Workshop Calculation & Science (NSQF) 2nd Year PDF

WORKSHOP CALCULATION & SCIENCE (NSQF Syllabus) 2nd Year Common for All Engineering Trades under CTS (All 2 Year Trades) DIRECTORATE GENERAL OF TRAINING MINISTRY OF SKILL DEVELOPMENT & ENTREPRENEURS...

WORKSHOP CALCULATION & SCIENCE (NSQF Syllabus) 2nd Year Common for All Engineering Trades under CTS (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) Workshop Calculation & Science (NSQF) - 2nd Year Common for All Engineering Trades Under CTS (All 2 year Trades) Developed & Published by National Instructional Media Institute Post Box No.3142 Guindy, Chennai - 32 INDIA Email: [email protected] Website: www.nimi.gov.in Printed by National Instructional Media Institute Chennai - 600 032 First Edition : November 2019 Copies : 1,000 First Reprint : February 2020 Copies : 5,000 Second Reprint : March 2020 Copies : 5,000 Rs.120/- (ii)  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 Director General Directorate General of Training Ministry of Skill Development & Entrepreneurship, Government of India. New Delhi - 110 001 (iii)  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 (Common 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 Chennai - 600 032 EXECUTIVE DIRECTOR (iv)  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 syllabus. 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)  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 : 80 Hrs Time allotment for each title of exercises 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) 1 Friction 2.1.01 - 2.1.03 10 2 Centre of Gravity 2.2.04 6 3 Area of cut out regular surfaces and area of irregular surfaces 2.3.05 - 2.3.07 14 4 Algebra 2.4.08 & 2.4.09 12 5 Elasticity 2.5.10 & 2.5.11 8 6 Heat Treatment 2.6.12 & 2.6.13 8 7 Profit and Loss 2.7.14 & 2.7.15 12 8 Estimation and Costing 2.8.16 & 2.8.17 10 80 Hrs 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)  CONTENTS Exercise No. Title of the Exercise Page No. Friction 2.1.01 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 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 2.3.07 Area of irregular surfaces and application related to shop problems 29 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 Elasticity - Elastic, plastic materials, stress, strain and their units and young’s modulus 41 2.5.11 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 Estimation and costing - Simple estimation of the requirement of material etc., as applicable to the trade 85 2.8.17 Estimation and costing - Problems on estimation and costing 92 (vii)  SYLLABUS 2nd Year Common for All Engineering Trades under CTS Duration: One Year (All 2 year Trades) S.No. Syllabus Time in Hrs I Friction 10 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 6 1 Centre of gravity and its practical application III Area of cut – out regular surfaces and area of irregular surfaces 14 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 12 1 Addition, Subtraction, Multiplication & Divisions 2 Algebra – Theory of indices, Algebraic formula, related problems V Elasticity 8 1 Elastic, plastic materials, stress, strain and their units and young’s modulus 2 Ultimate stress and working stress VI Heat Treatment 8 1 Heat treatment and advantages 2 Different heat treatment process – Hardening, Tempering, Annealing, Normalising, Case Hardening VII Profit and Loss 12 1 Simple problems on profit & loss 2 Simple and compound interest VIII Estimation and Costing 10 1 Simple estimation of the requirement of material etc., as applicable to the trade 2 Problems on estimation and costing Total 80 (viii) Friction - Advantages and disadvantages, Laws of friction, co-efficient of friction, angle of friction, simple problems related to friction Exercise 2.1.01 Introduction Forces acting on a body when a pulling force is When on a solid surface, another solid is rubbed a force is applied to move (Fig 2) created between the two solids which acts in the opposite Weight of the block acting vertically downward (W) direction of motion or tries to obstruct the motion of the object, this force is called frictional force. This phenomenon The normal reaction which acts upwards (R) is called friction. This happens due to roughness of the The applied pulling force (F) two surfaces. In other words, It is the force of resistance offered to motion, The frictional force (Ff) experienced by bodies which are in contact. It depends When the body is about to move W=R, F=P upon the normal reaction between the contacting surfaces and the nature of the surfaces. No surface is absolutely When pulling force is increased the body starts to move. friction less. 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) Frictional force is directly proportional to the normal reaction between contacting surfaces. Frictional force acts opposite to the direction of motion. Types of friction Frictional force depends on the nature of contacting surfaces. 1 Static friction Frictional force is independent over the area and shape 2 Dynamic friction 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. Coefficient of friction Rolling friction It is a ratio between the frictional force to the normal It is the friction that occurs when a circular object such as reaction when the body is just about to move but at a ball or roller rolls on a flat surface. Rolling friction is less equilibrium. It is represented by symbol . (read as ‘meu’) than sliding friction. (ball or roller bearing) 1 Therefore Limiting friction(or)force Co - efficient of friction = 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  P = F + W sin and R = W cos  forces 'R' and 'F' are compounded and we get the resultant  P = uR + W sin = u(W cos  W sin  force 'S'. The angle formed by 'S' with 'R' is the angle of  P = W x tan  cos + W cos  friction. Therefore Sin φ = Wx x cos θ + W cos θ Cos φ F tan  Sin φ + cos θ + Wcos φ x cos θ W =W x Cos φ tan  =W [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 Angle of repose (Fig 6) to move up the plane by the action of an applied force the 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 [Sin (θ + φ)] < P < = W [Sin (θ + φ)] Cos φ Cos φ A body placed on an inclined surface remains at rest till Advantages of friction the angle of inclination equals the angle of friction. When it exceeds the body starts sliding down. This is known as 1 Helps us to walk without slipping. angle of repose. 2 Used to stop vehicles when brakes are applied. Motion up the plane 3 Movement of vehicles due to friction between revolving When > a force must be applied to keep the body in wheels with tyres and the road. equilibrium. The applied force may be parallel to the plane, 4 Power transmission using gear drive or belt pulley drive. horizontal or at an angle to the plane itself. 5 Using friction we can sharp any object and also to hold When the body is at the point of motion up the plane the it. frictional force ‘F’ acts down the plane. 6 Nails and screws are held in wood by friction. Forces acting are W,R,P and F. The weight force ‘W’ is 7 Heat is produced when two rough surfaces are rubbed resolved into two components of W cos  perpendicular to against each other. the plane acting downwards and W sin  acting parallel to the plane downwards. Disadvantages of friction F 1 It causes wear and tear of the machine parts. = μ = tan φ R 2 It produces heat and may cause melting of machine parts. To avoid production of heat using of coolant is Where is the angle of friction. necessary. 2 Workshop Calculation & Science : (NSQF) Exercise 2.1.01 3 It reduces efficiency of a machine. W = 14500 kg = Weight 4 It reduces speed of the moving object. eg. spindle, shaft, F = ? = Force friction piston etc. F Friction can be reduced μ= W 1 By using suitable lubricants (oil, grease) between the F moving parts. 0.28 = 14500 2 By polishing the surface to make them smooth. F = 0.28 x 14500 3 By using ball bearings and roller bearings. F = 4060 kg. 4 By the use of wheel. 5 A force of 800 gram weight is needed to pull a Example block weighing 3200 gram. What is the co-efficient 1 A force of 40 kg is required to pull a weight of 400 of friction. kg on a horizontal plane. Determine the F = 800 gm = Force coefficient of friction. W = 3200 gm = Weight Force F Coefficient of friction = =  =? Weight W Co-efficient of friction = ? But F = P and R = W F  F Ff = = 40 Co-efficient of friction = μ = W R 400 W = 0.1 = 800 3200 2 A force for 30N is required to move a body of mass  = 0.25 35 kg on a flat surface horizontally at a constant velocity. Find the coefficient of friction. 6 A force of 40 kg is required to move a mass of 80 Mass of the body = 35 kg. = W (By taking kg on a flat surface horizontally at a constant velocity. Calculate its co-efficient of friction? The weight force=35 x 10 = 350 N 1kg = 10N) F = 40 kg = Force (By taking g = 10 meter/sec2) W = 80 kg = Weight F Ff 30 3 ∴μ= = = = = 0.086  =? W R 350 35 Co-efficient of friction = ? = 0.09 F Co-efficient of friction = μ = 3 A solid weights 20 kg. This is placed on a solid W surface. How much force does it require to come 40 in motion when co-efficient of friction is 0.24. = 80  = 0.24 = Co-efficient of friction W = 20 kg = Weight  = 0.5 F = ? = Force required 7 A weight of 10 kg is resting on a horizontal table and can just moved by a force of 2 kg. Find the F co-efficient of friction? μ= W W = 10 kg = Weight F F = 2 kg = Force 0.24 = 20  =? F = 20 x 0.24 Co-efficient of friction = ? F = 4.8 kg F μ= 4 A tanker with loaded weight of 14500 kg is running W on the road. If the co-efficient friction between 2 tyres and road surface is 0.28. Find out its force of = friction. 10  = 0.28 = Co-efficient of friction  = 0.2 Workshop Calculation & Science : (NSQF) Exercise 2.1.01 3 8 A body weighing 100kg is resting on a table. Find W = 150 kg = Work done the co-efficient of friction if a force of 30 kg makes  = 0.25 = Co-efficient of friction its just to move?   W = 100 kg = Weight F = 30 kg = Force F = ? = Force  =?  = tan  = 0.25 Co-efficient of friction = ? = 14º 2’ 20” F F μ= μ= W W 30 = 100 F 0.25 =  = 0.3 150 F = 0.25 x 150 9 A metal block weighing 10 kg rests on a horizontal table. A horizontal force of 2.5 kg can just slide F = 37.5 Kg. the block. Find the normal reaction, limiting 12 A body of mass 60kg rests on a horizontal plane. friction and co-efficient of friction? The value of co-efficient of friction between it and W = 10 kg = Weight the plane being 0.2. Find the work done in moving F = 2.5 kg = Force the body through a distance of 5 meters along the plane. R =?  = 0.2 = Co-efficient of friction Normal reaction = W Limiting friction = ? W = 60 kg = Weight  =? S = 5 m = Distance R = Normal reaction = 10 kg W = ? = Work done Limiting friction = F = 2.5 kg F μ= F W μ= W F 100 0.2 = 2.5 60 = 10 F = 60 x 0.2  = 0.25 = 12 kg Work done = Force x distance = F x S 10 A wooden block weights 100 kg. If the co-efficient = 12 x 5 of friction is 0.3, find out force required to move the block. = 60 Kg.m W = 10 kg = Weight (ie) Work done (or) Applied force = 60 Kg.m  = 0.3 = Co-efficient of friction 13 If a force of 30N is required to move a mass of F = ? = Force 35kg on a flat surface horizontally at constant velocity, what will be the co-efficient of friction? F μ= F = 30 N = Force W W = 35 kg = Weight 0.3 = F 1 kg = 9.8 N 100 35 Kg = 9.8 x 35 = 343 N F = 100 x 0.3 F Co-efficient of friction = μ= F = 30 kg W 30 N 11 Calculate the angle of inclination, if a weight of = 150 kg is in equilibrium, co-efficient of friction is 35 kg 0.25. Calculate the force of normal reaction also. 4 Workshop Calculation & Science : (NSQF) Exercise 2.1.01  = 2.2/cos 30º 30 N = = 2.2/0.8660 35 x 9.8 N Force acting at an angle of 30º = 2.54 kg  = 0.087 14 A block of ice weighing one quintal rests in 17 Calculate the angle of inclination, if a weight of equilibrium on a wooden plank inclined at 30º. 250 kg is in equilibrium. Coefficient of friction is Find the coefficient of friction between the ice and 0.36. Calculate the force of normal reaction also. wood.  = ? = angle of inclination W = 1 quintal = 100 kg = Weight W = 250kg = Weight F  = 30º    tan   = 0.36 W  = tan tan 30º F = ? = Force  = 0.5774 tan  =  tan  = 0.36 15 Calculate the force that is required to slide a mass of 980 kg on a guide, when the coefficient of  = 19º 48' friction between the surfaces is 0.09. F W = 980 kg = Weight μ= W  = 0.09 = Co-efficient of friction F = Force = ? F 0.36 = F 250 kg Co-efficient of friction = μ= W F = 0.36 x 250 kg F F = 90 kg. 0.09 = 980 kg 18 A body of mass 10 kg rests on a horizontal plane. F = 0.09 x 980 kg The co-efficient of friction between the body and Required force(F) = 88.2 kg plane is 0.15. Find the work done in moving the body through a distance of 10 meter. 16 A metal block weighing 10kg rests on a horizontal W = 10kg = Weight board and the coefficient of friction between the surfaces is 0.22. Find (a) the horizontal force which  = 0.15 = Co-efficient of friction will just move the block and (b) the force acting S = 10 meter = distance at an angle of 30º with the horizontal, which will just move the block. W = Work done = ? W = 10 kg = Weight F μ= Co-efficient of friction = = 0.22 W (a) F =? F 0.15 = (b) Force acting at an angle of 30º with the horizontal? 10 Kg F = 0.15 x 10 kg F (a) μ= F = 1.5 kg W Work done =W=FxS 0.22 = F = 1.5 kg x 10 m 10 kg Work done = 15 kg.m F = 2.2 Kg. F (b) Force acting at an angle of 30º= Cos θ Workshop Calculation & Science : (NSQF) Exercise 2.1.01 5 Assignment A 1 A force 50N is required to move a mass of 40kg on a 7 A body of mass 2000 kg moves a distance of 10 meters flat surface horizontally at a constant velocity. Find the in 5 sec. If the co-efficient of friction between the body coefficient of friction. (9.8N = 1kg) and floor is 0.3 find the horizontal force required to move 2 A vehicle having a weight of 800kg is moving on the the body and horsepower absorbed against friction. road. If the coefficient of friction between the tyres and 8 A vehicle is moving at 50kmph and the load on the road surface is 0.3, then calculate the force of friction. vehicle is 5000 kg. Find the H.P. required to move the 3 A solid weighing 50kg is place on a solid surface. How vehicle if = 0.2. much force is required to move the block when 9 Find out the power lost due to friction by a planer under coefficient of friction is 0.25 between the block and the the following conditions. surface. 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 Rate of moment of the table=0.5 m/sec force required to move the wagon. Co-efficient of friction between the table and the 5 A body of mass 100kg rests on a horizontal plane. The ways=0.06 angle of friction between the body and the plane being 10 A truck having weight 12000 kg is moving on the road. 0.025. Find the work done is moving the body through If the co-efficient of friction between the tyres and the a distance of 16m along the plane. 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 metres. Assignment B 1 F = 1800 N 4 F = 1.2 kN μ (static) = 0.16 d = 60 mm μ (dynamic) = 0.012 μ = 0.03 FR to overcome Frictional torque MR static friction = ______ Nm = ______ N (Frictional torque = FR to overcome Frictional force x dynamic friction radius) = ______ N 5 mass = 180 kg 2 mass = 250 kg μ = 0.15 FR = 160 N FR = ______ N μ = ______ 6 FR = 120 N μ = 0.032 Normal force F = ______ N 3 F = 5000 N μ (dry) = 0.03 7 m = 1000 kg μ (fluid friction) = 0.01 μ = 0.4 FR when dry Force required to move = ______ N FR = ______ N FR when lubricated = ______ N 6 Workshop Calculation & Science : (NSQF) Exercise 2.1.01 C MCQ 1 Which one of the following is useful friction A 0.0215 B 0.0152 A Rings in cylinders B Crankshaft bearings C 0.0125 D 0.0251 C Brake shoe linings D Wheel hole bearings 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 A 27 Kg B 28 Kg 3 Which direction of motion frictional force acts. C 29 Kg D 30 Kg A equal B opposite C inclined D forward 8 Determine the co-efficient of friction() between brass and steel when a brass slider was placed on the 4 What is the formula of angle of friction, if ‘F’ is the horizontal steel surface until it is just moving, if brass frictional force, R is the normal reaction and q is the slides (W) = 3 Kgf angle of friction. Brass slides (W) = 3 Kgf F F Force (F) required = 0.7 kgf A Tan q = B Cot q = R R A 0.033 B 0.133 F F C 0.233 D 0.333 C Sin q = D Cos q = R R 9 Which is necessary to avoid production of heat. 5 What is the formula for Co-efficient of friction (m). A sand B coolant C lubricant D salt A  R B  F F R 10 Which is using for reduce the friction. C F x R D F + R A lubricants B sand C coal D coolant 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 Key Answers A B C MCQ 7 F = 600 Kg 1 0.1275 1 288N, 21.6 N 1 C 6 C P = 16 HP 2 240 Kg 2 0.065 2 D 7 D 8 185.2 HP 3 12.5 Kg 3 150 N, 50 N 3 B 8 C 9 1.4 HP 4 36 N, 1.08 Nm 4 A 9 B 4 3.75 Tonne 10 3600 Kg 5 264.6 N 5 B 10 A 5 40 m-kg 6 3750 N 6 60 m-kg 7 3920 N Workshop Calculation & Science : (NSQF) Exercise 2.1.01 7 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 In other systems one of the rotating elements comes in The oil hole or grease point leading to each bearing is contact with that of the oil level and splash the whole fitted with a nipple, and by pressing the nose of the gun system with lubricating oil while working. (Fig 6) Such against this, the lubricant is forced to the bearing. Greases systems can be found in the headstock of a lathe machine are also force fed using grease cup. (Fig 3) 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 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 Lubricate bearings The moving parts experience some kind of resistance even when the surface of the parts seems to be very smooth. A shaft moving in a bearing is also subjected to frictional resistance. The shaft rotates in a bush bearing or in ball/ The resistance is caused by irregularities which cannot roller bearing, experiencing friction. be detected by the naked eyes. When the shaft is at rest on the bottom of the bush Without a lubricant the irregularities grip each other as bearing, there is hardly any lubricant between the shaft shown in the diagram. (Fig 10) and the bush. (Fig 14) 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) When the shaft starts rotating the lubricant maintains a The slideways are lubricated frequently by an oilcan. film between the shaft and the bush and an uneven ring of (Fig 12) lubricant builds up. (Fig 15) Workshop Calculation & Science : (NSQF) Exercise 2.1.02 9 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: - to cool the tool as well as the workpiece This lubrication ring decreases the frictional resistance very much and at the same time protects the mating - to reduce the friction between the chip and the tool face members against wear and changes. by lubricating - to prevent the chip from getting welded to the tool Some bush bearings have oil feeding holes over which the cutting edge oil or grease cup is mounted and the lubricant is fed through the holes into the bearing by gravity feed system.(Fig 17) - 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. 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. Hints for lubricating machines: As the chips are flushed away, the cutting zone will be neat. - identify the oiling and greasing points The machine or job will not get rusted because the coolant - select the right lubricants and lubricating devices prevents corrosion. - apply the lubricants. Properties of a good cutting fluid The manufacturer’s manual contains all the necessary A good cutting fluid should be sufficiently viscous. details for lubrication of parts in machine tools. Lubricants At cutting temperature, the coolant should not catch fire. are to be applied daily, weekly, monthly or at regular intervals at different points or parts as stipulated in the It should have a low evaporation rate. manufacturer’s manual. It should not corrode the workpiece or machine. These places are indicated in the maintenance manuals It must be stable and should not foam or fume. with symbols as shown in Fig 18. It should not create any skin problems to the operator. Cutting Fluids Should not give off bad smell or cause itching etc. which Cutting fluids and compounds are the substances used for are likely to irritate the operator, thus reducing his efficiency. efficient cutting while cutting operations take place. Should be transparent. 10 Workshop Calculation & Science : (NSQF) Exercise 2.1.02 Types of cutting fluids Compounded or blended oil The following are the common cutting fluids. These oils are used in automatic lathes. These oils are Straight mineral oil much cheaper and have more fluidity than fatty oil. Chemical solution (synthetic fluids) Fatty oil Compounded or blended oil Lard oil and vegetable oil are fatty oils. They are used on heavy duty machines with less cutting speed. They are Fatty oils also used on bench-works for cutting threads by taps and Soluble oil (Emulsified oil-suds) dies. Straight mineral oil Soluble oil (Emulsified oil) Straight mineral oils are the coolants which can be used Water is the cheapest coolant but it is not suitable undiluted. Use of straight mineral oil as a coolant has the because it causes rust to ferrous metals. An oil called following disadvantages. soluble oil is added to water which gets a non-corrosive It gives off a cloud of smoke. effect with water in the ratio of about 1: 20. It dissolves in water giving a white milky solution. Soluble oil is an oil It has little effect as a cutting fluid. blend mixed with an emulsifier. Hence straight mineral oils are poor coolants. But kerosene Other ingredients are mixed with the oil to give better which is a straight mineral oil is widely used as a coolant protection against corrosion, and help in the prevention of for machining aluminium and its alloys. skin irritations. Chemical solution (Synthetic oil) Soluble oil is generally used as a cutting fluid for centre These consist of carefully chosen chemicals in dilute lathes, drilling, milling and sawing. solution with water. They possess a good flushing and a Soft soap and caustic soda serve as emulsifying agents. good cooling action, and are non-corrosive and non- A chart showing coolants for different metals is given clogging. Hence they are widely used for grinding and below. sawing. They do not cause infection and skin trouble. They are artificially coloured. Recommended cutting fluids for various metals and different operations Material Drilling Reaming Threading Turning Milling Aluminium Soluble oil Soluble oil Soluble oil Soluble oil Soluble oil Kerosene Kerosene Kerosene Lard oil Kerosene and Mineral oil Lard oil Mineral oil lard oil Dry Brass Dry Dry Soluble oil Soluble oil Dry soluble oil soluble oil Lard oil soluble oil Mineral oil Lard oil Bronze Dry Dry Soluble oil Soluble oil Dry soluble oil soluble oil Lard oil soluble oil Mineral oil Mineral oil Mineral oil Lard oil Lard oil Lard oil Cast iron Dry Dry Dry Dry Dry Air jet soluble oil sulphurized oil soluble oil soluble oil Soluble oil Mineral lard oil Mineral lard oil Copper Dry Soluble oil Soluble oil Soluble oil Dry soluble oil Lard oil Lard oil soluble oil Mineral lard oil Kerosene Steel Soluble oil Soluble oil Sulphurized oil Soluble oil Soluble oil alloys Sulphurized oil Sulphurized oil Lard oil Mineral Mineral lard oil Mineral lard oil General Soluble oil Soluble oil Sulphurized oil Soluble oil Soluble oil purpose Sulphurized oil Sulphurized oil Lard oil Lard oil steel Lard oil Lard oil Mineral lard oil Workshop Calculation & Science : (NSQF) Exercise 2.1.02 11 Friction - Co -efficient of friction, application and effects of friction in workshop practice Exercise 2.1.03 Co-efficient of friction 2 An empty drum weighing 50kg is resting on a shop The ratio between the limiting frictional force and the normal floor. Find the coefficient of friction if a force of reaction is called co-efficient of friction. 15kg makes it just move. Suppose, by applying a force P kg, the object is just fit to W = 50 kg = Weight move, then limiting frictional force will be produced in F = 15 kg = Force between the two surfaces. The limiting frictional force will be equal to external force applied and will work in the F Co-efficient of friction = μ= opposite direction. W  F = P kg 15 kg = According to the second law of limiting frictional force, the 50 kg frictional force will be proportional to normal reaction.  = 0.3 F  R (sign is proportional to) 3 A machine crate weighing 1000kg moves distance F = R x constant of 5m in 5 sec. If the coefficient of friction between the crate and floor is 0.3, calculate the horizontal F force required to move the crate and horse power or = constant R absorbed against friction. This constant between objects is called co-efficient of Weight (W) = 1000 kg friction. This is represented by . Distance (S) = 5 meter F  or F  .R Time (t) = 5 second R i Co-efficient of friction () = 0.3 Limiting frictional force Co-efficient of friction = ii Force (F) =? Normal reaction Co-efficient of friction is always constant for any two objects Horse power (H.P.) = ? and it has no unit. F Example i μ= W 1 The sliding valve of a steam engine has dimensions 25cm by 45 cm and the steam pressure on the back F 0.3 = of the valve is 25 kg/cm2. If the co-efficient of friction 1000 Kg is 0.13. Calculate the force required to move the valve. Dimension of steam valve = 25 cm x 45 cm. F = 0.3 x 1000 kg Steam pressure = 25 kg/cm2 F = 300 kg (1 HP = 75 m.kg/sec) Co-efficient of friction = 0.13 FxS 1 Force required to move the valve = ? ii H.P = x t 75 F=? 1 Force of the steam = Pressure x Area H.P = 300 x 5 x = 4 H.P 5 75 = 25 x 25 x 45 25kg Horse power absorbed against friction = 4.H.P.  25cm  45cm = 28125 kg. cm2 4 A weight of 600 kg is kept on the inclined plane Force acts on the valve = 28125 kg at 300. Calculated the normal reaction and force rolling downwards. F μ= Solution: W Weight kept on the inclined plane (W) = 600kg F 0.13 = Angle of the inclined plane () = 300 28125  Normal reaction (R) = W. cos  F = 0.13 x 2812 Force required to move the valves = 3656.25 Kg 12  = 600 x cos 300 75 kgm/sec = 1 H.P = 600 (0.8660) 105  1 105 kgm/sec = = 1.4 H.P = 519.6 kg 75 Force rolling downwards = W. sin  Power lost due to friction = 1.4 H.P = 600 x sin 300 6 A planner table weighting 800 kg moves a distance of 2 metres in seconds on its bed. If co- = 600 (0.5000) efficient of friction between bed and table is 0.30 = 300 kg find the power required to move the table against the friction.  Normal reaction = 519.6 kg 7 On a milling machine table a component of 20 Force rolling downwards = 300 kg kgf is clamped with the help of three equidistant 5 Find out the power lost due to friction by a planner clamps. What force must be exerted by each under the following conditions. clamp to avoid slipping of the component when Mass of the planer table = 3500 kg the horizontal cutting force is 60 kgf and the coefficient of friction is equal to 0.2. Rate of movement of the table = 0.5m/sec 8 A machine weight of 14500 kg moving on the floor. Co-efficient of friction between the table and the ways = 0.06 If the co-efficient of friction between the machine and floor surface is 0.28 then calculate the force Solution: of friction. Weight of planer (W) = 3500 kg 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 Distance moved (d) = 0.5 m/sec horizontal force will be required to slide the tail Co-efficient of friction () = 0.06 stock? 10 An inclined surface makes an angle of 30 degrees F with the horizontal. An object weighting 5 tons is Co-efficient of friction = μ= W placed on the surface. Find out the normal reaction at the object and also the effective force F required to bring the object downwards. 0.06 = 3500 11 A glass block of 400 grams has been placed on F = 0.06 x 3500 = 210 kg the table. The glass is commuted by a string to a 40 grams scale pan. The string passes over pulley. Workdone = F x distance moved When a weight of 60 grams is placed on the scale = 210 x 0.5 = 105 kgm/sec pan, the block starts sliding. Find out the co- efficient of friction between wood and glass. Workshop Calculation & Science : (NSQF) Exercise 2.1.03 13 Centre of gravity - Centre of gravity and its practical application Exercise 2.2.04 Any object comprises of a large number of particles. Each per unit area, then the centroid is also the centre of gravity particle is pulled towards the earth due to the force of in a uniform gravitational field. gravity. Thus, the forces on the particles are equal, parallel and act in the same direction. These forces will have a Methods to calculate centre of gravity resultant which acts through a particular point ‘G’. This 1 By geometrical consideration. fixed point ‘G’ is called the centre of gravity. 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. 1 The centre of gravity of a circle is its centre. Centroid 2 The centre of gravity of a square, rectangle or a Different geometrical shapes such as the circle, triangle parallelogram is at the points where its diagonals meet and rectangle are plane figures having only 2-dimensions. each other. It is also the middle point of the length as They are also known as laminas. They have only area, but well as the width. no mass. The centre of gravity of these plane figures is 3 The centre of gravity of a triangle is at the point where called as the Centroid. It is also known as the geometrical the medians of the triangle meet. centre. The method of finding out the centroid of a plane 1 figure is the same as that of finding out the centre of gravity 4 The centre of gravity of a right circular Cone is at a 4 of a body. If the lamina is assumed to have uniform mass distance of from its base. 14 5 The centre of gravity of a hemisphere is at a distance wide base. The taller and more top-heavy an object, the 3r more likely it is to tip over when it is tilted by a force. This of from its base. figure demonstrates a bus driving on two different grades; 8 the second one is steep enough to cause the centre of 6 The centre of gravity of a segment of a sphere of radius gravity to fall outside of the base of the vehicle, which will ‘h’ is at a perpendicular distance of from the centre of cause it to topple over. the sphere. 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 'a'. 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 Equilibrium distance of from every point. A body is said to be in equilibrium if the resultant of all the Centre of gravity; An experiment forces acting on a body is zero and if there is no turning moment. Number of 2 pencil 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 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 1 Stable equilibrium Once the pencil is balanced, mark the location of the balancing point with a permanent marker. A body is said to be in a stable equilibrium if it returns to its original position when slightly displaced. (The C.G. is Step 3 as low as possible). Measure the distance between the ends of the pencil and the balancing point you have marked. Are the two lengths E.g : 1 A cone resting on its base equal? On my pencil, the length from the eraser to the 2 A ball on a concave surface balancing point was actually 1.25 inches less than the 3 Funnel resting on its base. (Fig 6) 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. 2 Unstable equilibrium Keeping up with that centre A body is said to be in an unstable equilibrium if it does The centre of gravity is an important concept in determining not return to its original position when slightly displaced. the stability of a structure. It’s the reason why a good Its centre of gravity falls taking it away from its original homeowner will keep the top branches of his trees trimmed. position. (CG is at high points) It’s also the reason why a pick-up truck might not be the E.g: 1 A cone resting on its tip best vehicle choice for a first time driver. Stability is maximized in objects with a lower centre of gravity and a 2 A ball on convex surface Workshop Calculation & Science : (NSQF) Exercise 2.2.04 15  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 While carrying load on back the man bends forward so 3 Neutral equilibrium that his and the load’s centre of gravity falls on his feet, if he walks erect, he will fall backward. A body is said to be in a neutral equilibrium if on being slightly displaced, it takes a new position similar to its 4 While climbing a mountain, a man bends forward and original one. The centre of gravity remains undisturbed. bends backward while descending so that the centre (CG is neither raised or lowered) of gravity of his load falls on his feet. Eg: 1 A cone resting on its side 5 In a double-decker, more passengers are accommodated in the lower deck and less on the upper 2 A ball on flat surface so that the centre of gravity of the bus and the 3 Funnel resting on its side (Fig 8) 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. Since BCD=45º then BD=DC=x Conditions of equilibrium As per Pythagoras theorem A body is said to be in a state of equilibrium under the BD2 + DC2 = CB2 action of forces when there is no motion of rotation or x2 + x2 = 402 translation of the body. There are three conditions of equilibrium of a body which are given below: 2x2 = 1600 i Algebraic sum of the horizontal components of all the 1600 x2 = = 800 forces acting on the body must be zero. 2 H = 0 x = 800 = 28.28 cm ii Algebraic sum of the vertical components of all the forces acting on the body must be zero. x 28.28 Centroid from BD = = = 9.43 cm V = 0 3 3 iii Algebraic sum of the moments of all the forces acting 2 A rectangular lamina has 10 cm and 8 cm. Find on the body must be zero. the centroid. (Centroid of rectangular = Diagonals M = 0 intersecting point.) Centroid of rectangular = Diagonals intersecting point Torque or twisting moment of a couple is given by the product of force applied and the arm of 10 Centre of AB = =5 the couple (i.e. Radius). In fact, moment means 2 the product of “force applied” and the “perpendicular distance of the point and the 8 Centre of AD = =4 line of the force”. 2 Centroid lying 4 cm from AB and 5 cm from AD 16 Workshop Calculation & Science : (NSQF) Exercise 2.2.04  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. 1 Area of right angled triangle (a1) = 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 from A (h1) = 12 - 4 = 8 cm Centroid of rectangle 1 Area of half circle (a2) =  r2 Area of rectangle= 5 x 10 = 50cm2 2 1 1 Area of triangle = bh 2 = x 3.14 x 5 x 5 2 1 = x 5 x 5 = 12.5 cm2 = 39.25 cm2 2 Centroid of semi circle = (Vertical distance from Total area = 50 + 12.5 = 62.5 cm2 centre of diagonal) The centre of gravity for rectangle is the point of intersection 4r of diagonal = 5 cm distance from AD (CG1) = 3π 1 Centre of gravity for triangle is distance from its height. 3 4x5 Centroid distance from E to D = 3 x 3.14 1 5 =5x = = 1.67 cm = 2.123 cm 3 3 (CG2) Centroid of plate is lying in between CG1 and CG2. ⎛ Height of ⎞ ⎛ Centroid of ⎞ From the figure torque is about AD. (h2 ) = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ ⎝ triangle ⎠ ⎝ half circle ⎠ 62.5x = 50 x 5 + 12.5 x 11.67 = 12 + 2.123 = 250+145.875 = 14.123 cm 62.5x = 395.875 To find centroid of lamina 395.875 x = = 6.334 cm a1 h1 + a2 h2 62.5 y= a1 + a2 Centre of gravity is 6.334 cm from AD, on the centre axis. Workshop Calculation & Science : (NSQF) Exercise 2.2.04 17  = 54 KNm + 90KNm 60 x 8 + 39.25 x 14.123 = 60 + 39.25 = 144 KNm B This moment is equal to moment of section A and 480 + 554.328 section B about 'O' (distance of action being D meter) = 99.25 That is =(36KN+20KN) x D(meter)=56 DKNm 1034.328 Again equating A and B = 99.25 144 KNm = 56 DKNm = 10.421 cm Centroid is lying at 10.421 cm from point A 144 KNm =D 56 KNm 5 A uniform rod weighing 50kg and 3m long carries loads as shown below. Find out the distance of 144 =D the CG of the system from the left hand end. 56 18 Therefore D = 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 7 A thin lamina is shown in the figure. Find centre of gravity. 110 x x = (50 x 1.5) + (15 x 1.75) + (20 x 2.25) + (25 x 2.75) = 75 + 26.25 + 45 + 68.75 = 215 215 Therefore x = = 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 Let a1= Area of rectangle CDBA = 45 x 40 = 1800 mm2 Moment of section B about O = 20 KN x 4.5 m h1 = Distance between centre of gravity of rectangle Adding both we get as below of point F = 40 = 20 mm Total moment about O=(36 KN x 1.5 m + (20KN x 4.5m) 2 18 Workshop Calculation & Science : (NSQF) Exercise 2.2.04  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 = 450 mm2 h2 = distance between centre of gravity of triangle of 30 Distance of CG2 from AB = 20 + mm point F. 2 =1/3rd height of triangle +width of rectangle = 20 + 15 mm 1 50 170 = 35 mm = (50) + 40 = + 40 = mm 3 3 3 1 Area of triangle = x 40 x 40 mm2 Applying formula 2 a1 h1 + a2 h2 = 800 mm2 y= a1 + a2  PTR - Isosceles triangle Draw perpendicular line PS on TR from P. ⎛ 170 ⎞ PSR - right angled triangle 1800(20) + 1125⎜ ⎟ = ⎝ 3 ⎠ By applying Pythagoras theorem, 1800 + 1125 800 5 x2 + x2 = 402 36000 + 63753.75 99753.75 2x2 = 1600 = = 2925 2925 x 2 = 800 y = 34.10 mm x = 800 The CG is at a distance of 34.1mm from point F = 28.28 mm the point of reference in the line AB. Distance of CG3 from TR = x = 28.28 mm = 9.43 mm 8 Find the CG of the lamina shown below. 3 3