Engineering Mechanics: Statics ENG2008 PDF

Engineering Mechanics: Statics ENG2008 Topic 2: Basis Vector Operations and The Cartesian System Lecturer: Tara-Sue Rhoden  2–D VECTOR ADDITION Objective: Students will be able to : a) Resolve a 2-D vector into components (also 3D) b) Add 2-D vectors using Cartesia...

Engineering Mechanics: Statics ENG2008 Topic 2: Basis Vector Operations and The Cartesian System Lecturer: Tara-Sue Rhoden  2–D VECTOR ADDITION Objective: Students will be able to : a) Resolve a 2-D vector into components (also 3D) b) Add 2-D vectors using Cartesian vector notations (also 3D). In-Class activities: o Application of adding forces o Resolution of a vector using Cartesian vector notation (CVN) o Addition using CVN o Attention quiz SCALARS AND VECTORS VECTOR OPERATIONS Scalar Multiplication and Division  VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law: (valid for TWO vectors, drawn to scale) 𝑎 𝑏 𝑐 Triangle method (always = = 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 𝑠𝑖𝑛𝐶 ‘tip to tail’): (valid for TWO vectors does not have to be drawn to scale) 𝑎2 = 𝑏2 + 𝑐 2 − 2𝑏𝑐 ∙ 𝑐𝑜𝑠𝐴 How do you subtract a vector? VECTOR ADDITION (example) In this instance we could use the triangle or parallelogram rule  VECTOR ADDITION How can you add more than two concurrent vectors graphically ?  RESOLUTION OF A VECTOR “Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse. Instead of “travelling” in one stage, “travel” instead in two stages, but along pre-determined directions. Resolution tells you how far to travel in each direction.  CARTESIAN VECTOR NOTATION - We ‘resolve’ vectors into components using the x and y axes system - Each component of the vector is shown as a magnitude (Fx or Fy) and a direction, i or j. The directions are based on the x and y axes. We use the “unit vectors” (direction only) i and j to designate the x and y axes. For example, F = Fx i + Fy j or F' = F'x i + F'y j The x and y axes are always perpendicular to each other. Together, they can be directed at any inclination. APPLICATION OF VECTOR ADDITION There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket ? ADDITION OF SEVERAL VECTORS - Step 1 is to resolve each force into its components - Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector. - Step 3 is to find the magnitude and angle of the resultant vector. Example of this process You can also represent a 2-D vector with a magnitude and angle.  EXAMPLE Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force. Collinear vectors – parallel Concurrent vectors – line of action intersect Coplanar vectors – lie in same plane Plan: a)Resolve the forces in their x-y components. b)Add the respective components to get the resultant vector. c)Find magnitude and angle from the resultant components.  EXAMPLE (continued) F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN CVN = { -24 i + 10 j } kN F3 = { 36 cos 30° i – 36 sin 30° j } kN = { 31.18 i – 18 j } kN  EXAMPLE (continued) Summing up all the i and j components respectively, we get, FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN = { 16.82 i + 3.49 j } kN F = ((16.82)2 + (3.49)2)1/2 = 17.2 kN R  = tan-1(3.49/16.82) = 11.7°   PROBLEM SOLVING Given: Three concurrent forces acting on a bracket Find: The magnitude and angle of the resultant force. Plan: a)Resolve the forces in their x-y components. b)Add the respective components to get the resultant vector. c)Find magnitude and angle from the resultant components.  PROBLEM SOLVING (continued) F1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i - 510 j } N F2 = { -625 sin(30°) i - 625 cos(30°) j } N CVN = { -312.5 i - 541.3 j } N F3 = { -750 sin(45°) i + 750 cos(45°) j } N { -530.3 i + 530.3 j } N  PROBLEM SOLVING (continued) Summing up all the i and j components respectively, we get FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N = { - 162.8 i - 521 j } N FR = ((162.8)2 + (521)2) ½ = 546 N  = tan–1(521/162.8) = 72.64° or From Positive x axis  = 180 + 72.64 = 253°  ATTENTION QUIZ 1. Resolve F along x and y axes and write it in vector form. F={ }N A)80 cos (30°) i - 80 sin (30°) j B)80 sin (30°) i + 80 cos (30°) j C)80 sin (30°) i - 80 cos (30°) j D)80 cos (30°) i + 80 sin (30°) j 2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N. A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N  3-D CARTESIAN VECTOR VECTOR – Position Vectors TERMINOLOGY POSITION A position vector is defined as a fixed vector that locates a point in space relative to another point. Consider two points, A & B, in 3-D space. Let their coordinates be (XA, YA, ZA) and ( XB, YB, ZB ), respectively. The position vector directed from A to B, rAB , is defined as r AB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m Please note that B is the ending point and A is the starting point. So ALWAYS subtract the “tail” coordinates from the “tip” coordinates!  3-D CARTESIAN VECTOR TERMINOLOGY Consider a box with sides AX, AY, and AZ meters long. The vector A can be defined as A = (AX i + AY j + AZ k) m CVN (3-D) The projection of the vector A in the x-y plane is A´. The magnitude of this projection, A´, is found by using the same approach as a 2-D vector: A´ = (AX2 + AY2)1/2 The magnitude of the position vector A can now be obtained as A = ((A´)2 + AZ2) ½ = (AX 2 + AY 2 + AZ2) ½  ADDITION/SUBTRACTION OF VECTORS Once individual vectors are written in Cartesian form, it is easy to add or subtract them. The process is essentially the same as when 2-D vectors are added. For example, if A = AX i + AY j + AZ k and B = BX i + BY j + BZ k ,then A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k or A – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k  IMPORTANT NOTES Sometimes 3-D vector information is given as: a)Magnitude and the coordinate direction angles, or b)Magnitude and projection angles. You should be able to use both these types of information to change the representation of the vector into the Cartesian form, i.e., F = {10 i – 20 j + 30 k} N  A UNIT VECTOR For a vector A with a magnitude of A, a unit vector is defined as UA = A / A. Characteristics of a unit vector: a) Its magnitude is 1. b) It is dimensionless. c) It points in the same direction as the original vector (A). The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and z axes respectively. 25 Engineering Mechanics: Statics ENG2008 END OF LECTURE and The Cartesian System Lecturer: Tara-Sue Rhoden