ECOR 1045 Lecture 3 PDF

ECOR 1045 – Statics Course Instructor: Dr. Thomas Walker Department of Civil and Environmental Engineering Lecture 3: Vectors II Hibbeler 15th edition, Chapter 2.5-2.8  More...

ECOR 1045 – Statics Course Instructor: Dr. Thomas Walker Department of Civil and Environmental Engineering Lecture 3: Vectors II Hibbeler 15th edition, Chapter 2.5-2.8  More Vectors ▪ Objectives: | Learn to express 3D vectors in Cartesian and spherical notation | Learn to construct unit vectors and to calculate coordinate directional angles (direction cosines) | Apply vectors in solving problems involving positions and forces ECOR 1045 – Lecture 3: Vectors II 2  3D Cartesian Vectors ▪ 3D Vectors are best represented in Cartesian notation ▪ Right-handed Coordinate System: | Thumb represents the positive z-axis | Fingers curl around z-axis | Sweep fingers from x-axis to y-axis ▪ Cartesian Unit Vectors: | Cartesian unit vectors i, j, k designate x, y and z directions | The positive directions of the units vectors are as shown ECOR 1045 – Lecture 3: Vectors II 3  Cartesian Vector Representation ▪ Resolution of A into Cartesian unit vectors requires two applications of the parallelogram law 𝐀 = 𝐴 𝑥 𝐢 + 𝐴 𝑦 𝐣 + 𝐴𝑧 𝐤 ECOR 1045 – Lecture 3: Vectors II 4  Cartesian Vector Representation ▪ The magnitude of a Cartesian vector is determined using the Pythagorean Theorem: 𝐀= 𝐴2𝑥 + 𝐴2𝑦 + 𝐴𝑧2 ECOR 1045 – Lecture 3: Vectors II 5  Cartesian Vector Representation ▪ Direction of a Cartesian vector | Defined by angles ,  and  with respect to the x, y, and z axes | ,  and  are called coordinate direction angles 𝐴𝑥 cos 𝛼 = 𝐴 𝐴𝑦 cos 𝛽 = 𝐴 𝐴𝑧 cos 𝛾 = 𝐴 ECOR 1045 – Lecture 3: Vectors II 6  Cartesian Vector Representation ▪ A vector A can be represented using unit vectors as: | 𝐀 = 𝐴𝐮𝐀 | where uA is a unit vector in the direction of A 𝐀 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐮𝐀 = 𝐴 = 𝐴 𝐢 + 𝐴 𝐣 + 𝐴 𝐤 𝐮𝐀 = cos 𝛼𝐢 + cos 𝛽𝐣 + cos 𝛾𝐤 ▪ The magnitude of a unit vector is 1 | If any two coordinate angles are known, we can find the third 𝑢𝐴 = cos2 𝛼 + cos2 𝛽 + cos2 𝛾 = 1 cos2 𝛼 + cos2 𝛽 + cos2 𝛾 = 1 ECOR 1045 – Lecture 3: Vectors II 7  Spherical Vector Representation ▪ The direction of A can alternatively be found from two angles,  and  𝐴𝑧 = 𝐀 cos 𝜙 𝐴′ = 𝐀 sin 𝜙 𝐴𝑥 = 𝐀′ cos 𝜃 = 𝐀 sin 𝜙 cos 𝜃 𝐴𝑦 = 𝐀′ sin 𝜃 = 𝐀 sin 𝜙 sin 𝜃 𝐀 = 𝐴 sin 𝜙 cos 𝜃 𝐢 + 𝐴 sin 𝜙 sin 𝜃 𝐣 + A 𝑐𝑜𝑠𝜙 𝐤 ECOR 1045 – Lecture 3: Vectors II 8  Cartesian Vector Addition ▪ Given two vectors A and B: | 𝐀 = 𝐴𝑥 𝐢 + 𝐴𝑦 𝐣 + 𝐴𝑧 𝐤 | 𝐁 = 𝐵𝑥 𝐢 + 𝐵𝑦 𝐣 + 𝐵𝑧 𝐤 ▪ We can add A and B using Cartesian components as follows: | 𝐑 = 𝐀 + 𝐁 = (𝐴𝑥 +𝐵𝑥 )𝐢 + (𝐴𝑦 +𝐵𝑦 )𝐣 + (𝐴𝑧 +𝐵𝑧 )𝐤 ▪ We can also subtract A and B using Cartesian components as follows: | 𝐑 = 𝐀 − 𝐁 = (𝐴𝑥 −𝐵𝑥 )𝐢 + (𝐴𝑦 −𝐵𝑦 )𝐣 + (𝐴𝑧 −𝐵𝑧 )𝐤 ▪ General formulation: | 𝐑 = ∑𝐅 = ∑𝐹𝑥 𝐢 + ∑𝐹𝑦 𝐣 + ∑𝐹𝑧 𝐤 ECOR 1045 – Lecture 3: Vectors II 9  Sample Problem Determine the resultant force acting on the hook. ECOR 1045 – Lecture 3: Vectors II 10  Sample Problem Determine the resultant force acting on the hook. Approach: 1. Determine F1 and F2 in Cartesian vector form 2. Sum the x and y components of the two forces ECOR 1045 – Lecture 3: Vectors II 11  Sample Problem ▪ Resolve F1 | 𝐹1𝑥 = 0 𝑙𝑏 4 | 𝐹1𝑦 = 500 𝑙𝑏 5 = 400 𝑙𝑏 3 | 𝐹1𝑧 = 500 𝑙𝑏 5 = 300 𝑙𝑏 ▪ In Cartesian vector form (remember the units): | 𝐅𝟏 = 0𝐢 + 400𝑙𝑏 𝐣 + 300𝑙𝑏 𝐤 | or, | 𝐅𝟏 = 0𝐢 + 400 𝐣 + 300 𝐤 𝑙𝑏 ECOR 1045 – Lecture 3: Vectors II 12  Sample Problem ▪ Resolve F2 | We only know two angles, so we need to resolve F2 into the z-axis and the xy-plane | 𝐹2𝑧 = 𝐹2 sin 45 = 800 sin 45 = 565.69 𝑙𝑏 | 𝐹2𝑥𝑦 = 𝐹2 𝑐𝑜𝑠45 = 800 cos 45 = 565.69 𝑙𝑏 | 𝐹2𝑥 = 𝐹2𝑥𝑦 cos 30 = 565.69 cos 30 = 489.90 𝑙𝑏 | 𝐹2𝑦 = 𝐹2𝑥𝑦 sin 30 = 565.69 sin 30 = 282.84 𝑙𝑏 | 𝐅𝟐 = 489.90𝐢 + 282.84𝐣 − 565.69𝐤 𝑙𝑏 ECOR 1045 – Lecture 3: Vectors II 13  Sample Problem ▪ Solution | 𝐅𝟏 = 0𝐢 + 400𝐣 + 300𝐤 𝑙𝑏 | 𝐅𝟐 = 489.90𝐢 + 282.84𝐣 − 565,69𝐤 𝑙𝑏 | 𝐅𝐑 = 𝐅𝟏 + 𝐅𝟐 | 𝐅𝐑 = ሼ 0 + 489.90 𝐢 + (400 + 282.84)𝐣 + 300 − 565.69 𝐤ሽ 𝑙𝑏 | 𝐅𝐑 = 490𝐢 + 683𝐣 − 266𝐤 𝑙𝑏 ECOR 1045 – Lecture 3: Vectors II 14  Position Vectors ▪ A position vector r is a fixed vector which defines a point in 3D space relative to another point e.g. a point P relative to the origin O | O (0, 0, 0) and P (x, y, z) 𝐫 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 ECOR 1045 – Lecture 3: Vectors II 15  Position Vectors ▪ If a position vector is directed from point A (xA, yA, zA) to B (xB, yB, zB) then: | 𝐫𝐀𝐁 = 𝐫𝐁 − 𝐫𝐀 | 𝐫𝐀𝐁 = 𝑥𝐵 𝐢 + 𝑦𝐵 𝐣 + 𝑧𝐵 𝐤 − 𝑥𝐴 𝐢 + 𝑦𝐴 𝐣 + 𝑧𝐴 𝐤 | 𝐫𝐀𝐁 = 𝑥𝐵 − 𝑥𝐴 ) 𝐢 + (𝑦𝐵 − 𝑦𝐴 )𝐣 + (𝑧𝐵 −𝑧𝐴 𝐤 ▪ Hint: Position vectors tells you how to get from A to B ECOR 1045 – Lecture 3: Vectors II 16  Position Vectors ▪ The magnitude of the position vector rAB is given as: 𝑟𝐴𝐵 = 𝑥𝐵 − 𝑥𝐴 2 + 𝑦𝐵 − 𝑦𝐴 2 + 𝑧𝐵 − 𝑧𝐴 2 ▪ The direction of the position vector rAB is described by the direction cosines of rAB, as specified by the unit vector: 𝐫𝐀𝐁 𝐮= = cos 𝛼𝐢 + cos 𝛽𝐣 + cos 𝛾𝐤 𝑟𝐴𝐵 ECOR 1045 – Lecture 3: Vectors II 17  Force Along a Line ▪ In 3-dimensional statics, the force F can be specified by two points, A and B, through which passes the line of action of F ▪ F can be represented by the position vector r from point A to B ▪ Hence: 𝐫 𝐅 = 𝐹𝐮 = 𝐹 𝑟 𝑥𝐵 − 𝑥𝐴 ) 𝐢 + (𝑦𝐵 − 𝑦𝐴 )𝐣 + (𝑧𝐵 −𝑧𝐴 𝐤 𝐅=𝐹 𝑥𝐵 − 𝑥𝐴 2 + 𝑦𝐵 − 𝑦𝐴 2 + 𝑧𝐵 − 𝑧𝐴 2 ECOR 1045 – Lecture 3: Vectors II 18  Problem F2-22 Express the force as a Cartesian vector. ECOR 1045 – Lecture 3: Vectors II 19  Problem F2-23 Determine the magnitude of the resultant force at A. ECOR 1045 – Lecture 3: Vectors II 20  Problem 2-95 At a given time, the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the distance d between A and B at this instant. Hint: To solve the problem, formulate a position vector, directed from A to B, and then determine its magnitude. ECOR 1045 – Lecture 3: Vectors II 21  Problem 2-105 (13th ed) The pipe is supported at its ends by a cord AB. If the cord exerts a force of 𝐹 = 12 𝑙𝑏 on the pipe at A, express this force as a Cartesian vector. ECOR 1045 – Lecture 3: Vectors II 22