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The Project deals with the development of the computerized system for maintaining the regular records and services that are undertaken in the furniture business. This project titled "Web Based Integrated Furniture Showroom Management System" has been aimed to design and computerized system that can handle various activities are been carried out at the Furniture Showroom. This application has been developed using PHP Programming Language as its front end and the back end is MYSQL Server In the existing system all the activities and record maintenance of the furniture showroom are done manually by the manager. The Project deals with the development of the computerized system for maintaining the regular records and services that are undertaken in this most important and large business oriented furniture business. This Project also enables the users to perform all the day to day business operations in the furniture showroom business most efficiently.
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In present era, the scopes of information technology growing with a very fast .We do not see any are untouched from this industry. The scope of information technology has become wider includes: Business and industry. Household Business, Communication, Education, Entertainment, Science, Medicine, Engineering, Distance Learning, Weather Forecasting. Carrier Searching and so on. My project named “Event Management System” is software that store and maintained all events coordinated in college. It also helpful to print related reports. My project will help to record the events coordinated by faculties with their Name, Event subject, date & details in an efficient & effective ways. In my system we have to make a system by which a user can record all events coordinated by a particular faculty. In our proposed system some more featured are added which differs it from the existing system such as security.
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NO1 Pandit Amil Baba In Bahawalpur, Sargodha, Sialkot, Sheikhupura, Rahim Yar Khan, Jhang, Dera Ghazi Khan, Gujrat +92322-6382012
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Laundry firms currently use a manual system for the management and maintenance of critical information. The current system requires numerous paper forms, with data stores spread throughout the laundry management infrastructure. Often information is incomplete or does not follow management standards. Records are often lost in transit during computation requiring a comprehensive auditing process to ensure that no vital information is lost. Multiple copies of the same information exist in the laundry firm data and may lead to inconsistencies in data in various data stores. A significant part of the operation of any laundry firm involves the acquisition, management and timely retrieval of great volumes of information. This information typically involves; customer personal information and clothing records history, user information, price of delivery and received date, users scheduling as regards customers details and dealings in service rendered, also our products package waiting list. All of this information must be managed in an efficient and cost wise fashion so that the organization resources may be effectively utilized. We present the design and implementation of a laundry database management system (LBMS) used in a laundry establishment. Laundry firms are usually faced with difficulties in keeping detailed records of customers clothing; this little problem as seen to most laundry firms is highly discouraging as customers are filled with disappointments, arising from issues such as customer clothes mix-ups and untimely retrieval of clothes. The aim of this application is to determine the number of clothes collected, in relation to their owners, as this also helps the users fix a date for the collection of their clothes. Also customer’s information is secured, as a specific id is allocated per registration to avoid contrasting information.
Laundry management system project report.pdf
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About Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol. • Remote control: Parallel or serial interface. • Compatible with MAFI CCR system. • Compatible with IDM8000 CCR. • Compatible with Backplane mount serial communication. • Compatible with commercial and Defence aviation CCR system. • Remote control system for accessing CCR and allied system over serial or TCP. • Indigenized local Support/presence in India. • Easy in configuration using DIP switches. Technical Specifications Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol. Key Features Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol. • Remote control: Parallel or serial interface • Compatible with MAFI CCR system • Copatiable with IDM8000 CCR • Compatible with Backplane mount serial communication. • Compatible with commercial and Defence aviation CCR system. • Remote control system for accessing CCR and allied system over serial or TCP. • Indigenized local Support/presence in India. Application • Remote control: Parallel or serial interface. • Compatible with MAFI CCR system. • Compatible with IDM8000 CCR. • Compatible with Backplane mount serial communication. • Compatible with commercial and Defence aviation CCR system. • Remote control system for accessing CCR and allied system over serial or TCP. • Indigenized local Support/presence in India. • Easy in configuration using DIP switches.
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The Project entitled “PHARMACY MANAGEMENT SYSTEM” is systematic approach made towards maintaining purchase details, stock details, order details, sales details, sales return details, payment details, buyer details, the record are manually maintained id’s provide lots of problem. Hence we need of data to be computerized.
Pharmacy management system project report..pdf
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The export maintenance system is a fully featured application that can help we manage fruit delivery business and achieve more control and information at a very low cost of total ownership. A fruit export maintains automatically monitors purchase, sales, supplier information. The system includes receiving fruit from the different supplier. Customer order is placed in the system, based on the order fruit has been sales to the customer. The report contains the details about product, purchase, sales, stock, and invoice. The main objective of this project is to computerize the company activities and to provide details about the production process at the fruit export maintenance system. The demand of fresh fruit fruits and processed food items in international and domestic market has shown a decent increase. This estimation is creating a necessity for growing more and more fruit fruits to cater the growing demand of domestic & international market. The customers effectively and hence help for establishing good relation between customer and fruit shop organization. It contains various customized modules for effectively maintaining fruit and stock information accurately and safely. When the fruits are sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting fruits for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item. The proposed project is developed to manage the fruit shop in the fruits for shop. The first module is the login. The admin should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
Fruit shop management system project report.pdf
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Kamal Acharya
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In this project presentation, I will show you the design and build a simple Car Speed Detector circuit using Arduino UNO and IR Sensors. This Arduino Car Speed Detector project can be used to detect speed of a moving car. this is a very simple project idea to present. dept. of electrical and electronic engineering, university of chittagong.
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A computer reservation system or central reservation system is a computerized system used to store and retrieve information and conduct transactions related to air travel, hotels, car rental, or activities. These systems typically allow users to book hotel rooms, rental cars, airline tickets as well as activities and tours. They also provide access to railway reservations and bus reservations in some markets, although these are not always integrated with the main system. For these systems to be accessible on mobile phones and computers outside the premises of the airport, cinema, train station or stadiums, they need to be on the internet or a network. This project focuses on the design and implementation of a web based cinema management system for the allocation of seat tickets online. The system would feature the registration of users, use of serial numbers and pins gotten from scratch cards sold and a printed slip. The system would have a store of all the seats and automate the generation of fresh serial numbers and pins.
A case study of cinema management system project report..pdf
A case study of cinema management system project report..pdf
Kamal Acharya
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C Sai Kiran
Teaching effects after 128 hours of Building Information Modeling course in Cracow, Poland. Natalia works in Revit, Navisworks and Dynamo for BIM Coordination position. More https://bim.edu.pl or https://bimedu.eu
Natalia Rutkowska - BIM School Course in Kraków
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HYDROPOWER - Hydroelectric power generation
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Robbie Edward Sayers
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PHP and MySQL project on Hall Booking System is a web based project and it has been developed in PHP and MySQL and we can manage Payment, Booking, Inventory, Booking Dates, Customers and Hall from this project. The main objective to develop Hall Booking System PHP, MySQL, JAVA SCRIPT and BOOTSRAP Project is to overcome the manual errors and make a computerized system. In this project, there are various type of modules available to manage Customers, Booking, Payment. We can also generate reports for Booking, Payment, Booking Dates, Hall. Here the Payment module manage all the operations of Payment, Booking module can manage Booking, Inventory module is normally developed for managing Inventory, Booking Dates module manages Booking Dates operations, Customers module has been implemented to manage Customers. In this project all the modules like Payment, Booking Dates, Booking are tightly coupled and we can track the information easily. Ifyou are looking for Free Hall Booking System Project in PHP and MySQL then you can visit our free projects section. We can easily get the list of wedding halls & lawns in Nagpur. Also we have detailed contact information for some particular hall. But we cannot get the availability about hall. So background behind this web portal is that it gives the area wise listing of wedding halls & lawns with the detailed information of individual and also display for particular date the hall is available or not. Just dial is the system in which we can only find the name of Hall and Lawns in city. In just dial we cannot find Halls in specific area. This system cannot show all information about any Hall. This system is not able to book the Halls online. The A Web Based Hall Booking Management System is designed to overcome the disadvantage of previous system.We can easily get the list of Wedding Halls. But we cannot get the availability about Hall. So background behind this web portal is that it gives the area wise listing of Wedding Halls with the detailed information of individual and also display for particular date the Hall is available or not. This is a special type of web portal to easily get the information of all Wedding Halls in Nagpur which display separate calendar for separate Hall. For particular date the Hall. We can availability of Hall as well as Lawns detailed information about individuals Hall in our web portal . It provides all facilities to clients with lowest cost and lowest maintenance problems.
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ASME IX(9) 2007 Full Version .pdf
ENERGY STORAGE DEVICES INTRODUCTION UNIT-I
ENERGY STORAGE DEVICES INTRODUCTION UNIT-I
Hall booking system project report .pdf
Hall booking system project report .pdf
Ch03
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VECTOR MECHANICS FOR
ENGINEERS: STATICS Eighth Edition Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 3 Rigid Bodies: Equivalent Systems of Forces
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Contents Introduction External and Internal Forces Principle of Transmissibility: Equivalent Forces Vector Products of Two Vectors Moment of a Force About a Point Varigon’s Theorem Rectangular Components of the Moment of a Force Sample Problem 3.1 Scalar Product of Two Vectors Scalar Product of Two Vectors: Applications Mixed Triple Product of Three Vectors Moment of a Force About a Given Axis Sample Problem 3.5 Moment of a Couple Addition of Couples Couples Can Be Represented By Vectors Resolution of a Force Into a Force at O and a Couple Sample Problem 3.6 System of Forces: Reduction to a Force and a Couple Further Reduction of a System of Forces Sample Problem 3.8 Sample Problem 3.10
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Introduction • Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. • Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. • Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. • moment of a force about a point • moment of a force about an axis • moment due to a couple • Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition External and Internal Forces • Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces • External forces are shown in a free-body diagram. • If unopposed, each external force can impart a motion of translation or rotation, or both.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Principle of Transmissibility: Equivalent Forces • Principle of Transmissibility - Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces. • Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. • Principle of transmissibility may not always apply in determining internal forces and deformations.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1. Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is 3. Direction of V is obtained from the right-hand rule. θsinQPV = • Vector products: - are not commutative, - are distributive, - are not associative, ( )QPPQ ×−=× ( ) 2121 QPQPQQP ×+×=+× ( ) ( )SQPSQP ××≠××
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Vector Products: Rectangular Components • Vector products of Cartesian unit vectors, 0 0 0 =×=×−=× −=×=×=× =×−=×=× kkikjjki ijkjjkji jikkijii • Vector products in terms of rectangular coordinates ( ) ( )kQjQiQkPjPiPV zyxzyx ++×++= ( ) ( ) ( )kQPQP jQPQPiQPQP xyyx zxxzyzzy −+ −+−= zyx zyx QQQ PPP kji =
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. • The moment of F about O is defined as FrMO ×= • The moment vector MO is perpendicular to the plane containing O and the force F. • Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment. • Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO. The sense of the moment may be determined by the right-hand rule. FdrFMO == θsin
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Force About a Point • Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure. • The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane. • If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. • If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. • Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F. ( ) +×+×=++× 2121 FrFrFFr
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Rectangular Components of the Moment of a Force ( ) ( ) ( )kyFxFjxFzFizFyF FFF zyx kji kMjMiMM xyzxyz zyx zyxO −+−+−= = ++= The moment of F about O, kFjFiFF kzjyixrFrM zyx O ++= ++=×= ,
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Rectangular Components of the Moment of a Force The moment of F about B, FrM BAB ×= / ( ) ( ) ( ) kFjFiFF kzzjyyixx rrr zyx BABABA BABA ++= −+−+−= −=/ ( ) ( ) ( ) zyx BABABAB FFF zzyyxx kji M −−−=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Rectangular Components of the Moment of a Force For two-dimensional structures, ( ) zy ZO zyO yFxF MM kyFxFM −= = −= ( ) ( )[ ] ( ) ( ) zBAyBA ZO zBAyBAO FyyFxx MM kFyyFxxM −−−= = −−−=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper. ( ) ( )( )in.12lb100 in.1260cosin.24 = =°= = O O M d FdM inlb1200 ⋅=OM
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 c) Horizontal force at A that produces the same moment, ( ) ( ) in.8.20 in.lb1200 in.8.20in.lb1200 in.8.2060sinin.24 ⋅ = =⋅ = =°= F F FdM d O lb7.57=F
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA. ( ) in.42 in.lb1200 in.42in.lb1200 ⋅ = =⋅ = F F FdMO lb50=F
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 d) To determine the point of application of a 240 lb force to produce the same moment, ( ) in.5cos60 in.5 lb402 in.lb1200 lb240in.lb1200 =° = ⋅ = =⋅ = OB d d FdMO in.10=OB
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.4 The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, FrM ACA ×=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.4 SOLUTION: 12896120 08.003.0 −− = kji M A ( ) ( ) ( )kjiM A mN8.82mN8.82mN68.7 ⋅+⋅+⋅−= ( ) ( ) jirrr ACAC m08.0m3.0 +=−= FrM ACA ×= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )kji kji r r FF DC DC N128N69N120 m5.0 m32.0m0.24m3.0 N200 N200 −+−= −+− = == λ
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© 2007 The
McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as ( )resultscalarcosθPQQP =• • Scalar products: - are commutative, - are distributive, - are not associative, PQQP •=• ( ) 2121 QPQPQQP •+•=+• ( ) undefined=•• SQP • Scalar products with Cartesian unit components, 000111 =•=•=•=•=•=• ikkjjikkjjii ( ) ( )kQjQiQkPjPiPQP zyxzyx ++•++=• 2222 PPPPPP QPQPQPQP zyx zzyyxx =++=• ++=•
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Scalar Product of Two Vectors: Applications • Angle between two vectors: PQ QPQPQP QPQPQPPQQP zzyyxx zzyyxx ++ = ++==• θ θ cos cos • Projection of a vector on a given axis: OL OL PP Q QP PQQP OLPPP == • =• == θ θ θ cos cos alongofprojectioncos zzyyxx OL PPP PP θθθ λ coscoscos ++= •= • For an axis defined by a unit vector:
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Mixed Triple Product of Three Vectors • Mixed triple product of three vectors, ( ) resultscalar=ו QPS • The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign, ( ) ( ) ( ) ( ) ( ) ( )SPQQSPPQS PSQSQPQPS ו−=ו−=ו−= ו=ו=ו ( ) ( ) ( ) ( ) zyx zyx zyx xyyxz zxxzyyzzyx QQQ PPP SSS QPQPS QPQPSQPQPSQPS = −+ −+−=ו • Evaluating the mixed triple product,
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O, FrMO ×= • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis, ( )FrMM OOL ו=•= λλ • Moments of F about the coordinate axes, xyz zxy yzx yFxFM xFzFM zFyFM −= −= −=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Force About a Given Axis • Moment of a force about an arbitrary axis, ( ) BABA BA BBL rrr Fr MM −= ו= •= λ λ • The result is independent of the point B along the given axis.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.5 a) about A b) about the edge AB and c) about the diagonal AG of the cube. d) Determine the perpendicular distance between AG and FC. A cube is acted on by a force P as shown. Determine the moment of P
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.5 • Moment of P about A, ( ) ( ) ( ) ( ) ( )jiPjiaM jiPjiPP jiajaiar PrM A AF AFA +×−= +=+= −=−= ×= 2 222 ( )( )kjiaPM A ++= 2 • Moment of P about AB, ( )( )kjiaPi MiM AAB ++•= •= 2 2aPM AB =
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.5 • Moment of P about the diagonal AG, ( ) ( ) ( ) ( ) ( )111 6 23 1 2 3 1 3 −−= ++•−−= ++= −−= −− == •= aP kji aP kjiM kji aP M kji a kajaia r r MM AG A GA GA AAG λ λ 6 aP M AG −=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.5 • Perpendicular distance between AG and FC, ( ) ( ) ( ) 0 110 63 1 2 = +−=−−•−=• P kjikj P P λ Therefore, P is perpendicular to AG. Pd aP M AG == 6 6 a d =
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Couple • Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. • Moment of the couple, ( ) ( ) FdrFM Fr Frr FrFrM BA BA == ×= ×−= −×+×= θsin • The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Moment of a Couple Two couples will have equal moments if • 2211 dFdF = • the two couples lie in parallel planes, and • the two couples have the same sense or the tendency to cause rotation in the same direction.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Addition of Couples • Consider two intersecting planes P1 and P2 with each containing a couple 222 111 planein planein PFrM PFrM ×= ×= • Resultants of the vectors also form a couple ( )21 FFrRrM +×=×= • By Varigon’s theorem 21 21 MM FrFrM += ×+×= • Sum of two couples is also a couple that is equal to the vector sum of the two couples
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Couples Can Be Represented by Vectors • A couple can be represented by a vector with magnitude and direction equal to the moment of the couple. • Couple vectors obey the law of addition of vectors. • Couple vectors are free vectors, i.e., the point of application is not significant. • Couple vectors may be resolved into component vectors.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Resolution of a Force Into a Force at O and a Couple • Force vector F can not be simply moved to O without modifying its action on the body. • Attaching equal and opposite force vectors at O produces no net effect on the body. • The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Resolution of a Force Into a Force at O and a Couple • Moving F from A to a different point O’ requires the addition of a different couple vector MO’ FrMO ×′=' • The moments of F about O and O’ are related, ( ) FsM FsFrFsrFrM O O ×+= ×+×=×+=×= '' • Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.6 Determine the components of the single couple equivalent to the couples shown. SOLUTION: • Attach equal and opposite 20 lb forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed. • Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions..
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.6 • Attach equal and opposite 20 lb forces in the +x direction at A • The three couples may be represented by three couple vectors, ( )( ) ( )( ) ( )( ) in.lb180in.9lb20 in.lb240in.12lb20 in.lb540in.18lb30 ⋅+=+= ⋅+=+= ⋅−=−= z y x M M M ( ) ( ) ( )k jiM in.lb180 in.lb240in.lb540 ⋅+ ⋅+⋅−=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.6 • Alternatively, compute the sum of the moments of the four forces about D. • Only the forces at C and E contribute to the moment about D. ( ) ( ) ( ) ( )[ ] ( )ikj kjMM D lb20in.12in.9 lb30in.18 −×−+ −×== ( ) ( ) ( )k jiM in.lb180 in.lb240in.lb540 ⋅+ ⋅+⋅−=
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© 2007 The
McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition System of Forces: Reduction to a Force and Couple • A system of forces may be replaced by a collection of force-couple systems acting a given point O • The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, ( )∑∑ ×== FrMFR R O • The force-couple system at O may be moved to O’ with the addition of the moment of R about O’ , RsMM R O R O ×+=' • Two systems of forces are equivalent if they can be reduced to the same force-couple system.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Further Reduction of a System of Forces • If the resultant force and couple at O are mutually perpendicular, they can be replaced by a single force acting along a new line of action. • The resultant force-couple system for a system of forces will be mutually perpendicular if: 1) the forces are concurrent, 2) the forces are coplanar, or 3) the forces are parallel.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Further Reduction of a System of Forces • System of coplanar forces is reduced to a force-couple system that is mutually perpendicular. R OMR and • System can be reduced to a single force by moving the line of action of until its moment about O becomes R OM R • In terms of rectangular coordinates, R Oxy MyRxR =−
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.8 For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant. Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium. SOLUTION: a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A. b) Find an equivalent force-couple system at B based on the force- couple system at A. c) Determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A.
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.8 SOLUTION: a) Compute the resultant force and the resultant couple at A. ( ) ( ) ( ) ( ) jjjj FR N250N100N600N150 −+−= = ∑ ( ) jR N600−= ( ) ( ) ( ) ( ) ( ) ( ) ( )ji jiji FrM R A 2508.4 1008.26006.1 −×+ ×+−×= ×= ∑ ( )kM R A mN1880 ⋅−=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.8 b) Find an equivalent force-couple system at B based on the force-couple system at A. The force is unchanged by the movement of the force-couple system from A to B. ( ) jR N600−= The couple at B is equal to the moment about B of the force-couple system found at A. ( ) ( ) ( ) ( ) ( )kk jik RrMM AB R A R B mN2880mN1880 N600m8.4mN1880 ⋅+⋅−= −×−+⋅−= ×+= ( )kM R B mN1000 ⋅+=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.10 Three cables are attached to the bracket as shown. Replace the forces with an equivalent force- couple system at A. SOLUTION: • Determine the relative position vectors for the points of application of the cable forces with respect to A. • Resolve the forces into rectangular components. • Compute the equivalent force, ∑= FR • Compute the equivalent couple, ( )∑ ×= FrM R A
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.10 SOLUTION: • Determine the relative position vectors with respect to A. ( ) ( ) ( )m100.0100.0 m050.0075.0 m050.0075.0 jir kir kir AD AC AB −= −= += • Resolve the forces into rectangular components. ( ) ( )N200600300 289.0857.0429.0 175 5015075 N700 kjiF kji kji r r F B BE BE B +−= +−= +− == = λ λ ( )( ) ( )N1039600 30cos60cosN1200 ji jiFD += += ( )( ) ( )N707707 45cos45cosN1000 ji jiFC −= −=
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McGraw-Hill Companies, Inc. All rights reserved. Vector Mechanics for Engineers: StaticsEighth Edition Sample Problem 3.10 • Compute the equivalent force, ( ) ( ) ( )k j i FR 707200 1039600 600707300 −+ +−+ ++= = ∑ ( )N5074391607 kjiR −+= • Compute the equivalent couple, ( ) k kji Fr j kji Fr ki kji Fr FrM DAD cAC BAB R A 9.163 01039600 0100.0100.0 68.17 7070707 050.00075.0 4530 200600300 050.00075.0 =−=× = − −=× −= − =× ×= ∑ kjiM R A 9.11868.1730 ++=
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