ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη)

Παρουσίαση με θέμα: "ΜΗΧΑΝΙΚΗ Ι - ΣΤΑΤΙΚΗ 1. Στατική Ισορροπία (επανάληψη)"— Μεταγράφημα παρουσίασης:

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

2 Άσκηση - 5 Τρία βάρη G1, G2 και G3 προσαρμόζονται σε δύο σχοινιά, όπως φαίνεται στο σχήμα, χωρίς να υφίστανται τριβές. Να υπολογίσετε τις γωνίες α1 και α2 στην κατάσταση ισορροπίας.

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

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

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

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 Associative Law for Scalar Multiplication Distributive Law for Vector Addition Distributive Law for Scalar Addition Identity Element for Scalar Multiplication: number 1 such that

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.

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,

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:

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

11 Vector Addition Vector Sum: Components

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

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

14 Cartesian Vectors (3) 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 °

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

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

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

18 Cartesian Vectors (7) Direction of a Cartesian Vector
- Angles α, β and γ can be determined by the inverse cosines - Given A = Axi + Ayj + AZk Then, uA = A /A = (Ax/A)i + (Ay/A)j + (AZ/A)k where

19 Cartesian Vectors (8) Direction of a Cartesian Vector
- uA can also be expressed as uA = cosαi + cosβj + cosγk - Since and magnitude of uA = 1, - A as expressed in Cartesian vector form A = AuA = Acosαi + Acosβj + Acosγk = Axi + Ayj + AZk

20 Παράδειγμα - 4 Express the force F as Cartesian vector

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

22 By inspection, α = 60° since Fx 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

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

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

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

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

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

29 Παράδειγμα - 6 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. The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles qx, qy, qz defining the direction of the force

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

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

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

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.

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.

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 Dot Product 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)

37 Dot Product 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 = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 0 j·k = 0

38 Dot Product Cartesian Vector Formulation
- Dot product of 2 vectors A and B A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk) = AxBx(i·i) + AxBy(i·j) + AxBz(i·k) + AyBx(j·i) + AyBy(j·j) + AyBz(j·k) + AzBx(k·i) + AzBy(k·j) + AzBz(k·k) = AxBx + AyBy + AzBz Note: since result is a scalar, be careful of including any unit vectors in the result

39 Dot Product 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-10= 90°, A is perpendicular to B

40 Dot Product 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

41 Dot Product 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

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.