ENGINEERING MECHANICS - PDF Free Download

ENGINEERING MECHNICS (For.E /.Tech Engineering Students) s Per Kerala Technological University New Syllabus Dr. S.Ramachandran, M.E., Ph.D., Professor and Research Head Dr.. nderson, M.E., Ph.D., Faculty of Mechanical Engineering Sathyabama University Jeppiaar Nagar, Chennai - 600 119 Prof. Karthick YVS. Prof. George V.J (Structural engineer - Civil Engineering) IR WLK PULICTIONS (Near ll India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai - 600 004. Ph.: 2466 1909, 94440 81904 Email: aishram2006@gmail.com www.airwalkpublications.com

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J bdul Kalam Technological University (KERL TECHNOLOGICL UNIVERSITY) ENGINEERING MECHNICS Course Plan Module Contents Hrs. Sem. Exam Marks I Statics: Fundamental concepts and laws 2 15% of mechanics - Rigid body - Principle of transmissibility of forces Coplanar force systems - Moment of a 2 force - Principle of moments Resultant of force and couple system 4 II Equilibrium of rigid body - Free body diagram - Conditions of equilibrium in two dimensions - Two force and three force members Types of supports - Problems involving point loads and uniformly distributed loads only Force systems in space - Degrees of freedom - Free body diagram - Equations of equilibrium - Simple resultant and Equilibrium problems FIRST INTERNL EXM 3 5 15% 4

III IV V VI Properties of planar surfaces - Centroid and second moment of area (Derivations 3 15% not required) - Parallel and perpendicular axis theorem - Centroid and Moment of Inertia of composite area Polar Moment of Inertia - Radius of 2 gyration - Mass moment of inertia of cylinder and thin disc (No derivations required) Product of inertia - Principal Moment of 3 Inertia (conceptual level) Theorems of Pappus and Guldinus 1 Friction - Characteristics of dry friction 6 15% - Problems involving friction of ladder, wedges and connected bodies Definition of work and virtual work - 4 Principle of virtual work for a system of connection bodies - Problems on determinate beams only. SECOND INTERNL EXM Dynamics: Rectangular and Cylindrical 1 20% co-ordinate system Combined motion of rotation and 4 translation - Concept of instantaneous centre - Motion of connecting rod of piston and crank of a reciprocating pump Rectilinear translation - Newton s second 4 law - D lembert s Principle - pplication to connected bodies (Problems on motion of lift only). Mechanical Vibrations - Free and forced 1 20% vibration - Degree of freedom

Simple Harmonic motion - Spring - Mass model - Period - Stiffness - Frequency - Simple numerical problems of single degree of freedom. END SEMESTER EXM 7

Contents C.1 Contents 1. Statics 1.1 Introduction.... 1.1 1.2 Classification of Engineering Mechanics... 1.2 1.3 Fundamental Concepts of Mechanics... 1.4 1.4 Scalar and Vector Quantities... 1.5 1.5 Units and Dimensions... 1.6 1.6 Laws of Mechanics... 1.14 1.6.1 Newton s Three Fundamental Laws... 1.14 1.6.2 Newton s Law of Gravitation... 1.16 1.6.3 The Parallelogram Law of Forces... 1.17 1.6.4 Triangular Law of Forces... 1.17 1.6.5 Polygon Law of Forces... 1.19 1.6.6 Law (or) Principle of Transmissibility of Forces...... 1.20 1.6.7 Law of conservation of energy... 1.21 1.6.8 Principle of work and energy... 1.22 1.7 Force and Force System... 1.22 1.7.2 Types of force system... 1.23 (I) Coplanar force system... 1.24 (a) Concurrent forces... 1.24 (b) Coplanar - concurrent force system... 1.24 (c) Non concurrent and non-parallel forces... 1.25 (d) Coplanar - Non concurrent forces... 1.25 (e) Collinear forces... 1.25 (f) Parallel forces... 1.26 II Non coplanar force system... 1.26 (a) Non coplanar concurrent forces... 1.26 (b) Non coplanar Non concurrent forces... 1.27 (c) Non coplanar parallel forces... 1.27

C.2 Engineering Mechanics (d) Non coplanar Non concurrent and Non parallel forces (skew forces)... 1.27 1.8 Moment of a Force... 1.28 Units of Moment... 1.29 Clockwise Moment... 1.29 nticlockwise Moment... 1.29 1.8.1 Principle of Moments (or Varignon s Principle)... 1.30 1.9 Resultant of Force System... 1.31 1.9.1 Resultant of two coplanar concurrent forces 1.31 I. nalytical method - parallelogram law of forces... 1.32 II nalytical Method Triangle Law of Forces. 1.34 III Graphical method - parallelogram law of forces... 1.35 IV Graphical method - Triangle law of forces... 1.35 1.9.2 Rectangular components of force system (Resolution)... 1.37 1.9.3 Resultant of several concurrent forces... 1.38 1.10 Parallel Force System... 1.58 (a) Like parallel forces... 1.58 (b) Unlike parallel forces... 1.59 1.11 Couple...... 1.63 rm of a Couple... 1.63 1.11.1 Moment of a Couple... 1.63 Clockwise Couple and nticlockwise Couple... 1.64 1.11.2 Equivalent couples... 1.64 1.11.3 Resolution of a given Force into a Force and a Couple : Force - Couple system... 1.65 1.12 Equivalent Force System... 1.66 1.13 Resultant of Force and Couple System... 1.67 1.14 Equilibrium of Particle... 1.91

Contents C.3 1.14.1 Equilibrant... 1.92 1.15 Equilibrium of Rigid ody... 1.94 1.16 Free ody Diagram... 1.94 1.17 Types of Forces on Rigid odies... 1.98 1.19 Conditions of Equilibrium in two Dimensions...1.101 1.20 Equilibrium of Two Force ody...1.104 1.21 Equilibrium of a Three Force ody...1.105 1.22 Condition for Three Forces in Equilibrium...1.105 1.22.1 Lami s Theorem...1.106 1.23 Graphical Method...1.142 2. Types of Supports - Force Systems in Space 2.1 Introduction.... 2.1 2.2 Types of Supports... 2.2 1. Frictionless (Simple) Support... 2.2 2. Roller and Knife Edge Supports... 2.2 3. Hinged (Pin) Support... 2.3 4. uilt-in Support or Fixed support... 2.3 2.2.1 Types of eams... 2.5 2.3 Various Types of Loadings... 2.6 1. Concentrated or Point Load... 2.6 2. Uniformly Distributed Load (UDL)... 2.6 3. Uniformly Varying Load... 2.7 2.4 Problems Involving Point Loads and Uniformly Distributed Loads...... 2.8 2.5 Review of Vectors.... 2.26 2.6 Force Systems in Space... 2.32 2.6.1 Forces in three dimensional space (Component of force in space)... 2.33 2.6.2 Resultant of several concurrent forces... 2.35 2.6.3 Equilibrium of particle in space... 2.36

C.4 Engineering Mechanics 2.6.4 Equivalent system of forces... 2.36 2.7 Degrees of Freedom... 2.46 2.8 Free ody Diagram... 2.62 2.9 Equilibrium of Rigid ody in Three Dimensions... 2.64 2.10 Moment of Force bout a Point... 2.77 Moment of force about any other point... 2.78 Varignon s Theorem... 2.78 Vectorial treatment... 2.79 2.11 Moment of a Force about an axis... 2.84 2.12 Equilibrium of Non-concurrent and Non-parallel Spatial Forces...... 2.98 3. Properties of Planar Surfaces and Solids 3.1 Centre of Gravity or Centroid... 3.1 3.1.1 First Moment of rea... 3.2 3.1.2 Centroid of a Uniform Lamina... 3.2 3.1.3 Centre of Gravity... 3.4 3.1.4 Solved problems on centroids... 3.8 3.2 Moment of Inertia (or) Second Moment of rea... 3.31 3.3 Parallel xis Theorem... 3.32 3.4 Perpendicular xis Theorem... 3.34 3.4.1 Polar Moment of Inertia... 3.34 3.5 Centroid and Moment of Inertia of Composite rea. 3.36 3.6 Polar Moment of Inertia... 3.69 3.7 Radius of Gyration of an rea... 3.69 3.8 Mass Moment of Inertia of Cylinder and Thin Disc. 3.73 3.8.1 Parallel xis Theorem... 3.73 3.9 Solved Problems on Mass Moment of Inertia... 3.76 3.10 Product of Inertia... 3.81 3.11 Principal Moment of Inertia... 3.82

Contents C.5 3.12 Solved Problems on Product of Inertia and Principal Moment of Inertia... 3.83 3.13 Theorems of Pappus and Guldinus... 3.91 3.13.1 Surface of Revolution... 3.91 3.13.2 Volume of Revolution... 3.91 3.13.3 Pappus-Guldinus Theorem I... 3.92 3.13.4 Pappus-Guldinus Theorem II... 3.92... 4. Friction 4.1 Introduction.... 4.1 4.2 Limiting Friction..... 4.1 4.3 Characteritics of Dry Friction (Coloumb s Law of Dry Friction)...... 4.2 4.4 ngle of Friction..... 4.3 4.5 Inclined Plane and ngle of Repose... 4.16 4.6 Problem Related to Inclined Plane... 4.17 4.7 Friction on Ladder... 4.31 4.8 elt Drives..... 4.40 4.8.1 Open elt Drive... 4.40 4.8.2 Cross belt drive... 4.41 4.8.3 Ratio of Tensions for Flat elts... 4.42 4.9 Wedge Friction..... 4.56 4.10 Virtual Work...... 4.68 4.10.1 Principle of virtual work... 4.71 4.10.2 pplications of virtual work... 4.72 4.10.3 Procedure for analysis using principle of virtual work... 4.72 4.10.4 Sign convention... 4.73 4.11 Degree of Freedom... 4.73 4.12 pplication of Principle of Virtual Work to eams. 4.84

C.6 Engineering Mechanics 5. Dynamics 5.1 Introduction.... 5.1 5.1.1 Kinematics... 5.1 5.1.2 Kinetics... 5.1 5.2 Rectangular Co-ordinate System... 5.1 5.2.1 Position... 5.2 5.2.2 Displacement... 5.2 5.2.3 verage velocity... 5.3 5.2.4 verage acceleration... 5.4 5.2.5 Instantaneous acceleration (or) Simply acceleration... 5.4 5.2.6 Motion Under Variable cceleration... 5.5 5.3 Velocity - Time Diagram... 5.18 5.4 Freely Falling ody... 5.20 5.5 Relative Motion... 5.23 5.6 Motion of a Projectile... 5.28 (i) Time of flight... 5.33 (ii) Horizontal range... 5.33 (iii) Maximum height... 5.33 5.7 Cylindrical Co-ordinate System... 5.37 5.7.1 Rotation Motion... 5.38 5.7.2 Equations of rotational motion:... 5.39 5.8 Summary of Translation and Rotational Motion... 5.42 5.9 Combined Motion of Rotation and Translation [General Plane Motion]... 5.47 5.10 Concept of Instantaneous Center... 5.61 5.10.1 Motion of connecting rod of piston and crank of a reciprocating pump... 5.62 5.11 Rectilinear Translation - Newton s Second Law... 5.76 5.12 D lemberts Principle... 5.80

Contents C.7 5.13 pplication to Connected odies... 5.83 5.14 Lift Motion... 5.85 5.16.1 Problems on motion of lift... 5.86 6. Mechanical Vibrations 6.1 Introduction.... 6.1 6.2 Commonly used Definitions in Vibratory Motions (or) asic Features of Vibrating System... 6.2 1. Period of Vibration... 6.2 2. Cycle... 6.2 3. Frequency... 6.2 4. mplitude... 6.2 5. Natural frequency... 6.3 6. Fundamental (or principal) mode of vibration... 6.3 7. Degree of freedom... 6.3 8. Damping... 6.3 9. Phase difference... 6.3 10. Resonance... 6.3 11. Mechanical system... 6.3 12. Discrete (or lumped) system... 6.3 6.3 Types of Vibratory Motion... 6.4 1. Free (or) Natural Vibrations... 6.4 2. Forced Vibrations... 6.4 3. Damped Vibrations... 6.4 6.3.1 Types of Free Vibrations... 6.5 1. Longitudinal Vibrations... 6.5 2. Transverse Vibrations:... 6.6 3. Torsional Vibrations... 6.6 6.4 asic Elements of Vibrating System... 6.7 (i) Inertia elements... 6.7 (ii) Restoring elements... 6.7

C.8 Engineering Mechanics (iii) Damping elements... 6.7 6.5 Degrees of Freedom... 6.8 (a) Single degree of freedom... 6.8 (b) Two degree of freedom... 6.8 (c) Discrete or lumped system (Finite degree of freedom)... 6.9 (d) Infinite degree of freedom... 6.9 6.6 Simple Harmonic Motion... 6.9 6.6.1 Introduction... 6.9 6.6.2 Velocity and cceleration of a Particle Moving with Simple Harmonic Motion... 6.10 6.6.3 Differential Equation of Simple Harmonic Motion...... 6.12 6.6.4 Terms Used in Simple Harmonic Motion... 6.12 6.7 Spring - Mass Model... 6.24 6.7.1 Equivalent Stiffness of Spring... 6.29 (i) Spring in series... 6.29 (ii) Springs in parallel... 6.29 6.8 Simple Pendulum.... 6.44 6.9 Compound Pendulum... 6.48

Module 1 STTICS Statics: Fundamental concepts and laws of mechanics - Rigid body - Principle of transmissibility of forces - Coplanar force systems - Moment of a force - Principle of moments - Resultant of force and couple system - Equilibrium of rigid body - Free body diagram - Conditions of equilibrium in two dimensions - Two force and three force members. 1.1 INTRODUCTION The different motions that we notice, everyday, like balls bouncing or wheels rolling, are interaction of different bodies and effect of forces acting on them - the study is called mechanics, which 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 Mechanics, when applied in engineering is called Engineering mechanics which concerns itself mainly with the application of the principles of mechanics to the solution of problems commonly encountered in the field of engineering practice. Thus, Engineering mechanics is the study of forces and motion of bodies in mechanisms.

1.2 Engineering Mechanics - www.airwalkpublications.com 1.2 CLSSIFICTION OF ENGINEERING MECHNICS Engineering mechanics may be classified based upon the type or nature of the body involved and is shown in Fig. 1.1. 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 particle is defined as an idealized model of actual physical body of real world, as an entity having only mass and location in space. Its dimensions are negligible when compared with the distances involved in the discussion of its motion. Rigid body Rigid body is one which can retain its shape, size or one which does not undergo any deformation under the loads. It is a combination of a large number of particles which occupy fixed positions with respect to each other both before and after applying a load.

Statics 1.3 rigid body is one which does not deform under load. However, actual bodies are not absolutely rigid and deform slightly. Since such slight deformations do not affect the conditions of equilibrium or motion, they are neglected in the study of rigid bodies. Rigid body mechanisms found a suitable basis for the analysis and design of structural, mechanical or electrical engineering devices and are divided into two areas - Statics and Dynamics. Dynamics is further divided into Kinetics and Kinematics. Statics is branch of mechanics which deals with the force and its effects on bodies at rest. The configuration of different forces is such that the resultant force on the system is zero. Dynamics is branch of mechanics which deals with the force and its effects on bodies in motion. 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.

1.4 Engineering Mechanics - www.airwalkpublications.com Fluid mechanics is branch of mechanics which deals with study of fluids both liquids and gases at rest or in motion. 1.3 FUNDMENTL CONCEPTS OF MECHNICS 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. 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. force is characterized by its point of application, its magnitude and its direction. force is represented by

Statics 1.5 a vector. Simply force is push or pull, 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 1.4 SCLR ND VECTOR QUNTITIES Scalar quantities are those physical quantities that have only magnitude but no direction. eg. mass, time, volume, etc. Vector quantities Magnitude (length) are those physical quantities that have both magnitude and direction. eg. Displacement, velocity, acceleration, momentum, force, etc. Sense (arrow) Tail of Vector Direction Line of ction Head of Vector Reference axis Fig.1.2 Graphical representation of a vector 1.2. vector is represented graphically as shown in Fig. For mathematical operations involving vectors, the rules of vector algebra should be applied.

1.6 Engineering Mechanics - www.airwalkpublications.com 1.5 UNITS ND DIMENSIONS Unit is defined as the numerical standard used to express the definite magnitude of a physical quantity. 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) cceleration (a)...(1.2) F ma or F mass Length Time 2 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 units of three 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 unit of force is a derived unit and is expressed in Newton (N). 1.5.1 SI units SI units are absolute system of units which are independent of the location where the measurements are made. International System of Units is abbreviated as SI units (syste rne International d unit es).

SI units have 7 base units, 2 supplementary units and many number of derived units. The various SI units are shown in the Table 1.1. Table 1.1 SI units Parameter Symbol Unit Statics 1.7 ase 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 rea m 2 Velocity v, u m/s Moment or couple M N m ngle rad ngular velocity rad/s cceleration a m/s 2 ngular 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

1.8 Engineering Mechanics - www.airwalkpublications.com 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 1.5.2 Dimensional nalysis ny physical variable can be described by using qualitative and quantitative approach. The qualitative description of physical variable is known as dimension.

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 bsolute system (MLT system) and Grativational system (FLT system). bsolute 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 Statics 1.9 The basic and derived dimensions of various quantities in MLT system are shown in Table 1.3

1.10 Engineering Mechanics - www.airwalkpublications.com Sl. No. Table 1.3 Dimensions of physical quantities in MLT system Physical quantity Unit MLT system 1. Length (l) metre L 2. Mass (m) kg M 3. Time (T) sec T 4. Force (F) N kg m s 2 MLT 2 2 T 2 5. cceleration (a) m s 2 LT 6. ngular acceleration rad s 2 7. ngular velocity rad T 1 s 8. rea () m 2 L 2 3 9. Density kg m 3 ML 10. Energy (E) N m kgm s 2 m ML 2 T 2 11. Impulse kg m s MLT 1 12. Moment of force (M) N-m ML 2 T 2 13. Moment of inertia (rea) m 4 L 4 14. Moment of inertia (Mass) kg m 2 ML 2 15. Momentum kg m s MLT 1

Statics 1.11 Sl. No. Physical quantity Unit MLT system 16. Modulus of Elasticity (E) N m 2 kg m s 2 1 m 2 ML 1 T 2 17. Modulus of rigidity (K) N m 2 ML 1 T 2 18. Power (P) Watts N m s ML 2 T 3 19. Pressure (p) N m 2 ML 1 T 2 20. Specific height (w) N m 3 ML 2 T 2 21. Stress N m 2 ML 1 T 2 22. Velocity (v, u) m s LT 1 23. Volume (V) m 3 L 3 24. Weight (w) N MLT 2 1.5.3 Dimensional Homogeneity n 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. Consider an equation y K X

1.12 Engineering Mechanics - www.airwalkpublications.com 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 ] [L] [L] [T 2 ]

Statics 1.13 [L] [L] [T 2 ] The equation is not dimensionally homogeneous. Problem 1.2: 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 F R 2 G M 1 M 2

1.14 Engineering Mechanics - www.airwalkpublications.com Substituting the various dimensions of the above equation (Table 1.3) G [MLT 2 ] [L 2 ] [M] [M] 1.6 LWS OF MECHNICS [M 1 L 3 T 2 ] 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 principle of transmissibility of forces (iv) The parallelogram law of addition of forces (v) The triangular law of forces (vi) The law of conservation of Energy (vii) The principle of work and Energy 1.6.1 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). Or in other words, Every particle continues in its state of rest or of motion in a straight line unless it

is compelled to change that state by an external force imposed on the body. So the first law is used 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 force and will move in the direction of this resultant force. F Resultant force or Net force F ma where m Mass of the body a cceleration 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 Mass cceleration Force mass acceleration Statics 1.15

1.16 Engineering Mechanics - www.airwalkpublications.com i.e., F d dv mv m dt 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 and have the same line of action. 1.6.2 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, also called the Constant of Gravitation. Refer Fig 1.3(a). M -F Example The attraction of the earth on a particle located on its surface. The force F Fig.1.3 exerted by the earth on the particle is known as weight W. y the law of gravitation r F m

Statics 1.17 we have, W GMm GM, Introducing the constant g 2 r 2, we have 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. 1.6.3 The Parallelogram Law of Forces Construct a parallelogram using a force P and a force Q as two sides of the parallelogram. Now the P diagonal passing through represents the resultant force. This is called as Q parallelogram law for Fig.1.3(b) addition of two forces. Refer Fig 1.3 (b). Or in other words If two forces P and Q acting at a point 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 1.6.4 Triangular Law of Forces If two forces acting simultaneously on a body are represented by the sides of a triangle taken in order, then

1.18 Engineering Mechanics - www.airwalkpublications.com their resultant is represented by the closing side of the triangle taken in the opposite order. 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. ddition of Two Vectors (Forces) P and Q [Refer Fig. 1.4 (a) and (b)] P Q Even if we P R esultant P+Q change the order we get same resultant Resultant 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 P P Q (a) P- Q -Q (b) subtraction Fig.1.5. Resultant P- Q

Statics 1.19 by adding to P with the negative force Q as shown in Fig. 1.5 (a) & (b). So, P Q P Q 1.6.5 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 the opposite order. Graphically arrange all the given forces in the tip-to-tail fashion in any order. Join the tail of the second force with the tip of the first force and so on. To obtain the resultant, draw vector joining tail of first force and tip of last force as shown in Fig. 1.6(b). 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 y using Polygon Law of Forces Vectorial ddition of F 1, F 2, F 3, F 4, and F 5 Resultant This means: We can add the forces as in Fig 1.6 (b)

1.20 Engineering Mechanics - www.airwalkpublications.com R esultant = F +F +F +F + F 1 2 3 4 5 o y using Polygon law of forces F 1 F 2 oth Resultant are equal Change the order : o R esultant = F +F +F +F +F F 4 1 2 3 4 5 F 2 F 5 F 4 F 5 F 3 F 1 F 3 Fig.1.6.(b) Fig.1.6.( c ) We can also add the forces as in Fig. 1.6 (c). F 2 F 1 F 4 F 3 F 5 Resultant. Consider Fig. 1.6 (b) and 1.6 (c). Here order is changed. Even then, we get the same resultant. Hence, the resultant R does not depend upon the order in which the forces are chosen to draw the polygon. 1.6.6 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 and 1.8 Line of action a F b = a Same Effect (Push) Fig.1.7. (Pull) F b

Statics 1.21 force F acting on a body at point is transferred to point 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 line of action Fig.1.8. = line of action F lternatively, 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. 1.6.7 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.

1.22 Engineering Mechanics - www.airwalkpublications.com 1.6.8 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 1.7 FORCE ND FORCE SYSTEM 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. force can also change or tend to change the size and shape of a deformable bodies. The characteristics of forces are (i) Magnitude (ii) Direction (slope) (iii) Sense (iv) Point of application. These characteristics distinguish one force from O M a gn itu de (F ) point of action F = inclination of force w ith horizontal Fig. 1.9 Force lin e of a ctio n

Statics 1.23 another. Representation of a force is shown in the Fig. 1.9. 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 =30N C =40N (a) Space diagram Fig 1.10 C (b) Force diagram Force diagram is the diagram in which the forces are drawn to scale, parallel to their respective line of action. 1.7.2 Types of force system When several forces are acting upon a body simultaneously, they constitute a system of forces. Force 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. Coplanar Force system Non-coplanar Fig. 1.11 Concurrent Parallel Non-concurrent, Non-parallel Collinear Concurrent Parallel Non-concurrent, Non-parallel

1.24 Engineering Mechanics - www.airwalkpublications.com 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 (I) Coplanar force system system of forces that are contained in a single plane or system of forces having their line of actions in the single plane is called coplanar force system. (Fig. 1.12) Forces F 1, F 2, F 3 are coplanar forces. F 1 F 2 F 3 Fig 1.12 Coplanar forces Plane Forces F 1 F 2 (a) 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. O F 3 Fig. 1.13 Concurrent Forces Plane (b) 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 force system (Fig. 1.14) Forces F 1, F 2, F 3 F 3 F 2 O 2 1 Fig 1.14 Coplanarconcurrent forces F 1

Statics 1.25 are in same plane and pass through common point O. (c) Non concurrent and non-parallel forces When the system of forces whose line of action does not pass through a common point and move of the forces are parallel is called non concurrent and non-parallel system. Fig: 1.15 Non-Concurrent Non-Parallel (d) 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. 1.16 is an example of coplanar - non concurrent forces. F 1 F 2 F 3 Plane Fig 1.16 Coplanar unlike parallel forces forces (e) Collinear forces When the system of forces acting in a single plane with a common line of action are called collinear forces. Fig. 1.17 shows collinear force system. O F 2 F 1 F 3 F 4 F 5 Plane Common line of action Forces Fig 1.17 Collinear forces

1.26 Engineering Mechanics - www.airwalkpublications.com (f) Parallel forces When the lines of action of all the forces of the system are parallel, then the system is called parallel force system. When the lines of action of all the forces are parallel and all of them act in the same direction, then the force system is called like parallel forces. Plane F 1 F 2 F 3 F 4 Fig 1.18 Coplanar like parallel forces Plane When the line of action F 1 F 3 of all the forces are parallel forces and some forces act in one direction while others in F opposite direction, then the 2 force system is called unlike Fig 1.16 Coplanar parallel force system. Fig. unlike parallel forces 1.18 shows an example of coplanar like parallel force system and Fig. 1.19 shows an example of coplanar unlike parallel force system. II Non coplanar force system system of forces in which the forces lie in different planes or in a three dimensional space is called a Non coplanar force system. (a) Non coplanar concurrent forces system of forces lying in different planes with their line of action intersecting at a common point are called Non

Statics 1.27 Z Planes Z Planes F 1 F 1 F 3 O O X O X F 2 F 2 Y Fig 1.20 Non Coplaner Concurrent forces Y Fig 1.21 Non Coplanar non concurrent forces coplanar concurrent force system. Fig. 1.20 shows such an example. (b) Non coplanar Non concurrent forces system of forces lying in different planes with different lines of actions are called non coplanar non concurrent forces (Fig. 1.21) (c) Non coplanar parallel forces system of forces lying in different plane but their line of action parallel to each other are called non coplanar parallel forces. In Fig. 1.22 forces F 1 and F 2 are non coplanar like parallel force and F 3 and F 4 are non coplanar unlike parallel forces. (d) Non coplanar Non concurrent and Non parallel forces (skew forces) 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 to each other are called

1.28 Engineering Mechanics - www.airwalkpublications.com Z Z F 1 F 2 F 1 F 3 F 4 O X F 3 O X F 2 Y Fig 1.22 Non Coplanar parallel forces Y Fig 1.23 Skew forces Non coplanar Non concurrent and non parallel forces. These forces are also called as skew forces. 1.8 MOMENT OF FORCE Let P Force acting on a body. l Perpendicular distance between the point O and line of action of the force P. M The moment of the force P about O P l Moment of a force is the turning effect produced by a force, on the body, on which it acts. The moment of a force O (a) l ody Line of ction of force P l Line of P ction of force Perpendicular distance O (b) Fig.1.24. Moment of Force.

Statics 1.29 is equal to the product of the force and the perpendicular distance between the point and the line of action of the force. The point O is called moment centre, l is the moment arm and the line perpendicular to the plane containing the force, P and the point, O and passing through O is called axis of moment. Units of Moment Since the moment of a force, is the product of force and distance, therefore the units of the moment will be Newton metre (briefly written as N-m). Similarly the units of moment may be kn-m and N-mm. Clockwise Moment It is the moment of a force, whose effect is to turn or rotate the body, in the clockwise sense Fig. 1.25(a). nticlockwise Moment O O l Fig.(a) P It is the moment of a force, whose effect is to turn or rotate the body, in the anticlockwise sense Fig. 1.25 (b). l Fig.(b) Fig.1.25 P Note: The general convention is to take clockwise moment as negative and anticlockwise moment as positive.

1.30 Engineering Mechanics - www.airwalkpublications.com 1.8.1 Principle of Moments (or Varignon s Principle) Principle of moments states that the moment of R cos the resultant of a number of forces about r 2 any point is equal to the algebraic sum of the O R sin R moments of all the forces r 1 of the system about the Fig.1.26.Varignon s Principle. same point. In otherwords the moment of a force about any point is equal to the algebraic sum of the moments of its components about that point. ccording to Varignon s principle: Refer 1.26 Moment of R about O must be equal to algebraic sum of moments of its components R sin & R cos about O. M o R r R sin r 1 R cos r 2 This principle of moments is very important in solving the problems of rigid bodies since it is often easier to determine the moments of force components rather than the moment of the applied force itself. If we want to find moments of several coplanar concurrent forces about a point, then the formula used is M F x r y F y r x where F x is component of each force in x direction F y is component of each force in y direction r y is the perpendicular distance of F x from the point r x is the perpendicular distance of F y from the point r

Statics 1.31 1.9 RESULTNT OF FORCE SYSTEM 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. 1.27 shows a system of three forces acting F 3 F 2 Particle on a particle with the resultant R which can produce the same effect. 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. Thus, the resultant is a representative force which has the same effect on the particle as the group of forces it replaces. R (Resultant ) F 1 = Fig 1.27 Resultant force 1.9.1 Resultant of two coplanar concurrent forces Resultant of two coplanar concurrent forces can be obtained by the following methods. 1. nalytical 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. nalytical 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. 1.28 (a) F 2 F 2 R R F 1 F 1 Hence, the relationship between F 1, F 2 and R can be derived as follows [Fig. 1.28 (b)]. Consider the Fig. 1.28(b) Resultant of two forces parallelogram OC. Let O and O 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 O D 2 CD 2 Fig 1.28(a) Parallelogram law O 2 2 O D D 2 CD 2 O 2 2 O D C 2 O F 1 D [... C 2 D 2 CD 2 ] R 2 F 1 2 2 F 1 F 2 cos F 2 2 [... C O F 2 ] F 2 R C [... D C cos F 2 cos ]

Statics 1.33 2 Hence R F 1 2 F2 2F1 F 2 cos lso tan CD O D 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 F 1 2 F2 2 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 1.1: 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.

1.34 Engineering Mechanics - www.airwalkpublications.com Solution: We know that Resultant of two coplanar concurrent forces is 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 is 50 kn, If the resultant of forces F 1, F 2 acting at an angle 50 2 30 2 20 2 2 30 20 cos cos 1 or 0 Thus F 1 and F 2 are collinear to each other when the resultant is 50 kn. II nalytical 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 F 2 F 1 F 1 Fig. 1.29 Triangle law of forces R F 2 C Fig. 1.29(a) : Representation of forces

Statics 1.35 side (closing side) represents the resultant in the opposite order. (Fig. 1.29) Thus by applying trigonometric relations can be applied. From Fig. 1.29 (a), we have sin sin C sin III Graphical method - parallelogram law of forces Consider two forces F 1 & F 2 acting at a point O as shown in Fig. 1.30 (a). Draw two sides of parallelogram representing the forces F 1 & F 2 to some scale. Fig. 1.30 (b) as O and O. Complete the parallelogram by drawing C, C equal to O and O to the same scale. Draw the diagonal OC representing the resultant of the two force system. F 2 F 2 C R O F 1 F 1 x axis O (a) (b) Fig 1.30 Graphical method- parallelogram law of forces The length of diagonal measured to the chosen scale represents the magnitude of the resultant acting at an angle from the x axis. IV Graphical method - Triangle law of forces Consider two forces F 1 and F 2 as shown in Fig. 1.31 (a). Draw O to some scale as shown in the Fig. 1.31 (b).

1.36 Engineering Mechanics - www.airwalkpublications.com F 2 R esultant(r ) F 2 O (a) F 1 O (b) F 1 x axis From the end, draw to same scale. Join O as the closing side of triangle. The length O 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. Problem 1.2: If two forces F 1 20 kn and F 2 15kN act on a particle as shown in Fig., find their resultant by (1) Parallelogram law and (2) Triangle law. (nna GE 1151, Nov 2009) Given F 1 20 kn, F 2 15 kn 15kN 70 Using parallelogram law 2 R F 1 2 F2 2 F 1 F 2 cos O 70 o Fig (a) 20kN R 20 2 15 2 2 20 15 cos 70 R 28.813 Using Triangle law [Take scale 1 cm = 1 kn]

Statics 1.37 Draw horizontal O of length 20 cm. From, draw a line of length 15 at an angle of 70. Now join O. The length of O measures 28.813 cm. This is the resultant. O R esultant Fig (b) 20kN 70 o15 kn ie R 28.813 kn 1.9.2 Rectangular components of force system (Resolution) The process of Y replacing a single force F acting on a particle by two F or more forces which F y together have the same effect of a single force is X called resolution of force O F x into components. Fig 1.32 Resolution of forces Theoretically a force can be resolved into an infinite number of component sets, however in practice a force is resolved into two components. Consider a force F acting at an angle from x axis as shown in Fig. 1.32. The two rectangular components of forces are resolved along x and y axis and are given as F x F cos ; F y F sin In the vector form Force F F x i F y j

1.38 Engineering Mechanics - www.airwalkpublications.com [Vectors will be explained in detail later] Considering resolution of force on an inclined plane as shown in Fig. 1.33. The force F is resolved into two components F n and F t Normal component, F n F cos Tangential F t F sin component, F F t F n x Fig 1.33 Resolution of forces on inclined plane Writing in vector form F F t Fnsub In simple words, Resolution of Forces into components single force can be resolved into two components which give the same effect on the particle. Refer Fig. 1.34 (a). Now a force F is Horizontal resolved as - Horizontal component of F component of F i.e. = F cos F cos and Vertical component of F i.e. F sin. F = F sin Fig.1.34.(a) 1.9.3 Resultant of several concurrent forces Consider a particle shown in Fig. 1.35 (a) subjected to four forces F 1, F 2, F 3, F 4. The resolution of each force is shown in Fig. 1.35 (b).

Statics 1.39 Y Y F 2 F 2 F 1 F 2y F 1 F 1y 2 1 X 3 o 4 F 3x F 2x o F 1x F 4x X F 3 F 3y F 3 F 4 F 4y F 4 Particle (a) (b) Fig 1.35 Several Concurrent forces The Resultant 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 later) The Resultant of the above forces is shown in Fig. 1.35 (c). The ngle made by Resultant R is given by Y R y R X O R x Fig 1.35 (c) Resultant tan R y R x F y F 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

1.40 Engineering Mechanics - www.airwalkpublications.com 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. 1.36) 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 2 Ry F 2 F 3 Y 2 3 4 F 1 1 F 4 Fig 1.36 Force system X ngle of inclination of Resultant tan 1 R y R x 1.9.4 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 Fig.1.37. Q

The resultant force, of a given system of forces, is found out by the method of resolution as followed: 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 ) 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 angle will vary depending upon the values of F y and F x as discussed below: (Refer Fig. 1.38) (a) (b) III Quadrant 180 o (-) x axis 90 o (-) y axis (+) y axis I Quadrant (+) x axis Statics 1.41 o o 0,360 When F y is ve, resultant will be in 1 st Quadrant or 2 nd Quadrant. (i.e. in between 0 to 180). II Quadrant 270 o IV Quadrant Fig.1.38. When F y is ve, resultant will be in III rd Quadrant or IV th Quadrant. (i.e. in between 180 to 360).

1.42 Engineering Mechanics - www.airwalkpublications.com (c) (d) 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). 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). 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 PROLEMS Problem 1.3: Four forces act on a bolt 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 Fsin 1 F 4 =100N Fig.(a) Fig.(b) F 3 =110N

Statics 1.43 Sl. No. Force in N F x in N F y in N 1. 150 cos 30 129.9 150 sin 30 75 30 o F =150N 1 150 sin 30 150 cos 30 2. 80 cos 70 27.36 80 sin 70 = 75.18 80 sin 70 F =80N 2 70 o -80 cos 70 ( sign indicates that force acts left side) 3. 0 Here, there is no [No F x] horizontal component and hence F x3 = 0 110 ( sign indicates force acts downward) F = -110N 3 4. 100 cos 15 = 96.593 100 sin 15 = 25.882 100 cos 15 15 o ( sign indicates force acts downward) -100 sin 15 Total F x = 199.133 F y = 14.298 N

1.44 Engineering Mechanics - www.airwalkpublications.com 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. dd all the x components and find F x. Similarly add all the y components and find F y. Refer Table. 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 2 199.6 N Resolution of forces in x components and y components are given in detail in the next page. R 199.6 N F =14.3N y 4.1 o Fig.(c) tan F y F x 14.3 199.1 0.07164 R=199.6N F=199.1N x 4.1 Problem 1.4: The following forces act as shown in figure. y Determine the resultant of this force system. 25N 50N Solution: Let F x lgebraic sum of the resolved 60 o 30 o 45 o 40N x 30N Fig.