

Disclaimer:
This document does not claim any originality and cannot be used as a substitute for prescribed textbooks. I would like to acknowledge various sources like freely available materials from internet particularly NPTEL/ SWAYAM course material from which the lecture note was prepared. The ownership of the information lies with the respective authors or institutions. Further, this document is not intended to be used for commercial purpose and the BlogSpot owner is not accountable for any issues, legal or otherwise, arising out of use of this document.
This
open resource is a collection of academic courses of under graduation program for
B.Tech (Civil Engineering) as per the syllabus of Dr. B.A.T University, Lonere, Raigad
(m.s), India prepared by Dr. Mohd.
Zameeruddin, Associate Professor, MGM's College of Engineering, Nanded for
use in the out-of-class activity. The content covers both theoretical and
analytical studies. There are six lessons as part of this document, and each
deal with an aspect related to Engineering Mechanics (Course Code BTES 103/203).
Unit 1: Basic Concepts
Unit 2: Equilibrium
Unit 3: Kinematics
Unit 4: Kinetics
Unit 5: Work, Power, Energy
Unit 1: Basic Concepts
Objectives
of Engineering Analysis and Design, Idealization of Engineering Problems, Simplification of real 3D problems to 2D and 1D domain, Basis of Assumptions, Types
of Supports, Types of Loads, Free Body Diagram, Laws
of Motion, Fundamental principles, Resolution and Composition
of a Forces, Resultant, Couple, Moment, Varignon's
theorem, Force Systems, Centroid of Composite Shapes, Moment of
Inertia of Planer Sections and Radius of Gyration.
Unit 2: Equilibrium
Static Equilibrium, analytical and graphical conditions of equilibrium, Lami's theorem, equilibrium of coplanar concurrent forces, coplanar nonconcurrent forces, parallel forces, beam reactions.
Simple trusses (Plane and space): method of joint for plane truss, method of section for plane trusses.
Friction: Columb law, friction angles, wedge friction, sliding friction and rolling resistance
Unit 3: Kinematics
Types of motions, kinematics of particles, rectilinear motion, constant and variable acceleration, relative motion, motion under gravity, study of motion diagrams, angular motion, tangential and radial acceleration projectile motion, kinematics of rigid bodies, concept of instantaneous center of rotation, concept of relative velocity
Unit 4: Kinetics
Mass moment of inertia, kinetics of particle, D' Alembert's principle, application of linear motion, kinetics of rigid bodies, applications in translation, applications in fixed axis rotation
Unit 5: Work, Power and Energy
Principle of virtual work, virtual displacements for particle and rigid bodies, work done by a force, spring, potential energy, kinetic energy of linear motion and rotation, work energy principle, conversion of energy, power, impulse moment principle, collision of elastic bodies
Course Objectives:
- Understand and apply the fundamental laws of engineering mechanics
- Understand and apply the condition of equilibrium to analyze the given force system
- Compute the center of gravity and moment of inertia of plane surfaces
- Evaluate the characteristics of a body/ particle in motion with or without the application of force
Course Outcome:Students' will be able to utilize the fundamental laws of mechanics and solve the engineering problems of rigid body in rest or motion under the action of force or force system
पाठ्यक्रम के उद्देश्य:
- इंजीनियरिंग यांत्रिकी के
मूलभूत नियमों को समझें और लागू करें
- दी गई बल प्रणाली का विश्लेषण करने के लिए संतुलन की स्थिति को समझें और लागू करें
- गुरुत्वाकर्षण के केंद्र और समतल सतहों की जड़ता के क्षण की गणना करें
- बल के अनुप्रयोग के साथ या उसके बिना गति में एक निकाय / कण की विशेषताओं का मूल्यांकन करें
पाठ्यक्रम परिणाम:
छात्र यांत्रिकी के मूलभूत नियमों का उपयोग करने और बल या बल प्रणाली की कार्रवाई के तहत आराम या गति में कठोर निकाय की इंजीनियरिंग समस्याओं को हल करने में सक्षम होंगेENGINEERING MECHANICS: WHAT AND WHY?
In today’s scientific
world, the various spheres of scientific activities are grouped commonly by
applying the principle of employing observation and experimentation. The branch
of science which deals or co-ordinate research work for practical utility and service
of mankind is known as "Applied
Sciences".
Engineering is an activity concerned with
the creation of a new system for the benefit of mankind. The process of
creating proceeds by way of research, design, and development. A new system
emerges from innovation and system may be constituted by mechanical, mechatronics,
hydraulic, thermal or other elements. "It may also be defined as an art of
executing a partial application of scientific knowledge." - Living
Webster Encyclopedia.
Let we try to understand
the difference between Science and Engineering. Science is concerned with a
systematic understanding and gathering of facts and laws and principles
governing natural phenomena. On the other hand, engineering is an art of the
utilization of established facts, laws and principles create a certain desired
phenomenon as shown in figure 1.

इंजीनियरिंग यांत्रिकी: क्या और क्यों ?
आज की वैज्ञानिक दुनिया में, वैज्ञानिक गतिविधियों के विभिन्न क्षेत्रों को आमतौर पर अवलोकन और प्रयोग को नियोजित करने के सिद्धांत को लागू करके समूहीकृत किया जाता है। विज्ञान की शाखा जो मानव जाति की व्यावहारिक उपयोगिता और सेवा के लिए अनुसंधान कार्य से संबंधित या समन्वय करती है, उसे "एप्लाइड साइंसेज" के रूप में जाना जाता है।
इंजीनियरिंग मानव जाति के लाभ के लिए एक नई प्रणाली के निर्माण से संबंधित एक गतिविधि है। अनुसंधान, डिजाइन और विकास के माध्यम से आगे बढ़ने की प्रक्रिया। नवाचार से एक नई प्रणाली उभरती है और प्रणाली यांत्रिक, मेक्ट्रोनिक्स,
हाइड्रोलिक, थर्मल या अन्य तत्वों द्वारा गठित की जा सकती है। "इसे वैज्ञानिक ज्ञान के आंशिक अनुप्रयोग को निष्पादित करने की कला के रूप में भी परिभाषित किया जा सकता है।
आइए हम विज्ञान और इंजीनियरिंग के बीच अंतर को समझने की कोशिश करें। विज्ञान प्राकृतिक घटनाओं को नियंत्रित करने वाले तथ्यों और कानूनों और सिद्धांतों की व्यवस्थित समझ और इकट्ठा करने से संबंधित है। दूसरी ओर इंजीनियरिंग स्थापित तथ्यों के उपयोग की एक कला है, कानून और सिद्धांत एक निश्चित वांछित घटना बनाते हैं जैसा कि चित्र 1 में दिखाया गया है।
The
branch of science which deals with the study of the action of forces on the
objects is called as Mechanics. Engineering Mechanics is that branch of applied
science, which deals with the laws and principles of mechanics, along with
their application to engineering problems.
विज्ञान की
वह
शाखा
जो
वस्तुओं पर
बलों
की
क्रिया
का
अध्ययन
करती
है,
यांत्रिकी कहलाती
है।
अभियांत्रिकी यांत्रिकी अनुप्रयुक्त विज्ञान की
वह
शाखा
है,
जो
यांत्रिकी के
नियमों
और
सिद्धांतों के
साथ-साथ इंजीनियरिंग समस्याओं में
उनके
अनुप्रयोग से
संबंधित है।
Engineering
mechanics can be broadly classified into two groups: Fluid Mechanics and Solid
Mechanics
इंजीनियरिंग
यांत्रिकी को मोटे तौर पर दो समूहों में वर्गीकृत किया जा सकता है: द्रव यांत्रिकी
और ठोस यांत्रिकी।

Fluid Mechanics:
It is the branch of science which deals with
the behavior of fluids [in gases or liquid state], and the forces on them. Fluid
mechanics is further classified as fluid statics and fluid dynamics. The study
of fluids at rest is known as fluid statics and the study of fluids in motion
considering the pressure forces is known as fluid dynamics, The fluid dynamics
is further divided into fluid kinematics and kinetics based on the effects of
pressure.
तरल
यांत्रिकी:
यह विज्ञान की वह शाखा है जो तरल पदार्थ [गैसों या तरल अवस्था में] के व्यवहार और उन पर लगने वाले बलों से संबंधित है। द्रव यांत्रिकी को आगे द्रव स्थैतिकी और द्रव गतिकी
के रूप में वर्गीकृत किया गया है। आराम पर तरल पदार्थ के अध्ययन को द्रव स्थैतिकी के
रूप में जाना जाता है और दबाव बलों पर विचार करते हुए गति में तरल पदार्थ का अध्ययन
द्रव गतिकी के रूप में जाना जाता है, द्रव गतिकी को दबाव के प्रभाव के आधार पर द्रव
कीनेमेटीक्स और कैनेटीक्स में विभाजित किया जाता है।यह विज्ञान की वह शाखा है जो तरल पदार्थ [गैसों या तरल अवस्था में] के व्यवहार और उन पर लगने वाले बलों से संबंधित है। द्रव यांत्रिकी को आगे द्रव स्थैतिकी और द्रव गतिकी
के रूप में वर्गीकृत किया गया है। आराम पर तरल पदार्थ के अध्ययन को द्रव स्थैतिकी के
रूप में जाना जाता है और दबाव बलों पर विचार करते हुए गति में तरल पदार्थ का अध्ययन
द्रव गतिकी के रूप में जाना जाता है, द्रव गतिकी को दबाव के प्रभाव के आधार पर द्रव
कीनेमेटीक्स और कैनेटीक्स में विभाजित किया जाता है।
Solid mechanics
It is the branch of engineering mechanics with deals with the
behavior of rigid and deformable bodies and the forces on them. Mechanics of
rigid bodies is further classified into static and dynamics. Statics deals with
the forces and their effects while acting on the bodies at rest. Dynamics deals
with the forces and their effects, on the bodies which are in state of motion.
Dynamics is further sub-divided into; (i) Kinematics deals with the bodies in
motion without any reference to force which are responsible for a motion and
(ii) Kinetics deals with the bodies which are in motion due to the application
of the force.
Mechanics of deformable bodies depends on
material behavior. The responses are evaluated by using the theories of
failures, theory of elasticity and theory of plasticity.
ठोस यांत्रिकी:
यह इंजीनियरिंग यांत्रिकी की शाखा है जो कठोर और विकृत निकायों के व्यवहार
और उन पर लगने वाले बलों से संबंधित है। कठोर निकायों के यांत्रिकी को आगे स्थिर और
गतिकी में वर्गीकृत किया गया है। स्टैटिक्स आराम की अवस्था में पिंडों पर कार्य करते
हुए बलों और उनके प्रभावों से संबंधित है। डायनेमिक्स गति की स्थिति में निकायों पर बलों
और उनके प्रभावों से संबंधित है। डायनेमिक्स को आगे उप-विभाजित किया गया है; (i)
किनेमैटिक्स बल के संदर्भ के बिना गति में निकायों से संबंधित है जो गति के लिए
ज़िम्मेदार हैं और (ii) कैनेटीक्स उन निकायों से संबंधित हैं जो बल के आवेदन के
कारण गति में हैं।डायनेमिक्स गति की स्थिति
में निकायों पर बलों और उनके प्रभावों से संबंधित है। डायनेमिक्स को आगे उप-विभाजित
किया गया है; (i) किनेमैटिक्स बल के संदर्भ के बिना गति में निकायों से संबंधित है
जो गति के लिए ज़िम्मेदार हैं और (ii)
कैनेटीक्स उन निकायों से संबंधित हैं जो बल के आवेदन के कारण गति में हैं।
विकृत निकायों के यांत्रिकी भौतिक व्यवहार पर निर्भर करते हैं। विफलताओं के सिद्धांत, लोच के सिद्धांत और प्लास्टिक के सिद्धांत का उपयोग करके प्रतिक्रियाओं
का मूल्यांकन किया जाता है।
Employability - The opportunity to employ all one's abilities.
रोज़गार
योग्यता
- अपनी
सभी
क्षमताओं
को
नियोजित
करने
का
अवसर।
Mechanics and its
relevance to Engineering
Mechanics is a physical science, concern with the dynamical behavior of bodies in presence of the mechanical disturbances which is a point of interest in disciplines of Mechanical, Electrical, Civil, Chemical, Aeronautical, textile, metallurgical and mining Engineers. This is a reason that makes it to appear in the core disciplines of the engineering analysis.
Mechanics
is the physical science concerned with the dynamical behavior of material
bodies in presence of mechanical disturbances which are interest point of
Mechanical, Electrical, Civil, Chemical, Aeronautical, textile,
metallurgical and mining Engineers. This makes it appear in the core of all
engineering analysis. Engineering Mechanics refers to a course in mechanics
tailored exclusively for engineers with features such as:
- The subject matter is not presented as rigorously
as a course in analytical or axiomatic mechanics may demand.
- It provides through grounding of the basic principles with the engineering applications.
- It provides a basic knowledge for other courses to be built-up on the concept of engineering mechanics.
यांत्रिकी
और
इंजीनियरिंग
के
लिए
इसकी
प्रासंगिकता
यांत्रिकी
एक
भौतिक
विज्ञान
है,
जो
यांत्रिक
गड़बड़ी
की
उपस्थिति
में
निकायों
के
गतिशील
व्यवहार
से
संबंधित
है
जो
मैकेनिकल,
इलेक्ट्रिकल,
सिविल,
केमिकल,
एयरोनॉटिकल,
टेक्सटाइल,
मेटलर्जिकल
और
माइनिंग
इंजीनियर्स
के
विषयों
में
रुचि
का
एक
बिंदु
है।
यह
एक
कारण
है
जो
इसे
इंजीनियरिंग
विश्लेषण
के
मुख्य
विषयों
में
दिखाई
देता
है। इंजीनियरिंग मैकेनिक्स यांत्रिकी
में एक पाठ्यक्रम को संदर्भित करता है जो विशेष रूप से इंजीनियरों के लिए इस तरह की
विशेषताओं के साथ तैयार किया गया है:
- विषय
वस्तु
को
उतनी
कठोरता
से
प्रस्तुत
नहीं
किया
जाता
है
जितना
कि
विश्लेषणात्मक
या
स्वयंसिद्ध
यांत्रिकी
में
एक
पाठ्यक्रम
की
मांग
हो
सकती
है।
- यह
इंजीनियरिंग
अनुप्रयोगों
के
साथ
बुनियादी
सिद्धांतों
की
ग्राउंडिंग
प्रदान
करता
है।
- यह
इंजीनियरिंग
यांत्रिकी
की
अवधारणा
पर
निर्मित
होने
वाले
अन्य
पाठ्यक्रमों
के
लिए
एक
बुनियादी
ज्ञान
प्रदान
करता
है।
State-of -Art-of-Development
Modernization of mankind started with the invention of wheels. The study of civilization of Babylonians, Egyptians, Greeks and Romans reveals that they used water wheels and windmills to maintain their basic amenities. The word Mechanics was put forth by a Greek philosopher "Aristotle (BC 322-384)", for the problems of lever and concept of center of gravity. First mathematical concept was given by Archimedes (BC 212-287) in the form of law of hydrostatics. Leonardo Da Vici (1452-1519) a great engineer and painter, gave many ideas for the study of mechanism, friction and the motion of bodies on inclined planes. Galileo (1564-1642) established the theory of projectiles and rudimentary idea of inertia. Huygens (1629-1695) developed the analysis of motion of pendulum. Sir Isaac Newton (1643-1727) introduced the concept of force and mass and stated the laws of motions. John Bernoulli (1667-1748) has given the concept of principle of virtual work. James Watt (1736-1819) introduced the term Horsepower for comparing performances of his engines. With the passage of time engineering mechanics coupled with knowledge of other specialized subjects for example, strength of materials, theory of machines, calculus, vector and algebra.
Review of Engineering Mechanics (Video Lecture 1)
Review of Engineering Mechanics (Video Lecture 2)
अत्याधुनिक
विकास
मानव जाति का आधुनिकीकरण पहियों के आविष्कार के साथ शुरू हुआ। बेबीलोन, मिस्र, यूनानियों और रोमनों की सभ्यता के अध्ययन से पता चलता है कि उन्होंने अपनी बुनियादी सुविधाओं को बनाए रखने के लिए पानी के पहियों और पवन चक्कियों का इस्तेमाल किया। यांत्रिकी शब्द को एक ग्रीक
दार्शनिक "अरस्तू (बीसी 322-384)" द्वारा लीवर की समस्याओं और गुरुत्वाकर्षण
के केंद्र की अवधारणा के लिए रखा गया था। पहली गणितीय अवधारणा आर्किमिडीज (बीसी 212-287) द्वारा जलस्थैतिकी के नियम के
रूप में दी गई थी। एक महान इंजीनियर और चित्रकार
लियोनार्डो दा विकी (1452-1519) ने तंत्र, घर्षण और झुकाव वाले सतह पर निकायों
की गति के अध्ययन के लिए कई विचार दिए। गैलीलियो (1564-1642) ने प्रक्षेप्य के सिद्धांत और
जड़ता के अल्पविकसित विचार की स्थापना की। ह्यूजेंस
(1629-1695) ने लोलक
की
गति
का
विश्लेषण
विकसित
किया। सर इस्साक न्यूटन (1643-1727) ने बल और द्रव्यमान की अवधारणा पेश की और गति के नियमों को बताया। जॉन बर्नौली (1667-1748) ने आभासी कार्य के सिद्धांत
की अवधारणा दी है। जेम्स वाट (1736-1819) ने अपने इंजनों के प्रदर्शन की तुलना करने के लिए हॉर्सपावर शब्द की शुरुआत की। समय बीतने के साथ इंजीनियरिंग
यांत्रिकी अन्य विशिष्ट विषयों के ज्ञान के साथ युग्मित हो गया, उदाहरण के लिए, सामग्री
की ताकत, मशीनों का सिद्धांत, कलन, वेक्टर और बीजगणित।
Humans learn best when they learn in a community
featuring social, cognitive, and teaching presence - D. Randy Garrison.
मनुष्य
सबसे
अच्छा
सीखते
हैं
जब
वे
सामाजिक,
संज्ञानात्मक
और
शिक्षण
उपस्थिति
वाले
समुदाय
में
सीखते
हैं। - डी रैंडी गैरीसन
Simplification
of real 3-D problems to 2-D and 1-D domain
The state of solid body
is defined with its volume and mass occupied in that volume. The dimensional
proportion can be traced through coordinates system either Cartesian coordinates
(x, y and z) or polar coordinates (r, q and f). When a
solid body is subjected to forces, effects may be seen on all three dimensions.
In order to avoid complex analysis, in structural analysis effects of forces
are neglected on one or more direction, this type of idealization is said to be
simplification of real 3-Dproblems.
The idealization in
which one dimension say length is considered to be very large compared to other
two dimensions, the effects along the length dominates surpassing the effects
on other two dimension is said to be 1-D
domain. For Example: Beams, frames, trusses, cables, arch, and bars.
The idealization in
which length and breadth are large compared to thickness, the effects along the
length and breadth dominates surpassing the effects on the thickness is said to
be 2-D domain. For Example: Slabs,
Deep beams, Shells or Walls.
2-डी
और
1-डी
डोमेन
के
लिए
वास्तविक
3-डी
समस्याओं
का
सरलीकरण
ठोस शरीर की स्थिति को इसके आयतन और द्रव्यमान से परिभाषित किया जाता है वह मात्रा। निर्देशांक प्रणाली के माध्यम से आयामी अनुपात का पता लगाया जा सकता है या तो कार्तीय निर्देशांक (x, y और z) या ध्रुवीय निर्देशांक (r, q and f)। जब एक ठोस पिंड पर बल लगाया जाता है, तो प्रभाव तीनों आयामों पर देखा जा सकता है। जटिल विश्लेषण से बचने के लिए, संरचनात्मक विश्लेषण में बलों के प्रभावों को एक या एक से अधिक दिशाओं में उपेक्षित किया जाता है, इस प्रकार के आदर्शीकरण को वास्तविक 3-डी समस्याओं का सरलीकरण कहा जाता है।
आदर्शीकरण
जिसमें
एक
आयाम
कहता
है
कि
लंबाई
अन्य
दो
आयामों
की
तुलना
में
बहुत
बड़ी
मानी
जाती
है,
लंबाई
के
साथ
प्रभाव
अन्य
दो
आयामों
पर
प्रभाव
को
पार
करते
हुए
1-डी
डोमेन
कहा
जाता
है।
उदाहरण
के
लिए:
बीम,
फ्रेम,
ट्रस,
केबल,
आर्क
और
बार।
आदर्शीकरण
जिसमें
मोटाई
की
तुलना
में
लंबाई
और
चौड़ाई
बड़ी
होती
है,
मोटाई
पर
प्रभाव
को
पार
करते
हुए
लंबाई
और
चौड़ाई
के
साथ
प्रभाव
को
2-डी
डोमेन
कहा
जाता
है।
उदाहरण
के
लिए:
स्लैब,
डीप
बीम,
शेल
या
दीवारें।
Fundamental units
All fundamental units relating to Engineering Mechanics are expressed
in terms of three fundamental quantities.
1. Length
2. Mass 3. Time
The
other fundamental units are
4.
Electric Current 5. Temperature
6. Luminous Intensity
मौलिक
इकाइयाँ
इंजीनियरिंग
यांत्रिकी
से
संबंधित
सभी
मूलभूत
इकाइयों
को
तीन
मूलभूत
मात्राओं
के
रूप
में
व्यक्त
किया
जाता
है।
1. लम्बाई 2. द्रव्यमान 3. समय
अन्य
मूलभूत
इकाइयाँ
हैं
4. विद्युत
प्रवाह 5. तापमान 6. चमकदार
तीव्रता
Derived Units
These
are derived from fundamental units, for example.
Area
= Length x Breadth = m x m =m2
Volume= Length
x Breadth x Height = m x m x m = m3
Velocity=Distance/Time
= m/sec
व्युत्पन्न इकाइयाँ
ये मौलिक इकाइयों से प्राप्त
होते हैं, उदाहरण के लिए।
क्षेत्रफल = लंबाई x चौड़ाई
= मी x मी = मी2
आयतन = लंबाई x चौड़ाई x
ऊँचाई = मी x मी x मी = मी3
वेग=दूरी/समय = मी/सेकंड
System of Units
- C.G.S Unit: Centimeter, Gram, Second...
-
F.P.S Units: Foot, Pound, Second.......
-
M.K.S Unit: Meter, Kilogram, Seconds......
-
SI Unit: General Conference of Weight and Measures (C.G.M.W)
इकाइयों की प्रणाली
- सीजीएस यूनिट: सेंटीमीटर, ग्राम, सेकेंड...
- एफपीएस इकाइयां: फुट, पाउंड, सेकंड .......
- एमकेएस यूनिट: मीटर, किलोग्राम, सेकेंड...
- SI इकाई: वजन और माप का सामान्य सम्मेलन (C.G.M.W)
Fundamental Quantities -
Video Lecture
Some basic units in the SI system
एसआई प्रणाली में कुछ बुनियादी इकाइयां
Length-Meter-m
Mass-Kilogram-Kg
Time-Seconds-s
Electric
current-Ampere-A
Temperature-Kelvin-K
Luminous
intensity-Candela-cd
Plane
angle-Radian-rad
Solid
angle-ste-radian-sr
|
लंबाई-मीटर-एम
मास-किलोग्राम-किग्रा
समय-सेकंड-एस
विद्युत धारा-एम्पीयर-ए
तापमान-केल्विन-के
चमकदार तीव्रता-कैंडेला-सीडी
समतल कोण-रेडियन-रेड
ठोस कोण-स्टे-रेडियन-एसआर
|
Fundamental Concepts and
Principles
(Assumptions made in Engineering Mechanics)
मौलिक अवधारणाएं और सिद्धांत
(इंजीनियरिंग यांत्रिकी में
की गई धारणाएँ)
The basic concepts used in mechanics are
space, time, mass and force.
यांत्रिकी में उपयोग की जाने वाली बुनियादी अवधारणाएं अंतरिक्ष, समय, द्रव्यमान और बल हैं।
Mass
Mass is the quantity of matter in a body
regardless of its volume or of any forces acting on it. Mass is always constant
at any place and at any time. An object on the moon would weigh less than it
does on the earth because of lower gravity, but it would still have the same
mass. This is because weight is a force, while mass is the property that (along
with gravity) determines the strength of this force [Wikipedia, Dec 2018].
Mass is expressed in Kilogram (kg), Gram (gm), and Milligram (mg). There are
several distinct phenomena which can be used to measure mass:
- Inertial mass measures an object's resistance
to being accelerated by a force (F=ma)
- Active gravitational mass measures the
gravitational for exerted by an object
- Passive gravitational mass measures the
gravitational force exerted on an object in a known gravitational field.
द्रव्यमान
द्रव्यमान एक शरीर में पदार्थ की मात्रा है, भले ही इसकी मात्रा या उस पर कार्य करने वाले किसी भी बल की परवाह किए बिना। द्रव्यमान हमेशा किसी भी स्थान पर और किसी भी समय स्थिर होता है। चंद्रमा पर एक वस्तु का वजन कम गुरुत्वाकर्षण के कारण पृथ्वी की तुलना में कम होगा, लेकिन इसका द्रव्यमान अभी भी समान होगा। ऐसा इसलिए है क्योंकि वजन एक बल है, जबकि द्रव्यमान वह गुण है जो (गुरुत्वाकर्षण के साथ) इस बल की ताकत निर्धारित करता है [विकिपीडिया, दिसंबर 2018] ।
द्रव्यमान किलोग्राम (किलो), ग्राम (ग्राम), और मिलीग्राम (मिलीग्राम) में व्यक्त किया जाता है। कई अलग-अलग घटनाएं हैं जिनका उपयोग द्रव्यमान को मापने के लिए किया जा सकता है:
- जड़त्वीय द्रव्यमान एक बल द्वारा त्वरित होने के लिए एक वस्तु के प्रतिरोध को मापता है (एफ = एमए)।
- सक्रिय गुरुत्वाकर्षण द्रव्यमान किसी वस्तु द्वारा लगाए गए गुरुत्वाकर्षण को मापता है।
- निष्क्रिय गुरुत्वाकर्षण द्रव्यमान एक ज्ञात गुरुत्वाकर्षण क्षेत्र में किसी वस्तु पर लगाए गए गुरुत्वाकर्षण बल को मापता है।
Newton's Law of Gravitational Constant
It states that, every particle attracts every other particle in the universe with a force that is directly proportional to their masses and inversely proportional to the square
of the distance between them.
Let us consider two
bodies of mass m1 and m2 respectively, which are at a
distance d from each other and is under the force of attraction as shown in
figure 2.
 |
Figure 2 |
The
force of attraction is expressed as,
F= G (m1m2/d2)
where,
G is the constant of proportionality and is known as constant of gravity
G = Fd2/m1m2 =
Nm2/kg.kg = Nm2/kg2= 6.673 x 10-11
For
example
1
kg mass on earth surface experience a force of,
F
= (6.673 x 10-11x 1 x 5.96504 x 1024) / (6371 x 103)2
F
= 9.80665 N
where,
Mass of earth = 5.96504 x 1024
Radius
of earth = 6371 x 103
This
force of attraction is always directed towards the center of earth.
Newtons Law of Gravity - video Lecture
State
and explain Newton’s laws of motion and gravitation?
न्यूटन का गुरुत्वाकर्षण स्थिरांक का नियम
यह बताता है कि, प्रत्येक कण ब्रह्मांड में हर दूसरे कण को एक बल के साथ आकर्षित करता है जो सीधे उनके द्रव्यमान के आनुपातिक होता है और उनके बीच की दूरी के वर्ग के व्युत्क्रमानुपाती होता है।
आइए हम क्रमशः द्रव्यमान m1 और m2 के दो पिंडों पर विचार करें, जो एक दूसरे से d की दूरी पर हैं और आकर्षण बल के अधीन हैं जैसा कि चित्र 2 में दिखाया गया है।
आकर्षण
बल
को
इस
प्रकार
व्यक्त
किया
जाता
है;
F= G (m1m2/d2)
जहां, G आनुपातिकता का स्थिरांक है और इसे गुरुत्वाकर्षण के स्थिरांक के रूप में जाना जाता है
उदाहरण के लिए
पृथ्वी
की
सतह
पर
1 किलो
द्रव्यमान
किसके
बल
का
अनुभव
करता
है?
F = (6.673 x 10-11x 1 x 5.96504 x 1024) / (6371 x 103)2
F = 9.80665 N
जहां
पृथ्वी
का
द्रव्यमान = 5.96504 x 1024
पृथ्वी
की
त्रिज्या = 6371 x 103
आकर्षण
का
यह
बल
हमेशा
पृथ्वी
के
केंद्र
की
ओर
निर्देशित
होता
है।
Weight (W)
The gravitational attraction of the earth on a body with quantity of
mass (m) is said to be weight (W). The weight of an object depends on the
gravity of that place. Weight is expressed in Newton (N). The weight of the
particle having a mass (m) is stated as,
W = m x g
A body of mass 1kg on the earth surface weighs 1 x 9.81 = 9.81 N
For example:
A body of mass 56 kgs weighs
- On earth = 56 x 9.81 = 549.36 kgs
- On Mercury = 0.378 x 549.36= 207.7 kgs
- On Venus = 0.907 x 549.36 = 499.9 kgs
- On Mars = 0.377 x 549.36 = 208.2 kgs
- On Jupiter = 2.364 x 549.36 = 1285.5 kgs
- On Saturn = 0.910 x 549.36 = 510.9 kgs
- On Uranus = 0.889 x 549.36 = 505.4 kgs
- On Neptune = 1.125 x 549.36 = 615.30 kgs
- On Moon = 0.165 x 549.36 = 90.64 kgs
वजन (W)
द्रव्यमान (m) की मात्रा के साथ एक पिंड पर पृथ्वी के गुरुत्वाकर्षण आकर्षण को वजन (W) कहा जाता है। किसी वस्तु का वजन उस स्थान के गुरुत्वाकर्षण पर निर्भर करता है। वजन न्यूटन (एन) में व्यक्त किया गया है। द्रव्यमान (m) वाले कण का वजन इस प्रकार बताया गया है,
W = m x g
पृथ्वी
की
सतह
पर
1 किग्रा
द्रव्यमान
के
पिंड
का
वजन
1 x 9.81 = 9.81 N होता है
उदाहरण
के
लिए:
56 किग्रा
द्रव्यमान
के
पिंड
का
भार
है
- पृथ्वी
पर
= 56 x 9.81 = 549.36 किग्रा
- बुध
पर
= 0.378 x 549.36 = 207.7 किग्रा
- शुक्र
पर
= 0.907 x 549.36 = 499.9 किग्रा
- मंगल
पर
= 0.377 x 549.36 = 208.2 किग्रा
- बृहस्पति
पर
= 2.364 x 549.36 = 1285.5 किग्रा
- शनि
पर
= 0.910 x 549.36 = 510.9 किग्रा
- यूरेनस
पर
= 0.889 x 549.36 = 505.4 किग्रा
- वरुण
पर
= 1.125 x 549.36 = 615.30 किलोग्राम
- चंद्रमा
पर
= 0.165 x 549.36 = 90.64 किग्रा
Space and Time
It is a region in all directions encompassing
the universe. It is a geometric position occupied by bodies. These positions
are described by linear or angular measurements with reference to a defined
system of co-ordinates.
The time is the measure of the succession of events. The selected successive
event is the rotation of the earth about its own axis, which is called
a day. The day is divided into 24 hours, an hour is divided in to 60 minutes and
a minute in 60 seconds.
The concept of space is associated with the notion of the position of point
"P" by three-length measured from Cartesian Reference Point, or
Origin, in three mutually perpendicular directions known as Coordinate of a
point whereas the time indicates its position in the space.
स्थान
और
समय
यह ब्रह्मांड को शामिल करने वाली सभी दिशाओं में एक क्षेत्र
है। यह निकायों द्वारा कब्जा की गई एक ज्यामितीय स्थिति है। निर्देशांक की एक
परिभाषित प्रणाली के संदर्भ में इन स्थितियों का वर्णन रैखिक या कोणीय मापों
द्वारा किया जाता है।
समय
घटनाओं
के
उत्तराधिकार
का
माप
है।
चयनित
क्रमिक
घटना
पृथ्वी
का
अपनी
धुरी
पर
घूमना
है,
जिसे
एक
दिन
कहा
जाता
है।
दिन
को
24 घंटे,
एक
घंटे
को
60 मिनट
और
एक
मिनट
को
60 सेकंड
में
बांटा
गया
है।
अंतरिक्ष की अवधारणा बिंदु "पी" की स्थिति की धारणा से जुड़ी है, कार्टेसियन संदर्भ बिंदु, या मूल से मापी गई तीन-लंबाई, तीन परस्पर लंबवत दिशाओं में एक बिंदु के समन्वय के रूप में जानी जाती है, जबकि समय अंतरिक्ष में इसकी स्थिति को इंगित करता है।.
Length
It is a concept to measure linear distance. The standard unit of length
is a meter (m). A meter is defined as the length of the standard bar of
platinum-radium kept at the international bureau of weight and measure. The other
units are millimeter (mm), centimeter (cm), and kilometer (km).
for example:
- The diameter of the needle is 1.5 mm
- The diameter of a cylinder is 30 cm
- The
height of the building is 15 m
- The
distance between two cities is 400 km
1
km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm.
लंबाई
यह
रैखिक
दूरी
को
मापने
की
अवधारणा
है।
लंबाई
की
मानक
इकाई
एक
मीटर
(एम)
है।
एक
मीटर
को
वजन
और
माप
के
अंतरराष्ट्रीय
ब्यूरो
में
रखे
प्लेटिनम-रेडियम
के
मानक
बार
की
लंबाई
के
रूप
में
परिभाषित
किया
गया
है।
अन्य
इकाइयां
मिलीमीटर,
सेंटीमीटर
और
किलोमीटर
हैं।यह
रैखिक
दूरी
को
मापने
की
अवधारणा
है।
लंबाई
की
मानक
इकाई
एक
मीटर
(एम)
है।
एक
मीटर
को
वजन
और
माप
के
अंतरराष्ट्रीय
ब्यूरो
में
रखे
प्लेटिनम-रेडियम
के
मानक
बार
की
लंबाई
के
रूप
में
परिभाषित
किया
गया
है।
अन्य
इकाइयां
मिलीमीटर,
सेंटीमीटर
और
किलोमीटर
हैं।
उदाहरण के लिए:
- सुई का व्यास 1.5 मिमी है
- एक बेलन का व्यास 30 सेमी है
- भवन की ऊंचाई 15 मीटर है
- दो शहरों के बीच की दूरी 400 किमी है
1 किमी = 1000 मीटर, 1 मीटर
= 100 सेमी, 1 सेमी = 10 मिमी।
mass, weight, length, time and space - video lecture
Continuum
The
body is treated as a continuum. The body can be sub-divided into molecules,
atoms, and electrons. It is very difficult to solve any engineering problem by
treating a body as a conglomeration of such discrete particles.
सातत्य
शरीर को एक निरंतरता के रूप में माना जाता है। शरीर को अणुओं, परमाणुओं और इलेक्ट्रॉनों में उप-विभाजित किया जा सकता है। ऐसे असतत कणों के समूह के रूप में शरीर मानते हुए किसी भी इंजीनियरिंग समस्या को हल करना बहुत मुश्किल है।
Rigid
Body
A body is said to be
rigid, if the relative positions of any two-particle do not change under the
action of forces acting on it.
Consider a body as shown in
figure 3, Point A and B are the original positions in a body. After application
of the forces F1, F2, F3, and F4 the body takes a new position, the point A'
and B' represent a new position, if the body is rigid the relative positions of AB
and A'B' measures are same.
 |
Figure 3 |
कठोर शरीर
एक शरीर को कठोर कहा जाता है, यदि किसी दो-कण की सापेक्ष स्थिति उस पर कार्य करने वाली शक्तियों की कार्रवाई के तहत नहीं बदलती है।
चित्र 3 में दर्शाए अनुसार
किसी पिंड पर विचार करें, बिंदु A और B किसी पिंड की मूल स्थिति हैं। F1, F2, F3, और
F4 बलों के आवेदन के बाद शरीर एक नई स्थिति लेता है, बिंदु A' और B' एक नई स्थिति का
प्रतिनिधित्व करते हैं, यदि शरीर कठोर है तो AB और A'B' उपायों की सापेक्ष स्थिति समान
होती है .
Particle
A particle may be
defined as an object which has only mass and no size. Theoretically, such a
body cannot exist. However, in dealing with problems involving distances
considerably larger compared to the size of the body, the body is treated as a
particle, taking care of accuracy [ Bhavikatti S. S].
For example
- A bomber aircraft is a
particle for a gunner operating from the ground.
- A ship in mid-sea is a
particle in the study of its relative motion from the control tower.
- In the study of the
movement of the earth in the celestial sphere, earth is treated as a particle.
कण
एक कण को एक वस्तु के रूप में परिभाषित किया जा सकता है जिसका केवल द्रव्यमान और कोई आकार नहीं है। सैद्धांतिक रूप से, ऐसा शरीर मौजूद नहीं हो सकता। हालांकि, शरीर के आकार की तुलना में काफी बड़ी दूरियों से जुड़ी समस्याओं से निपटने में, शरीर को एक कण के रूप में माना जाता है, सटीकता का ख्याल रखते हुए [भविकट्टी एस.एस.]।

उदाहरण के लिए:
- एक बमवर्षक विमान जमीन से संचालित होने वाले गनर के लिए एक कण है।
- मध्य समुद्र में एक जहाज नियंत्रण टॉवर से अपनी सापेक्ष गति के अध्ययन में एक कण है।
- आकाशीय मंडल में पृथ्वी की गति के अध्ययन में पृथ्वी को एक कण के रूप में माना जाता है
Deformable
body
When a body deforms due
to A force or A torque it is said deformable body. Material generates stresses
against deformation.
विकृत शरीर
जब कोई पिंड A बल या A बल आघूर्ण के कारण विकृत होता है तो उसे विरूपित पिंड कहा जाता है। सामग्री विरूपण के खिलाफ तनाव उत्पन्न करती है।
Define
the following terms: Continuum, rigid body and particle.
Force -बल
Force maybe defined as an agent
which, produces or tends to produce, destroys or tend to destroy motion in a
body.
बल
को एक एजेंट के रूप में परिभाषित किया जा सकता है, जो निकाय में गति उत्पन्न करता है
या उत्पन्न करता है, नष्ट करता है या नष्ट करता है।
Force is a physical quantity that
changes or tends to change the state of rest or motion of a body or changes the
dimensions of a body.
बल
एक भौतिक मात्रा है जो निकाय के आराम या गति की स्थिति को बदलता है या बदलता है या
शरीर के आयामों को बदलता है।
Effects of Force
A force has the following effects on
a body:
- External effect
- Internal effect
Force may produce the following
external effects on a body (a) It may produces or stop the motion of body (b)
It may balance the forces already acting on the body, thus bringing it to rest
or equilibrium as shown in Figure 4.
एक बल का शरीर पर निम्नलिखित प्रभाव होते हैं:
- बाहरी प्रभाव
- आंतरिक प्रभाव
बल किसी पिंड पर निम्नलिखित
बाहरी प्रभाव उत्पन्न कर सकता है (a) यह पिंड की गति उत्पन्न या रोक सकता है (b) यह
पिंड पर पहले से ही कार्य कर रहे बलों को संतुलित कर सकता है, इस प्रकार इसे आराम या
संतुलन में ला सकता है जैसा कि चित्र 4 में दिखाया गया है।
 |
Figure 4 |
The internal effect of a force is to produce
deformation (change of size or shape) in the body. If the body is rigid it will
not undergo any deformation under the action of forces acting on it (as
illustrated in Figure 5).
किसी बल का आंतरिक प्रभाव
शरीर में विरूपण (आकार या आकार में परिवर्तन) उत्पन्न करना है। यदि पिंड कठोर है तो
उस पर कार्यरत बलों की कार्रवाई के तहत कोई विरूपण नहीं होगा (जैसा कि चित्र 5 में
दिखाया गया है)
What is force: it can be simply stated as “Action of one body on another”
Force and its effects – Video Lecture
Define the term Force and state the various characteristics of
force
Unit of Force
The
following are two systems of units of force
1.
Gravitational
units
2.
Absolute
units
The
gravitational or engineering unit of a force is defined as, “the force with
which the earth attracts a unit mass of the body. The gravitational units of force
are lb wt., gm wt, kg wt in FPS, CGS and SI system respectively.
The absolute or scientific units of
force are based on Newton's Second Law of Motion. Force is defined as the rate
of change of momentum. For an unchanging mass, this is equivalent to mass x
acceleration.
So,
1 N = 1 kg m s-2, or 1 kg m/s2.
The
absolute units of force are Poundal, dyne, Newton in the FPS, CGS and SI
system.
Definition of force according to various kinds of systems.
- C.G.S: 1 dyne is that amount of force which causes a one-gram mass to move with an acceleration of 1cm/s2.
- F.P.S: 1 slag is that amount of force which gives an acceleration of 1ft/s2 when acted by a force of one pound.
- M.K.S: kilo-gram weight is the force required to move a mass of one kilogram with an acceleration equal to gravitational acceleration g=9.8m/s2; i.e. 1 kg-wt = 9.8 N
- SI Unit: 1 N is the force which gives an acceleration of 1m/s2 to move a mass of 1 kg.
Characteristic of a force
A force is
characterized by its point of application, magnitude, and direction.
Magnitude: The quantity of a force is called its magnitude
Direction: The line along which the force acts is called its
direction
Sense: The way in which the force acts along the line of action is called the
sense of force
Point of Application: The Point at which the force acts on the
body is called the point of application.
Define the term Force and state the various characteristics of force - Video Lecture
Representation of Force
A force is a vector quantity possessing both magnitude and direction. it
is represented in two ways by a vector or bow's notation. A vector is
geometrically represented by a straight line having an arrow at the endpoint is
called a "head", and the initial point is called tail; the length of
a straight line is proportional to the magnitude of a vector. A vector is
represented by putting an arrow over the symbol for example: Force [FAB]
is represented by vector AB, the length of arrow gives the magnitude of the force and arrow shows
the direction and line of action of the force.
In Bow's notation a force is represented by putting capital alphabets on either side of the line of action of the force. The placing of letters is
either done clockwise or anti-clockwise [as shown in figure].
Scalar quantities
Physical quantities that are completely described by magnitude having a
specific unit and can be manipulated by ordinary algebra are called as scalar
quantities.
For example: Mass, Time, Energy, Area and Temperature are scalar
quantities that obey ordinary algebraic law of addition and multiplication.
(1) Cumulative law:
a + b = b + a
a x b
= b x a
(2) Associative law:
a + b
+ c = (a + b) + c = (c + b) + a = (a + c) + b
a x b
x c = (a x b) x c = (c x b) x a = (a x c) x b
(3) Distributive Law:
a x
(b + c) = (a x b) + (a x c)
Vectors
Physical quantities which have magnitude with specified unit and
direction and could be manipulated by vector algebra are called as vector
quantities.
Example: Displacement, Force, Velocity, Acceleration, Momentum, Couple
etc.
Distinguish between scalars and vectors. Give examples
System
of Forces
When several forces act
simultaneously on a body, they constitute a system of forces. When all the
forces acting on system do not lie in a single plane they constitute the system
of forces in space. The other forms of force system can be classified as;
System of Force
When several forces act simultaneously on a body,
they constitute a system of forces. When all the forces acting on system do not lie in a single plane they constitute the
system of forces in space. The other forms of force system can be classified as;
1. Col-linear forces- The line of action of all the forces act along the same line.
2. Co-planar forces- All forces are parallel to each other and lie in a single plane.
Like forces and Unlike forces
The forces are said to be like if they have the same magnitude and direction. The forces are said to be unlike if they have the same magnitude and not in the same direction.
3. Concurrent forces- Line of action of all forces pass through a single point.
4. Co-planar concurrent forces- Line of action of all forces pass through a single point and the forces lie in the same plane.
5. Co-planar non-concurrent forces- All forces do not meet at the same point but lie in a single plane.
6. Non-co-planar non-concurrent forces- All forces do not lie in the same plane, but their line of action does not pass through a single point.
7. Non-co planar concurrent forces-All forces do not lie in the same plane, but their lines of action pass through a single point.

Definesystem of forces and classify the system of forces with neat sketches -Video lecture
Type of Forces
Forces can be divided into the following types:
- Tensile Forces
- Compression Forces
- Shear Forces
- Bending Forces
- Torsional Forces
Tension Forces
Compression Forces
Bending Forces
Exercise No. 1:
(Reference RK Dhawan (2011), Applied Mechanics, S. Chand and Company Ltd, New Delhi)
- Define Mechanics and Engineering Mechanics. What are the different classes of Engineering Mechanics?
- Describe the various systems of units. Which systems of units are being followed in India these days and why?
- Define Mass and Weight. How do they differ from each other?
- What do you understand by 1 Newton force? state how force is expressed in gravitational units.
- Define force and give units in which it is measured.
- Define the term force and state clearly the effect of force.
- What do you mean by a force system? Explain the various force systems.
Exercise No. 2:
(Reference RK Dhawan (2011), Applied Mechanics, S. Chand and Company Ltd, New Delhi)
Fill in the blanks with appropriate words and rewrite
- The branch of Engineering Mechanics which deals with the action of forces on bodies at..............is called statics.
- The body in which the distance between any two particles remains constant is known as .......body
- The force with which the body is attracted to the earth's center is called.............................
- The quantity of matter contained in a body is called ...................
- The
action in the form of ............. or ....... which produces or tends
to produce, stops or tends to stop the motion of a body is
called................
- The unit of force in the SI system is .................
- One newton is that much force produces an acceleration of 1 m/sec^2 in a mass of ................
Newton's First Law of Motion
Statement
An
object at rest stays at rest and an object in motion stays in motion
with the same speed and in the same direction unless acted upon by an
unbalanced force.
This statement composes of two parts;
- which predicts the behavior of stationary objects (v= 0 m/s; a = 0 m/s^2) stay rest.
- which predicts the behavior of moving loads (v≉ 0 m/s; a = 0 m/s^2) stay in motion with same speed and direction.
For example:
The
water would have a tendency to spill from the container during specific
locations on the track. In general, the water gets spilled when;
- the container was at rest and you attempted to move it.
- the container was in motion and you attempted to stop it.
- the container was moving in one direction and you attempted to change its direction
Newtons Law of Motions - (Video Lecture)
Newton's Second Law of Motion
Statement
The
acceleration of an object as produced by a net force is directly
proportional to the magnitude of the net force, in the same direction as
the net force, and inversely to the mass of the object.
Mathematically,
Fnet = m ⨯ a; a = Fnet / m
where Fnet is applied force, m is the mass of the body, and a is an acceleration
If
the body is subjected to multiple forces at the same time, then the
acceleration produced is proportional to the vector sum all the
individual forces.
F1+F2+F3+..............+Fn = ma
Fnet = ma = m (dv/dt) = d(mv)/dt = dM/dt
Hence, Newtons second law also states that the net force acting on body is equal to the rate of change of momentum of the body.
Newton's Third Law
Statement
For every action, there is an equal and opposite reaction
This statement means that in every interaction, there is a pair of forces
acting on the two interacting objects. The size of the forces on the
first object equals to the size of the force on the second objects. The
direction of the first force on the object is opposite to the direction
of the force on the second object. Forces always come in pairs-equal and
opposite action-reaction force pairs.
For example:
The flying motion of birds
A
bird flies by use of its wings, The wings of bird push air downwards.
Since forces results from mutual interactions, the air must also be
pushing the bird upwards. The size of the force on the air equals to the
size of the force on bird; the direction of the force on the air
(downwards) is opposite to the direction of the force on the bird
(upwards). For every action, there is an equal ( in size) and opposite
(in direction) reaction. Action-reaction for pairs make it possible for
bird to fly.
Law of Transmissibility
Statement
The
state of rest or uniform motion of the rigid body is unaltered if a
force acting on the body is replaced by another force of the same
magnitude and direction but acting anywhere on the rigid body along the
line of action of the replaced force.
Let
F be a force acting on the rigid body on point A, as shown in figure 6.
According to the law of transmissibility, this force has the same effect on
the state of the body as the force F applied at point B along the line
of action of the force F.
Resultant Force
The
overall or combined effect of a system of forces on a body is called the resultant (R). The resultant is a single force. If it is zero the body remains at rest as in its original state, if it was originally in motion, the motion remains unaltered. The body is said to be in equilibrium. If the resultant is not zero the body will a varying state of motion.
Equilibrant
It
is a force that is equal in magnitude and opposite in direction to the
resultant force. The equilibrant brings the body to the condition that
existed when the forces were not acting.
Free Body Diagram (FBD)
A diagram that shows the forces on the body free of other bodies is called the free body diagram.
Addition of Two Forces:
Composition of forces
The
process of finding out the resultant forces, of a number of given
forces are called as the composition of forces or compounding of forces.
Methods for the resultant force
- Analytical Method - Parallelogram law of forces, Triangle law of forces, Polygon law of forces
- Method of Resolution
Resolution of forces
The single force acting on a particle can be replaced by two or more forces
which (together) have the same effect on the particle as that produced
by single force (alone). These forces are called a component of the
original force and the process are called a resolution of forces.
Principle of Resolution
The algebraic sum of the resolved parts of a number of forces, in a given the direction is equal to the resolved part of their resultant in the same
direction.
Methods of resolution
- Resolution of forces into two components in any direction
- Resolution of forces into two components in the right angle to each other (rectangular)
Resolution of forces into two components in any direction
∠AOC =𝞪
∠BOC = 𝛃
Now using sine rule;
(OA/sin𝛃)+(AC/sin𝞪)+(F/sin(𝞪+𝛃);
(P/sin𝛃)+(Q/sin𝞪)+(F/sin(𝞪+𝛃);
Hence,
P = (F/sin(𝞪+𝛃)/ sin𝛃
Q = (F/sin(𝞪+𝛃)/ sin𝞪
Resolution of forces into two components in the right angle to each other (Rectangular)
∠AOC = 𝛉
Now using triangle rule;
cos𝛉 = (OA/OC); OA = OC cos𝛉; P = F cos𝛉
sin𝛉 = (OB/OC); OB = OC sin𝛉; Q = F sin𝛉
Sign Conventions
Example
N-1: Resolve the forces shown in figure in two component parts
Solution:
FAX = 5cos30 = -4.33N FCX = 10cos70 = 3.42N
FAY = 5sin30 = -2.50N FCY = 10sin70 = 1.73N
FBX = 7cos40 = 5.36N FDX = 8cos20 = -7.51N
FBY = 7sin40 = -4.5N FDY = 8sin20 = 2.73N
N-2: Resolve the force of 2000N in two components P and Q as shown in figure
Solution:
∠FOP =𝞪
∠QOF = 𝛃
Now using sine rule;
(P/sin𝛃)+(Q/sin𝞪)+(F/sin(𝞪+𝛃);
Hence,
P = (F/sin(𝞪+𝛃)/ sin𝛃 = [2000 x sin38⁰]/ sin (38⁰+15⁰) = 1541.78 N
Q = (F/sin(𝞪+𝛃)/ sin𝞪 = [2000 x sin15⁰]/ sin (38⁰+15⁰) = 648.15 N
Analytical Method
Triangle Law
Statement
If two forces acting on a particle are represented in magnitude and
direction by the two adjacent sides of a triangle taken in order, then their resultant
will be given by the third side of a triangle taken in the opposite direction.
Figure 6: Triangle Law of Forces
Let, force P and Q act at angle 𝛉 as shown in the figure 6 space diagram. Now applying the head to the tail method, to construct the triangle. In figure 6 vector diagram ab and bc, represent P and Q in magnitude and direction. The closing side of the
triangle taken in reversed order represents the resultant R of the
forces P and Q. The magnitude and direction of R is found by using Sine
and cosine laws of triangle.
R2 = P2 + 2PQcos𝛃 + Q2
[P/sin𝝰 ]+[Q/sin𝝲]+[R/sin𝝱]
Parallelogram Law
Statement:
If two forces acting at a point are represented in
magnitude and direction by the adjacent side of parallelogram, the diagonal
passing through their point of intersection represents the resultant both in
magnitude and direction [Varignon and Newton, 1687].
ANALYTICAL METHOD
Figure 7: Parallelogram Law of Forces
Let two forces P and Q act at a point as shown in
figure 7. The forces P and Q are represented in magnitude and direction by OC
and OA. Let the angle between the two forces be θ. A parallelogram is drawn
with OA and OC as its adjacent sides. The resultant R is represented by OB in
magnitude and direction. Its value can be calculated as follows;
Let D be the foot of perpendicular dropped from
point B on extension of line OC.
Let ‹AOC = ‹BCD =θ
In triangle BCD
BC = AO = Q
OC = AB = P
CD = BCcosθ = Q cosθ
BD = BCsinθ = Q sinθ
In triangle OBC
OB2=OD2 + BD2
R2 = (P + Q cosθ)2 +(Q sinθ)2
R2 = P2 + 2PQcosθ + Q2cos2θ
+Qsin2θ
R2 = P2 + 2PQcosθ + Q2(cos2θ
+sin2θ)
R2 = P2 + 2PQcosθ + Q2
The angle between the horizontal and resultant is
found as;
𝛼
=tan⁻¹ (BD/OD) = tan⁻¹ (BD/OC+CD) = tan⁻¹ (Q sinθ/P+Q cosθ)
Polygon Law of Force
Statement
It is an extension of triangle law of force for more than two forces which states that, "
if a number of forces acting simultaneously on a particle, be
represented in magnitude and direction, by the sides of a polygon taken
in order; then the resultant of all these forces may be represented in
magnitude and direction by closing side of the polygon, taken in
opposite order".
Exercise No. 3:
- State and explain Newton's law of gravitational attraction.
- State and explain Newton's Three Fundamental Laws
- What do you understand about law of transmisibility? Explain it.
- Define Resultant Force and Equillibriant Force
- What do you understand about composition and resolution of forces
- State and explain parallelogram law of forces
Graphical Solution of Law of Forces
Experiment No. 1
Verification of triangle law of forces and Parallelogram law of forces
Objective
To
verify the triangle law of forces with the help of Gravesend's apparatus.
Apparatus
Gravesend's apparatus,
paper sheet, weights, thread, drawing pine, mirror strip, three scale pans, set
square, pencil, etc.
Procedure
- Fix
the paper sheet with drawing pins on the board set in a vertical plane.
- Pass
one thread over the pulleys carrying a pan (for adding weight during the performance of the experiment) at its end.
- Take
a second thread and tie its one end in the middle of the first thread and tie a
pan at its other end.
- Add weights in the pans in such a manner that
the small knot comes approximately in the center.
- Displace slightly the pans from their position of equilibrium
and note if they come to their original position of rest. This will ensure the free
movement of pulleys.
- Take the mirror strip and place under each thread lengthwise and
mark two points keeping the eye, the thread and its image in the same line
without disturbing the system.
- Mark
lines of forces and write the magnitude of forces.
- Remove
the paper from the board and produce the lines to meet at O.
- Use
Bow’s notation to name the forces P, Q, and R as AB,
BC and CA.
- Select a suitable scale and draw the line ab parallel to
force P and cut it equal to the magnitude of P. From b draw
the line bc parallel to force Q and cut it equal to the magnitude
of Q (Fig. 8). Calculate the magnitude of ca i.e., R1 which
will be equal to the third force R which proves the triangle law of
forces. If R1 differs from the original magnitude of R, then
percentage error is found as follows;

Figure 8: Gravesand's Apparatus (Parallelogram Law setup)

Figure 9: Gravesand's Apparatus (Polygon Law setup)
Parallelogram
law of forces
Cut OA = P and
OB = Q. From A draw AC’ parallel to OB’ and BC
parallel to OA. The resultant of forces P and Q will
be shown by R1 which will be equal to R.(Note that R and R1 are in the
opposite direction)
Observations
Scale.........
Sr. No
|
Total weight of pan “P”
|
Total weight of pan “Q”
|
Total weight of pan “R”
|
Calculate resultant R1
|
Avg. R1 |
|
|
|
|
|
|
Precautions
- Pans/weights
should not touch the vertical board.
- There
should be only one central knot on the thread which should be small.
- While
calculating the total force in each case the weight of the pan should be added
to the weights put into the pan.
- Make
sure that all the pans are at rest when the lines of action of forces are
marked.
- All
the pulleys should be free from friction.
Experiment No. 2
Verification of Lami's Theorem
Objective
To
verify the Lami's Theorem of forces with the help of Gravesend's apparatus.
Apparatus
Gravesend's apparatus,
paper sheet, weights, thread, drawing pine, mirror strip, three scale pans, set
square, pencil, etc.
Theory
According to Lami's Theorem,
"If
a body is in equilibrium under the action of three forces, each force
is proportional to the sine of the angle between the other two forces"
Thus, if three forces P, Q, and R acting at a point O are in equilibrium, Let the angle between their line of actions be α, β, and γ as shown in the figure.then according to lamis theorem
P:Q:R::sin𝛂 :sin𝛃:sin𝛄
OR
[P/sin𝝰 ]+[Q/sin𝛃]+[R/sin𝛄]
These three forces are represented completely in magnitude and direction by OA, OB, and OD respectively. Complete the parallelogram OACB. The diagonal OC represents the resultant of force P and Q. The resultant OC must be equal and opposite to the third force OD (R), as the points are in equillibrium.
Vector OA = BC = P
Vector OB = AC = Q
Vector OC = OD =R
from △ OBC
∠BOC = 180°-𝛼
∠OCB = 180°-β
In △ OBC
∠OBC = 180°-(180°-𝛼)-(180°-β)
= 𝛼+β - 180°
But 𝛼+β+ɣ = 360°
∠OBC = 180°-ɣ
BC/sin ∠BOC = OB/sin∠OCB = OC/sin∠OBC
P/sin (180°-𝛼) = Q/sin (180°-β) = R/sin (180°-ɣ)
P/sin𝛼 = Q/sinβ = R/sinɣ
Procedure
- Fix
the paper sheet with drawing pins on the board set in a vertical plane.
- Pass
one thread over the pulleys carrying a pan (for adding weight during the performance of the experiment) at its end.
- Take
a second thread and tie its one end in the middle of the first thread and tie a
pan at its other end.
- Add weights in the pans in such a manner that
the small knot comes approximately in the center.
- Displace slightly the pans from their position of equilibrium
and note if they come to their original position of rest. This will ensure the free
movement of pulleys.
- Take the mirror strip and place under each thread lengthwise and
mark two points keeping the eye, the thread and its image in the same line
without disturbing the system.
- Mark
lines of forces and write the magnitude of forces.
- Remove
the paper from the board and produce the lines to meet at O.
- Use
Bow’s notation to name the forces P, Q, and R as AB,
BC and CA.
- Select a suitable scale and draw the line ab parallel to
force P and cut it equal to the magnitude of P. From b draw
the line bc parallel to force Q and cut it equal to the magnitude
of Q.
- measure the angle α, β, and γ. Now by using Lami’s theorem calculate the magnitude of the third force R . If R1 differs from the original magnitude of R, then
percentage error is found as follows;
Observations
Scale.........
Sr. No
|
Total weight of pan “P”
|
Total weight of pan “Q”
|
Total weight of pan “R”
|
Calculate resultant R1
|
Avg. R1 |
|
|
|
|
|
|
Precautions
- Pans/weights
should not touch the vertical board.
- There
should be only one central knot on the thread which should be small.
- While
calculating the total force in each case the weight of the pan should be added
to the weights put into the pan.
- Make
sure that all the pans are at rest when the lines of action of forces are
marked.
- All
the pulleys should be free from friction.
Numerical
N-2:
A body is supported to pull of 50 N and 100 N, if the angle between
them is 70⁰. Determine the resultant magnitude and direction
solution:
R= √(
P2 + 2PQcosθ + Q2)
R= √(
502 + 2×50×100 cos(70)+ 1002)
R= 126.175 N
𝛼 = tan⁻¹ (Q sinθ/P+Q cosθ)
𝛼 = tan⁻¹ (100 sin(70)/50+100 cos(70))
𝛼 =48.15⁰
N-3:
Two forces of 100 N and 80 N respectively having included angle 135⁰
are acting on a particle. Find the resultant in magnitude and direction,
when:
- Both the forces are pull
- The force 100 N is pull but that of 80 N is push
Solution:
Case 1:
R= √(P2 + 2PQcosθ + Q2)
R= √(
1002 + 2×80×100 cos(135)+ 802)
R= 71.318 N
𝛼 = tan⁻¹ (Q sinθ/P+Q cosθ)
𝛼 = tan⁻¹ (80 sin(135)/100+80 cos(135))
𝛼 =52.48⁰
Case 2:
R= √(P2 + 2PQcosθ + Q2)
R= √(
1002 + 2×80×100 cos(45)+ 802)
R= 166.47 N
𝛼 = tan⁻¹ (Q sinθ/P+Q cosθ)
𝛼 = tan⁻¹ (80 sin(45)/100+80 cos(45))
𝛼 =19.86⁰
N-4: Determine the resultant of two equal forces P and Q when angle between them is;
- 0⁰ 5. 90⁰
- 30⁰ 6. 120⁰
- 45⁰ 7. 150⁰
- 60⁰ 8. 180⁰
Solution:
Case 1:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(0)+ P2)
R= √(4
P2)
R= 2P N
Case 2:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(30)+ P2)
R= √(3.732
P2)
R= 1.932P N
Case 3:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(45)+ P2)
R= √(3.
414P2)
R= 1.847P N
Case 4:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(60)+ P2)
R= √(3
P2)
R= 1.732P N
Case 5:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(90)+ P2)
R= √(2
P2)
R= 1.414P N
Case 6:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(120)+ P2)
R= √(
P2)
R= P N
Case 7:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(150)+ P2)
R= √(0.268
P2)
R= 0.517P N
Case 8:
R= √(P2 + 2PQcosθ + Q2)
R= √(
P2 + 2×P×P cos(180)+ P2)
R= √(
0P2)
R= 0 N
N-5:
Find the magnitude of two forces, P and Q such that if they act at
right angle their resultant is √34 N. If they act at 60⁰ their resultant
is 7N.
Solution:
R2 = P2 + 2PQcosθ + Q2
(√34)² = P2 + 2PQcosθ + Q2
34 = P2 + Q2
R2 = P2 + 2PQcosθ + Q2
72 = P2 + 2PQcos60⁰ + Q2
49 = P2 + Q2+ PQ 49 = 34 + PQ
15 = PQ
Q = 15/P
P2 + (15/P)2= 34
P=5N; Q=3N
N-6:
Find the greatest and smallest resultant of two forces whose magnitude
are 20N and 15N respectively by making the suitable assumptions.
Solution:
R= √(
P2 + 2PQcosθ + Q2)
R= √(
202 + 2×20×15 cos(θ)+ 152)
For R to be maximum, cosθ should be maximum, θ = 0
R= √(202 + 2×20×15 cos(0)+ 152)
R = 34 N
For R to be minimum, cosθ should be maximum, θ = 180
R= √(202 + 2×20×15 cos(180)+ 152)
R = 5 N
N-7:
Determine the angle between two like forces P. When their resultant is; (a) P/4, (b) P/2, and (c) P.
Solution:
Case a:
R2 = P2 + 2PQcosθ + Q2
(P/4)2 = P2 + P2cosθ + P2
(P/4)2 = 2P2 (1+ cosθ)
(1/32) = (1+ cosθ)
cosθ = -(1-1/32) = -31/32
θ = 165.63⁰
Case b:
R2 = P2 + 2PQcosθ + Q2
(P/2)2 = P2 + P2cosθ + P2
(P/2)2 = 2P2 (1+ cosθ)
(1/8) = (1+ cosθ)
cosθ = -(1-1/8) = -7/8
θ = 151.044⁰
Case c:
R2 = P2 + 2PQcosθ + Q2
(P)2 = P2 + P2cosθ + P2
(P)2 = 2P2 (1+ cosθ)
(1/2) = (1+ cosθ)
cosθ = -(1-1/2) = -1/2
θ = 180⁰
N-8:
The sum of the two forces is 9N their resultant which is perpendicular
to the smaller is 6N. Find the magnitude of the two forces and angle
between them.
Solution:
P+Q = 9N
tan𝛼 = (Q sinθ/P+Q cosθ)
(sin𝛼 /cos𝛼) = (Q sinθ/P+Q cosθ)
(sin 90⁰ /cos 90⁰) = (Q sinθ/P+Q cosθ)
P+Q cosθ = 0
cosθ = -P/Q
R2 = P2 + 2PQcosθ + Q2
R2 = P2 + 2PQ(-P/Q ) + Q2
36 = Q2-P2
36 = (Q+P)(Q-P)
36 = 9(Q-P)
Q-P = 4N
from equations
P+Q = 9N
Q-P = 4N
we get P = 2.5N and Q = 6.5N
cosθ = -P/Q = -(2.5/6.5)
θ = 112.62⁰
N-9:
The resultant of the two forces is 180 kN and its direction is inclined
at 22⁰ with one of the force whose magnitude is 125 kN. Determine the
other force and angle between the two forces.
Solution
tan𝛼 = (Q sinθ/P+Q cosθ)
(sin𝛼 /cos𝛼) = (Q sinθ/P+Q cosθ)
cos 22⁰ = P+Q cosθ
Q cosθ = cos 22⁰-P
Q cosθ = cos 22⁰- 125
Q cosθ = -124.07
R2 = P2 + 2PQcosθ + Q2
1802 = 1252 + 2 x125 x -124.07 + Q2
Q = 218.61 kN
cosθ = -124.07/218.61
θ = 124.54⁰
N-10:
Add the following forces in a plane; 150N at 30⁰ and -90N at 90⁰.
Solution
R2 = P2 + 2PQcosθ + Q2
R2 = 1502 + 2×150×90cos120⁰ + 902
R = 130.76 N
tan𝛼 = (Q sinθ/P+Q cosθ)
𝛼 = tan⁻¹ (90 sin120⁰/150+90cos120⁰)
𝛼 = 36.58⁰
Resultant of ;
Co-linear Forces
R = ∑ Fⱼ = F1+F2+F3+F4
R = 7+9+10-12 = 14 kN
Coplanar Concurrent Forces
- Resolve
all forces in components that coincide with two arbitrary chosen axes
which are mutually perpendicular (in general x-axes and y axes)
- Take summation of all component for in respective axes i.e ∑ Fₓ and ∑ Fy
- Resultant force equals to √(∑ F²ₓ + ∑ F²y )
- Direction 𝛼 = tan⁻¹ ( ∑ Fy / ∑ Fₓ)
Example:
∑ Fₓ = P cosθ1 + Q cosθ2 + S cosθ3
∑ Fy = P sinθ1 + Q sinθ2 + S sinθ3
N-11: The following forces act at a point;
- 30 N inclined at 30⁰ towards North-East
- 25 N towards North
- 30 N towards North-West and
- 35 N inclined at 40⁰ towards South-West
Find the magnitude and direction of the resultant force.
Force θx Fₓ = Fcos θx Fy = Fsinθx
20N 30 17.32N 10N
25N 90 00.00N 25N
30N 45 - 21.21N 21.21N
35N 40 -26.81N -22.50N
∑Fₓ =-30.70N ∑Fy =+33.71N
R = √(∑ F²ₓ + ∑ F²y )
R = √ (-30.70)² + (33.715)²
R = 45.60 N
𝛼 = tan⁻¹ ( ∑ Fy / ∑ Fₓ)
𝛼 = tan⁻¹ ( 33.715 / -30.70)
𝛼 = -47.67º
actual 𝛼 = 180-47.67º = 132.30º
N-12: Determine the resultant of the following force system shown in
figure
∑ Fₓ =
+5cos30͐º+10cos60͐º+12cos40͐º-4-15cos60͐º-12sin40͐º = -0.69 N
∑Fy =
+5sin30͐º+10sin60͐º+12sin40͐º-4-15sin60͐º-12cos40͐º = -8.35 N
R = √(∑ F²ₓ
+ ∑ F²y )
R = √ (-0.69)² + (-8.35)²
R = 8.37 N
𝛼 = tan⁻¹
( ∑ Fy / ∑ Fₓ)
𝛼 = tan⁻¹
( -8.35 /
-0.69)
𝛼 = 85.287º
N-13: Determine (a) angle𝞪 for which resultant of three forces is vertical, (b) also find the corresponding magnitude of resultant.
For resultant to be vertical;
let ∑ Fₓ =0
∑ Fₓ =
+40 + 40sin𝛼 -80cos⍺ = 0
40(1 + sin𝛼 ) = 80cos⍺
(1 + sin𝛼 ) = 2cos⍺
(1 + sin𝛼 )² = (2cos⍺)²
(1 + 2sin𝛼 + sin²𝛼) = 4(1-sin²𝛼)
5sin²𝛼 + 2sin
𝛼-3 = 0
𝛼 = 36.87° or 90°
∑Fy =
40cos𝛼 -80sin⍺ = -R
40cos36.87° -80
sin36.87° =
-R
R = 80 N (↓)
Moment (M)
It is the turning effect produced by a force. It is the product of the magnitude of the force and perpendicular distance about that point. The point about which moment is taken is known as the moment center.
Moment (M) = Force x Perpendicular distance = F x d
Units are kNm, Nm, etc.
Sign Convention: Clockwise (+ve) and anti-clockwise (-ve)
Varignon's Theorem (Law of Moments)
For any system of forces acting in the space, the algebraic summation of the moment of all forces about a point is equal to the moment of resultant about the same point
Let,
P and Q are the force acting at a point "A"
R be the resultant of P and Q
Moment of 'P' about "O" is
Mp = P x d₁
Moment of 'Q' about "O" is
Mq = Q x d₃
Moment of 'R' about "O" is
Mr = R x d₃
Mp + Mq = (P x d₁) + (Q x d₃)
= (P x OA cosθ₁) + (Q x OA cosθ₃)
= [(P cosθ₁) + (Q cosθ₃)] x OA
= [Rcosθ₃] x OA
= Rd₃
Mp + Mq = Mr
Gravity Axis of Body:
(Reference Sanju Unadkat (2017), Engineering Mechanics, Tech-Max Publication, Pune)
The gravity axis of the body is the line of action of the gravitational force that acts on the body.
Center of gravity:
The center of gravity of the body is the point of intersection of all gravity axes of the body
Center
of gravity of a body is the point through which results of the
distributed gravity forces act irrespective of the orientation of the
body
Consider body be made up of an infinite number of particles having weights W1, W2, W3,......, Wn. Let the system be observed in the x-y plane. Let (x₁,y₁), (x₂, y₂), (x₃ y₃),....., (xₙ,yₙ). Let (x,y) be the coordinates of resultant gravity for Wr of all these gravity forces acting as parallel forces.
CG about x-axis and y-axis is given by:
⃑⃑x = (W₁x₁ + W₂x₂ + W₃x₃+....+Wₙxₙ)/ (W₁ + W₂ + W₃+....+Wₙ)
⃑y = (W₁y₁ + W₂y₂ + W₃y₃+....+Wₙyₙ)/ (W₁ + W₂ + W₃+....+Wₙ)
Centroid
The point through which whole area of the plane lamina is supposed to act is termed as centriod. Consider a lamina is made of infinite number of elemental areas A₁ (x₁,y₁), A₂ (x₂,y₂), A₃ (x₃,y₃),.......,Aₙ (xₙ,yₙ),
⃑⃑x = (A₁x₁ + A₂x₂ + A₃x₃+....+Aₙxₙ)/ (A₁ + A₂ + A₃+....+Aₙ)
⃑y = (A₁y₁ + A₂y₂ + A₃y₃+....+Aₙyₙ)/ (A₁ + A₂ + A₃+....+Aₙ)
Methods
of Determining Centro of Gravity
The center of gravity of a lamina can be determined
by the following method
- By
Geometrical Method
- By
the Method of Moments
- By
Graphical Method
- By
Integration Method
The position of center of gravity of many bodies can be found by inspection, that is geometric center the method is called as geometric method. Below are some standard examples of plane lamina's with their geometric centroid.
In the methods of
moment for determining the center of gravity principle of moments is applied.
According to this principle,
“ If a lamina is a
combination of a number of basic section, the algebraic sum of the moments of
the individual areas of sections about any axis is equal to moment of whole
area of the lamina about the same axis”.
The following steps
should be taken into consideration while solving the problems on center of
gravity of composite section
- Choose the two mutually perpendicular
axis OX and OY. The axes should be so selected that they touch or pass through
the nearest line of the figure.
- Check whether the section is symmetric
about any X-X axis or Y-Y axis, the CG will lie on that axis
- The composite figure is divided into
small areas such as square, rectangle, circles, etc.
- Then find the position of CG by
calculating the X and Y
- If the area is removed from the figure
such as hole, it should be subtracted from the original area and the moment
should be treated as negative
C1: Centroid of
Triangle
Consider a triangular
section of base “b” and height “h” as shown in figure. Consider elemental strip
of thickness ‘dx” and at a distance “y” from apex.
Width of elemental
strip “x” = (by/h)
Area of elemental strip
dA = (b/h) ydy
Moment of area of
elemental strip about apex = dA * y
= (b/h)*y*y*dy
= (b/h)*y2*dy
for the moment of area of area about the centroid of section, integrate for limit "0" to "h"
C 2: Centroid of Semi-circle
Consider a semi-circle of radius "r". Let POQ be a elemental of an angle "dθ" inclined at an angle θ with respect to positive direction of X - axis and ∠POQ =dθ. Length of the elemental sector PQ = ds = rdθ.
As dθ is very small so ds is treated as straight line and sector is considered as right angle traingle.
Area of elemental sector OPQ is,
dA = 1/2*r*rdθ
Distance of center of gravity of elemental sector along PQ = 2/3*rsinθ
Moment of area of elemental sector dA y = r²/2 *dθ*2/3*rsinθ
= r³/3*sinθ*dθ
for moment of area of the complete semi-circle integrate between 0 to 𝜋
Experiment No. 3
Center of Gravity of Regular Lamina
Objective
To find the center of gravity of regular lamina
Apparatus
Paper sheet, drawing pin or tape. set squares, scale cards, etc
Theory
Every body is attracted towards the center of the earth due to gravity. The force with which the body is attracted towards earth is called weight of the body. the whole weight of the body is supposed to act at a point where the whole area of a lamina is supposed to act is also known as centroid or center of gravity
Procedure
The center of gravity of any lamina may be founded by adopting the following steps:
- Choose the two mutually perpendicular
axis OX and OY. The axes should be so selected that they touch or pass through
the nearest line of the figure.
- Check whether the section is symmetric
about any X-X axis or Y-Y axis, the CG will lie on that axis
- The composite figure is divided into
small areas such as square, rectangle, circles, etc.
- Then find the position of CG by
calculating the X and Y
- If the area is removed from the figure
such as hole, it should be subtracted from the original area and the moment
should be treated as negative
Observation:
Sr. No
|
Areas in “mm2”
|
Distance from OX axis in mm
|
Distance from OY axis in mm
|
Position of CG from OX
|
Position of CG from OY
|
|
|
|
|
⃑⃑x = (A₁x₁ + A₂x₂ + A₃x₃+....+Aₙxₙ)/
(A₁ + A₂ + A₃+....+Aₙ)
|
⃑y = (A₁y₁ + A₂y₂ + A₃y₃+....+Aₙyₙ)/
(A₁ + A₂ + A₃+....+Aₙ)
|
Precautions
- Calculate the areas accurately
- Find the CG position from standard relationships
Types of Loads
Concentrated Load or Point Load
When the load is acting at a single point on the structure is called point load or concentrated load
Uniformly Distributed Load
When the intensity of load is uniformly distributed over a certain length of structure is said to be uniformly distributed load. Expressed as load per unit run ( w/run)
Uniformly Varying Load
When the intensity of load varies over the certain length of the structure is said to be uniformly distributed load. Expressed as load per unit run ( w/run)
Couple
when two equal, opposite, parallel and non-colinear forces act on body forms couple. Expressed as force (F) multiplied by the perpendicular distance (d)
Couple = F x d
WORK,
ENERGY AND POWER
Work
The work done by a force on a
moving body is defined as the product of the force and the distance moved in
the direction of the force. It is abbreviated with a symbol (W). Figure
below illustrates the work done by a force.

Mathematically, work done by a
force is;
F = force × distance = F x S
The unit of work depends upon the
units of the force and distance. In SI units work done is represented as Nm,
which is also termed as Joule (J). One
joule may be defined as the work done by 1 N force, when it displaces the body
through 1m; mathematically 1 joule (J) = 1Nm.
The other commonly used units are;
Kilo joules (KJ) kNm, mille joules (MJ) N-mm, etc.
Work
done by a constant force
The work done
by a constant force may be defined as the
product of the component of the force in the direction of displacement and the
distance moved. Mathematically;
W = Fcosθ x d
Figure (a), demonstrates the work done by a
constant force. When a force acts on a body in the direction of
motion, the work done by the force is taken to be positive, but if the force
acts in opposite to the direction of motion, the work done by the force is
taken to be negative. Figure (b), depicts that as long as the person shown in picture
does not lift or lower the bag of groceries, he is doing no work on it. The
force he exerts, has no component in the direction of motion. 
Graphical
representation of the work done
The work done,
during any operation, can also be represented by a graph, by plotting distance
along x-axis and the force along y-axis. The work done, is equal to the product
of the force and distance, therefore the area under the figure (a) and (b),
represents the work done to some scale, such diagrams are called force-distance
diagrams. Figure (a), represents the work done by a constant force, force
diagram appears to be rectangular. Figure (b), represents the work done by a
variable force, force diagram appears to be trapezium.
Exercise
1. A
trolley of mass 200 kg moves on a level track for a distance of 500 m. If the
resistance of the track is 100 N, find the work done in moving the trolley.
Solution:
Given data; m = 200 kg; d = 500 m, Track
resistance = 100 N
Work done =
Force x distance
= Track resistance x distance
= 100 x 500 = 50,000 Nm or 50
kJ
2.
A horse pulling a cart
exerts a steady horizontal pull of 250 N and walks at the rate of 5 km/h. How
much work is done by the horse in 7 min.?
Solution: Given
data; P = 250 kN, velocity (v) = 5 km/h, = 83.33 m/min, time (t) = 7 min.
The distance
traveled in 7 min = 83.33 x 7 =583.33 m
Work done = Force x distance
= 250 x 583.33 = 145833.33 N-m = 145.833
kNm or kJ
3. A
spring is stretched by 50 mm by the application of a force. Find the work done, if
the force required to stretch 1 mm of the spring is 8 N.
Solution: Force
required stretching the spring by 50 mm = 50 x 8 = 400 N.
Average force = 400/2 = 200 mm
Work done = Average force x distance
= 200 x 50 = 10000 Nm or 10 J
4. A
force
acts on a particle. Find the work done by the
force during the displacement of the particle from x = 0 m to x = 3
m. Given that force is in N.
Exercise
(Home Work)
1. How
much work is done when a force of 5 kN moves its point of application 550 mm in
the direction of force?
2. Find the work done in raising 100 kg of water
through a vertical distance of 2 m.
3. A
block weighing 400 N is resting on a smooth horizontal surface. What work will
be done if the block is to be moved through 2 m distance by applying (a) A horizontal force of
150 N and (b) A force of 200 N whose
line of action makes an angle of 300 with the horizontal
4.
When is work done by a
force is positive and when it is negative
5. A
man carrying a bucket of water is walking on a level road with a uniform
velocity. Does he do any work on the bucket while carrying it?
Energy
Energy is defined as the
ability to do the work. It exists in many forms as shown in figure