Transcript Chapter 2
Engineering Mechanics: STATICS Fifth Edition in SI Units Chapter 2: Vectors By: Anthony Bedford and Wallace Fowler Learning Objective If an object is subjected to several forces that have different magnitudes and act in different directions, how can the magnitude and direction of the resulting total force on the object be determined? This chapter reviews vector operations, express vectors in terms of components, and present examples of engineering applications of vectors. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 2 Chapter Outline • Scalars & Vectors • Components in Two Dimensions • Components in Three Dimensions • Dot Products • Cross Products Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 3 Scalars & Vectors • Scalar – a physical quantity that is completely described by a real number – E.g. Time, mass • Vector – both magnitude (nonnegative real number) & direction – E.g. Position of a point in space relative to another point, forces – Represented by boldfaced letters: U, V, W, … – Magnitude of vector U = |U| Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 4 Scalars & Vectors – Graphical representation of vectors: arrows • Direction of arrow = direction of vector • Length of arrow magnitude of vector • Example: – rAB = position of point B relative to point A – Direction of rAB = direction from point A to point B Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 5 Scalars & Vectors – |rAB| = distance between 2 points Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 6 Scalars & Vectors • Vector Addition: – When an object undergoes a displacement (moves from 1 location in space to another) – Displacement vector: U – Direction of U = direction of displacement – |U| = distance the book moves Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 7 Scalars & Vectors – 2nd displacement V – Final position of book is the same whether we give it displacement U then V, or vice versa – U and V equivalent to a single displacement W: U + V=W Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 8 Scalars & Vectors • Definition of Vector Addition: – Vector from tail of U to head of V • Triangle rule – Sum is independent of the order in which the vectors are placed head to tail • Parallelogram rule Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 9 Scalars & Vectors – Vector addition is commutative: U+V=V+U (2.1) – Vector addition is associative: (U + V) + W = U + (V + W) (2.2) – If the sum of 2 or more vectors = 0, they form a closed polygon Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 10 Scalars & Vectors – Example: • Vector rAC from A to C is the sum of rAB & rBC Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 11 Scalars & Vectors • Product of a Scalar & a Vector: – Product of scalar (real number) a & vector U = vector aU – Magnitude = |a||U| , where |a| is the absolute value of the scalar a – Direction of aU is the same as direction of U when a is positive – Direction of aU is opposite to direction of U when a is negative Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 12 Scalars & Vectors – Division of a vector U by a scalar a: U 1 U a a Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 13 Scalars & Vectors – The product is associative with respect to scalar multiplication: a(bU) = (ab)U (2.3) – The product is distributive with respect to scalar addition: (a + b)U = aU + bU (2.4) – The product is distributive with respect to vector addition: a(U + V) = aU + aV (2.5) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 14 Scalars & Vectors • Vector Subtraction: U – V = U + (1)V (2.6) • Unit Vectors: –Magnitude = 1 –Specifies a direction –If a unit vector e & a vector U have the same direction: U = |U|e Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 15 Example 2.1 Vector Operations The magnitudes of the vectors shown are |U| = 8 and |V| = 3. The vector V is vertical. Graphically determine the magnitude of the vector U + 2V. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 16 Example 2.1 (continued) Strategy By drawing the vectors to scale and applying the triangle rule for addition, we can measure the magnitude of the vector U + 2V. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 17 Example 2.1 (continued) Solution Drawing the vectors U and 2V to scale, place them head to tail. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 18 Example 2.1 (continued) Solution The measured value of |U + 2V| is 13.0. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 19 Example 2.1 (continued) Practice Problem The magnitudes of the vectors shown are |U| = 8 and |V| = 3. The vector V is vertical. Graphically determine the magnitude of the vector U 2V. Answer: |U 2V| = 5.7 Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 20 Example 2.2 Adding Vectors (refer to textbook) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 21 Components in Two Dimensions • Vectors are much easier to work with when expressed in terms of mutually perpendicular vector components: – Consider vector U: – Place a cartesian coordinate system so that the vector U is parallel to the x-y plane – U = sum of perpendicular vector components Ux & Uy that are parallel to the x & y axes: U = Ux + Uy Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 22 Components in Two Dimensions – Introduce a unit vector i defined to point in the direction of the positive x axis & a unit vector j defined to point in the direction of the positive y axis: U = Uxi + Uyj where Ux & Uy are scalar components of U (2.7) – Magnitude of U is given in terms of its components by the Pythagorean theorem: U U 2x U 2y Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.8) Engineering Mechanics: STATICS Fifth Edition Page 23 Components in Two Dimensions • Manipulating Vectors in Terms of Components: – Sum of 2 vectors U & V: U + V = (Uxi + Uyj) + (Vxi + Vyj) = (Ux + Vx)i + (Uy + Vy)j (2.9) – Graphically: Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 24 Components in Two Dimensions • Manipulating Vectors in Terms of Components: – Product of number a & vector U: aU = a(Uxi + Uyj) = aUxi + aUyj Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 25 Components in Two Dimensions • Position Vectors in Terms of Components: – Consider point A with coordinates (xA, yA) & point B with coordinates (xB, yB) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 26 Components in Two Dimensions • Position Vectors in Terms of Components: – Let rAB be the vector that specifies the position of B relative to A: rAB = (xB xA)i + (yB yA)j Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.10) Engineering Mechanics: STATICS Fifth Edition Page 27 Example 2.3 Determining Components The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F. Express F in terms of components using the coordinate system shown. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 28 Example 2.3 (continued) Strategy We will determine the components of the vector F in two ways. In the first method, we will determine the angle between F and the y axis and use trigonometry to determine the components. In the second method, we will use the given slope of the cable AB and apply similar triangles to determine the components of F. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 29 Example 2.3 (continued) Solution First Method Determine the angle between F and the y axis: 40 α arctan 26.6 80 Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 30 Example 2.3 (continued) Solution Use trigonometry to determine F in terms of its components: F F sin i F cosj 900sin 26.6o i 900cos 26.6o j N 402i 805j N Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 31 Example 2.3 (continued) Solution Second Method Using the given dimensions, calculate the distance from A to B: 40 m 80 m 2 2 Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd 89.4 m Engineering Mechanics: STATICS Fifth Edition Page 32 Example 2.3 (continued) Solution Use similar triangles to determine the components of F: Fx 40 m F 89.4m and Fy 80 m F 89.4m so 40 40 900 N i 900 N j F 89.4 89.4 402i 805j N Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 33 Example 2.3 (continued) Practice Problem The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F. Suppose that you change the placement of point B so that the magnitude of the y component of F is three times the magnitude of the x component of F. Express F in terms of its components. How far along the x axis from the origin of the coordinate system should B be placed? Answer: F = 285i - 854j (N). Place point B at 26.7 m from the origin. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 34 Example 2.4 Determining Components in Terms of an Angle (refer to textbook) Example 2.5 Determining an Unknown Vector Magnitude (refer to textbook) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 35 Components in Three Dimensions • Review of drawing objects in 3 dimensions: (a) A cube viewed with the line of sight – perpendicular to a face (b) An oblique view of the cube (c) A cartesian coordinate system aligned with the edges of the cube (d) 3-D representation of the coordinate system Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 36 Components in Three Dimensions • Right-handed coordinate system: – Express vector U in terms of vector components Ux, Uy & Uz parallel to the x, y & z axes respectively: U = Ux + Uy + Uz Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.11) Engineering Mechanics: STATICS Fifth Edition Page 37 Components in Three Dimensions – Introducing unit vectors i, j & k that point in the positive x, y & z directions, U can be expressed in terms of scalar components: U = Uxi + Uyj + Uz k (2.12) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 38 Components in Three Dimensions • Magnitude of a Vector in Terms of Components: – Consider a vector U & its vector components: Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 39 Components in Three Dimensions • Magnitude of a Vector in Terms of Components: – From the right triangles formed by vectors Uy, Uz & their sum Uy + Uz: |Uy + Uz|2 = |Uy|2 + |Uz|2 (2.13) – The vector U is the sum of the vectors Ux & Uy + Uz. The 3 vectors form a right triangle: |U|2 = |Ux|2 + |Uy + Uz|2 Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 40 Components in Three Dimensions • Magnitude of a Vector in Terms of Components: – Substituting Eqn (2.13): |U|2 = |Ux|2 + |Uy| + |Uz|2 = Ux2 + Uy2 + Uz2 – Thus, magnitude of vector U: U U x2 U y2 U z2 Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.14) Engineering Mechanics: STATICS Fifth Edition Page 41 Components in Three Dimensions • Direction Cosines: – One way to describe the direction of a vector is by specifying the angles x, y & z between the vector & the positive coordinate axes: Ux = |U| cos x, Uy = |U| cos y, Uz = |U| cos z Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.15) Engineering Mechanics: STATICS Fifth Edition Page 42 Components in Three Dimensions • Direction Cosines: Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 43 Components in Three Dimensions • Direction Cosines: – Direction cosines: cos x, cos y & cos z – Direction cosines satisfy the relation: cos2 x + cos2 y + cos2 z = 1 – Suppose that e is a unit vector with the same direction as U: (2.16) U = |U| e – In terms of components: Uxi + Uyj + Uzk = |U| (exi + eyj + ezk) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 44 Components in Three Dimensions • Direction Cosines: – Thus: Ux = |U| ex, Uy = |U| ey, Uz = |U| ez – By comparing these equations to Eqn (2.15): cos x = ex, cos y = ey, cos z = ez – The direction cosines of a vector U are the components of a unit vector with the same direction as U Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 45 Components in Three Dimensions • Position Vectors in Terms of Components: – Consider point A with coordinates (xA, yA, zA) & point B with coordinates (xB, yB, zB) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 46 Components in Three Dimensions • Position Vectors in Terms of Components: – The position vector rAB from A to B: rAB = (xB xA)i + (yB yA)j + (zB zA)k Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd (2.17) Engineering Mechanics: STATICS Fifth Edition Page 47 Components in Three Dimensions • Components of Vector Parallel to a Given Line: – In 3-D applications, the direction of a vector U is often defined by specifying the coordinates of 2 points A & B on a line that is parallel to U – Determine position vector rAB using Eqn (2.17) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 48 Components in Three Dimensions • Components of Vector Parallel to a Given Line: – Divide rAB by its magnitude unit vector eAB that points from A to B – eAB has the same direction as U – Determine U as the product of its magnitude & eAB: U = |U| eAB Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 49 Example 2.6 Direction Cosines The coordinates of point C of the truss are xC = 4 m, yC = 0, zC = 0, and the coordinates of point D are xD = 2 m, yD = 3 m, zD = 1 m. What are the direction cosines of the position vector rCD from point C to point D? Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 50 Example 2.6 (continued) Strategy Knowing the coordinates of points C and D, we can determine rCD in terms of its components. Then we can calculate the magnitude of rCD (the distance from C to D) and use Eqn (2.15) to obtain the direction cosines. Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 51 Example 2.6 (continued) Solution Determine the position vector rCD in terms of its components. rCD = (xD xC)i + (yD yC)j + (zD zC)k = (2 4)i + (3 0)j + (1 0) k (m) = 2i + 3j + k (m) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 52 Example 2.6 (continued) Solution Calculate the magnitude of rCD. rCD rCD 2x rCD 2y rCD 2z 2 m 2 3 m 2 1 m 2 3.74 m Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 53 Example 2.6 (continued) Solution Determine the direction cosines. cos x cos y cos z Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd rCDx rCD rCD y rCD rCDz rCD 2m 0.535, 3.74 m 3m 0.802, 3.74 m 1m 0.267 3.74 m Engineering Mechanics: STATICS Fifth Edition Page 54 Example 2.6 (continued) Practice Problem The coordinates of point B of the truss are xB = 2.4 m, yB = 0, zB = 3 m. Determine the components of a unit vector eBD that points from point B toward point D. Answer: eBD = 0.110i + 0.827j 0.551k Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 55 Example 2.7, 2.8 & 2.9 Determining Components in Three Dimensions (refer to textbook) Example 2.10 Determining Components of a Force (refer to textbook) Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 56 Refer To Textbook Chapter 2 For Further Details Chapter 2 Copyright 2008 Pearson Education South Asia Pte Ltd Engineering Mechanics: STATICS Fifth Edition Page 57