lecture_01 - PART A: STATICS 1. INTRODUCTION 1.1 What is...

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Unformatted text preview: PART A: STATICS 1. INTRODUCTION 1.1 What is Mechanics? Mechanics is that branch of science which describes and predicts the conditions of rest or motion of bodies under the action of forces Mechanics Mechanics of rigid bodies Mechanics of deformable bodies Mechanics of fluids Statics deals with bodies at rest Dynamics deals with bodies in motion EG1109 STATICS and MECHANICS AND MATERIALS 1.2 Fundamental Quantities of Mechanics x t m F space time mass force In Newtonian mechanics, these are absolute quantities, independent of each other. But force is dependent on the three. Particle – A body which may be assumed to occupy a single point in space Rigid Body – Combination of large amount of particles occupying fixed position with respect to each other EG1109 STATICS and MECHANICS AND MATERIALS 1.4 Fundamental Principles of Elementary Mechanics (a) Parallelogram Law for Addition of Force F2 F1 (b) The Principle of Transmissibility F1+F2 FF F F FF FFFFF (c) Newton’s First Law (Law of Inertia) Every body continues in its state of rest, or in uniform motion in a right (straight) line unless it is compelled to change that state by forces impressed upon it. EG1109 STATICS and MECHANICS AND MATERIALS (d) Newton’s Second Law F = ma Work Energy Principle dv F = mv ds Impulse-Momemtum Principle F =m dv dt ∫ Fds = ∫ mvdv Work done = 1 1 2 mv2 − mv12 2 2 = change in kinetic energy ∫ Fdt = ∫ mdv Impulse = mv 2 − mv1 = change in momemtum (e) Newton’s Third Law "To every action there is always opposed an equal reaction” Newton, third law of motion, Philosophiae Naturalis Principia Mathematica (1686) EG1109 STATICS and MECHANICS AND MATERIALS (f) Newton’s Law of Gravitation Mm F = G 2 , where G is the universal gravitational constant r r m F F M This principle leads to the concept of weight: W = mg where g = G Me re2 = gravitational acceleration, M e and re are the mass and the radius of the Earth, respectively. EG1109 STATICS and MECHANICS AND MATERIALS These quizzes are based on ideas from Eric Mazur, These Peer Instruction, Prentice Hall, Saddle River, N.J. 1997 Peer collision 1. A collision occurs between a bus and a small sports car. During impact: impact: a) There is no force between the bus and the car b) The bus exerts a force on the car, but the car does not exert any force on the bus c) The bus exerts a much larger force on the car than the car does on the bus d) The force that the bus exerts on the car is equal to the force the car exerts on the bus e) The car exerts a larger force on the bus than the bus on the car EG1109 STATICS and MECHANICS AND MATERIALS These quizzes are based on ideas from Eric Mazur, Peer Instruction, Prentice Hall, Saddle River, N.J. 1997 If your 2. If you moved to a planet of the same radius but a greater mass, your weight would weight a) decrease b) increase c) remain the same our 3. If you moved to a planet of the same mass but a larger radius, your mass would mass a) decrease b) increase c) remain the same EG1109 STATICS and MECHANICS AND MATERIALS 2. STATICS OF PARTICLES 2.1 Definition of a Force Force is a vector. It possesses both magnitude and direction. F In SI Units, the force unit is Newton (N). This is the force which, applied to a body of mass 1 kilogram, causes an acceleration of 1 meter per second2 (m/s2) in the direction of application of the force. EG1109 STATICS and MECHANICS AND MATERIALS 2.2 Addition of Forces The addition of forces obeys the parallelogram law of parallelogram law vector addition Magnitude and direction of c may be obtained by the law of cosines α b b a+ c2 = a2 + b2 – 2abcos γ c= γ or the law of sines β a b c a = = s in α s in β s in γ The addition of 3 or more forces is carried out by arranging the given vectors in tip-to-tail fashion and connecting the tail of the first vector with the tip of the last one. - polygon rule for addition of vectors b a c d EG1109 STATICS and MECHANICS AND MATERIALS d=a+b+c 2.3 Force Components The component of a force along a given axis may be obtained using the dot (or scalar) product of vectors: a ● b = a b cos θ For example, to find the component of force P along the direction of vector Q P cos θ θ O B P Q A Q The component of P along Q direction is OB = OA cos θ = P cos θ As , λQ θ P Q = P Q cos θ PQ Q P co s θ = =P Q Q . . . P But Q = λ Q is a unit vector along Q. Therefore the component of P along Q = P λ Q Q . EG1109 STATICS and MECHANICS AND MATERIALS Any force in space can be represented as the sum of its component force vectors along three convenient axes, normally mutually perpendicular (rectangular Cartesian coordinates system). y j k θz z i θy θx F Fy x Fx Fz In terms of the unit vectors i, j, k the force F may be expressed as F = Fx i + Fy j + Fz k EG1109 STATICS and MECHANICS AND MATERIALS in which the Cartesian components Fx = F . i = F cos θx , Fy = F . j = F cos θy , Fz = F . k = F cos θz are the projections of F onto the coordinate axes. The unit vector in the direction of F is given by λF = cos θx i + cos θy j + cos θz k It follows that the unit vector has the direction cosines of that vector as its Cartesian components. Note that cos2 θx + cos2 θy + cos2 θz = 1 EG1109 STATICS and MECHANICS AND MATERIALS Addition of forces by summing up their Cartesian components Earlier, the addition of vectors was done graphically using the polygon rule. The same addition may be done analytically by summing algebraically the corresponding scalar components of the vectors. For example, R = P+Q+ S R x i + R y j + Rz k = Px i + Py j + Pz k + Q x i + Q y j + Qz k + S x i + S y j + Sz k + ( Pz + Qz + S z ) k = ( Px + Q x + S x ) i + Py + Q y + S y j ( ) ⇒ R x = Px + Q x + S x , R y = Py + Q y + S y , R z = Pz + Q z + S z The magnitude of the resultant vector R is given by 2 2 R = Rx + R y + Rz2 EG1109 STATICS and MECHANICS AND MATERIALS 2.4 Forces on a Particle and Equilibrium Forces passing through a single point (or a particle) are called concurrent forces. The resultant of concurrent forces, may be obtained by the polygon rule (graphically) or by the addition of the scalar components of the forces. y Fn F2 R Rx = ∑ Fx ,i R y = ∑ F y ,i i =1 n n 2D 3D 0 z R = ∑Fi i =1 n F1 x Rz = ∑ Fz ,i i =1 i =1 n 2 2 R = Rx + R 2 + Rz y EG1109 STATICS and MECHANICS AND MATERIALS zero If the resultant force on a particle is ______, the particle is equilibrium said to be in _______________. R x = ∑ Fx , i = 0 R = ∑Fi = 0 i =1 n n R y = ∑ F y ,i = 0 Rz = ∑ Fz ,i = 0 i =1 i =1 n i =1 n Note: With 3 equations, 3 unknowns maybe solved (3-D). However, only 2 equations are available for 2-D. The concept of equilibrium will be used not to solve for the resultant, which is zero, but to solve for the required forces that act on the particle and maintain its equilibrium. EG1109 STATICS and MECHANICS AND MATERIALS 2.5 Free Body Diagram A sketch showing the physical conditions of the problem is known as a space diagram [see Figure (a) below] FB FA FB 40o 30 o FA A θ = 40o A W To calculate the tensile forces in the chains, consider a free body diagram of the joint A as shown in Figure (b). (b) Free Body Diagram FB W FA For equilibrium, FA FB 500 = = sin 40o sin 30o sin 110o FA = 342 N and FB = 266 N (c) Force Triangle (a) Space Diagram W = 500 N EG1109 STATICS and MECHANICS AND MATERIALS is A free-body diagram is a sketch of an object or a connected free group of objects, modeled as a single particle/rigid body that is group completely completely isolated from its environment or surrounding bodies and and represents the interactions of its environment by appropriate appropriate external forces (and moments). body Drawing a free-body diagram is an art, and can be learned only Drawing body by practice. If a correct free-body diagram is constructed, then by the balance of the forces can be carried out in a very systematic the manner. manner. No No equilibrium problem should be solved without first drawing body the free-body diagram, so as to account for all the forces and the couple moments that act on the free-body. couple EG1109 STATICS and MECHANICS AND MATERIALS body Construct a free-body diagram for the object shown object α R W R R R α 3R B α N A P 3R / 2 2 R cos α Free-Body Diagram of Member AB EG1109 STATICS and MECHANICS AND MATERIALS body Construct a free-body diagram for the object shown object D E Ball and socket support F B G C Cables T1 T2 Ay AX AZ B G Free body diagram of rigid member ABC T3 C W EG1109 STATICS and MECHANICS AND MATERIALS ...
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