Recommended Content

Page 1 :
WORKSHOP CALCULATION, & SCIENCE, (As Per NSQF), 2nd Year, Common for All Engineering Trades under CTS, (For all 2 Year Trades), , DIRECTORATE GENERAL OF TRAINING, MINISTRY OF SKILL DEVELOPMENT & ENTREPRENEURSHIP, GOVERNMENT OF INDIA, , NATIONAL INSTRUCTIONAL, MEDIA INSTITUTE, CHENNAI, Post Box No. 3142, CTI Campus, Guindy, Chennai - 600 032, (i), , Copyright free, under CC BY Licence

Page 2 :
Workshop Calculation & Science (NSQF) - 2nd Year, Common for All Engineering Trades Under CTS, (For All 2 year Trades), , Copyright © 2018 National Instructional Media Institute, Chennai, First Edition : September 2019, , Copies :, , Rs. /-, , All rights reserved., No part of this publication can be reproduced or transmitted in any form or by any means, electronic or mechanical,, including photocopy, recording or any information storage and retrieval system, without permission in writing from the, National Instructional Media Institute, Chennai., , Published by:, NATIONAL INSTRUCTIONAL MEDIA INSTITUTE, P. B. No.3142, CTI Campus, Guindy Industrial Estate,, Guindy, Chennai - 600 032., Phone : 044 - 2250 0248, 2250 0657, 2250 2421, Fax : 91 - 44 - 2250 0791, email : [email protected], [email protected], Website: www.nimi.gov.in, (ii), , Copyright free, under CC BY Licence

Page 3 :
FOREWORD, , The Government of India has set an ambitious target of imparting skills to 30 crores people, one out of every, four Indians, by 2020 to help them secure jobs as part of the National Skills Development Policy. Industrial, Training Institutes (ITIs) play a vital role in this process especially in terms of providing skilled manpower., Keeping this in mind, and for providing the current industry relevant skill training to Trainees, ITI syllabus, has been recently updated with the help of comprising various stakeholder's viz. Industries, Entrepreneurs,, Academicians and representatives from ITIs., The National Instructional Media Institute (NIMI), Chennai, has now come up with instructional material to, suit the revised curriculum for Workshop Calculation & Science 2nd Year NSQF Commom for all 2 year, engineering trades under CTS will help the trainees to get an international equivalency standard where their, skill proficiency and competency will be duly recognized across the globe and this will also increase the, scope of recognition of prior learning. NSQF trainees will also get the opportunities to promote life long, learning and skill development. I have no doubt that with NSQF the trainers and trainees of ITIs, and all, stakeholders will derive maximum benefits from these IMPs and that NIMI's effort will go a long way in, improving the quality of Vocational training in the country., The Executive Director & Staff of NIMI and members of Media Development Committee deserve appreciation, for their contribution in bringing out this publication., Jai Hind, , RAJESH AGGARWAL, Director General/ Addl. Secretary, Ministry of Skill Development & Entrepreneurship,, Government of India., , New Delhi - 110 001, , (iii), , Copyright free, under CC BY Licence

Page 4 :
PREFACE, The National Instructional Media Institute(NIMI) was set up at Chennai, by the Directorate General of Training,, Ministry of skill Development and Entrepreneurship, Government of India, with the technical assistance, from the Govt of the Federal Republic of Germany with the prime objective of developing and disseminating, instructional Material for various trades as per prescribed syllabus and Craftsman Training Programme(CTS), under NSQF levels., The Instructional materials are developed and produced in the form of Instructional Media Packages (IMPs),, consisting of Trade Theory, Trade Practical, Test and Assignment Book, Instructor Guide, Wall charts,, Transparencies and other supportive materials. The above material will enable to achieve overall improvement, in the standard of training in ITIs., A national multi-skill programme called SKILL INDIA, was launched by the Government of India, through a, Gazette Notification from the Ministry of Finance (Dept of Economic Affairs), Govt of India, dated 27th, December 2013, with a view to create opportunities, space and scope for the development of talents of, Indian Youth, and to develop those sectors under Skill Development., The emphasis is to skill the Youth in such a manner to enable them to get employment and also improve, Entrepreneurship by providing training, support and guidance for all occupation that were of traditional, types. The training programme would be in the lines of International level, so that youths of our Country can, get employed within the Country or Overseas employment. The National Skill Qualification Framework, (NSQF), anchored at the National Skill Development Agency(NSDA), is a Nationally Integrated Education, and competency-based framework, to organize all qualifications according to a series of levels of Knowledge,, Skill and Aptitude. Under NSQF the learner can acquire the Certification for Competency needed at any, level through formal, non-formal or informal learning., The Workshop Calculation & Science 2nd Year (Comon for All 2 year Engineering Trades under CTS) is, one of the book developed by the core group members as per the NSQF syllabus., The Workshop Calculation & Science (Common for All 2 year Engineering Trades under CTS as per, NSQF) 2nd Year is the outcome of the collective efforts of experts from Field Institutes of DGT, Champion, ITI’s for each of the Sectors, and also Media Development Committee (MDC) members and Staff of NIMI., NIMI wishes that the above material will fulfill to satisfy the long needs of the trainees and instructors and, shall help the trainees for their Employability in Vocational Training., NIMI would like to take this opportunity to convey sincere thanks to all the Members and Media Development, Committee (MDC) members., , R. P. DHINGRA, EXECUTIVE DIRECTOR, , Chennai - 600 032, , (iv), , Copyright free, under CC BY Licence

Page 5 :
ACKNOWLEDGEMENT, The National Instructional Media Institute (NIMI) sincerely acknowledge with thanks the co-operation and, contribution of the following Media Developers to bring this IMP for the course Workshop Calculation & Science, (2nd Year) as per NSQF., , MEDIA DEVELOPMENT COMMITTEE MEMBERS, Shri. M. Sangara pandian, , -, , Training Officer (Retd.), CTI, Guindy, Chennai., , Shri. G. Sathiamoorthy, , -, , Jr.Training Officer (Retd.), Govt I.T.I, DET - Tamilnadu., , NIMI CO-ORDINATORS, Shri. Nirmalya nath, , -, , Deputy General Manager,, NIMI, Chennai - 32., , Shri. G. Michael Johny, , -, , Assistant Manager,, NIMI, Chennai - 32., , NIMI records its appreciation of the Data Entry, CAD, DTP Operators for their excellent and devoted services in, the process of development of this IMP., NIMI also acknowledges with thanks, the efforts rendered by all other staff who have contributed for the development of this book., , (v), , Copyright free, under CC BY Licence

Page 6 :
INTRODUCTION, The material has been divided into independent learning units, each consisting of a summary of the topic and an, assignment part. The summary explains in a clear and easily understandable fashion the essence of the mathematical, and scientific principles. This must not be treated as a replacment for the instructor’s explanatory information to be, imparted to the trainees in the classroom, which certainly will be more elaborate. The book should enable the, trainees in grasping the essentials from the elaboration made by the instructor and will help them to solve independently, the assignments of the respective chapters. It will also help them to solve the various problems, they may come, across on the shop floor while doing their practical exercises., The assignments are presented through ‘Graphics’ to ensure communications amongst the trainees. It also assists, the trainees to determine the right approach to solve the problems. The required relevent data to solve the problems, are provided adjacent to the graphics either by means of symbols or by means of words. The description of the, symbols indicated in the problems has its reference in the relevant summaries., At the end of the exercise wherever necessary assignments, problems are included for further practice., Duration:, 2nd Year, , Time allotment : 84 Hrs, , Time allotment for each module has been given below. Common for all 2 year Engineering Trades. Instructors are, here with informed to make use of the same., S.No, , Title, , Exercise No., , Time allotment (Hrs), , 2.1.01 - 2.1.03, , 14, , 2.2.04, , 6, , 1, , Friction, , 2, , Centre of Gravity, , 3, , Area of cut out regular surfaces, and area of irregular surfaces, , 2.3.05 - 2.3.07, , 16, , 4, , Algebra, , 2.4.08 & 2.4.09, , 12, , 5, , Elasticity, , 2.5.10 & 2.5.11, , 9, , 6, , Heat Treatment, , 2.6.12 & 2.6.13, , 3, , 7, , Profit and Loss, , 2.7.14 & 2.7.15, , 12, , 8, , Estimation and Costing, , 2.8.16 & 2.8.17, , 12, , LEARNING / ASSESSABLE OUTCOME, On completion of this book you shall be able to, • Demonstrate basic mathematical concept and principles to perform, practical operations., • Understand and explain basic science in the field of study including, simple machine., , (vi), , Copyright free, under CC BY Licence

Page 7 :
CONTENTS, Exercise No., , Title of the Exercise, , Page No., , Friction, Friction - Advantages and disadvantages, Laws of friction, co-efficient of, friction, angle of friction, simple problems related to friction, , 1, , 2.1.02, , Friction - Lubrication, , 8, , 2.1.03, , Friction - Co- efficient of friction, application and effects of friction in workshop, practice, , 12, , 2.1.01, , Centre of Gravity, 2.2.04, , Centre of gravity - Centre of gravity and its practical application, , 14, , Area of cut out regular surfaces and area of irregular surfaces, 2.3.05, , Area of cut out regular surfaces - circle, segment and sector of circle, , 24, , 2.3.06, , Related problems of area of cut out regular surfaces - circle, segment and, sector of circle, , 27, , Area of irregular surfaces and application related to shop problems, , 29, , 2.3.07, , Algebra, 2.4.08, , Algebra - Addition , subtraction, multiplication & division, , 32, , 2.4.09, , Algebra - Theory of indices, algebraic formula, related problems, , 36, , Elasticity, 2.5.10, 2.5.11, , Elasticity - Elastic, plastic materials, stress, strain and their units and, young’s modulus, , 42, , Elasticity - Ultimate stress and working stress, , 53, , Heat Treatment, 2.6.12, , Heat treatment and advantages, , 56, , 2.6.13, , Heat treatment - Different heat treatment process – Hardening, tempering,, annealing, normalising and case hardening, , 58, , Profit and Loss, 2.7.14, , Profit and loss - Simple problems on profit & loss, , 67, , 2.7.15, , Profit and loss - Simple and compound interest, , 73, , Estimation and Costing, 2.8.16, 2.8.17, , Estimation and costing - Simple estimation of the requirement of material etc.,, as applicable to the trade, , 85, , Estimation and costing - Problems on estimation and costing, , 92, , (vii), , Copyright free, under CC BY Licence

Page 8 :
SYLLABUS, 2nd Year, , Common for All Engineering Trades under CTS, (For All 2 year Trades), , S.no., I, , Duration: One Year, , Syllabus, Friction, , Time, , Marks, , 14, , 7, , 6, , 4, , 16, , 9, , 12, , 8, , 9, , 4, , 3, , 3, , 12, , 8, , 12, , 7, , 84, , 50, , 1 Advantages and disadvantages, Laws of friction, co- efficient of friction, angle of, friction, simple problems related to friction, 2 Friction – Lubrication, 3 Co- efficient of friction, application and effects of friction in workshop practice, II, , Centre of Gravity, 1 Centre of gravity and its practical application, , III, , Area of cut – out regular surfaces and area of irregular surfaces, 1 Area of cut – out regular surfaces – circle, segment and sector of circle, 2 Related problems of area of cut – out regular surfaces – circle, segment and, sector of circle, 3 Area of irregular surfaces and application related to shop problems, , IV, , Algebra, 1 Addition, Subtraction, Multiplication & Divisions, 2 Algebra – Theory of indices, Algebraic formula, related problems, , V, , Elasticity, 1 Elastic, plastic materials, stress, strain and their units and young’s modulus, 2 Ultimate stress and working stress, , VI, , Heat Treatment, 1 Heat treatment and advantages, 2 Different heat treatment process – Hardening, Tempering, Annealing, Normalising,, Case Hardening, , VII, , Profit and Loss, 1 Simple problems on profit & loss, 2 Simple and compound interest, , VIII, , Estimation and Costing, 1 Simple estimation of the requirement of material etc., as applicable to the trade, 2 Problems on estimation and costing, Total, , (viii), , Copyright free, under CC BY Licence

Page 9 :
Friction - Advantages and disadvantages, Laws of friction, co-efficient of, friction, angle of friction, simple problems related to friction Exercise 2.1.01, Introduction, When on a solid surface, another solid is rubbed a force is, created between the two solids which acts in the opposite, direction of motion or tries to obstruct the motion of the, object, this force is called frictional force. This phenomenon, is called friction. This happens due to roughness of the, two surfaces., In other words, It is the force of resistance offered to motion,, experienced by bodies which are in contact. It depends, upon the normal reaction between the contacting surfaces, and the nature of the surfaces. No surface is absolutely, friction less., , Forces acting on a body when a pulling force is, applied to move (Fig 2), •, , Weight of the block acting vertically downward (W), , •, , The normal reaction which acts upwards (R), , •, , The applied pulling force (F), , •, , The frictional force (Ff), , When the body is about to move W=R, F=P, When pulling force is increased the body starts to move., , Friction plays an important role in our daily life. It would, not be possible to walk without friction between our foot, and floor. Vehicles are able to run on roads because of the, friction between the wheels and road., , Laws of friction (Fig 3 & 4), , Types of friction, 1 Static friction, 2 Dynamic friction, , •, , Frictional force is directly proportional to the normal, reaction between contacting surfaces., , •, , Frictional force acts opposite to the direction of motion., , •, , Frictional force depends on the nature of contacting, surfaces., , •, , Frictional force is independent over the area and shape, of contacting surfaces., , 1 Static friction, The friction between two solid objects when at rest is called, static friction., Eg. Static friction can prevent an object from sliding down, on a sloped surface., Limiting friction, When the frictional force (F) is equal to the applied pulling, force (P) then the friction between two surfaces is known, as limiting friction. (i.e F=P), 2 Dynamic friction, It is the friction between two objects, when are in motion, is called dynamic friction. It is also called kinetic friction., Sliding friction, It is the friction experienced by an object when its slides, over another object. Sliding friction is always less than, limiting friction., Rolling friction, It is the friction that occurs when a circular object such as, a ball or roller rolls on a flat surface. Rolling friction is less, than sliding friction. (ball or roller bearing), , Coefficient of friction, It is a ratio between the frictional force to the normal, reaction when the body is just about to move but at, equilibrium. It is represented by symbol . (read as ‘meu’), , 1, , Copyright free, under CC BY Licence

Page 10 :
Therefore, , Co - efficient of friction =, =, , Limiting friction(or)force, normal reaction(or) weight, , F, F, (or ) f, W, R, , Angle of friction (Fig 5), The forces acting on a body when it is just about to move, by the application of a pulling force are W, R, P and F. The, forces 'R' and 'F' are compounded and we get the resultant, force 'S'. The angle formed by 'S' with 'R' is the angle of, friction., Therefore, ,  P = F + W sin and R = W cos ,  P = uR + W sin = u(W cos  W sin ,  P = W x tan  cos + W cos , = Wx, , F, W, tan , tan , , =W x, =W, , Sin φ, x cos θ + W cos θ, Cos φ, , Sin φ + cos θ + Wcos φ x cos θ, Cos φ, , [Sin (θ + φ)], Cos φ, , Similarly when the body is about to slide down the plane, the applied force P must be equal to, =W, , [Sin (θ + φ)], Cos φ, , To keep the body under equilibrium when the body is about, to move up the plane by the action of an applied force the, , Angle of repose (Fig 6), , applied force P = W, , [Sin (θ + φ)] and the force necessary, , Cos φ, to be applied to the body to prevent it from sliding down, the plane will be P = W, , [Sin (θ + φ)], , Cos φ, Note : Under all circumstances, =W, A body placed on an inclined surface remains at rest till, the angle of inclination equals the angle of friction. When, it exceeds the body starts sliding down. This is known as, angle of repose., Motion up the plane, When > a force must be applied to keep the body in, equilibrium. The applied force may be parallel to the plane,, horizontal or at an angle to the plane itself., When the body is at the point of motion up the plane the, frictional force ‘F’ acts down the plane., Forces acting are W,R,P and F. The weight force ‘W’ is, resolved into two components of W cos  perpendicular to, the plane acting downwards and W sin  acting parallel to, the plane downwards., F, = μ = tan φ, R, , Where is the angle of friction., 2, , [Sin (θ + φ)] < P < = W [Sin (θ + φ)], Cos φ, , Cos φ, , Advantages of friction, 1 Helps us to walk without slipping., 2 Used to stop vehicles when brakes are applied., 3 Movement of vehicles due to friction between revolving, wheels with tyres and the road., 4 Power transmission using gear drive or belt pulley drive., 5 Using friction we can sharp any object and also to hold, it., 6 Nails and screws are held in wood by friction., 7 Heat is produced when two rough surfaces are rubbed, against each other., Disadvantages of friction, 1 It causes wear and tear of the machine parts., 2 It produces heat and may cause melting of machine, parts. To avoid production of heat using of coolant is, necessary., , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence

Page 11 :
3 It reduces efficiency of a machine., , W = 14500 kg = Weight, , 4 It reduces speed of the moving object. eg. spindle, shaft,, piston etc., , F, , = ? = Force friction, , F, , μ=, , Friction can be reduced, , W, , 1 By using suitable lubricants (oil, grease) between the, moving parts., , 0.28 =, , 2 By polishing the surface to make them smooth., , F, , 14500, = 0.28 x 14500, , 3 By using ball bearings and roller bearings., , F, , = 4060 kg., , 4 By the use of wheel., Example, 1 A force of 40 kg is required to pull a weight of 400, kg on a horizontal plane. Determine the, coefficient of friction., , Coefficient of friction =, , Force, Weight, , =, , F, W, , But F = P and R = W, , 5 A force of 800 gram weight is needed to pull a, block weighing 3200 gram. What is the co-efficient, of friction., F, , W = 3200 gm = Weight, , , =, , 2 A force for 30N is required to move a body of mass, 35 kg on a flat surface horizontally at a constant, velocity. Find the coefficient of friction., Mass of the body = 35 kg. = W, The weight force=35 x 10 = 350 N, , (By taking, 1kg = 10N), , (By taking g = 10 meter/sec2), , W, , =, , Ff, R, , =, , 30, 350, , =, , 3, 35, , = 0.086, ,  = 0.24 = Co-efficient of friction, W = 20 kg = Weight, , 800, 3200, ,  = 0.25, 6 A force of 40 kg is required to move a mass of 80, kg on a flat surface horizontally at a constant, velocity. Calculate its co-efficient of friction?, F, , = 40 kg = Force, , , , =?, , Co-efficient of friction = μ =, =, , F, W, , 40, 80, ,  = 0.5, 7 A weight of 10 kg is resting on a horizontal table, and can just moved by a force of 2 kg. Find the, co-efficient of friction?, , F = ? = Force required, , F, W, F, , W = 10 kg = Weight, F, , = 2 kg = Force, , 20, , , , =?, , F = 20 x 0.24, , Co-efficient of friction = ?, , F = 4.8 kg, 4 A tanker with loaded weight of 14500 kg is running, on the road. If the co-efficient friction between, tyres and road surface is 0.28. Find out its force of, friction., , , W, , Co-efficient of friction = ?, , 3 A solid weights 20 kg. This is placed on a solid, surface. How much force does it require to come, in motion when co-efficient of friction is 0.24., , 0.24 =, , F, , W = 80 kg = Weight, , = 0.09, , μ=, , =?, , Co-efficient of friction = ?, , = 0.1, , ∴μ=, , = 800 gm = Force, , Co-efficient of friction = μ =, , 40, F Ff, , = =, 400, W R, , F, , F, , = 0.28 = Co-efficient of friction, , μ=, =, , F, W, 2, 10, ,  = 0.2, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 3

Page 12 :
8 A body weighing 100kg is resting on a table. Find, the co-efficient of friction if a force of 30 kg makes, its just to move?, W = 100 kg = Weight, F = 30 kg = Force,  =?, Co-efficient of friction = ?, , μ=, =, , , , F, , = ? = Force, , , , = tan  = 0.25, , , , tan-1 0.25, , 30, , F, W, , 100, F, , 0.25 =, , R =?, Normal reaction = W, Limiting friction = ?, =?, , R = Normal reaction = 10 kg, , 150, , F, , = 0.25 x 150, , F, , = 37.5 Kg., , 12 A body of mass 60kg rests on a horizontal plane., The value of co-efficient of friction between it and, the plane being 0.2. Find the work done in moving, the body through a distance of 5 meters along the, plane., , , = 0.2 = Co-efficient of friction, , W, , = 60 kg = Weight, , S, , = 5 m = Distance, , W = ? = Work done, , Limiting friction = F = 2.5 kg, , F, , μ=, , W, , 100, 2.5, , F, W, , 0.2 =, , F, 60, , 10, F, ,  = 0.25, , = 60 x 0.2, = 12 kg, , 10 A wooden block weights 100 kg. If the co-efficient, of friction is 0.3, find out force required to move, the block., W = 10 kg = Weight, , Work done = Force x distance = F x S, = 12 x 5, = 60 m - Kg., , , , = 0.3 = Co-efficient of friction, , (ie) Work done (or) Applied force = 60 m - Kg., , F, , = ? = Force, , 13 If a force of 30N is required to move a mass of, 35kg on a flat surface horizontally at constant, velocity, what will be the co-efficient of friction?, , μ=, 0.3 =, , F, W, F, , F, , = 100 x 0.3, , F, , = 30 kg, , F, , = 30 N = Force, , W, , = 35 kg = Weight, , 1 kg = 9.8 N, , 100, , 35 Kg = 9.8 x 35 = 343 N, Co-efficient of friction, , 11 Calculate the angle of inclination, if a weight of, 150 kg is in equilibrium, co-efficient of friction is, 0.25. Calculate the force of normal reaction also., 4, , , , μ=, , = 2.5 kg = Force, , =, , = 0.25 = Co-efficient of friction, , W, , W = 10 kg = Weight, , μ=, , , , = 14º 2’, , 9 A metal block weighing 10 kg rests on a horizontal, table. A horizontal force of 2.5 kg can just slide, the block. Find the normal reaction, limiting, friction and co-efficient of friction?, , , , = 150 kg = Work done, , F, ,  = 0.3, , F, , W, , =, , 30 N, 35 kg, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , = μ=, , F, W

Page 13 :
=, , , = 2.2/cos 30º, , 30 N, 35 x 9.8 N, , = 2.2/0.8660, = 2.54 kg, , = 0.087, , 14 A block of ice weighing one quintal rests in, equilibrium on a wooden plank inclined at 30º., Find the coefficient of friction between the ice and, wood., W, , 17 Calculate the angle of inclination, if a weight of, 150 kg is in equilibrium. Coefficient of friction is, 0.25. Calculate the force of normal reaction also.,  = ? = angle of inclination, , = 1 quintal = 100 kg = Weight, , , , = 30º, , , , = tan tan 30º, , , , = 0.5774, ,  = 0.25, F = ? = Force, tan  = , , 15 Calculate the force that is required to slide a mass, of 980 kg on a guide, when the coefficient of, friction between the surfaces is 0.09., , tan  = 0.25,  = 14º 2'20", , μ=, , W, , = 980 kg = Weight, , , , = 0.09 = Co-efficient of friction, , F, , = Force = ?, = μ=, , F = 0.25 x 150 kg, , W, , F = 37.5 kg., 18 A body of mass 10 kg rests on a horizontal plane., The co-efficient of friction between the body and, plane is 0.15. Find the work done in moving the, body through a distance of 10 meter., , F = 0.09 x 980 kg, Required force(F) = 88.2 kg, , W = 10kg = Weight, , 16 A metal block weighing 10kg rests on a horizontal, board and the coefficient of friction between the, surfaces is 0.22. Find (a) the horizontal force which, will just move the block and (b) the force acting, at an angle of 30º with the horizontal, which will, just move the block., W, (a), , F, , W = Work done = ?, , = = 0.22, 0.15, , =?, , μ=, 0.22 =, F, , S = 10 meter = distance, , μ=, , (b) Force acting at an angle of 30º with the horizontal?, (a), ,  = 0.15 = Co-efficient of friction, , = 10 kg = Weight, , Co-efficient of friction, , W, , 150 Kg, , F, , F, 980 kg, , F, , F, , 0.25 =, , Co-efficient of friction, 0.09 =, , W = 150kg = Weight, , F,    tan , W, , =, , F, W, F, 10 Kg, , F = 0.15 x 10 kg, , F, , F = 1.5 kg, , W, , Work done = W = F x S, = 1.5 kg x 10 m, , F, 10 kg, , = 15 m - kg, , = 2.2 Kg., , (b) Force acting at an angle of 30º=, , F, Cos θ, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 5

Page 14 :
Assignment A, 1 A force 50N is required to move a mass of 40kg on a, flat surface horizontally at a constant velocity. Find the, coefficient of friction. (9.8N = 1kg), 2 A vehicle having a weight of 800kg is moving on the, road. If the coefficient of friction between the tyres and, road surface is 0.3, then calculate the force of friction., 3 A solid weighing 50kg is place on a solid surface. How, much force is required to move the block when, coefficient of friction is 0.25 between the block and the, surface., , 7 A body of mass 2000 kg moves a distance of 10 meters, in 5 sec. If the co-efficient of friction between the body, and floor is 0.3 find the horizontal force required to move, the body and horsepower absorbed against friction., 8 A vehicle is moving at 50kmph and the load on the, vehicle is 5000 kg. Find the H.P. required to move the, vehicle if = 0.2., 9 Find out the power lost due to friction by a planer under, the following conditions., Mass of the planer table = 3500 kg, , 4 A railway wagon weighs 1250 tonnes. If the coefficient, of friction between it and the rails is 0.003, find the, force required to move the wagon., 5 A body of mass 100kg rests on a horizontal plane. The, angle of friction between the body and the plane being, 0.025. Find the work done is moving the body through, a distance of 16m along the plane., , Rate of moment of the table=0.5 m/sec, Co-efficient friction between the table and the, ways=0.06, 10 A truck having weight 12000 kg is moving on the road., If the co-efficient of friction between the tyres and the, road surface is 0.3, then calculate the force of friction., , 6 A body of mass 20kg rests on a horizontal plane the, co-efficient of friction between the body and plane is, 0.3. Find the work done in moving the body through a, distance of 10 meters., , Assignment B, 1, , F = 1800 N, , F = 1.2 kN, , μ (static) = 0.16, , d = 60 mm, , μ (dynamic) = 0.012, , μ = 0.03, , FR to overcome, static friction, = ______ N, , Frictional torque MR, = ______ Nm, , FR to overcome, dynamic friction, = ______ N, 2, , 4, , (Frictional torque =, Frictional force x, radius), 5, , mass = 180 kg, , mass = 250 kg, , μ = 0.15, , FR = 160 N, , FR = ______ N, , μ = ______, 6, , FR = 120 N, μ = 0.032, , 3, , Normal force F, = ______ N, , F = 5000 N, μ (dry) = 0.03, μ (fluid friction), = 0.01, , 7, , FR when dry, = ______ N, FR when lubricated =, ______ N, 6, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , m = 1000 kg, μ = 0.4, Force required to move, FR = ______ N

Page 15 :
C MCQ, 1 Which one of the following is useful friction, A Rings in cylinders, , B Crankshaft bearings, , C Brake shoe linings, , D Wheel hole bearings, , A 0.0215, , B 0.0152, , C 0.0125, , D 0.0251, , 7 Calculate the pulling force required for the figure shown., , 2 Which is in between the wheels and road, if vehicles, are able to run on roads., A erosion, , B motion, , C corrosion, , D friction, , 3 Which direction of motion frictional force acts., A equal, , B opposite, , C inclined, , D forward, , 4 What is the formula of angle of friction, if ‘F’ is the, frictional force, R is the normal reaction and q is the, angle of friction., A Tan q =, , F, R, , B Cot q =, , F, R, , C Sin q =, , F, R, , D Cos q =, , F, R, , B  F, R, , C F x R, , D F + R, , B 28 Kg, , C 29 Kg, , D 30 Kg, , 8 Determine the co-efficient of friction() between brass, and steel when a brass slider was placed on the, horizontal steel surface until it is just moving, if brass, slides (W) = 3 Kgf, Brass slides (W) = 3 Kgf, Force (F) required = 0.7 kgf, A 0.033, , B 0.133, , C 0.233, , D 0.333, , 9 Which is necessary to avoid production of heat., , 5 What is the formula for Co-efficient of friction (m)., A  R, F, , A 27 Kg, , A sand, , B coolant, , C lubricant, , D salt, , 10 Which is using for reduce the friction., , 6 A loaded truck weighs 2400 kg and it can be moved by, a force of 30 kg. Determine the co-efficient of rolling, friction, , A lubricants, , B sand, , C coal, , D coolant, , Key Answers, A, 1 0.1275, 2 240 Kg, 3 12.5 Kg, 4 3.75 Tonne, 5 40 m-kg, 6 60 m-kg, , 7 F = 600 Kg, , B, , C MCQ, 1 288N, 21.6 N, , 1 C, , 6, , C, , 2 0.065, , 2 D, , 7, , D, , 3 150 N, 50 N, , 3 B, , 8, , C, , 9 1.4 HP, , 4 36 N, 1.08 Nm, , 4 A, , 9, , B, , 10 3600 Kg, , 5 264.6 N, , 5 B, , 10 A, , P = 16 HP, 8 185.2 HP, , 6 3750 N, 7 3920 N, , Workshop Calculation & Science : (NSQF) Exercise 2.1.01, , Copyright free, under CC BY Licence, , 7

Page 16 :
Friction - Lubrication, , Exercise 2.1.02, , There are 3 systems of lubrication., •, , Gravity feed system, , •, , Force feed system, , •, , Splash feed system, , Gravity feed, The gravity feed principle is employed in oil holes, oil cups, and wick feed lubricators provided on the machines., (Figs 1 & 2), , Oil pump method, In this method an oil pump driven by the machine delivers, oil to the bearings continuously, and the oil afterwards, drains from the bearings to a sump from which it is drawn, by the pump again for lubrication., Splash lubrication, In this method a ring oiler is attached to the shaft and it, dips into the oil and a stream of lubricant continuously, splashes around the parts, as the shaft rotates. The rotation, of the shaft causes the ring to turn and the oil adhering to, it is brought up and fed into the bearing, and the oil is then, led back into the reservoir. (Fig 5) This is also known as, ring oiling., , Force feed/Pressure feed, Oil, grease gun and grease cups, The oil hole or grease point leading to each bearing is, fitted with a nipple, and by pressing the nose of the gun, against this, the lubricant is forced to the bearing. Greases, are also force fed using grease cup. (Fig 3), , In other systems one of the rotating elements comes in, contact with that of the oil level and splash the whole, system with lubricating oil while working. (Fig 6) Such, systems can be found in the headstock of a lathe machine, and oil engine cylinder., , Oil is also pressure fed by hand pump and a charge of oil, is delivered to each bearing at intervals once or twice a, day by operating a lever provided with some machines., (Fig 4) This is also known as shot lubricator., 8, , Copyright free, under CC BY Licence

Page 17 :
Types of grease guns, The following types of grease guns are used for lubricating, machines., •, , ‘T’ handle pressure gun (Fig 7), , •, , Automatic and hydraulic type pressure gun (Fig 8), , After cleaning the open gears, oil them and repeat, lubrication regularly. (Fig 13), , •, , Lever-type pressure gun (Fig 9), , Lubrication to exposed slideways, The moving parts experience some kind of resistance even, when the surface of the parts seems to be very smooth., The resistance is caused by irregularities which cannot, be detected by the naked eyes., Without a lubricant the irregularities grip each other as, shown in the diagram. (Fig 10), , With a lubricant the gap between the irregularities fills up, and a film of lubricant is formed in between the mating, components which eases the movement. (Fig 11), The slideways are lubricated frequently by an oilcan., (Fig 12), , Lubricate bearings, A shaft moving in a bearing is also subjected to frictional, resistance. The shaft rotates in a bush bearing or in ball/, roller bearing, experiencing friction., When the shaft is at rest on the bottom of the bush, bearing, there is hardly any lubricant between the shaft, and the bush. (Fig 14), , When the shaft starts rotating the lubricant maintains a, film between the shaft and the bush and an uneven ring of, lubricant builds up. (Fig 15), , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence, , 9

Page 18 :
When the shaft is rotating at full speed a full ring of, lubricating film surrounds the shaft (Fig 16) which is known, as hydro dynamic lubrication., , Functions, The functions of cutting fluids are:, This lubrication ring decreases the frictional resistance, very much and at the same time protects the mating, members against wear and changes., Some bush bearings have oil feeding holes over which the, oil or grease cup is mounted and the lubricant is fed through, the holes into the bearing by gravity feed system.(Fig 17), , -, , to cool the tool as well as the workpiece, , -, , to reduce the friction between the chip and the tool face, by lubricating, , -, , to prevent the chip from getting welded to the tool, cutting edge, , -, , to flush away the chips, , -, , to prevent corrosion of the work and the machine., , Advantages, As the cutting fluid cools the tool, the tool will retain its, hardness for a longer period; so the tool life is more., Because of the lubricating function, the friction is reduced, and the heat generated is less. A higher cutting speed can, be selected., , Hints for lubricating machines:, , As the coolant avoids the welding action of the chip to the, tool-cutting edge, the built up edge is not formed. The tool, is kept sharp and a good surface finish is obtained., , -, , identify the oiling and greasing points, , As the chips are flushed away, the cutting zone will be neat., , -, , select the right lubricants and lubricating devices, , -, , apply the lubricants., , The machine or job will not get rusted because the coolant, prevents corrosion., Properties of a good cutting fluid, , The manufacturer’s manual contains all the necessary, details for lubrication of parts in machine tools. Lubricants, are to be applied daily, weekly, monthly or at regular, intervals at different points or parts as stipulated in the, manufacturer’s manual., These places are indicated in the maintenance manuals, with symbols as shown in Fig 18., , A good cutting fluid should be sufficiently viscous., At cutting temperature, the coolant should not catch fire., It should have a low evaporation rate., It should not corrode the workpiece or machine., It must be stable and should not foam or fume., , Cutting Fluids, , It should not create any skin problems to the operator., , Cutting fluids and compounds are the substances used for, efficient cutting while cutting operations take place., , Should not give off bad smell or cause itching etc. which, are likely to irritate the operator, thus reducing his efficiency., Should be transparent., , 10, , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence

Page 19 :
Types of cutting fluids, , Compounded or blended oil, , The following are the common cutting fluids., •, , Straight mineral oil, , These oils are used in automatic lathes. These oils are, much cheaper and have more fluidity than fatty oil., , •, , Chemical solution (synthetic fluids), , Fatty oil, , •, , Compounded or blended oil, , •, , Fatty oils, , •, , Soluble oil (Emulsified oil-suds), , Lard oil and vegetable oil are fatty oils. They are used on, heavy duty machines with less cutting speed. They are, also used on bench-works for cutting threads by taps and, dies., , Straight mineral oil, , Soluble oil (Emulsified oil), , Straight mineral oils are the coolants which can be used, undiluted. Use of straight mineral oil as a coolant has the, following disadvantages., It has little effect as a cutting fluid., , Water is the cheapest coolant but it is not suitable, because it causes rust to ferrous metals. An oil called, soluble oil is added to water which gets a non-corrosive, effect with water in the ratio of about 1: 20. It dissolves in, water giving a white milky solution. Soluble oil is an oil, blend mixed with an emulsifier., , Hence straight mineral oils are poor coolants. But kerosene, which is a straight mineral oil is widely used as a coolant, for machining aluminium and its alloys., , Other ingredients are mixed with the oil to give better, protection against corrosion, and help in the prevention of, skin irritations., , Chemical solution (Synthetic oil), , Soluble oil is generally used as a cutting fluid for centre, lathes, drilling, milling and sawing., , It gives off a cloud of smoke., , These consist of carefully chosen chemicals in dilute, solution with water. They possess a good flushing and a, good cooling action, and are non-corrosive and nonclogging. Hence they are widely used for grinding and, sawing. They do not cause infection and skin trouble., They are artificially coloured., , Soft soap and caustic soda serve as emulsifying agents., A chart showing coolants for different metals is given, below., , Recommended cutting fluids for various metals and different operations, Material, , Drilling, , Reaming, , Threading, , Turning, , Milling, , Aluminium, , Soluble oil, Kerosene, Kerosene and, lard oil, , Soluble oil, Kerosene, Mineral oil, , Soluble oil, Kerosene, Lard oil, , Soluble oil, , Soluble oil, Lard oil, Mineral oil, Dry, , Brass, , Dry, soluble oil, Mineral oil, Lard oil, , Dry, soluble oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, , Bronze, , Dry, soluble oil, Mineral oil, Lard oil, , Dry, soluble oil, Mineral oil, Lard oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, Mineral oil, Lard oil, , Cast iron, , Dry, Air jet, Soluble oil, , Dry, soluble oil, Mineral lard oil, , Dry, sulphurized oil, Mineral lard oil, , Dry, soluble oil, , Dry, soluble oil, , Copper, , Dry, soluble oil, Mineral lard oil, Kerosene, , Soluble oil, Lard oil, , Soluble oil, Lard oil, , Soluble oil, , Dry, soluble oil, , Steel, alloys, , Soluble oil, Sulphurized oil, Mineral lard oil, , Soluble oil, Sulphurized oil, Mineral lard oil, , Sulphurized oil, Lard oil, , Soluble oil, , Soluble oil, Mineral, , General, purpose, steel, , Soluble oil, Sulphurized oil, Lard oil, Mineral lard oil, , Soluble oil, Sulphurized oil, Lard oil, , Sulphurized oil, Lard oil, , Soluble oil, , Soluble oil, Lard oil, , Workshop Calculation & Science : (NSQF) Exercise 2.1.02, , Copyright free, under CC BY Licence, , 11

Page 20 :
Friction - Co -efficient of friction, application and effects of friction in workshop, practice, Exercise 2.1.03, Co-efficient of friction, The ratio between the limiting frictional force and the normal, reaction is called co-efficient of friction., , 2 An empty drum weighing 50kg is resting on a shop, floor. Find the coefficient of friction if a force of, 15kg makes it just move., , Suppose, by applying a force P kg, the object is just fit to, move, then limiting frictional force will be produced in, between the two surfaces. The limiting frictional force will, be equal to external force applied and will work in the, opposite direction., , , W, , = 50 kg = Weight, , F, , = 15 kg = Force, , Co-efficient of friction, , F = P kg, , =, , According to the second law of limiting frictional force, the, frictional force will be proportional to normal reaction., F  R (sign is proportional to), F = R x constant, F, = constant, R, , or, , , , , , R, , or F   .R, , Co-efficient of friction =, , Limiting frictional force, Normal reaction, , = 1000 kg, , Distance (S), , = 5 meter, , Time (t), , = 5 second, , Co-efficient of friction () = 0.3, , ii, , Force (F), , =?, , Horse power (H.P.) = ?, , μ=, , 1 The sliding valve of a steam engine has dimensions, 25cm by 45 cm and the steam pressure on the back, of the valve is 25 kg/cm2. If the co-efficient of friction, is 0.13. Calculate the force required to move the, valve. Dimension of steam valve = 25 cm x 45 cm., , F, W, , 0.3 =, , = 25 kg/cm2, , F, 1000 Kg, , F, , = 0.3 x 1000 kg, , F, , = 300 kg, , Co-efficient of friction = 0.13, , H.P =, , ii, , Force required to move the valve = ?, F=?, Force of the steam, , cm2, , 4 A weight of 600 kg is kept on the inclined plane, at 300. Calculated the normal reaction and force, rolling downwards., , Force acts on the valve = 28125 kg, , F, , Solution:, , W, , 0.13 =, , FxS, 1, x, t, 75, , Horse power absorbed against friction = 4.H.P., ,  25cm  45cm = 28125 kg., , μ=, , (1 HP = 75 m.kg/sec), , 1, = 4 H.P, H.P = 300 x 5 x, 75, 5, , = Pressure x Area, = 25 x 25 x 45, , 25kg, , = 0.3, , Weight (W), , i, , Steam pressure, , W, , 15 kg, 50 kg, , i, , Co-efficient of friction is always constant for any two objects, and it has no unit., Example, , F, , 3 A machine crate weighing 1000kg moves distance, of 5m in 5 sec. If the coefficient of friction between, the crate and floor is 0.3, calculate the horizontal, force required to move the crate and horse power, absorbed against friction., , This constant between objects is called co-efficient of, friction. This is represented by ., F, , = μ=, , Weight kept on the inclined plane (W) = 600kg, , F, , Angle of the inclined plane () = 300, , 28125, F = 0.13 x 2812, , , , Normal reaction (R) = W . cos , , Force required to move the valves = 3656.25 Kg, 12, , Copyright free, under CC BY Licence

Page 21 :
= 600 x cos 300, = 600 (0.8660), = 519.6 kg, Force rolling downwards = W . sin , = 600 (0.5000), = 300 kg, Normal reaction, , = 519.6 kg, , Force rolling downwards, , = 300 kg, , 5 Find out the power lost due to friction by a planner, under the following conditions., Mass of the planer table, , = 3500 kg, , Rate of movement of the table, , = 0.5m/sec, , , , Co-efficient of friction between, the table and the ways, = 0.06, Solution:, Weight of planer (W), , = 3500 kg, , Distance moved (d), , = 0.5 m/sec, , Co-efficient of friction () = 0.06, Co-efficient of friction, , = μ=, , F, W, , F, 3500, , 0.06, , =, , F, , = 0.06 x 3500 = 210 kg, , Workdone, , = 1 H.P, , 105 kgm/sec, , =, , 105  1, = 1.4 H.P, 75, , Power lost due to friction = 104 H.P, , = 600 x sin 300, , , , 75 kgm/sec, , = F x distance moved, = 210 x 0.5 = 105 kgm/sec, , 6 A planner table weighting 800 kg moves a, distance of 2 metres in seconds on its bed. If coefficient of friction between bed and table is 0.30, find the power required to move the table aganist, the friction., 7 On a milling machine table a component of 20, kgf is clamped with the help of three equidistant, clamps. What force must be exerted by each, clamp to avoid slipping of the component when, the horizontal cutting force is 60 kgf and the, coefficient of friction is equal to 0.2., 8 A machine weight of 14500 kg moving on the, floor.If the co-efficient of friction between the, machine and floor surface is 0.28 then calculate, the force of friction., 9 A tail stock of a lathe has a mass of 21.5 kg and, co-efficient of friction at the slides is 0.122. What, horizontal force will be required to slide the tail, stock?, 10 An inclined surface makes an angle of 30 degrees, with the horizontal. An object weigting 5 tons is, placed on the surface. Find out the normal, reaction at the object and also the effective force, required to bring the object downwards., 11 A glass block of 400 grams has been placed on, the table. The glass is commuted by a string to a, 40 grams scale pan. The string passes over pulley., When a weight of 60 grams is placed on the scale, pan, the block starts sliding. Find out the coeffiecient of friction between wood and glass., , Workshop Calculation & Science : (NSQF) Exercise 2.1.03, , Copyright free, under CC BY Licence, , 13

Page 22 :
Centre of gravity - Centre of gravity and its practical application, Exercise 2.2.04, Any object comprises of a large number of particles. Each, particle is pulled towards the earth due to the force of, gravity. Thus, the forces on the particles are equal, parallel, and act in the same direction. These forces will have a, resultant which acts through a particular point ‘G’. This, fixed point ‘G’ is called the centre of gravity., , of a body. If the lamina is assumed to have uniform mass, per unit area, then the centroid is also the centre of gravity, in a uniform gravitational field., Methods to calculate centre of gravity, 1 By geometrical consideration., 2 By moments., Principle : The total moment of a weight about any axis, = The sum of the moments of the various parts about the, same axis., 3 By graphical method., The first two methods are generally used to find out the, centre of gravity or centroid, as the third method can, become tedious., Centre of gravity by geometrical consideration, , Concept of Centre of gravity, In physics, an imaginary point in a body of matter where,, for convenience in certain calculations, the total weight of, the body may be thought to be concentrated. The concept, is sometimes useful in designing static structures (e.g.,, buildings and bridges) or in predicting the behaviour of a, moving body when it is acted on by gravity., In a uniform gravitational field the centre of gravity is, identical to the centre of mass, a term preferred by, physicists., , Gravitation, The mutual attractive force of bodies due to which they, attract each other is called gravitation., 1 Gravity, The attractive force of the earth due to which it attracts all, bodies towards its centre is called gravity., The value of gravity varies from place to place on the ground, surface. Its general value is 9.81 m/s2., Centroid, Different geometrical shapes such as the circle, triangle, and rectangle are plane figures having only 2-dimensions., They are also known as laminas. They have only area, but, no mass. The centre of gravity of these plane figures is, called as the Centroid. It is also known as the geometrical, centre. The method of finding out the centroid of a plane, figure is the same as that of finding out the centre of gravity, 14, , 1 The centre of gravity of a circle is its centre., 2 The centre of gravity of a square, rectangle or a, parallelogram is at the points where its diagonals meet, each other. It is also the middle point of the length as, well as the width., 3 The centre of gravity of a triangle is at the point where, the medians of the triangle meet., 4 The centre of gravity of a right circular Cone is at a, distance of from its base., , Copyright free, under CC BY Licence

Page 23 :
5 The centre of gravity of a hemisphere is at a distance, of from its base., 6 The centre of gravity of a segment of a sphere of radius, h is at a perpendicular distance of from the centre of, the sphere., , more likely it is to tip over when it is tilted by a force. This, figure demonstrates a bus driving on two different grades;, the second one is steep enough to cause the centre of, gravity to fall outside of the base of the vehicle, which will, cause it to topple over., , 7 The centre of gravity of a semicircle is at a perpendicular, distance of from its centre., 8 The centre of gravity of a trapezium with parallel side, 'a' and 'b' is at a distance of measured from the base, 'b'., 9 The centre of gravity of a cube of side L is at a distance, of from every face., 10 The centre of gravity of a Sphere of diameter 'd' is at a, distance of from every point., , Equilibrium, , •, , Number of 2 pencil, , A body is said to be in equilibrium if the resultant of all the, forces acting on a body is zero and if there is no turning, moment., , •, , A fine edge like a ruler or a credit card, , There are three states of equilibrium (Fig 5), , •, , A permanent marker, , 1 Stable equilibrium, , •, , A ruler, , 2 Unstable equilibrium, , Centre of gravity; An experiment, , Step 1, , 3 Neutral equilibrium, , Attempt to balance the pencil on the edge you have, selected, Balancing the pencil may take some trial and error. The, point at which the pencil balances may not be where you, first thought. If it begins to tip in one direction, move the, pencil back slowly in the opposite direction until it will, stay there on its own., Step 2, Once the pencil is balanced, mark the location of the, balancing point with a permanent marker., Step 3, Measure the distance between the ends of the pencil and, the balancing point you have marked. Are the two lengths, equal? On my pencil, the length from the eraser to the, balancing point was actually 1.25 inches less than the, length from the pencil tip to the balancing point. Why, would this be the case?, In our experiment, the balancing point was another word, for the centre of gravity of this pencil. In other words, if we, cut the pencil in two at the mark we made in the experiment,, the two parts would be equal in weight. However, they are, not equal in length. As you may have already figured out,, the metal piece that houses the eraser contributes more, to the weight of the pencil, so the CG is closer to that side, of the pencil., Keeping up with that centre, The centre of gravity is an important concept in determining, the stability of a structure. It’s the reason why a good, homeowner will keep the top branches of his trees trimmed., It’s also the reason why a pick-up truck might not be the, best vehicle choice for a first time driver. Stability is, maximized in objects with a lower centre of gravity and a, wide base. The taller and more top-heavy an object, the, , 1 Stable equilibrium, A body is said to be in a stable equilibrium if it returns to, its original position when slightly displaced. (The C.G. is, as low as possible)., E.g :, , 1 A cone resting on its base, 2 A ball on a concave surface, 3 Funnel resting on its base. (Fig 6), , 2 Unstable equilibrium, A body is said to be in an unstable equilibrium if it does, not return to its original position when slightly displaced., Its centre of gravity falls taking it away from its original, position. (CG is at high points), E.g:, , 1 A cone resting on its tip, 2 A ball on convex surface, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 15

Page 24 :
3 Funnel standing on its tube end. (Fig 7), , Some example of equilibrium in daily life, 1 The lower decks of the ships are loaded with heavy, cargoes. This makes the centre of gravity of the whole, ship lower and its equilibrium becomes more stable., 2 A man carrying a bucket full of water in one hand extends, his opposite arm and bends his body towards it., , 3 Neutral equilibrium, A body is said to be in a neutral equilibrium if on being, slightly displaced, it takes a new position similar to its, original one. The centre of gravity remains undisturbed., (CG is neither raised or lowered), Eg:, , 1 A cone resting on its side, 2 A ball on flat surface, 3 Funnel resting on its side (Fig 8), , 3 While carrying load on back the man bends forward so, that his and the load’s centre of gravity falls on his, feet, if he walks erect, he will fall backward., 4 While climbing a mountain, a man bends forward and, bends backward while descending so that the centre, of gravity of his load falls on his feet., 5 In a double-decker, more passengers are, accommodated in the lower deck and less on the upper, so that the centre of gravity of the bus and the, passengers is kept low to eliminate any chance of, turning., Example, 1 Find the centroid of the isosceles triangular plate, as shown in the figure., , Model 1, Conditions for stable equilibrium, •, , The CG should be as low as possible., , •, , It should have a broad base., , •, , The vertical line passing through the CG should fall, within the base., , As per Pythagoras theorem, , Conditions of equilibrium, A body is said to be in a state of equilibrium under the, action of forces when there is no motion of rotation or, translation of the body. There are three conditions of, equilibrium of a body which are given below:, i, , Since BCD=45º then BD=DC=x, , Algebraic sum of the horizontal components of all the, forces acting on the body must be zero., H = 0, , ii Algebraic sum of the vertical components of all the, forces acting on the body must be zero., V = 0, iii Algebraic sum of the moments of all the forces acting, on the body must be zero., M = 0, Torque or twisting moment of a couple is given, by the product of force applied and the arm of, the couple (i.e. Radius). In fact, moment means, the product of “force applied” and the, “perpendicular distance of the point and the, line of the force”., , BD2 + DC2 = CB2, x2 + x2, , = 402, , 2x2 = 1600, x2, , =, , 1600, 2, , x, , 800 = 28.28 cm, , =, , Centroid from BD =, , = 800, , x, 3, , =, , 28.28, 3, , = 9.43 cm, , 2 A rectangular lamina has 10 cm and 8 cm. Find, the centroid. Centroid of rectangular = Diagonals, intersecting point., Centroid of rectangular = Diagonals intersecting point, Centre of AB =, Centre of AB =, , 10, =5, 2, , 8, , =4, , 2, Centroid lying 4 cm from AB and 5 cm from AD, , 16, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

Page 25 :
4 A thin lamina consists of an isosceles triangle of, height 120mm and base 100mm placed on a, semicircle of diameter 100mm. find the location, of its centre of gravity., , 3 A thin lamina is shown in the figure below. Find, the centre of gravity., Area of right angled triangle (a1) =, =, , 1, , bh, , 2, 1, , x 10 x 12, , 2, = 60 cm2, , 1, , Centroid of right angled triangle =, , h from base, , 3, 1, , =, , x 12, , 3, Centroid from E = 4 cm, , Centroid of rectangle, Area of rectangle= 5 x 10 = 50cm2, Area of triangle =, =, Total area, , 1, 2, , Centroid from A (h1), , = 12 - 4 = 8 cm, , Area of half circle (a2), , =, , bh, , 1, 2, , 1, , =5x, , Centroid distance from E to D, 1, , 3, , =, , 5, = 1.67 cm, 3, , (CG2) Centroid of plate is lying in between CG1 and CG2., From the figure torque is about AD., 62.5x, , = 50 x 5 + 12.5 x 11.67, = 395.875, , x =, , 395.875, , x 3.14 x 5 x 5, , =, , 4r, 3π, , =, , 4x5, 3 x 3.14, , = 2.123 cm, ⎛ Height of ⎞ ⎛ Centroid of, ⎟⎟ + ⎜⎜, (h2 ) = ⎜⎜, ⎝ triangle ⎠ ⎝ half circle, , ⎞, ⎟⎟, ⎠, , = 12 + 2.123, , = 250+145.875, 62.5x, , 2, , Centroid of semi circle = (Vertical distance from, centre of diagonal), , distance from its height., , 3, , 1, , = 39.25 cm2, , x 5 x 5 = 12.5 cm2, , The centre of gravity for rectangle is the point of intersection, of diagonal = 5 cm distance from AD (CG1), Centre of gravity for triangle is, ,  r2, , 2, =, , = 50 + 12.5 = 62.5 cm2, , 1, , = 14.123 cm, To find centroid of lamina, , = 6.334 cm, , 62.5, , y=, , Centre of gravity is 6.334 cm from AD, on the centre, axis., , a1 h1 + a2 h2, a1 + a2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 17

Page 26 :
=, , 60 x 8 + 39.25 x 14.123, 60 + 39.25, , =, , 480 + 554.328, 99.25, , =, , = 54 KNm + 90KNm, = 144 KNm, B This moment is equal to moment of section A and, section B about 'O' (distance of action being D meter), That is =(36KN+20KN) x D(meter)=56 DKNm, , 1034.328, 99.25, , Again equating A and B, 144 KNm = 56 DKNm, , = 10.421 cm, Centroid is lying at 10.421 cm from point A, 5 A uniform rod weighing 50kg and 3m long carries, loads as shown below. Find out the distance of, the CG of the system from the left hand end., , 144 KNm, =D, 56 KNm, 144, 56, , =D, , Therefore D =, , 18, 7, , = 2.57 meters, The distance of CG of the shaft from left hand is 2.57, meters., , Distance of CG from A = x, Total weight = 50 + 15 + 20 + 25 = 110 kg, 110 x x = (50 x 1.5) + (15 x 1.75) + (20 x 2.25) + (25 x, 2.75), , 7 A thin lamina is shown in the figure. Find centre, of gravity., , = 75 + 26.25 + 45 + 68.75 = 215, Therefore x =, , 215, = 1.96 m, 110, , Distance of CG of the system from A = 1.95 m, 6 A long shaft is composed of two section A and B, each 3 meter long and weight 36KN and 20KN, respectively. Find out the position of centre of, gravity of the shaft., , Solution, Let G1 be the c.g. point of section A, , As the body is symmetrical about y-axis centre of, gravity lies on this axis., , Let G2 be the common c.g. of the shaft and its distance is, D from left hand end., , Let AB is the axis of reference, , Now, take moments about 'O', , Let y = The distance between centre of gravity and point, F, the point of reference as shown in the figure., , A Moment of section A about O = 36 KN x 1.5 m, Moment of section B about O, , = 20 KN x 4.5 m, , Adding both we get as below, Total moment about O=(36 KN x 1.5 m + (20KN x 4.5m), 18, , Let a1= Area of rectangle CDBA = 45 x 40 = 1800 mm2, h1 = Distance between centre of gravity of rectangle, of point F = 40 = 20 mm, 2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

Page 27 :
Let a2 = Area of triangle ECD=1/2 x base x height, , Area of B1, , = 30 x 15 mm2, , =1/2 x 45 x 50 = 1125 square mm, h2 = distance between centre of gravity of triangle of, point F., , = 450 mm2, Distance of CG2 from AB, , = 20 +, , =1/3rd height of triangle +width of rectangle, =, , 1, , (50) + 40 =, , 3, , 170, 50, + 40 =, mm, 3, 3, , Applying formula, , y=, , = 20 + 15 mm, = 35 mm, Area of triangle =, , 1, , x 40 x 40 mm2, , 2, = 800 mm2, , a1 h1 + a2 h2, ,  PTR - Isosceles triangle, , a1 + a2, , Draw perpendicular line PS on TR from P., , ⎛ 170 ⎞, 1800(20) + 1125⎜, ⎟, ⎝ 3 ⎠, =, 1800 + 1125, 800, 5, 36000 + 63753.75, 99753.75, =, =, 2925, 2925, y, , 30, mm, 2, , PSR - right angled triangle, By applying Pythagoras theorem,, , x2 + x2 = 402, 2x2, , = 1600, 2, , x, , = 34.10 mm, , = 800, , x = 800, , The CG is at a distance of 34.1mm from point F, the point of reference in the line AB., 8 Find the CG of the lamina shown below., , = 28.28 mm, Distance of CG3 from TR = x = 28.28 mm = 9.43 mm, 3, 3, Dist. Of CG3 from AB = 20 + 30 + 9.43 mm = 59.43 mm, Total area =1200 + 450 + 800 mm2 = 2450 mm2, Distance from AB = Ymm, Taking moment at AB 2450 x y = 1200 x 10 + 450 x 35 +, 800 x 9.43, =12000 +15750 + 7544, =35294, y, , =, , 35294, = 14.41 mm, 2450, , Distance of CG is on the line PQ from side, AB = 14.41 mm., 9 A rectangular sheet of cardboard of uniform, thickness measuring 20 cm by 15 cm is divided, into four parts by drawing the diagonals. One of, the triangles formed on a 15 cm side is removed., Find the position of the C.G. of the remainder., CG is in PQ, CG1, CG2 and CG3 - centres of centre of gravity., Area of A1, , = 60 x 20 mm2, =1200 mm2, , Distance of CG1, from AB, , =, , 20, mm, 2, , = 10 mm, Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 19

Page 28 :
= r2h unit3, , ABCD - rectangle hard board, AC, BD - diagonals, O, - meeting point, AOD, - removed portion, , Volume of circle, ,  AOB = CG1,  BOC = CG2,  COD = CG3, , Remaining volume, , =  x 3 x 3 x 12 cm3, = 339.3 cm3, = 14400 - 339.8 cm3, = 14060.7 cm3, C.G is on PQ, , C.G, = Centre of gravity of the hard board, Take CG is on PQ, Area of  AOB, , = 1/2 bh unit2, = 1/2 x 20 x 7.5 cm2, = 75 cm2, , Area of BOC, , = 1/2 x 15 x 10 cm2, = 75 cm2, , Area of COD, , = 1/2 x 20 x 7.5 cm2, , = 75 cm, Total area, = 75 + 75 + 75 cm2, = 225 cm2, Taking moment at side BC, 2, , Distance of CG1 =, , 20, in = 10 cm, 2, , Distance of CG2 =, , 10, in = 3.33 cm, 3, , Distance of CG3 =, , 20, in = 10 cm, 2, , CG in before drilling = C.G2, CG in after drilling, , = C.G, , Calculating the moment on side AB, 100, = 50 cm, 2, , Distance of CG1, , =, , Distance of CG2, , = 10 cm, , Distance of C.G, , =x, , 14060.7 x x + 339.3 x 10 =14400 x 50, 14060.7x + 3393, , = 720000, , 14060.7x, , = 720000 - 3393, = 716607, , x, , =, , 716607, 14060.7, , = 50.97 cm, C.G. of the strip, , Distance of C.G = x, 225 x x, , CG in before drilling = C.G1, , = 75 x 10 + 75 x 3.33 + 75 x 10, = 750 + 249.75 + 750, = 1749.75, , = 50.97 cm from side AB., , 11 Centre of gravity in a lamina (Area), Find the position of c.g of the area shown in Fig11., (All dimensions are in mm.), , x = 1749.75 = = 7.777, 225, , = 7.777, Distance of CG is on the line PQ from side BC=7.78cm, 10 A steel rod 100x12x12cm has a hole dia 6cm, drilled in it as shown in the figure. Find the, position of the C.G. of the strip., , Solution, Taking moments of area about the line ab, we get the, equation as below., Moment of area abcd+moment of area efgh=moment of, area of complete figure., Volume of rod, , Now to calculate the areas, , = a2h unit3, = 12 x 12 x 100 cm3, = 14400 cm3, , 20, , 1 Area of abcd = 120 mm x 50 mm, = 6000 mm2, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

Page 29 :
Area of efgh = 200 x 75 mm2, = 15000 mm, Total area, , = (6000+15000) mm2, , (abcd+efgh) = 21000 mm2, 2 (6000 m2 x 25) + (15000 mm2 + 150 mm), , = 1000 cm3, Mass of copper cube, , = Volume x Density, = 1000 x 8.9, , = (21000mm2) x (x mm), , = 8900 gms, , 150000mm2 + 2250000mm2 = (21000 mm2) x (x mm), 240000mm2 = (21000 mm2) x (x mm), Therefore x =, , 21000 mm 2, , 8900, Kg, 1000, , =, , 2400000 mm 2, , = 8.9 Kg., (g=Acceleration due to gravity=10m/sec2), , 2400, mm, =, 21, =, , 10 6 mm 3, 10 3 mm 3, , =, , 2, , Weight of copper cube = 8.9 kg x 10 m/sec2, Similarly, , 800, mm, 7, , = 89 N, , 1000 cm3 x 8.5 x 10, 1000, 2, (Take g = 10m/sec ) = 8.500 x 10 = 85N, , Weight of brass cube =, , = 114.3 mm, Hence c.g. point of composite figure is 114.3 mm from, A on the line ab., 12 Centre of gravity point of a composite body can, be found out by using a variation of principle of, moments., Example (Fig 12), , Let cg of separate cubes be Gc and GB as shown, in figure., The distance between Gc and GB=100mm or, 0.1 m, Let c.g of the total object be at G which is 'P', meter to the right of GC or (0.1-P) meter to the, left of GB., Take moments about G, Clock moments, , = WB x (0.1 - P), = 85 x (0.1 - P)), = 8.5 - 85P, , Anti clock moments = WC x P, = 89 x P[Nm], = 89P, By principle of moments, Moment of part "A" about O+ Moment of part "B" about, O=Moment of (A+B) about O., The moment of the (A+B) acting through, point G., A copper cube of 100mm side is attached to brass cube, of 100 mm side, as sketched in the figure. Calculate the, position of c.g of composite object. Take densities of copper, and brass as 8.9 gms/cm3 and 8.5 gms/cm3., Solution, , 89P = 8.5 - 85P [Equating clock moments with anti-clock, moments], 89P + 85P = 8.5, 174P, P=, , = 8.5, , 8.5, meter or 0.049 m or 49 mm, 174, , Centre of gravity of the composite object lies, 49 mm from point of Gc. Hence it lies within, copper cube., , Volume of copper cube = 100 x 100 x 100 cm3, = 106 mm3, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 21

Page 30 :
Assignment A, 1 Find the position of centre of gravity of the figure shown., (All dimensions in mm), , 2 A lamina consists of a square of 60mm side, on one, side of which an equilateral triangle is constructed. Find, the position of centroid of the composite., , 3 A steel strip 50x12x12cm has a hole of 8cm dia. drilled, through it at a distance of 10cm from end. Find out the, c.g of the strip., , 4 Find out the C.G. of the four sided figure ABCD when, A = B=90º and the side AB=15cm, BC=12cm and, AD=5cm., , Assignment B, 1 What is the centre of gravity of a semi-circle is at a, perpendicular distance from its centre?, A, , 3r, 4π, , B, , 4r, 3π, , A, , h, 2, , B, , h, 3, , C, , 8r, 3, , D 3r, 8, , C, , h, 4, , D, , h, 5, , 2 What is the centre of gravity of a hemisphere is at a, distance from its space., , 5 Centre of gravity is usually located where., A more weight is concentrated, , A, , B 3r, 8, , B less weight is concentrated, , 4r, C, 8, , 5r, D, 8, , D more mass is concentrated, , C less mass is concentrated, , 3 What is the centre of gravity of a triangle is at the point, where the medians of the triangle meet?, , 22, , 4 What is the centre of gravity of a right circular cone is, at a distance from its base., , A, , h, 2, , B, , h, 3, , C, , h, 4, , D, , h, 5, , 6 Centre of gravity of an object depends on it's., A weight, , B mass, , C density, , D shape, , 7 Point where whole weight of body acts vertically is, called., A centre of mass, , B mid point, , C centre of gravity, , D none of above, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence

Page 31 :
8 A simple method to find centre of gravity of a body is, usage of., , 10 Which of the following laminas do not have centroid at, its geometrical centre?, , A stop watch, , B plumbline, , A Circle, , B Equilateral triangle, , C pendulum, , D screw gauge, , C Right angled triangle, , D Isosceles triangle, , 9 If a material has no uniform density throughout the body,, then the position of centroid and centre of mass are., A identical, B not identical, C independent upon the density, D unpredictable, , Key Answers, A, , B, , 1 90.6 mm, , 1 B, , 2 44.3 mm, , 2 B, , 3 26.3 cm, , 3 B, , 4 The C.G. lies at a point at a distance of 4.49 cm, from, the line AB and at a distance of 6.47 from the line BC., , 4 C, 5 D, 6 B, 7 C, 8 B, 9 B, 10 C, , Workshop Calculation & Science : (NSQF) Exercise 2.2.04, , Copyright free, under CC BY Licence, , 23

Page 32 :
Area of cut out regular surfaces - circle, segment and sector of circle, Exercise 2.3.05, Circle (Fig 1), It is the path of a point which is always equal from its, centre is called a circle., r = radius of the circle, d = diametre of the circle, Area of the circle = r2, (or) =, , π, 4, , Area of the smaller segment, = Area of the sector - Area of  ABC, , d2 unit2, , Area of the greater segment, , Circumference of the circle = 2r (or) d unit, , = Area of the circle - Area of smaller segment, Semi Circle (Fig 4), •, , A semi circle is a sector whose central angle is 180°., , •, , Length of arc of semi circle, , Sector of a circle (Fig 2), The area bounded by an arc is called the sector of a circle., In the figure given ABC is the sector of a circle., , l = 2πr ×, , r = radius of the circle, = Angle of sector in degrees, , = r unit, , Area of sector ABC, =, , πr, , 2, , 360, , Area of sector =, , xθ, , Area of semi circle =, , unit2, , 0, , Length of arc of sector  radius, , Length of the arc  = 2r x, , 2, θ, 0, , 360, , 180, 1, = 2πr ×, 360, 2, , unit2, , r 2, 2, , unit2, , Perimeter of a semi circle, , ==, , 2πr, + 2r, 2, , = r + 2r, , unit, , = r ( + 2) unit, Quadrant of a circle (Fig 5), , Perimeter of the sector =  + 2r unit, , •, , r = radius, , A quadrant of a circle is a sector whose central angle, is 90°., , Segment of a circle (Fig 3), When a circle is divided into two by drawing a line, the, bigger part is called segment of the circle and the smaller, part is also called segment of the circle., 24, , Copyright free, under CC BY Licence

Page 33 :
•, ,  Circumference of circle = d, , Length and area of a quadrant of a circle, , l = 2πr ×, = 2πr ×, =, , 90, 360, ,  44 = d, , 1, 4, , d=, , πr, 2, , Area of quadrant of a circle =, , Perimeter of a quadrant =, , =, , πr, , 2 πr, 4, , 2, , 44, = 44 ÷ π, π, , = 44 ÷, , 22, 7, , = 44 ×, , 7, 22, , unit2, , 4, , = 14 cm, ,  Diameter of circle (d) = 14 cm, , + 2r, ,  Area of circle =, , πr, + 2r, 2, , 1, =π × d 2, 4, 22 1, = × × 14 ×14, 7 4, , ⎛ π +2 ⎞ unit, ⎟, ⎝2 ⎠, , =r ⎜, Examples :, , 1 Find the area of a sector of a circle whose radius, is 14 cm and the length of the arc of the sector is, 28 cm., Radius of sector r = 14 cm, , π 2, d unit2, 4, , = 154 cm2, Area of circle, , = 154 cm2, , Length of arc of sector = 28 cm, , 3 Find the remaining areas of circles of 10 cm dia, after inscribing triangles of 5 cm base and 10 cm, height., , Length of arc of sector () =, , Solution, , 28, , (i) Area of the circle, , =, , =, =, , , , =, , π 2, d, 4, , 22 × 10 × 10, 7×4, , = 114.550, , =, ,  Angle of sector  = 114.550, , 550, 7, , (ii) Area of the triangle inscribed in this circle, ,  Area of sector, , =, , =, =, , cm2, , = 196 cm, Area of sector, , 2, , = 196 cm2, , 2 If the circumference of a circle is 44 cm, find its, area. (Take π =, , 22, ), 7, , =, , Remaining area, , 1, 2, , × base × height, , 10 × 5, 2, , = 25 sq.cm, , =, , Remaining area of circle = 53, , 4, , Sq.cm, , 7, , Solution, ,  Let (d) = diamter of circle, , Workshop Calculation & Science : (NSQF) Exercise 2.3.05, , Copyright free, under CC BY Licence, , 25

Page 34 :
4 A rectangular sheet of metal measures 8 cm and, 6 cm. Four quadrants of circles each of radius 2 cm, are cut away at corners. Find the area of the, remaining portion., , = 48 - 12.57, = 35.428 cm2, = say 35.43 cm2, Area of remaining portion = 35.43 sq.cm, 5 Find the perimeter of the given circular disc., Sector :, , Area of rectangular sheet, , =8x6, , r, , = 100 mm, , , , = 360° - 45° = 315°, , l=, , = 48 cm2, There are four quadrants of a circle, each of radius 2 cm cut, away at the corners. Quadrant of circle means 1/4th of, circle., 4 quadrant of circles, , =4x, , =, , θ, 360, , 315, 360, , × 2π r unit, , × 2 × π × 100 mm, ,  = 550 mm, , of circle = 1 circle, , Area of 4 quadrant circles = Area of one circle, =  r2, =, , 22, 7, , × 2× 2, , Perimeter of the given circular Disc =  + 2r, , = 12.57 cm2, , = 550 + 200 = 750 mm, , Area of remaining portion =, Area of rectangular sheet - Area of four quadrant circles cut, at corners., , 26, , Perimeter of the given circular Disc = 750 mm, , Workshop Calculation & Science : (NSQF) Exercise 2.3.05, , Copyright free, under CC BY Licence

Page 35 :
Related problems of area of cut out regular surfaces - circle, segment and, sector of circle, Exercise 2.3.06, 1, , 2, , 7, d t = 21 mm, , d = 32 mm, , A t = ______ mm2, , Cross sectional area, = _____ mm2, , l, , = 750 mm, , b = 400 mm, , 8, , d = 180 mm, ,  = 60°, 9, s = 9.2 mm, , 4, , d = 580 mm, Angle of cut off, sector a = 135°, , A = Area of sector, = 140 mm2, , Area of the remaining portion, A, = _____ mm2, 10, , Equilateral triangle, of side a = 6 cm, ,  = ______°, , Radius of circle, = 1.732 cm, , d = 380 mm, , Shaded area, _______________, , No. of sectors of, equal area = 8, Area of each sector, = ______ mm2, , 11, , Two plugs having, diameters 2 cm and, 5 cm are placed on a, surface plate touching, each other. calculate, the distance ‘L’ in the, figure., , 12, , 90° vee block is 26 mm, wide at the top of the, vee block. What dia., of soft when laid in the, vee block will, have its top surface just, level with the top of the, vee block., ,  = ______ °, length of arc of each, sector = ______ mm, 6, , D = 880 mm, , A of sector, = _______ mm2, , d of the circle, = 30 mm, , 5, , Av (Area of shaded, part) = ______ mm2, Av = % of (Area of, rectangle) A1, , Area of sheet, = _______, 3, , D = 38 mm, ,  = 160°, A = 0.893 m2, d = ______ mm, , 27, , Copyright free, under CC BY Licence

Page 36 :
13, , 17, From a sheet of 5m , 3m how many circular, pieces of 12.5 cm dia, can be cut., , The arrangement of a, band saw blade is, shown in the figure given, below. Find out the, length of the saw blade., 18 Calculate the area covered by 3 equal circles of radius, 2.8 cm touches one another., , 14, Find out ‘L’ from the, given sketch., , 19, = 155°, d = 350 mm, b = ____mm, , 15, Find the value of ‘x’ in, the following fig., , 20, Find the area of shaded, portion., , 16, , 28, , Area of the shaded, portion, ______________mm2., , Workshop Calculation & Science : (NSQF) Exercise 2.3.06, , Copyright free, under CC BY Licence

Page 37 :
Area of irregular surfaces and application related to shop problems, Exercise 2.3.07, Area of irregular surface, , (b) Trapezoidal rule, , Surface area of irregular figures can be obtained by, applying either i) simpson’s rule or ii) trapezoidal rule. Area, found by simpson’s rule is more accurate than trapezoidal, rule. However accurate area can be obtained if the number, of ordinates are more i.e interval between ordinates is so, small as possible. (Fig 1), , A=, , h, [y1 + y8 + 2(y2 + y3 + y4 + y5 + y6 + Y7)] unit2, 3, , A=, , 5 [4 + 5 + 2(3 + 2 + 5 + 1 + 2 + 3)] m2, 2, , 5, x 41 m2, 2, = 102.5 m2, , A=, , Calculation of the area of an irregular surface, In this Calculation the area of an irregular surface may be, determined as follows., In this method of calculation a chain line known as base line, to be laid through the centre of the area of the surface., i, , Area as per simpson’s rule, , The offset are taken to the boundary points in the order of, their chainages on both the sides of the base line., , h, [y1 + y7 + 4(y2 + y4 + y6) + 2 (y3 + y5)], 3, , Area =, , The chain line and offsets are noted down., , h = interval between ordinates, , With reference to the notes the boundary points are plotted, and the area to be divided into number of triangles and, trapezium according to the shape., , ii Area as per trapezoidal rule, , Example, , where, , Area= [, , Now apply the geometrical formulae for calculating the, according to the shape of the figures. (Fig 2), , h, (first ordinate + last ordinate) + sum of, 2, remaining ordinate], , Calculate the area enclosed between the chain line,, the edge and the end offsets by, The offsets were taken from a chain line to a edge., Distance (M) 0, , 5, , 10, , 15, , 20, , 25, , 30, , 35, , Off set (M), , 3, , 2, , 5, , 1, , 2, , 3, , 5, , 4, , (a) Simpson’s rule, , (b) Trapezoidal rule, , (a) Simpson’s rule, A=, A=, , Chainline = AB, , h, [y1 + y8 + 4(y2 + y4 + y6) + 2 (y3 + y5 + Y7)] unit2, 3, 5, [4 + 5 + 4(3 + 5 + 2) + 2 (2 + 1 + 3)] m2, 3, , = 101.7 m2, , ., , Offsets, , = C,E, , 1 Area of triangle, ½ x base x height, 2 Area of trapezium, base (a  b), 2, ,  height, , 29, , Copyright free, under CC BY Licence

Page 38 :
Example, Plot the following details of a field and calculate its area all measurements are in metres (Fig 3), , Serial No. 1 In ABI, , Sl. No. 3, , Chainage in metres 0 and 20m., , Area of triangle KCD, , Offsets in metres 0 and 36m., In, ,  30  35, ,  30 ,  2, , 2, = 32.5m x 30m = 975 sq.m, , SI. No. 2, , Area of Trapezium LFGJ =, , Area of trapezium IBCK, Chainage in metres = 20m and 55m = 35m, Offsets in metres 36m and 20m = 28m, (a  b), 2, ,  36  20, ,  height = , , , 2, , Figure Chainline in, metres, , 1, , 2, , 1, , ABI, , 2, , Trapezium, , 3, , (a  b), ,  height = , , Sl. No. 6, , , , , Area of trapezium AJGH =, ,  35 , , 35  45, 2, ,  35 , , Base in, Metres, 4, , Offsets in, metres, , Mean offsets, in metres, , 5, , 6, , Area in square, Metres, +ve, -ve, 7, 8, , 0 and 20, , 20, , 0 and 36, , 18, , 360, , --, , 20 and 55, , 35, , 36 and 20, , 28, , 980, , --, , IBCK, 3, , KCD, , 55 and 100, , 45, , 0 and 20, , 10, , 450, , --, , 4, , Rectangle, , 100 and 75, , 25, , 0 and 30, , 15, , 750, , --, , 75 and 45, , 30, , 30 and 35, , 32.50, , 975, , --, , 45 and 0, , 45, , 45 and 35, , 40, , 1400, , --, , DEFL, Trapezium, , LFGJ, Trapezium, , JGHA, , 80,  35, 2, , = 40 x 35 = 1400 sq.m, , = 28 x 35 = 980 sq.m, , 30, ,  20  45, , Sl. No. 5 (LFGH), , =360 sq.m, , 6, , 2, , Area of rectangle DEFL = 25 x 30 = 750 sq.m, , =1/2 x 20 x 36, , 5, , 1, , Sl. No. 4, , Area = ½ x base x height, , S., No., , 2, , xbxh=, , = 45m x 10m = 450 Sq.m, ,  ABI, , =, , 1, , =, , Total, , 4915, , Workshop Calculation & Science : (NSQF) Exercise 2.3.07, , Copyright free, under CC BY Licence, , Remarks, , 9

Page 39 :
Assignment, Calculate the area of the irregular surfaces given below., , Note : All dimension are in mm., , Workshop Calculation & Science : (NSQF) Exercise 2.3.07, , Copyright free, under CC BY Licence, , 31

Page 40 :
Algebra - Addition , subtraction, multiplication & division, Exercise 2.4.08, Introduction, , – 72, , Algebra is a form of mathematics in which letters may be, used in place of unknown. In this mathematics numbers, are also used in addition to the letters and the value of, number depends upon its place. For example in 3x and x3,, the place of x is different. In 3x =3 is multiplied with x,, whereas in x3 - 3 is an Index of x., Positive and negative numbers, Positive numbers have a + sign in front of them, and, negative numbers have – sign in front of them. The same, applies to letters also., , +9, – 96, –6, , = –8, = +16, , When an expression contains addition,, subtraction, multiplication and division,, perform the multiplication and division, operations first and then do the addition and, subtraction., Example, 12 x 8 – 6 + 4 x 12 = 96 – 6 + 48 = 138, , Example + x, – y., , 102  6 – 6 x 2 + 3 = 17 – 12 + 3 = 8, , +8 or simply 8 positive number., – 8 negative number., , Parantheses and grouping symbols, , Addition and subtraction, , (, , ) Brackets, , Two positive numbers are added, by adding their absolute, magnitude and prefix the plus sign., , {, , } Braces, , To add two negative numbers, add their absolute magnitude and prefix the minus sign., , 6 x (8–5) = 6 x 3 = 18, , To add a positive and a negative number, obtain the, difference of their absolute magnitudes and prefix the sign, of the number having the greater magnitude., +7 + 22, –8 – 34, –27 + 19, 44 + (–18), 37 + (–52), , =, =, =, =, =, , +29, – 42, –8, +26, –15, , Multiplication of positive and negative numbers, The product of two numbers having like signs is positive and, the product of two numbers with unlike signs is negative., Note that, where both the numbers are negative, their, product is positive., Ex., , –20 x –3, 5x8, 4 x –13, –5 x 12, , =, =, =, =, , 60, 40, – 52, –60, , The number that is divided is the dividend, the number by, which we are dividing is the divisor and the answer is the, quotient. If the signs of the dividend and the divisor are the, same then the quotient will have a + sign. If they are unlike, then the quotient will have a negative sign., = +7, , +4, + 56, 32, , –4, , Parentheses, These are symbols that indicate that certain addition and, subtraction operations should precede multiplication and, division. They indicate that the operations within them, should be carried out completely before the remaining, operations are performed. After completing the grouping,, the symbols may be removed., In an expression where grouping symbols immediately, preceded or followed by a number but with the signs of, operation omitted, it is understood, that multiplication, should be performed., Grouping symbols are used when subtraction and multiplication of negative numbers is done., To remove grouping symbols which are preceded by, negative signs, the signs of all terms inside the grouping, symbols must be changed (from plus to minus and minus, to plus)., Parentheses which are preceded by a plus sign may be, removed without changing the signs of the terms within the, parentheses., , Division, , + 28, , 7 + (6–2) = 7 + 4 = 11, , When one set of grouping symbols is included within, another set, remove the innermost set first., When several terms connected by + or – signs contain a, common quantity, this common quantity may be placed in, front of a parentheses., 8 + 6(4–1) = 8 + 6 x 3 = 26, (6+2) (9–5) = 8 x 4 = 32, , = –14, , Plus 4 less negative 7 is written as 4 – (–7)., , Copyright free, under CC BY Licence

Page 42 :
Each member of an equation may be multiplied or divided, by the same number or symbol without changing its, equality., The equality of an equation is not altered when the numbers, or symbols are added or subtracted from both sides., Multiplication and division by the same numbers or symbols on both sides also will not affect the equality., Transposition of the terms of the equations, = equals to, + plus, – minus, , We must always perform the same operation on both sides, of the equation to keep the equilibrium. Add or subtract the, same amount from both sides. 5 + x = 9 By adding 3 on, both sides, the equation becomes 5 + x + 3 = 9 + 3 or x, + 8 = 12., 5 + x = 9 Subtract 5 from both sides then 5 + x – 5 = 9– 5., x = 4., 5 is transposed from left side to the right side by changing, its sign from + to –., x, x, = 20. Multiply both sides by 4. Then, x 4 = 20 x 4., 4, 4, x = 80,, , x multiply, , 5x = 25., ,  divided by, , Divide both sides by 5 then, , Concept of equality (Fig 1), , 5x, , 5, x = 5., , =, , 25, 5, , When transposing numbers or letter symbols from one, side to the other side multiplication becomes division and, the division becomes multiplication., The equality of an equation remains unchanged, when both sides of the equation are treated in, the same way. When transposing from one, side to the other side,, a plus quantity becomes minus quantity., a minus quantity becomes a plus quantity, a multiplication becomes a division, a division becomes a multiplication., To solve simple equations isolate the unknown, quantity which is to be found on the left side of, the equation., Example, •, An equation can be compared to a pair of scales which, always remain in equilibrium. The two sides of the equation, can fully be transposed. 9 = 5 + x may also be written as, 5 + x = 9., , Solve for x if 4x = 3(35 – x ), 4x = 105 – 3x (brackets removed), 4x + 3x = 105 (By transposing –3x on the right side, to the left side), 7x = 105, x = 15 (dividing both sides by 7), , 34, , Workshop Calculation & Science : (NSQF) Exercise 2.4.08, , Copyright free, under CC BY Licence

Page 44 :
Algebra – Theory of indices, Algebraic formula, related problems, Exercise 2.4.09, Calculations involving powers, , 2, , Power : Concept, , 2, , a.a.a... upto n times is = an, When a number, say 2 is multiplied by itself 4 times, we, write it as 24 (two to the power of 4) and it is equal to, 2 x 2 x 2 x 2 = 16., The exponent denotes how many times the base number, is multiplied by itself., Powers with a positive base have a positive result., Powers with a negative base and with an exponent that is, even will have a positive result., The sign, =, =, =, =, =, , 2, , =2, , 3–2, , 2x2x2, , a is the base, n is the exponent., , (+a) n, (– a) 2n, (2) 2, (– 2)2, (– 2)3, , 3, , an, a2n, 2 x 2 = 4 and, – 2 x – 2 = +4 but, –2x–2x–2=–8, , =, , 2x2, , 1, =2 =2, , 8, , =2, , 4, , Powers with the same exponents are divided by involving, the quotient of the bases by the common exponent., , n, , n, , 2, , 2, , ⎛a⎞, n =⎜ ⎟, b, ⎝b⎠, a, , ⎛2⎞ 2x2 4, =⎜ ⎟ =, =, 2, ⎝3⎠ 3x3 9, 3, 2, , Only like powers can be added or subtracted., Examples, (The exponent 1 is usually not written.), a1 = a, , Addition and subtraction of powers, , 21 =2, , Powers with the same base and exponents can be added, or subtracted by addition or subtraction of the coefficients., , 2a2 + 3a2 = 5a2, , x.an + y.an = an (x + y), , (Any number raised to the power of 0 is 1.), a0 = 1, , x.an – y.an = an (x – y), Ex .4x2 + x2 – 3x2 = x2 (4 + 1 – 3) = 2x2., , 20 = 1, A number raised to a negative power corresponds to its, reciprocal with the exponent's sign changed to +., , Multiplication, Powers with the same bases are multiplied by involving the, common base raised to the power of sum of the exponents., , a, , –n, , 1, = n, a, , am x an = am + n., 2 3 x 2 2 = 2 3 + 2 = 25, (2 x 2 x 2) x (2 x 2) = 2 x 2 x 2 x 2 x 2 = 25, , 1, 2–2 = 2, 2, , Powers are involved by multiplying the exponents., , 8 x 4 = 32., Powers with the same exponent of different base numbers, are multiplied by involving the product of the base numbers, raised to the common exponent., an x bn = (a x b)n, , (an)m = anm, (22) 3 = 22.3 = 26, Powers can be transposed without affecting the result., , 22 x 32 = (2 x 3)2, , a  = a , , 2 x 2 x 3 x 3 = 6 x 6 = 36, , (22) 3 = (23) 2, , n m, , m n, , Division, , (2 x 2) x (2 x 2) x (2 x 2) = (2 x 2 x 2) (2 x 2 x 2), , Powers with like bases are divided by involving the base, raised to the difference between the exponents., , 4 x 4 x 4 = 64, , a, , m, , 36 a, , n =a, , m–n, , 8 x 8 = 64, A mixed number raised to a power is first converted into an, improper fraction and then the result is evaluated., , Copyright free, under CC BY Licence

Page 47 :
v) a3 - b3, x -y, 3, , 3, , x3 - y3, vi), , =, , (a - b) (a2 + b2 + ab), , =, , (x - y) (x + y + xy), , =, , 1 (41 + 20), , =, , 1 x 61, , =, , 61, , =, , 61, , 2, , =, , 10, 2, , =, , 28561, , x4 + y4, , =, , 28561 - 7200, , x4 + y4, , =, , 21361, , 2, , 5. x + y = 5 ; x - y =, , 2, 2, 3 Find the value of 8 x y (x + y ), , x + y = 5; x - y =, , x+y = 9, x-y = 1, ---------------------2x, = 10, ---------------------x, , x4 + y4 + 7200, , 3 (take square on both sides), , (x + y)2 = 5; (x - y)2 = 3, Solve the equations, (x + y)2, , =, , (x - y)2, , = x2 + y2 - 2xy = 3, ---------------------------------------2 (x2 + y2), =, 8, , =5, , If x = 5, 5 + y = 9, y=9-5=4, , (x2 + y2), , x = 5; y = 4, , x2 + y2 + 2xy, , 8, , =, =, , 2. Solve (x + 5)2 - (x - 5)2, , 2, , =, , =4, , x2 + y2 + 2xy, , =, , a2 - b2 = (a + b) (a - b), (x + 5)2 - (x - 5)2 = [(x + 5) + (x - 5)] [(x + 5) - (x - 5)], , 2, , = (x + 5 + x - 5) (x + 5 - x + 5), , 2, , 2, , xy, , =, , =, , 8x, , =, , 4 x 4 = 16, , 4, , = (2x) (10), 8 xy (x2 + y2), , 3. If (x - y) = 4 and xy = 12, find the value of (x2 + y2), (x - y)2 =, , x2 + y2 - 2xy, , (4)2 =, , x2 + y2 - 2 x 12, , 16, , x + y - 24, , =, , x2 + y2 - 24 =, , 2, , 16, , x +y =, , 16 + 24, , x2 + y2 =, , 40, , 2, , 2, , 6. If (a -, , 2, , 4. If x - y = 7 and xy = 60 then find the value of x + y, (x - y)2, , =, , x2 + y2 - 2xy = 72, , x2 + y2 - 2 x 60, , =, , 49, , x2 + y2, , =, , 169, , (x2 + y2)2 =, x4 + y4 + 2x2y2, x + y + 2(xy), , =, , 28561, , x4 + y4 + 2(60)2, , =, , 28561, , x + y + 2(3600) =, 4, , 4, , a, , a2 +, , 28561, , 1, 2, , 1, a, , a2 +, , 2, , x4, , 1, , 2, , 1, a, , 2, , 2, , =, , 6, , =, , 62 (take square on both sides), , 2, , ⎛ 1 ⎞ - 2 (a), ⎜ ⎟, ⎝a⎠, , a, , a2 +, , 2, , 1, , a, , 2, , a2 +, , 1, , =, , ) = 6. Find the value of a2 +, , ⎛ 1⎞, ⎜a - ⎟, ⎝ a⎠, , (169)2, , =, , 4, , 4, , (169)2 (take square on both side), , 2, , 4, , 1, , ⎛ 1⎞, ⎜a - ⎟, ⎝ a⎠, , 4, , 5, , x + y - 2xy = 3, (-) (-) (+), (-), ------------------------------------------4xy, = 2, ------------------------------------------=, , If x + 5 = a and x - 5 = b, , = 20 x, , 5, , ⎛ 1 ⎞ = 36, ⎜ ⎟, ⎝a⎠, , - 2 = 36, , = 36 + 2, , = 38, , Workshop Calculation & Science : (NSQF) Exercise 2.4.09, , Copyright free, under CC BY Licence, , 39

Page 50 :
Elasticity - Elastic, plastic materials, stress, strain and their units and, young’s modulus, Exercise 2.5.10, Elastic material, , 11 Rubber Epichlorohydrin, , The Elastic materials are those materials that have the, ability to resist a distorting or deforming influence or force,, and then return to their original shape and size when the, same force is removed., , 12 Nylon, , Linear elasticity is widely used in the design and analysis, of structures such as beams, plates and sheets., , 15 Poilbutadiene, , Elastic materials are of great importance to society since, many of them are used to make clothes, tires, automotive, spare parts, etc., , 17 Vinyl stretch, , Characteristics of elastic materials, , 13 Terpene, 14 Isoprene Rubber, 16 Nitrile Rubber, 18 Thermoplastic elastomer, 19 Silicone rubber, 20 Ethylene-propylene-diene rubber (EPDM), , When an elastic material is deformed with an external, force, it experiences an internal resistance to the, deformation and restores it to its original state if the, external force is no longer applied., , 21 Ethylvinylacetate (EVA or foamy gum), , To a certain extent, most solid materials exhibit elastic, behavior, but there is a limit of the magnitude of the force, and the accompanying deformation within this elastic, recovery., , Plastic Material, , A material is considered as elastic if it can be stretched, up to 300% of its original length., For this reason there is an elastic limit, which is the, greatest force or tension per unit area of a solid material, that can withstand permanent deformation., For these materials, the elasticity limit marks the end of, their elastic behavior and the beginning of their plastic, behavior. For weaker materials, the stress or stress on, its elasticity limit results in its fracture., The elasticity limit depends on the type of solid considered., For example, a metal bar can be extended elastically up, to 1% of its original length., However, fragments of certain gummy materials may, undergo extensions of up to 1000%. The elastic properties, of most solid intentions tend to fall between these two, extremes., Maybe you might be interested How to Synthesize an, Elastolic Material?, , 22 Halogenated butyl rubbers (CIIR, BIIR), 23 Neoprene, Plastic Material Classification, “Plastic material” is a term that refers to a large class of, polymers, separated into various groups and sub-groups., Before starting the chapter on the uses and subsequent, recycling of plastic, let us establish a general classification, of these thermosetting resins or thermo-plastics (the two, big groups into which we include elastomers) by detailing, their properties, their make-up, their aspect, and their, final uses, while explaining which ones are recyclable., Thermoplastics, Remember that thermoplastic is a material whose, structure and viscosity can be modified both ways through, heating or cooling. This large family of materials is, commonly used by many industries and is easily, integrated into France’s recycling cycles., The following polymers are some examples of plastic, material:, 1 Polyolefins, 2 Vinyl polymers, 3 Polystyrenes, , Examples of elastic materials, , 4 Acrylate and methacrymate polymers, , 1 Natural gum, , 5 Polyamide, , 2 Spandex or lycra, , 6 Polycarbonates, , 3 Butyl Rubber (GDP), , 7 Celluloid, , 4 Fluoroelastomer, , 8 Linear polyesters, , 5 Elastomers, 6 Ethylene-propylene rubber (EPR), 7 Resilin, 8 Styrene-butadiene rubber (SBR), 9 Chloroprene, 10 Elastin, 42, , 9 Polyfluorethane, 10 Polyacetal, 11 Polysulfone, 12 Polyphenylene sulfide, 13 Modified polyphenylene oxide (PPO), , Copyright free, under CC BY Licence