FORCE VECTORS. Chapter Objectives

Transcription

1 CHAPTER Engineering Mechanics: Statics ORCE VECTORS Tenth Edition College of Engineering Department of Mechanical Engineering a by Dr. Ibrahim A. Assakkaf SPRING 007 ENES 110 Statics Department of Mechanical Engineering University of Maryland Baltimore County Chapter Objectives Slide No. 1 To show how to add forces and resolve them into components using the Parallelogram Law. To express force and position in Cartesian vector form and explain how to determine the vector s magnitude and direction. To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another. 1

2 Two-Dimensional Vectors Objectives Students will be able to: a) Resolve a -D vector into components b) Add -D vectors using Cartesian vector notations. Slide No. Two-Dimensional Vectors Slide No. 3 Application There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket?

3 Scalar and Vectors Slide No. 4 Scalar A quantity characterized by a positive or negative number. Vector A quantity that has both a magnitude and direction. Scalar and Vectors Slide No. 5 Notations for Vectors or handwritten work, a vector is generally represented by a letter with an arrow written over it, such as A r The magnitude is designated A r or A. In your textbook vectors are symbolized in boldface type; for example, A is used to designate the vector A. Its magnitude is symbolized by A, or simply A. 3

4 Scalar and Vectors Slide No. 6 Scalars Vectors Examples: mass, volume force, velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Cartesian Components Special Notation: None Bold font, a line, an arrow or a carrot Scalar and Vectors Graphical Representation of a Vector Slide No. 7 The arrow represents the vector graphically and used to define the vector magnitude, direction, and sense. The magnitude is the length of the arrow, and the direction is defined by the angle between a reference axis and the arrow s line of action. The sense is indicated by the arrowhead. 4

5 Vectors Operations Slide No. 8 Multiplication or Division of a Vector by a Scalar aa magnitude : aa = a A 1 A a 1 1 magnitude : A = a a A Vectors Operations Slide No. 9 Vector Addition Two vectors A and B such as force or position, ig. 4a, may be added to form a "resultant" vector R = A+ B by using the parallelogram law. ig. 4 5

6 Vectors Operations Slide No. 10 Vector Subtraction The resultant difference between two vectors A and B of the same type may be expressed as Vectors Operations Slide No. 11 Resolution of a Vector A vector may be resolved into components having known lines of action by using the parallelogram law. 6

7 Vector Addition of orces Slide No. 1 Addition of orces Application: Chains Vector Addition of orces Addition of orces (cont d) A force is vector quantity since it has a specified magnitude and direction. Two common problems in statics involve either finding the resultant force given its components or resolving a known force into components. The law of cosines is often used to find the magnitude, while the law of sines is used to find the direction. Slide No. 13 7

8 Vector Addition of orces Slide No. 14 Law of Sines and Cosines Vector Addition of orces Slide No. 15 Example 1 The screw eye in the figure is subjected to two forces 1 and. Determine the magnitude and direction of the resultant force. 8

9 Vector Addition of orces Slide No. 16 Example 1 (cont d) Parallelogram Law Trigonometry Vector Addition of orces Slide No. 17 Example 1 (cont d) R = ( 100) + ( 150) ( 100)( 150) cos115 = 1.6 = 13 N sin115 = θ = sin sinθ sin φ = o = o o = 54.8 o 39.8 o 9

10 Slide No. 18 orces (D) In the plane, a force can be resolved into two rectangular components. There are two separate notations for doing this: Scalar Notation: we write the force as ( x, y ) where x and y are scalar components of the force in the directions of the positive x- and y-axes, respectively. If x and y are negative, it means that x and y are directed along the negative x- and y-axes. orces (D) Slide No. 19 Cartesian Vector Notation: we write the force as = i + x y j Where i and j represent the positive directions of the x- and y-axes, respectively. = cosθ x y = sinθ θ 10

11 orces (D) Slide No. 0 The resultant of several coplanar forces can easily be determined if an x, y- coordinate system is established and the forces are resolved along the axis. or example, 1 3 n = i + 1x = = = x 3x M nx 1y i + i + i + ny j y 3 y j j j Slide No. 1 orces (D) Then the resultant is given by R = + + L+ = = 1 ( K ) i + ( K ) 1x ( ) i + ( )j Rx x 3 Ry 3x n In general case, the x and y components of the resultant of any number of coplanar forces can be represented symbolically by the algebraic sum of the x and y components of all the forces, that is nx 1y y 3 y ny j 11

12 orces (D) Rx = x x and y components Ry = of Resultant y Slide No. The magnitude and direction of the resultant force are given by R = R θ = tan 1 = Ry Rx Rx + Ry Magnitude & Direction orces (D) Slide No. 3 The resultant force of the four cable forces acting on the supporting bracket can be determined by adding algebraically the separate x and y components of each cable force. This resultant R produces the same pulling effect on the bracket as all four cables. 1

13 orces (D) Example Slide No. 4 The Three concurrent forces are acting on a bracket. ind the magnitude and the angle of the resultant force. Slide No. 5 orces (D) Example (cont d) Plan: a) Resolve the forces in their x-y components. b) Add the respective components to get the resultant vector. c) ind magnitude and angle from the resultant components. 13

14 Slide No. 6 orces (D) Example (cont d) 1 = { 15 sin 40 i + 15 cos 40 j } kn = { 9.64 i j } kn = { -(1/13)6 i + (5/13)6 j } kn = { -4 i + 10 j } kn 3 = { 36 cos 30 i 36 sin 30 j } kn = { i 18 j } kn θ x = cosθ = sinθ y orces (D) Example (cont d) Summing up all the i and j components respectively, we get Slide No. 7 R = { ( ) i + ( ) j } kn R θ = tan = { 16.8 i j } kn = R 1 = Rx + Ry = 11.7 o = ( 16.8) + ( 3.49) y θ = 17. kn R x 14

15 orces (D) Example 3 To be discussed and solved in class Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three forces acting on the ring A. Given: 1 = 500 N, = 400 N, 3 = 600 N θ = 0 o, φ = 30 o c = 3, d = 4 Slide No. 8 orces (D) Example 4 Slide No. 9 To be discussed and solved in class Determine the magnitude and direction θ of A so that the resultant force is directed along the positive x axis and has a magnitude 150 N. o = 30 = 800 N 15