Lect Vectors Assign.ppt PDF

Teaching Slides Chapter 2: Vectors (C) 2005 Pearson Education South Asia Pte Ltd 1  Chapter Outline Scalars & Vectors Rules for Manipulating Vectors Components in 2 Dimensions Components in 3 Dimensions Dot Prod...

Teaching Slides Chapter 2: Vectors (C) 2005 Pearson Education South Asia Pte Ltd 1  Chapter Outline Scalars & Vectors Rules for Manipulating Vectors Components in 2 Dimensions Components in 3 Dimensions Dot Products Cross Products Mixed Triple Products (C) 2005 Pearson Education South Asia Pte Ltd 2  2.1 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 (C) 2005 Pearson Education South Asia Pte Ltd 3 30 METERS 30 METERS @ 45O @ 90O 50 METERS @ 0O VECTOR ARROWS MAY BE DRAWN = 10 METERS ANYWHERE ON THE PAGE AS LONG AS THE PROPER LENGTH AND SCALE DIRECTION ARE MAINTAINED  2.2 Rules for Manipulating 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 (C) 2005 Pearson Education South Asia Pte Ltd 5  ALL VECTORS MUST BE DRAWN TO SCALE & POINTED IN THE PROPER DIRECTION A R D B C B C A D A+ B + C +D = R  Vector A 30 METERS @ 45O To add the vectors Place them head to tail Vector B 50 METERS @ 0O C Vector C 30 METERS B @ 90O A Angle is measured at 40o = 10 METERS SCALE Resultant = 9 x 10 = 90 meters  2.1 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 (a) to point A  Direction of rAB = direction from point A to point B  |rAB| = distance between 2 points (b) (C) 2005 Pearson Education South Asia Pte Ltd 8  2.2 Rules for Manipulating 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 (C) 2005 Pearson Education South Asia Pte Ltd 9  2.2 Rules for Manipulating 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 (C) 2005 Pearson Education South Asia Pte Ltd 10  2.2 Rules for Manipulating 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  Example: Vector rAC from A to C is the sum of rAB & rBC (C) 2005 Pearson Education South Asia Pte Ltd 11  2.2 Rules for Manipulating 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  Division of a vector U by a scalar a: U  1    U a  a (C) 2005 Pearson Education South Asia Pte Ltd 12  2.2 Rules for Manipulating 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) (C) 2005 Pearson Education South Asia Pte Ltd 13  2.2 Rules for Manipulating 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 Unit Vector Vector Subtraction (C) 2005 Pearson Education South Asia Pte Ltd 14 Resultant of Two Vectors  The resultant is the sum or the combined effect of two vector quantities Vectors in the same direction: 6N 4N = 10 N 6m = 10 m 4m Vectors in opposite directions: 6 m s-1 10 m s-1 = 4 m s-1 6N 10 N = 4N Up = + Down = - Right = + Left = - Rectangular Coordinates 90 o North y Quadrant II + Quadrant I - + x 0 East o West 180 o 360 o Quadrant III Quadrant IV - 270 o South  90O North +y 120O -240O West 180O +x 0O East -x 360O 30O West of North 30O Left of +y MEASURING THE SAME DIRECTION 60O North of West -y IN DIFFERENT WAYS 60O Above - x 270O South  RADIANS = ARC LENGTH / RADIUS LENGTH CIRCUMFERENCE OF A CIRCLE = 2  x RADIUS RADIANS IN A CIRCLE = 2  R / R 1 CIRCLE = 2  RADIANS = 360O 1 RADIAN = 360O / 2  = 57.3O /2 radians y Quadrant II + Quadrant I - + x 0 radians  radians 2 radians Quadrant III Quadrant IV - 3/2  radians The Parallelogram Law  When two vectors are joined tail to tail  Complete the parallelogram  The resultant is found by drawing the diagonal  When two vectors are joined head to tail  Draw the resultant vector by completing the triangle Problem: Resultant of 2 Vectors Two forces are applied to a body, as shown. What is the magnitude and direction of the resultant force acting on the body? Solution:  Complete the parallelogram (rectangle)  The diagonal of the parallelogram ac represents the resultant force  The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc 12 N a d 2 2 Magnitude ac  12  5 θ 13 5N N 5 ac 13 N 12 b 12 c Direction of ac : tan   5 12   tan  1 67  Resultant displacement is 13 N 5 67º with the 5 N force Problem: Resultant of 3 Vectors Find the magnitude (correct to two decimal places) and direction of the resultant of the three forces shown below. Solution:  Find the resultant of the two 5 N forces first (do right angles first) ac  52  52  50 7.07 N 5 d 5 c tan   1   45 5 N 5N Now find the resultant of the 10 N and 5 07  7. 7.07 N forces  The 2 forces are in a straight line (45º + 90º 45º θ a N b 135º = 180º) and in opposite directions 5N 93 135º 2. N  So, Resultant = 10 N – 7.07 N = 2.93 10 N in the direction of the 10 N force Calculating the Magnitude of the Perpendicular Components If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are: x = v Cos θ v y = v Sin θ y=v Sin θ y θ x=v x Cos θ  Proof: x y Cos  Sin  v v x vCos y vSin  Problem: Calculating the magnitude of perpendicular components A force of 15 N acts on a box as shown. What is the horizontal component of the force? Solution: Horizontal Component  x 15Cos 60 7.5 N Component 12.99 N N Vertical 15 Vertical Component  y 15Sin 60 12.99 N 60º Horizontal 7.5 N Componen t  Example 2.1 Adding Vectors Fig. 2.11 is an initial design sketch of part of the roof of a sports stadium to be supported by the cables AB & AC. The forces the cables exert on the pylon to which they are attached are represented by the vectors FAB & FAC. The magnitude of the forces are |FAB| = 100 kN & | FAC| = 60 kN. Determine the magnitude & direction of the sum of the forces exerted on the pylon by the cables (a) graphically & (b) using trigonometry. Figure 2.11 (C) 2005 Pearson Education South Asia Pte Ltd 24  Example 2.1 Adding Vectors Strategy (a) By drawing the parallelogram rule for adding the 2 forces with the vectors drawn to scale, we can measure the magnitude & direction of their sum. (b) We will calculate the magnitude & direction of the sum of the forces by applying the laws of sines & cosines to the triangles formed by the parallelogram rule. (C) 2005 Pearson Education South Asia Pte Ltd 25  Example 2.1 Adding Vectors Solution (a) Graphically construct the parallelogram rule with FAB & FAC proportional to their magnitudes: By measuring the figure, we estimate the magnitude of the vector FAB + FAC to be 155 kN & its direction to be 19° above the horizontal. (C) 2005 Pearson Education South Asia Pte Ltd 26  Example 2.1 Adding Vectors Solution (b) Consider the parallelogram rule: Since  + 30° = 180°,  = 150° Applying law of cosines to shaded triangle: 2 2 2 FAB  FAC  FAB  FAC  2 FAB FAC cos α 2 2 100   60   2100 60 cos150 Magnitude |FAB + FAC| = 155 kN (C) 2005 Pearson Education South Asia Pte Ltd 27  Example 2.1 Adding Vectors Solution (b) To determine the angle  between FAB + FAC & the horizontal, apply law of sines to shaded triangle: sin  sin α  FAB FAB  FAC The solution is  FAB sin α   100 sin 150  β arcsin   arcsin   18.8  FAB  FAC   155  (C) 2005 Pearson Education South Asia Pte Ltd 28  A person in a wheelchair is moving up a ramp at constant speed. Their total weight is 900 N. The ramp makes an angle of 10º with the horizontal. Calculate the force required to keep the wheelchair moving at constant speed up the ramp. (You may ignore the effects of friction). Solution: If the wheelchair is moving at constant speed (no acceleration), then the force that moves it up the ramp must be the same as the component of it’s weight parallel to the ramp. Complete the parallelogram. Component of weight 28 N 156. 10º parallel to ramp: 900Sin10 156.28 N 80º 10º Component of weight perpendicular to ramp: 886.33 N 900 N 900Cos10 886.33 N  2.4 Components in 3 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 (C) 2005 Pearson Education South Asia Pte Ltd 30  2.4 Components in 3 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 (2.11) (C) 2005 Pearson Education South Asia Pte Ltd 31  2.4 Components in 3 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) (C) 2005 Pearson Education South Asia Pte Ltd 32  2.4 Components in 3 Dimensions Direction Cosines: (C) 2005 Pearson Education South Asia Pte Ltd 33  2.4 Components in 3 Dimensions Direction Cosines:  Direction cosines: cos x, cos y & cos z  Direction cosines satisfy the relation: cos2 x + cos2 y + cos2 z = 1 (2.16)  Suppose that e is a unit vector with the same direction as U: U = |U| e  In terms of components: Uxi + Uyj + Uzk = |U| (exi + eyj + ezk) (C) 2005 Pearson Education South Asia Pte Ltd 34  2.4 Components in 3 Dimensions Position Vectors in Terms of Components:  Consider point A with coordinates (xA, yA, zA) & point B with coordinates (xB, yB, zB) (C) 2005 Pearson Education South Asia Pte Ltd 35  2.4 Components in 3 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 (2.17) (C) 2005 Pearson Education South Asia Pte Ltd 36  Example 2.6 Magnitude & Direction Cosines of a Vector The coordinates of point C of the truss in Fig. 2.26 are xC = 4 m, yC = 0, zC = 0 & the coordinates of point D are xD = 2 m, yD = 3 m, zD = 1 m. Let rCD be the position vector from C to D. (a) What is the magnitude of rCD? Fig. 2.26 (C) 2005 Pearson Education South Asia Pte Ltd 37  Example 2.6 Magnitude & Direction Cosines of a Vector Strategy (a) We can obtain the components of rCD by subtracting the coodinates of C from the coordinates of D (C) 2005 Pearson Education South Asia Pte Ltd 38  Example 2.6 Magnitude & Direction Cosines of a Vector Solution (a) The components of rCD are given by: rCD = (xD  xC)i + (yD  yC)j + (zD  zC)k = (2  4) i + (3  0)j + (1  0)k = 2i + 3j + k (C) 2005 Pearson Education South Asia Pte Ltd 39  Example 2.6 Magnitude & Direction Cosines of a Vector Solution (a) The magnitude of rCD is: rCD  rCD 2x  rCD 2y  rCD 2z 2 2 2        2 m  3 m  1m  3.74 m (C) 2005 Pearson Education South Asia Pte Ltd 40  2.5 Dot Products Definition:  Consider 2 vectors U & V:  Denoted by U∙V  Defined to be the product of the magnitude of U, the magnitude of V & the cosine of the angle  between U & V when they are placed tail to tail U∙V = |U||V| cos  (2.18) (C) 2005 Pearson Education South Asia Pte Ltd 41  2.5 Dot Products Definition:  Result is a scalar  sometimes called scalar product  Unit = product of the units of the 2 vectors (C) 2005 Pearson Education South Asia Pte Ltd 42  2.5 Dot Products  Notice that the dot product of two nonzero vectors = 0 if & only if the vectors are perpendicular  The dot product is commutative: U∙V = V∙U (2.19) (C) 2005 Pearson Education South Asia Pte Ltd 43  2.5 Dot Products  The dot product is associative with respect to scalar multiplication: a(U∙V) = (aU)∙V = U∙ (aV) (2.20)  The dot product is associative with respect to vector addition: U∙ (V + W) = U∙V + U∙W (2.21) (C) 2005 Pearson Education South Asia Pte Ltd 44  2.5 Dot Products Dot Products in Terms of Components:  Determine the dot products formed from the unit vectors i, j & k: i∙i = |i||i| cos (0) = (1)(1)(1) = 1 i∙j = |i||j| cos (90°) = (1)(1)(0) = 0  Continuing this way: i∙i = 1, i∙j = 0, i∙k = 0, j∙i = 0, j∙j = 1, j∙k = 0, k∙i = 0, k∙j = 0, k∙k = 1. (2.22) (C) 2005 Pearson Education South Asia Pte Ltd 45  2.5 Dot Products  Dot product of U & V expressed in terms of their components: U∙V = (Uxi + Uyj + Uzk) ∙ (Vxi + Vyj + Vzk) = UxVx(i∙i) + UxVy(i∙j) + UxVz(i∙k) + UyVx(j∙i) + UyVy(j∙j) + UyVz(j∙k) + UzVx(k∙i) + UzVy(k∙j) + UzVz(k∙k)  Substituting Eqs. (2.22): U∙V = UxVx + UyVy + UzVz (2.23) (C) 2005 Pearson Education South Asia Pte Ltd 46  2.5 Dot Products  Equate the expression for dor product given by Eq. (2.23) to the definition of dot product, Eq. (2.18) to solve for cos  : U V U xVx  U yV y  U zVz cos θ   UV UV (2.24) (C) 2005 Pearson Education South Asia Pte Ltd 47  Example 2.12 Using the Dot Product to Determine an Angle What is the angle  between the lines AB & AC in Fig. 2.35? Fig. 2.35 Strategy Coordinates of A, B & C  components of vector rAB & rAC Use Eq. (2.24) to determine  (C) 2005 Pearson Education South Asia Pte Ltd 48  Example 2.12 Using the Dot Product to Determine an Angle Solution Vectors rAB & rAC : rAB = (6  4) i + (1  3)j + (2  2)k = 2i  2j  4k (m) rAC = (8  4) i + (8  3)j + (4  2)k = 4i + 5j + 2k (m) Magnitudes: 2 rAB  2 m    2 m 2   4 m 2 4.90 m rAC  4 m 2  5 m 2  2 m 2 6.71 m (C) 2005 Pearson Education South Asia Pte Ltd 49  Example 2.12 Using the Dot Product to Determine an Angle Solution Dot product of rAB & rAC : rAB∙rAC = (2 m) (4 m) + (2 m) (5 m) + (4 m) (2 m) = 10 m2 Therefore, 2 rAB rAB  10 m cos    0.304 rAB rAC 4.90 m 6.71 m   arccos 0.304 107.7 (C) 2005 Pearson Education South Asia Pte Ltd 50  2.6 Cross Products Definition:  Consider 2 vectors U & V:  Denoted by U V  Defined by: U  V = |U||V| sin  e (2.28) where  = angle between U & V e = unit vector defined to be perpendicular to both U & V (C) 2005 Pearson Education South Asia Pte Ltd 51  2.6 Cross Products Definition:  U,V & e are defined to be a right- handed system Right-hand rule  Result is a vector  sometimes called vector product (C) 2005 Pearson Education South Asia Pte Ltd 52  2.6 Cross Products  Unit = product of the units of the 2 vectors  Notice that the cross product of 2 nonzero vectors = 0 if & only if the vectors are parallel  The dot product is not commutative: U  V = V  U (2.29) (C) 2005 Pearson Education South Asia Pte Ltd 53  2.6 Cross Products The dot product is associative with respect to scalar multiplication: a(U  V) = (aU)  V = U  (aV) (2.30) The dot product is distributive with respect to vector addition: U  (V + W) = (U  V) + (U  W) (2.31) (C) 2005 Pearson Education South Asia Pte Ltd 54  2.6 Cross Products Cross Products in Terms of Components:  Determine the cross products formed from the unit vectors i, j & k: i  i = |i||i| sin (0) e = 0 i  j = |i||j| sin (90°) e = e e = unit vector perpendicular to i & j e = k or e = k Applying right-hand rule, e = k ij=k (C) 2005 Pearson Education South Asia Pte Ltd 55  2.6 Cross Products  Continuing this way: i  i = 0, i  j = k, i  k = j, j  i = k, j  j = 0, j  k = i, k  i = j, k  j = i, k  k = 0. (2.32) These results can be easily remembered by arranging the unit vectors in a circle: (C) 2005 Pearson Education South Asia Pte Ltd 56  2.6 Cross Products  Cross product of U & V expressed in terms of their components: U  V = (Uxi + Uyj + Uzk)  (Vxi + Vyj + Vzk) = UxVx(i  i) + UxVy(i  j) + UxVz(i  k) + UyVx(j  i) + UyVy(j  j) + UyVz(j  k) + UzVx(k  i) + UzVy(k  j) + UzVz(k  k) (C) 2005 Pearson Education South Asia Pte Ltd 57  2.6 Cross Products  Substituting Eqs. (2.32): U  V = (UyVz  UzVy)i (UxVz UzVx)j + (UxVy UyVx)k (2.33)  This result can be compactly written as the determinant: Ux Uy Uz U V  V x Vy Vz Wx Wy Wz (2.34) (C) 2005 Pearson Education South Asia Pte Ltd 58  2.6 Cross Products Evaluating a 3  3 Determinant:  Repeat its first 2 columns & evaluate the products of the terms along the 6 diagonal lines i j k i j Ux Uy UzUx Uy Vx Vy Vz Vx Vy () () () (+) (+) (+) (C) 2005 Pearson Education South Asia Pte Ltd 59  2.6 Cross Products  Value of the determinant: i j k U yV zi  U zV xj  U xV yk Ux Uy Uz   U yV xk  U zV yi  U xV zj Vx Vy Vz (C) 2005 Pearson Education South Asia Pte Ltd 60  2.6 Cross Products  Alternatively: i j k Uy Uz Ux Uz Ux Uy U x U y U z i  j k Vy Vz Vx Vz Vx Vy Vx Vy Vz U yV z U zV y i  U xV z U zV x j  U xV y U yV x k (C) 2005 Pearson Education South Asia Pte Ltd 61  2.7 Mixed Triple Products Definition: U∙ (V  W) (2.35)  In terms of scalar components: i j k U V W  U x i  U y j  U z k V x Vy Vz Wx Wy Wz U x i  U y j  U z k [V yWz  VzWy i  VxWz  VzWx j  VxWy  V yWx k ] U x V yWz  VzWy  U y VxWz  VzWx   U z VxWy  V yWx  (C) 2005 Pearson Education South Asia Pte Ltd 62  2.7 Mixed Triple Products  This result can be expressed as the determinant: Ux Uy Uz U V W   V x V y V z (2.36) Wx Wy Wz  Interchanging any 2 vectors in the mixed triple product changes the sign but not the absolute value of the result E.g. U∙ (V  W) = W∙ (V  U) (C) 2005 Pearson Education South Asia Pte Ltd 63  (submission as an Assignment) Physics for Scientist and Engineers. Fundamental problems Chapter 01 Do Problem 25-35 Chapter 03 Pb 18-21 Pb 33-37,53 All problem done so far in lect.) Due date 7x days. (C) 2005 Pearson Education South Asia Pte Ltd 64