Η παρουσίαση φορτώνεται. Παρακαλείστε να περιμένετε

Η παρουσίαση φορτώνεται. Παρακαλείστε να περιμένετε

ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη) 2. Διανύσματα (Vectors) 3. Δυνάμεις στο χώρο 1.

Παρόμοιες παρουσιάσεις


Παρουσίαση με θέμα: "ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη) 2. Διανύσματα (Vectors) 3. Δυνάμεις στο χώρο 1."— Μεταγράφημα παρουσίασης:

1 ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη) 2. Διανύσματα (Vectors) 3. Δυνάμεις στο χώρο 1

2 Άσκηση - 5 Τρία βάρη G 1, G 2 και G 3 προσαρμόζονται σε δύο σχοινιά, όπως φαίνεται στο σχήμα, χωρίς να υφίστανται τριβές. Να υπολογίσετε τις γωνίες α 1 και α 2 στην κατάσταση ισορροπίας. 2

3 Coordinate System 1.An origin as the reference point 2.A set of coordinate axes with scales and labels 3.Choice of positive direction for each axis 4.Choice of unit vectors at each point in space Coordinate system: used to describe the position of a point in space and consists of Cartesian Coordinate System 3

4 Vector A vector is a quantity that has both direction and magnitude. Let a vector be denoted by the symbol The magnitude of is denoted by 4

5 Vector Addition Let and be two vectors. Define a new vector,the “vector addition” of and by the geometric construction shown in either figure 5

6 Summary: Vector Properties Addition of Vectors 1. Commutativity 2. Associativity 3.Identity Element for Vector Addition such that 4.Inverse Element for Vector Addition such that Scalar Multiplication of Vectors 1.Associative Law for Scalar Multiplication 2.Distributive Law for Vector Addition 3.Distributive Law for Scalar Addition 4.Identity Element for Scalar Multiplication: number 1 such that 6

7 Application of Vectors (1) Vectors can exist at any point P in space. (2) Vectors have direction and magnitude. (3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located. 7

8 Vector Decomposition Choose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis, for example along the x,y, and z-axes of a Cartesian coordinate system. A vector at P can be decomposed into the vector sum, 8

9 Unit Vectors and Components The idea of multiplication by real numbers allows us to define a set of unit vectors at each point in space with Components: 9

10 Vector Decomposition in Two Dimensions Consider a vector x- and y components: Magnitude: Direction: 10

11 Vector Addition Vector Sum: Components 11

12 Cartesian Vectors (1) Right-Handed Coordinate System A rectangular or Cartesian coordinate system is said to be right-handed provided: - Thumb of right hand points in the direction of the positive z axis when the right-hand fingers are curled about this axis and directed from the positive x towards the positive y axis 12

13 Cartesian Vector Representations - Three components of A act in the positive i, j and k directions A = A x i + A y j + A Z k *Note the magnitude and direction of each components are separated, easing vector algebraic operations. Cartesian Vectors (2) 13

14 Direction of a Cartesian Vector - Orientation of A is defined as the coordinate direction angles α, β and γ measured between the tail of A and the positive x, y and z axes - 0° ≤ α, β and γ ≤ 180 ° Cartesian Vectors (3)

15 Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A Cartesian Vectors (4) 15

16 Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A Cartesian Vectors (5) 16

17 Direction of a Cartesian Vector - For angles α, β and γ (blue colored triangles), we calculate the direction cosines of A Cartesian Vectors (6) 17

18 Direction of a Cartesian Vector - Angles α, β and γ can be determined by the inverse cosines - Given A = A x i + A y j + A Z k Then, u A = A /A = (A x /A)i + (A y /A)j + (A Z /A)k where Cartesian Vectors (7) 18

19 Direction of a Cartesian Vector - u A can also be expressed as u A = cosαi + cosβj + cosγk - Since and magnitude of u A = 1, - A as expressed in Cartesian vector form A = Au A = Acosαi + Acosβj + Acosγk = A x i + A y j + A Z k Cartesian Vectors (8) 19

20 Express the force F as Cartesian vector Παράδειγμα

21 Since two angles are specified, the third angle is found by Two possibilities exit, namely or Λύση

22 By inspection, α = 60° since F x is in the +x direction Given F = 200N F = Fcosαi + Fcosβj + Fcosγk = (200cos60°N)i + (200cos60°N)j + (200cos45°N)k = {100.0i j k}N Checking:

23 Rectangular Components in Space The vector is contained in the plane OBAC. Resolve into horizontal and vertical components. Resolve into rectangular components 23

24 Rectangular Components in Space With the angles between and the axes, is a unit vector along the line of action of and are the direction cosines for 24

25 Rectangular Components in Space Direction of the force is defined by the location of two points, 25

26 26 Να βρεθεί το μέτρο και η διεύθυνση της F. Οι δυνάμεις βρίσκονται σε ισορροπία. 1. Το σχήμα είναι και διάγραμμα ελευθέρου σώματος. 2. Παράδειγμα - 5

27 27 For the components: COUNT EQUATIONS AND UNKNOWNS!!!!!!! Now, what? (1) (2) Two equations, (1) and (2), and two unknowns, F and. 3. Collect and equate components.

28 28 From (1) and (2), (3) (4) Now, what? For the magnitude, For the angle, divide (4) by (3) to get

29 Παράδειγμα - 6 The tension in the guy wire is 2500 N. Determine: a) components F x, F y, F z of the force acting on the bolt at A, b) the angles  x,  y,  z  defining the direction of the force STEPS : Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. Apply the unit vector to determine the components of the force acting on A. Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. 29

30 Determine the unit vector pointing from A towards B. Determine the components of the force. Λύση 30

31 Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. 31

32 Άσκηση – 6 Να προσδιορίσετε το μέτρο και τις γωνίες των συντεταγμένων κατεύθυνσης της συνισταμένης δύναμης που δρα στο μεταλλικό δακτύλιο. 32

33 Άσκηση - 7 Ο άνδρας τραβά το σχοινί με δύναμη 350 Ν. Να αναπαραστήσετε τη δύναμη αντίδρασης στο σημείο Α ως ένα Καρτεσιανό διάνυσμα και να βρείτε τη διεύθυνσή της. 33

34 Άσκηση - 8 Μία κατασκευή αποτελείται από 2 ράβδους, 1 και 2, και ένα σχοινί 3(τα βάρη θεωρούνται αμελητέα). Στο σημείο Α φορτώνεται ένα βάρος G. Το σύστημα ισορροπεί. Να προσδιορίσετε τις δυνάμεις στις ράβδους 1 και 2 καθώς και στο σχοινί 3. 34

35 Dot Product Dot product of vectors A and B is written as A·B (Read A dot B) Define the magnitudes of A and B and the angle between their tails A·B = AB cosθ where 0°≤ θ ≤180° Referred to as scalar product of vectors as result is a scalar

36 Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D) Dot Product

37 Cartesian Vector Formulation - Dot product of Cartesian unit vectors Eg: i·i = (1)(1)cos0° = 1 and i·j = (1)(1)cos90° = 0 - Similarly i·i = 1j·j = 1k·k = 1 i·j = 0i·k = 0j·k = 0 Dot Product

38 Cartesian Vector Formulation - Dot product of 2 vectors A and B A·B = (A x i + A y j + A z k)· (B x i + B y j + B z k) = A x B x (i·i) + A x B y (i·j) + A x B z (i·k) + A y B x (j·i) + A y B y (j·j) + A y B z (j·k) + A z B x (k·i) + A z B y (k·j) + A z B z (k·k) = A x B x + A y B y + A z B z Note: since result is a scalar, be careful of including any unit vectors in the result Dot Product

39 Applications The angle formed between two vectors or intersecting lines θ = cos -1 [(A·B)/(AB)] 0°≤ θ ≤180° Note: if A·B = 0, cos -1 0= 90°, A is perpendicular to B Dot Product

40 Applications - If A ║ is positive, A ║ has a directional sense same as u - If A ║ is negative, A ║ has a directional sense opposite to u - A ║ expressed as a vector A ║ = A cos θ u = (A·u)u Dot Product

41 Applications For component of A perpendicular to line aa’ 1. Since A = A ║ + A ┴, then A ┴ = A - A ║ 2. θ = cos -1 [(A·u)/(A)] then A ┴ = Asinθ 3. If A ║ is known, by Pythagorean Theorem Dot Product

42 Άσκηση - 9 Ο σκελετός του σχήματος υπόκειται σε μια οριζόντια δύναμη F 300 N. Να προσδιορίσετε τις συνιστώσες αυτής της δύναμης παράλληλα και κάθετα στη διεύθυνση του τμήματος ΑΒ. 42


Κατέβασμα ppt "ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη) 2. Διανύσματα (Vectors) 3. Δυνάμεις στο χώρο 1."

Παρόμοιες παρουσιάσεις


Διαφημίσεις Google