Chapter Objectives To show how to add forces and resolve them into components using the Parallelogram Law. To express force and position in Cartesian vector form and explain how to determine the vector s magnitude and direction. To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another.
In-Class Activities 1. Reading Quiz. Applications 3. Scalars and Vectors 4. Vector Operations 5. Vector Addition of Forces 6. Addition of a System of Coplanar Forces 7. Cartesian Vectors 8. Addition and Subtraction of Cartesian Vectors 9. Position Vectors 10. Force Vector Directed along a Line 11. Dot Product 1. Concept Quiz
READING QUIZ 1. Which one of the following is a scalar quantity? a) Force b) Position c) Mass d) Velocity
READING QUIZ (cont). If a dot product of two non-zero vectors is 0, then the two vectors must be to each other. a) Parallel (pointing in the same direction) b) Parallel (pointing in the opposite direction) c) Perpendicular d) Cannot be determined.
APPLICATIONS
SCALARS AND VECTORS Scalar A quantity characterized by a positive or negative number Indicated by letters in italic such as A e.g., mass, volume and length
SCALARS AND VECTORS (cont) Vector A quantity that has magnitude and direction e.g., position, force and moment Represented by a letter with an arrow over it Magnitude is designated by In this subject, vector is presented as A and its magnitude (positive quantity) as A A A
VECTOR OPERATIONS Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = aa - Magnitude = aa - Law of multiplication applies e.g. A/a = ( 1/a ) A, a 0
VECTOR OPERATIONS (cont) Vector Addition - Addition of two vectors A and B gives a resultant vector R by the parallelogram law - Result R can be found by triangle construction - Communicative e.g. R = A + B = B + A - Special case: Vectors A and B are collinear (both have the same line of action)
VECTOR OPERATIONS (cont) Vector Subtraction - Special case of addition e.g. R = A B = A + ( - B ) - Rules of Vector Addition Applies
VECTOR ADDITION OF FORCES Finding a Resultant Force Parallelogram law is carried out to find the resultant force Resultant, F R = ( F 1 + F )
VECTOR ADDITION OF FORCES (cont) Procedure for Analysis Parallelogram Law Make a sketch using the parallelogram law component forces add to form the resultant force Resultant force is shown by the diagonal of the parallelogram The components are shown by the sides of the parallelogram
VECTOR ADDITION OF FORCES (cont) Procedure for Analysis Trigonometry Redraw half portion of the parallelogram Magnitude of the resultant force can be determined by the law of cosines Direction of the resultant force can be determined by the law of sines Magnitude of the two components can be determined by the law of sines
EXAMPLE 1 The screw eye is subjected to two forces, F 1 and F. Determine the magnitude and direction of the resultant force. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 (cont) Parallelogram Law Unknown: magnitude of F R and angle θ Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 (cont) Solution Trigonometry Law of Cosines 100N 150N 100N 150N cos115 500 30000 0.46 1.6N N F R 10000 13 Law of Sines 150N 1.6N sin sin115 150N sin 0.9063 1.6N 39.8 Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 (cont) Solution Trigonometry Direction Φ of F R measured from the horizontal 39.8 54.8 15 Copyright 011 Pearson Education South Asia Pte Ltd
ADDITION OF A SYSTEM OF COPLANAR FORCES Scalar Notation x and y axes are designated positive and negative Components of forces expressed as algebraic scalars F F F x x F F y cos and F F sin y
ADDITION OF A SYSTEM OF COPLANAR FORCES (cont) Cartesian Vector Notation Cartesian unit vectors i and j are used to designate the x and y directions Unit vectors i and j have dimensionless magnitude of unity ( = 1 ) Magnitude is always a positive quantity, represented by scalars F x and F y F F i F j x y
ADDITION OF A SYSTEM OF COPLANAR FORCES (cont) Coplanar Force Resultants To determine resultant of several coplanar forces: Resolve force into x and y components F F i F j 1 1x 1y F F i F x F F i F j 3 3x 3 y Addition of the respective components using scalar algebra y j
ADDITION OF A SYSTEM OF COPLANAR FORCES (cont) Coplanar Force Resultants Vector resultant is therefore F F F F R 1 3 F i F j If scalar notation is used Rx Ry F F Rx Ry F F 1x 1y F F x y F F 3x 3y
ADDITION OF A SYSTEM OF COPLANAR FORCES (cont) Coplanar Force Resultants In all cases we have F F Rx Ry F F x y * Take note of sign conventions Magnitude of F R can be found by Pythagorean Theorem F R F Rx F Ry and tan -1 F F Ry Rx
EXAMPLE Determine x and y components of F 1 and F acting on the boom. Express each force as a Cartesian vector. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE (cont) Solution Scalar Notation F F 1x 1y 00sin 30 00cos 30 N 100N 100N N 173N 173N Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE (cont) Solution By similar triangles we have F F x y 1 60 13 60 5 13 Scalar Notation: 40N 100N F F x y 100N Cartesian Vector Notation: 40N F F 1 100N 100i 173 j 40i 100 jn N Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 3 The link is subjected to two forces F1 and F. Determine the magnitude and orientation of the resultant force. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 3 (cont) Scalar Notation: Solution I F F F F Rx Rx F 36.8N Ry Ry F x y : 600cos 30 : 600sin 30 58.8N N 400sin 45 N 400cos 45 N N Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 3 (cont) Resultant Force F R 69N 36.8N 58.8N Solution I From vector addition, direction angle θ is tan 1 67.9 58.8N 36.8N Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 3 (cont) Cartesian Vector Notation F1 = { 600 cos 30 i + 600 sin 30 j } N F = { -400 sin 45 i + 400 cos 45 j } N Solution II Thus, F R = F1 + F = (600 cos 30º N 400 sin 45º N) i + (600 sin 30º N + 400 cos 45º N) j = {36.8i + 58.8j} N The magnitude and direction of F R are determined in the same manner as before. Copyright 011 Pearson Education South Asia Pte Ltd
CARTESIAN VECTORS Right-Handed Coordinate System A rectangular or Cartesian coordinate system is said to be right-handed provided: Thumb of right hand points in the direction of the positive z axis z-axis for the D problem would be perpendicular, directed out of the page.
CARTESIAN VECTORS (cont) Rectangular Components of a Vector A vector A may have one, two or three rectangular components along the x, y and z axes, depending on orientation By two successive applications of the parallelogram law, A = A + A z A = A x + A y Combing the equations, A can be expressed as A = A x + A y + A z
CARTESIAN VECTORS (cont) Unit Vector Direction of A can be specified using a unit vector Unit vector has a magnitude of 1 If A is a vector having a magnitude of A 0, unit vector having the same direction as A is expressed by u A = A / A. So that A = A u A
CARTESIAN VECTORS (cont) Cartesian Vector Representations 3 components of A act in the positive i, j and k directions A = A x i + A y j + A Z k *Note the magnitude and direction of each components are separated, easing vector algebraic operations.
Magnitude of a Cartesian Vector From the colored triangle, From the shaded triangle, Combining the equations gives magnitude of A CARTESIAN VECTORS (cont) z y x A A A A ' y x A A A ' z A A A
CARTESIAN VECTORS (cont) Direction of a Cartesian Vector Orientation of A is defined as the coordinate direction angles α, β and γ measured between the tail of A and the positive x, y and z axes 0 α, β and γ 180 The direction cosines of A are cos A x A cos A z A cos A y A
CARTESIAN VECTORS (cont) Direction of a Cartesian Vector Angles α, β and γ can be determined by the inverse cosines Given A = A x i + A y j + A Z k then, u A = A /A = (A x /A)i + (A y /A)j + (A Z /A)k where A A x A y A z
CARTESIAN VECTORS (cont) Direction of a Cartesian Vector u A can also be expressed as u A = cosαi + cosβj + cosγk Since A A A A and u A = 1, we have cos x A as expressed in Cartesian vector form is A = Au A = Acosαi + Acosβj + Acosγk = A x i + A y j + A Z k y cos z cos 1
ADDITION AND SUBTRACTION OF CARTESIAN VECTORS Addition and Subtraction of Forces - Concurrent Force Systems: Force resultant is the vector sum of all the forces in the system F R = F = F x i + F y j + F z k
EXAMPLE 4 Express the force F as Cartesian vector. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 4 (cont) Since two angles are specified, the third angle is found by cos cos cos cos cos 1 Two possibilities exist, namely 60 cos 1 0. 5 cos 0. 5 cos 60 0 1 45. 707 1 ± 0. 5 Solution cos 1 0.5 10 Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 4 (cont) Solution By inspection, α = 60º since Fx is in the +x direction Given F = 00N F = Fcosαi + Fcosβj + Fcosγk = (00cos60ºN)i + (00cos60ºN)j + (00cos45ºN)k = {100.0i + 100.0j + 141.4k}N Checking: F F x F y F z 100.0 100.0 141.4 00N Copyright 011 Pearson Education South Asia Pte Ltd
POSITION VECTORS x,y,z Coordinates Right-handed coordinate system Positive z axis points upwards, measuring the height of an object or the altitude of a point Points are measured relative to the origin, O.
POSITION VECTORS (cont) Position Vector Position vector r is defined as a fixed vector which locates a point in space relative to another point. E.g. r = xi + yj + zk
POSITION VECTORS (cont) Position Vector Vector addition gives r A + r = r B Solving r = r B r A = (x B x A )i + (y B y A )j + (z B z A )k or r = (x B x A )i + (y B y A )j + (z B z A )k
POSITION VECTORS (cont) Length and direction of cable AB can be found by measuring A and B using the x, y, z axes Position vector r can be established Magnitude r represents the length of cable Angles, α, β and γ represent the direction of the cable Unit vector, u = r/r
EXAMPLE 5 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 5 (cont) Position vector r = [-m 1m]i + [m 0]j + [3m (-3m)]k = {-3i + j + 6k}m Solution Magnitude = length of the rubber band 3 6 m r 7 Unit vector in the director of r u = r /r = -3/7i + /7j + 6/7k Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 5 (cont) Solution Copyright 011 Pearson Education South Asia Pte Ltd
FORCE VECTOR DIRECTED ALONG A LINE In 3D problems, direction of F is specified by points, through which its line of action lies F can be formulated as a Cartesian vector: F = F u = F (r/r) Note that F has the unit of force (N) unlike r, with the unit of length (m)
FORCE VECTOR DIRECTED ALONG A LINE (cont) Force F acting along the chain can be presented as a Cartesian vector by - Establish x, y, z axes - Form a position vector r along length of chain Unit vector, u = r/r that defines the direction of both the chain and the force We get F = Fu
EXAMPLE 6 The man pulls on the cord with a force of 350 N. Represent this force acting on the support A, as a Cartesian vector and determine its direction. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 6 (cont) End points of the cord are: A (0m, 0m, 7.5m) and B (3m, -m, 1.5m) Solution r = (3m 0m)i + (-m 0m)j + (1.5m 7.5m)k = {3i j 6k}m Magnitude = length of cord AB 3m m 6m m r 7 Unit vector, u = r /r = 3/7i - /7j - 6/7k Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 6 (cont) Force F has a magnitude of 350 N, direction specified by u. Solution F = Fu = 350N(3/7i - /7j - 6/7k) = {150i - 100j - 300k}N α = cos -1 (3/7) = 64.6 β = cos -1 (-/7) = 107 γ = cos -1 (-6/7) = 149 Copyright 011 Pearson Education South Asia Pte Ltd
DOT PRODUCT Dot product of vectors A and B is written as A B (Read A dot B) Define the magnitudes of A and B and the angle between their tails A B = AB cosθ where 0 θ 180 Referred to as scalar product of vectors as result is a scalar
DOT PRODUCT (cont) Laws of Operation 1. Commutative law A B = B A. Multiplication by a scalar a(a B) = (aa) B = A (ab) = (A B)a 3. Distributive law A (B + D) = (A B) + (A D)
DOT PRODUCT (cont) Cartesian Vector Formulation - Dot product of Cartesian unit vectors i i = (1)(1)cos0 = 1 i j = (1)(1)cos90 = 0 - Similarly i i = 1 j j = 1 k k = 1 i j = 0 i k = 1 j k = 1
DOT PRODUCT (cont) Cartesian Vector Formulation Dot product of vectors A and B is: A B = A x B x + A y B y + A z B z Applications The angle formed between two vectors or intersecting lines. θ = cos -1 [(A B)/(AB)] 0 θ 180 The components of a vector parallel and perpendicular to a line. A a = A cos θ = A u a
EXAMPLE 7 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB. Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 7 (cont) Solution Since r i j k u B 6 3 B rb 0.86i 0.857 j 0.49k Thus FAB F. u B F 57.1N 6 3 cos 300 j 0.86i 0.857 j 0.49k (0)(0.86) (300)(0.857) (0)(0.49) Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 7 (cont) Solution Since result is a positive scalar, F AB has the same sense of direction as u B. Expressed in Cartesian form: Perpendicular component: Copyright 011 Pearson Education South Asia Pte Ltd
EXAMPLE 7 (cont) Solution Magnitude can be determined from F or from Pythagorean Theorem, F F 300N 57.1N 155N F AB Copyright 011 Pearson Education South Asia Pte Ltd
CONCEPT QUIZ (cont) 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 y 30 x F = 80 N
CONCEPT QUIZ (cont) ) Determine the magnitude of the resultant (F1 + F) force in N when: F1 = {10i + 0j} N F = {0i + 0j} N a) 30 N b) 40 N c) 50 N d) 60 N e) 70 N
CONCEPT QUIZ (cont) 3) Vector algebra, as we are going to use it, is based on a coordinate system. a) Euclidean b) Left-handed c) Greek d) Right-handed e) Egyptian
CONCEPT QUIZ (cont) 4) The symbols,, and designate the of a 3-D Cartesian vector. a) Unit vectors b) Coordinate direction angles c) Greek societies d) X, Y and Z components
CONCEPT QUIZ (cont) 5) What is not true about a unit vector, u A? a) It is dimensionless. b) Its magnitude is one. c) It always points in the direction of the positive X-axis. d) It always points in the direction of vector A.
CONCEPT QUIZ (cont) 6) If F = { 10 i + 10 j + 10 k } N and G = { 0 i + 0 j + 0 k } N, then F + G = { } N a) 10 i + 10 j + 10 k b) 30 i + 0 j + 30 k c) 10 i 10 j 10 k d) 30 i + 30 j + 30 k
CONCEPT QUIZ (cont) 7) A position vector, r PQ, is obtained by a) Coordinates of Q minus coordinates of P a) Coordinates of P minus coordinates of Q a) Coordinates of Q minus coordinates of the origin a) Coordinates of the origin minus coordinates of P
CONCEPT QUIZ (cont) 8) A force of magnitude F, directed along a unit vector U, is given by F =. a) F (U) b) U / F c) F / U d) F + U e) F U
CONCEPT QUIZ (cont) 9) P and Q are two points in a 3-D space. How are the position vectors r PQ and r QP related? a) r PQ = r QP b) r PQ = - r QP c) r PQ = 1/r QP d) r PQ = r QP
CONCEPT QUIZ (cont) 10) Two points in 3 D space have coordinates of P (1,, 3) and Q (4, 5, 6) meters. The position vector r QP is given by a) {3 i + 3 j + 3 k} m b) { 3 i 3 j 3 k} m c) {5 i + 7 j + 9 k} m d) { 3 i + 3 j + 3 k} m e) {4 i + 5 j + 6 k} m
CONCEPT QUIZ (cont) 13) Force vector F directed along a line PQ is given by a) (F/ F) r PQ b) r PQ /r PQ c) F(r PQ /r PQ ) d) F(r PQ /r PQ ) 14) The dot product of two vectors P & Q is defined as a) P Q cos b) P Q sin c) P Q tan d) P Q sec P Q
CONCEPT QUIZ (cont) 15) The dot product can be used to find all of the following except. a) sum of two vectors b) angle between two vectors c) component of a vector parallel to another line d) component of a vector perpendicular to another line
CONCEPT QUIZ (cont) 16) Find the dot product of the two vectors P and Q. P = { 5 i + j + 3 k} m Q = {- i + 5 j + 4 k} m a) -1 m b) 1 m c) 1 m d) -1 m e) 10 m