ECOR 1045 Lecture 2: Vectors I PDF

ECOR 1045 – Statics Course Instructor: Dr. Thomas Walker Department of Civil and Environmental Engineering Lecture 2: Vectors I Hibbeler 15th Edition, Chapter 2.1-2.4  Scalars and Vectors...

ECOR 1045 – Statics Course Instructor: Dr. Thomas Walker Department of Civil and Environmental Engineering Lecture 2: Vectors I Hibbeler 15th Edition, Chapter 2.1-2.4  Scalars and Vectors ▪ Objectives: | Learn the difference between scalar and vector quantities | Learn to add and subtract vectors | Learn to express 2D vectors in Cartesian notation ECOR 1045 – Lecture 2: Vectors I 2  Scalars and Vectors ▪ Scalar: a value described by its magnitude alone | Mass, volume, length, time ▪ Vector: a value with more than one characteristic to describe it | Magnitude (Size) | Direction (Angle) | Point of application (Location) | Sense (Tension/Compression) ECOR 1045 – Lecture 2: Vectors I 3  Vectors ▪ Magnitude = Size ▪ Direction = Angle ▪ Point of Application = Location ▪ Sense = Tension/Compression Point of Application Angle = Direction Reference ECOR 1045 – Lecture 2: Vectors I 4  Vector Notation ▪ Vectors can be represented with different notations: | Bold face type (A) | Letter with arrow above (A) | Underlined letter (A) | Two letters denoting origin and end with arrow above (AB) ECOR 1045 – Lecture 2: Vectors I 5  Vector Operations Multiplication Division Positive scalar - Increases magnitude - Decreases magnitude - Direction unchanged - Direction unchanged Negative scalar - Increases magnitude - Decreases magnitude - Reverses direction - Reverses direction ECOR 1045 – Lecture 2: Vectors I 6  Vector Operations P 3P P/3 -3P P/(-3) ECOR 1045 – Lecture 2: Vectors I 7  Vector Operations – Parallelogram law ▪ Vectors obey the parallelogram law of addition ▪ When two vectors, A and B, are added, they form a resultant vector C ▪ The general rule is: | Join the tails of A and B at point O (this will also be the tail of C) | From the head of B, draw a line parallel to A and from the head of A draw a line parallel to B | The point at which the lines intersect is the head of C O O C P ECOR 1045 – Lecture 2: Vectors I 8 C  Vector Operations – Triangle Rule ▪ The triangle rule is an alternative to the parallelogram law for vector addition ▪ For two vectors A and B: | Draw vector A | Draw vector B at the end of A | The resultant vector C extends from the beginning of A to the end of B C C C C ECOR 1045 – Lecture 2: Vectors I 9  Vector Operations – Collinear Vectors ▪ Collinear vectors have the same direction (line of action) ▪ Collinear vectors can be added algebraically | The parallelogram law reduces to the algebraic sum of their magnitudes ECOR 1045 – Lecture 2: Vectors I 10  Vector Operations - Subtraction ▪ The difference (R’) between two vectors A and B is expressed as: 𝐑′ = 𝐀 − 𝐁 𝐑′ = 𝐀 + (−𝐁) ▪ R’ is found by reversing the direction of B and adding it to A, using one of the methods previously shown ECOR 1045 – Lecture 2: Vectors I 11  Forces ▪ Forces are vector quantities | They have magnitude, sense, direction and point of application ▪ In statics, we must often find the resultant of two or more forces ▪ Sometimes, we are given a force and must resolve it into components ECOR 1045 – Lecture 2: Vectors I 12  Addition of Forces ▪ Forces can be added as vectors | Can use parallelogram law or triangle rule ▪ When adding more than two forces, multiple iterations of the parallelogram law may be used 𝐅𝐑 = 𝐅𝟏 + 𝐅𝟐 + 𝐅𝟑 = ( 𝐅𝟏 + 𝐅𝟐 + 𝐅𝟑 ) 𝐅𝟏 + 𝐅𝟐 + 𝐅𝟑 = 𝐅𝟐 + 𝐅𝟑 + 𝐅𝟏 ECOR 1045 – Lecture 2: Vectors I 13  Sample Problem ▪ Given vectors Q, R and S, show that: | 𝐒 + 𝐐 + 𝐑 = 𝐑 + 𝐐 + 𝐒 = (𝐐 + 𝐑 + 𝐒) ECOR 1045 – Lecture 2: Vectors I 14  Sample Problem ▪ Given vectors Q, R and S, show that: | 𝐒 + 𝐐 + 𝐑 = 𝐑 + 𝐐 + 𝐒 = (𝐐 + 𝐑 + 𝐒) P Q R O S ECOR 1045 – Lecture 2: Vectors I 15  System of Coplanar Forces ▪ Any force can be resolved into coplanar components ▪ Using the parallelogram law or triangle rule, any two axes (u and v) may be used to resolve a force F into Fu and Fv 𝐅 = 𝐅𝑢 + 𝐅𝑣 ECOR 1045 – Lecture 2: Vectors I 16  Sine and Cosine Laws ▪ The sine and cosine laws can be used with the triangle rule or parallelogram law ▪ Can determine magnitude and direction of resultant force ECOR 1045 – Lecture 2: Vectors I 17  Scalar Notation ▪ Rectangular components: forces resolved along x and y axes ▪ Force components can be expressed in Scalar Notation using the parallelogram law ▪ The magnitudes of the force components can be calculated using the angle or slope of the force 𝑎 𝐹𝑥 = 𝐹 cos 𝜃 = 𝐹 𝑐 𝑏 𝐹𝑦 = 𝐹 sin 𝜃 = 𝐹 𝑐 ECOR 1045 – Lecture 2: Vectors I 18  Cartesian Vector Notation ▪ Force components can be expressed in Cartesian Notation using unit vectors ▪ Force F can be expressed as: | 𝐅 = 𝐹𝑥 𝐢 + 𝐹𝑦 𝐣 ▪ Fx and Fy are scalar quantities | Magnitude of the components of F ▪ Unit vectors i and j | Designate direction along x and y axes ECOR 1045 – Lecture 2: Vectors I 19  Resultant of Coplanar Forces ▪ Coplanar forces lie in the same plane ▪ Using Scalar or Cartesian Notation, the resultant of forces can be determined | Resolve forces into rectangular components | Take algebraic sum of collinear vectors 𝐹𝑅𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥 Scalar 𝐹𝑅𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦 𝐹𝑅 = 𝐹𝑅𝑥 + 𝐹𝑅𝑦 𝐹𝑅 = (𝐹1𝑥 +𝐹2𝑥 + 𝐹3𝑥 )𝐢 + (𝐹1𝑦 +𝐹2𝑦 + 𝐹3𝑦 )𝐣 Cartesian 𝐹𝑅 = ∑𝐹𝑥 𝐢 + ∑𝐹𝑦 𝐣 ECOR 1045 – Lecture 2: Vectors I 20  Resultant of Coplanar Forces ▪ The magnitude of the resultant force is given as: | 𝐹𝑅 = 2 + 𝐹2 𝐹𝑅𝑥 𝑅𝑦 ▪ While the direction is: 𝐹𝑅𝑦 | 𝜃 = tan−1 𝐹𝑅𝑥 ECOR 1045 – Lecture 2: Vectors I 21  Sample Problem ▪ The contact point between the femur and tibia bones of the leg is at A. If a vertical force of 175 lb is applied at this point, determine the components of the force acting along the x and y axes. ECOR 1045 – Lecture 2: Vectors I 22  Sample Problem 𝐹𝑥 5 5 = ⟹ 𝐹𝑥 = 𝐹 = 67.3 𝑙𝑏 𝐹 13 13 𝐹𝑦 12 12 = ⟹ 𝐹𝑦 = 𝐹 = 162 𝑙𝑏 Fy 𝐹 13 13 5 12 13  12  5 13 F Fx x ECOR 1045 – Lecture 2: Vectors I 23  Problem 2-27 ▪ Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force acting on the ring at O if FA = 750 N and  =45o. ECOR 1045 – Lecture 2: Vectors I 24  Cartesian Vector Notation 𝐅 = 𝐹𝐴𝑥 + 𝐹𝐵𝑥 𝐢 + 𝐹𝐴𝑦 + 𝐹𝐵𝑦 𝐣 𝐅 = 530.33 + 692.82 𝐢 + 530.33 − 400,0 j FAy = FAcos45o = 530.33 𝐅 = 1223.3 𝐢 + 130.33 𝐣 FAx = FAsin45o = 530.33 𝐹 = 1223.152 + 130.332 FBx = FB cos30o = 692.82 𝐹 = 1230.1 𝑁 = 1.23 × 103 𝑁 𝐹 = 1.23 𝑘𝑁 FBy = FBsin30o = 400 130.33 𝛽 = tan−1 1223.15 𝛽 = 6.08° ECOR 1045 – Lecture 2: Vectors I 25 25  Scalar Notation 𝐹 = 𝐹𝐴2 + 𝐹𝐵2 − 2𝐹𝐴 𝐹𝐵 cos 105° 𝐹 = 1230.1 𝑁 = 1.23 𝑘𝑁 o 45o 60 𝐹𝐵 𝐹 FB = sin 𝐵 sin 105 −1 800 sin 105 F 𝐵= sin 1230.1 𝐵 = 38.92° 𝛽 = 45 − 38.92 𝛽 = 6.08° ECOR 1045 – Lecture 2: Vectors I 26