )

Basics and Statics of Particles 1.41 3. The resultant R of the given forces is given by the equation: R F y 2 F x 2 4. The resultant force will be inclined at an angle, with the horizontal. tan F y F x Note: The value of the 90 o angle will vary depending upon the values of F y and F x as III Quadrant I Quadrant o o discussed below: (Refer 180 o 0,360 (-) x axis (+) x axis Fig. 1.34) II Quadrant IV Quadrant (a) When F y is ve, resultant will be in 1 st Quadrant or 2 nd 270 o Fig.1.34. Quadrant. (i.e. in between 0 to 180). (b) When F y is ve, resultant will be in III rd Quadrant or IV th Quadrant. (i.e. in between 180 to 360). (c) When F x is ve, the resultant will be in I st Quadrant or IV th Quadrant (i.e. in between 0 to 90 or 270 to 360). (d) When F x is ve, the resultant will be in II nd Quadrant or III rd Quadrant. (i.e. in between 90 to 180 or 180 to 270). (-) y axis (+) y axis

1.42 Engineering Mechanics - www.airwalkpublications.com The following sign conventions are followed for solving statics problems. Sign conventions for the direction of force: x axis Right side ve. x axis Left side ve y axis Upside ve y axis Downward ve SOLVED PROBLEMS Problem 1.5: Four forces act on a bolt A as shown in Fig (a). Determine the resultant of the forces on the bolt. Given diagram F 2 =80N y F 1 =150N 70 o 30 o o 15 o x =30 o Fcos 1 F 1 F 1 sin F 4 =100N Fig.(a) Fig.(b) F 3 =110N Solution: The force F 1 can be resolved into F 1 sin and F 1 cos as shown in Fig. (b). Similarly we can resolve all the forces F 2, F 3 and F 4. Add all the x components and find F x. Similarly add all the y components and find F y Refer Table.

Basics and Statics of Particles 1.43 F x 150 cos 30 80 cos 70 0 100 cos 15 199.1 F y 150 sin 30 80 sin 70 0 100 sin 15 14.3 The resultant R F 2 2 x F y 199.1 2 14.3 2 Resolution of forces in x components and y components are given in detail in the next page. 199.6 N R 199.6 N tan F y F x 14.3 199.1 0.07164 4.1 F =14.3N y 4.1 o Fig.(c) R=199.6N F=199.1N x Note: Conditions for a system to be in equilibrium When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium For a system to be in equilibrium, the following conditions should be satisfied. 1. Algebraic sum of all horizontal components of forces should be equal to zero, i.e F x 0 2. Algebraic sum of all vertical components of forces should be equal to zero, i.e. F y 0 3. Sum of all moments about any point in a system should be equal to zero, i.e M 0 [Moments will be studied in detail in the next chapter]