1. MEC 0011 Statics Lecture 1 Prof. Sanghee Kim Fall_ 2012

2. Chapter 1. General Principles Basic quantities and idealizations of mechanics Newton’s Laws of Motion and Gravitation Principles for applying the SI system of units Standard procedures for performing numerical calculations General guide for solving problems

3. 1.1 Mechanics Mechanics can be divided into 3 branches: - Rigid-body Mechanics - Deformable-body Mechanics - Fluid Mechanics - Statics : Equilibrium of bodies  At rest  Move with constant velocity  special case of dynamics with zero acceleration - Dynamics : Accelerated motion of bodies

4. Basic Quantities (used throughout in mechanics) 1.Length - locate the position of a point in space - describe the size of physical system 2.Mass - measure of a quantity of matter 3.Time - succession of events - principles of statics are time independent 4.Force - a “push” or “pull” exerted by one body on another - gravitational, electrical, and magnetic forces - characterized by magnitude, direction and point of application 1.2 Fundamentals Concepts

5. Idealizations 1.Particles ( 질점 ) - has a mass and size can be neglected 2.Rigid Body ( 강체 ) - a combination of a large number of particles 3.Concentrated Force ( 집중력 ) - the effect of a loading assumed to be act at a point (contract force btw wheel and the ground)

6. Newton ’ s Three Laws of Motion First Law “ A particle originally at rest, or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force ” Second Law “ A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force ” Third Law “ The mutual forces of action and reaction between two particles are equal and, opposite and collinear ”

7. Copyright © 2010 Pearson Education South Asia Pte Ltd Exercise #1 1. The subject of mechanics deals with what happens to a body when ______ is / are applied to it. A) magnetic field B) heat C) forces D) neutrons E) lasers 2. ________________ still remains the basis of most of today’s engineering sciences. A) Newtonian Mechanics B) Relativistic Mechanics C) Greek Mechanics C) Euclidean Mechanics

8. Newton ’ s Law of Gravitational Attraction Weight: Letting yields F = force of gravitation between two particles G = universal constant of gravitation G= 66.73 (10 -12 ) m 3 / (kg·s 2 ) m 1,m 2 = mass of each of the two particles r = distance between the two particles g, acceleration due to gravity, is determined at standard location (at sea level and at a altitude of 45°) The weight of body is not an absolute quantity

9. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. Exercise #2

10. 1.3 Units of Measurement SI Units Stands for Système International d’Unités SI system specifies length in meters (m), time in seconds (s) and mass in kilograms (kg) Force unit, Newton (N), is derived from F = ma : a force required to give 1 kg of mass an acceleration of 1m/s 2 NameLengthTimeMassForce International Systems of Units (SI) Meter (m) Second (s) Kilogram (kg) Newton (N) At the standard location, g = 9.806 65 m/s 2 For calculations, we use g = 9.81 m/s 2 Thus, W = mg(g = 9.81m/s 2 ) Hence, a body of mass 1 kg has a weight of 9.81 N, a 2 kg body weighs 19.62 N

11. 1.4 The International System of Units Prefixes For a very large or small numerical quantity, units can be modified by using a prefix Each represent a multiple or sub-multiple of a unit Eg: 4,000,000 N = 4000 kN (kilo-newton) = 4 MN (mega- newton) 0.005m = 5 mm (milli-meter)

12. - Rules for use several units which are multiples of one another are separated by dot : m·s (meter-second) vs ms (millisecond) avoid the use of a prefix in the denominator of composite : N/mm  kN/m, m/mg  Mm/kg converting all prefixes to power of 10 and use single prefix : (50 kN) (60 nm) = [50(10) 3 N] [60(10 -9 ) m] = 3000(10 -6 ) N·m = 3 mN·m exponential power on a unit applied to both the unit and prefix : µN 2 =(µN) 2 = µN·µN, mm 2 = (mm) 2 = mm·mm

13. 1.5 Numerical Calculations Dimensional Homogeneity Each term must be expressed in the same units Regardless of how the equation is evaluated, it maintains its dimensional homogeneity All terms can be replaced by a consistent set of units Significant Figures Accuracy of a number is specified by the number of significant figures it contains A significant figure is any digit including zero e.g. 5604 and 34.52 have four significant numbers When numbers begin or end with zero, we make use of prefixes to clarify the number of significant figures e.g. 400 as one significant figure would be 0.4(10 3 )

14. Calculations Retain a greater number of digits for accuracy Work out computations so that numbers that are approximately equal Round off final answers to three significant figures Rounding Off Numbers Accuracy obtained would never be better than the accuracy of the problem data Calculators or computers involve more figures in the answer than the number of significant figures in the data Calculated results should always be “ rounded off ” to an appropriate number of significant figures If digit preceding the 5 is even number  not round up (75.25  ) If digit preceding the 5 is odd number  round up (0.1275  )

15. Copyright © 2010 Pearson Education South Asia Pte Ltd 5. For a static’s problem your calculations show the final answer as 12345.6 N. What will you write as your final answer? A) 12345.6 N B) 12.3456 kN C) 12 kN D) 12.3 kN E) 123 kN Exercise #3

16. 1.6 General Procedure for Analysis To solve problems, it is important to present work in a logical and orderly way as suggested: 1.Correlate actual physical situation with theory 2.Draw any diagrams and tabulate the problem data 3.Apply principles in mathematics forms 4.Solve equations which are dimensionally homogenous 5.Report the answer with significance figures 6.Technical judgment and common sense

17. Convert to 2 km/h to m/s. Solution Exercise #4

18. Evaluate each of the following and express with SI units having an appropriate prefix (a) (50 mN) (6 GN) (b) (400 mm) (0.6 MN) 2 (c) 45 MN 3 /900 Gg Solution Exercise #6

19. Copyright © 2010 Pearson Education South Asia Pte Ltd Chapter 2. Force Vectors Parallelogram Law Cartesian vector form Dot product and angle between 2 vectors

20. 2.1 Scalars and Vectors Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as A e.g. Mass, volume and length Vector – A quantity that has magnitude and direction e.g. Position, force and moment – Represent by a letter with an arrow over it, – Magnitude is designated as – In this subject, vector is presented as A and its magnitude (positive quantity) as A

21. 2.2 Vector Operations Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = aA - Magnitude = - Law of multiplication applies e.g. A/a = ( 1/a ) A, a≠0 Vector Addition - Addition of two vectors A and B gives a resultant vector R by the parallelogram law - Result R can be found by triangle construction - Communicative e.g. R = A + B = B + A - Special case: Vectors A and B are collinear (both have the same line of action) Vector Subtraction - Special case of addition e.g. R ’ = A – B = A + ( - B ) - Rules of Vector Addition Applies

22. 2.3 Vector Addition of Forces Finding a Resultant Force Parallelogram law is carried out to find the resultant force Resultant, F R = ( F 1 + F 2 ) Procedure for Analysis Parallelogram Law – Make a sketch using the parallelogram law – 2 components forces add to form the resultant force – Resultant force is shown by the diagonal of the parallelogram – The components is shown by the sides of the parallelogram

23. Procedure for Analysis Trigonometry – Redraw half portion of the parallelogram – Magnitude of the resultant force can be determined by the law of cosines – Direction if the resultant force can be determined by the law of sines – Magnitude of the two components can be determined by the law of sines

24. Example 2.1 The screw eye is subjected to two forces, F 1 and F 2. Determine the magnitude and direction of the resultant force.

25. Solution Parallelogram Law Unknown: magnitude of F R and angle θ

26. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Trigonometry Law of Cosines Law of Sines

27. Copyright © 2010 Pearson Education South Asia Pte Ltd Solution Trigonometry Direction Φ of F R measured from the horizontal

28. Resolve the 1000 N ( ≈ 100kg) force acting on the pipe into the components in the (a)x and y directions, (b) x’ and y directions. Solution (a)Parallelogram Law From the vector diagram, Exercise #7

29. Solution (b) Parallelogram Law Solution (b) Law of Sines

30. Resolve F2 into components along the u and axes and determine the magnitudes of these components. Exercise #8

31. 2.4 Addition of a System of Coplanar Forces Scalar Notation – x and y axes are designated positive and negative – Components of forces expressed as algebraic scalars Cartesian Vector Notation – Cartesian unit vectors i and j are used to designate the x and y directions – Unit vectors i and j have dimensionless magnitude of unity ( = 1 ) – Magnitude is always a positive quantity, represented by scalars F x and F y

32. Coplanar Force Resultants To determine resultant of several coplanar forces: – Resolve force into x and y components – Addition of the respective components using scalar algebra – Resultant force is found using the parallelogram law – Cartesian vector notation: Coplanar Force Resultants – Vector resultant is therefore – If scalar notation are used

33. Coplanar Force Resultants – In all cases we have – Magnitude of F R can be found by Pythagorean Theorem * Take note of sign conventions

34. Example 2.5 Determine x and y components of F 1 and F 2 acting on the boom. Express each force as a Cartesian vector.

35. Solution Scalar Notation Hence, from the slope triangle, we have

36. By similar triangles we have Scalar Notation: Cartesian Vector Notation:

37. The contact point between the femur and tibia bones of the leg is at A. If a vertical force of 875 N is applied at this point, determine the components along the x and y axes. Note that the y component represents the normal force on the load-bearing region of the bones. Both the x and y components of this force cause synovial fluid to be squeezed out of the bearing space. Exercise #9