Engineering Mechanics: Statics in SI Units, 1e orce Vectors 1
Chapter Objectives Parallelogram Law Cartesian vector form Dot product and angle between vectors
Chapter Outline 1. Scalars and Vectors. Vector Operations 3. Vector Addition of orces 4. Addition of a System of Coplanar orces 5. Cartesian Vectors 6. Addition and Subtraction of Cartesian Vectors 7. Position Vectors 8. orce Vector Directed along a Line 9. Dot Product 3
.1 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 4
.1 Scalars and Vectors Vector A quantity that has magnitude and direction e.g. Position, force and moment Represent by a letter with an arrow over it, A Magnitude is designated as A In this subject, vector is presented as A and its magnitude (positive quantity) as A 5
. 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 6
. Vector Operations 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) 7
. Vector Operations Vector Subtraction - Special case of addition e.g. R = A B = A + ( - B ) - Rules of Vector Addition Applies 8
.3 Vector Addition of orces inding a Resultant orce Parallelogram law is carried out to find the resultant force Resultant, R = ( 1 + ) 9
.3 Vector Addition of orces Procedure for Analysis Parallelogram Law Make a sketch using the parallelogram law components forces add to form the resultant force Resultant force is shown by the diagonal of the parallelogram The components is shown by the sides of the parallelogram 10
.3 Vector Addition of orces Procedure for Analysis Trigonometry Redraw half portion of the parallelogram Magnitude of the resultant force can be determined by the law of cosines Direction if the resultant force can be determined by the law of sines Magnitude of the two components can be determined by the law of sines 11
Example.1 The screw eye is subjected to two forces, 1 and. Determine the magnitude and direction of the resultant force. 1
Solution Parallelogram Law Unknown: magnitude of R and angle θ 13
Solution Trigonometry Law of Cosines ( 100N ) + ( 150N ) ( 100N )( 150N ) cos115 + 500 30000( 0.46) = 1.6N = N R = = 10000 13 Law of Sines 150N 1.6N = sinθ sin115 150N sinθ = 1.6N θ = 39.8 ( 0.9063) 14
Solution Trigonometry Direction Φ of R measured from the horizontal φ = 39.8 = 54.8 + 15 φ 15
.4 Addition of a System of Coplanar orces Scalar Notation x and y axes are designated positive and negative Components of forces expressed as algebraic scalars = x x = + y cosθ and = sinθ y 16
.4 Addition of a System of Coplanar orces 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 x and y = i x + y j 17
.4 Addition of a System of Coplanar orces Coplanar orce Resultants To determine resultant of several coplanar forces: Resolve force into x and y components Addition of the respective components using scalar algebra Resultant force is found using the parallelogram law Cartesian vector notation: = i + j 1 3 = = 1x 3x x 1y i + i 3y y j j 18
.4 Addition of a System of Coplanar orces Coplanar orce Resultants Vector resultant is therefore R = 1 + + 3 = ( ) i + ( )j Rx Ry If scalar notation are used Rx Ry = = 1x 1y + x y + 3x 3y 19
.4 Addition of a System of Coplanar orces Coplanar orce Resultants In all cases we have Rx Ry = = x y * Take note of sign conventions Magnitude of R can be found by Pythagorean Theorem R = + and θ = tan Rx Ry -1 Ry Rx 0
Example.5 Determine x and y components of 1 and acting on the boom. Express each force as a Cartesian vector. 1
Solution Scalar Notation 1x 1y = 00sin30 = 00cos30 N = 100N = 100N N = 173N = 173N Hence, from the slope triangle, we have θ = tan 1 5 1
Solution By similar triangles we have 1 x = 60 = 40N 13 5 y = 60 = 100N 13 Scalar Notation: = x y 40N = 100N Cartesian Vector Notation: 1 = 100N { 100i + 173 j} = N = { 40i 100 j}n 3
Solution Scalar Notation 1x 1y = 00sin30 = 00cos30 N = 100N = 100N N = 173N = 173N 1 Hence, from the slope triangle, we have: θ = tan 1 5 1 Cartesian Vector Notation { 100i + 173 j} = N = { 40i 100 j}n 4
Example.6 The link is subjected to two forces 1 and. Determine the magnitude and orientation of the resultant force. 5
Solution I Scalar Notation: Rx Rx = Σ = = 36.8N Ry Ry = Σ = x y : 600cos30 : 600sin30 = 58.8N N 400sin 45 N + 400cos 45 N N 6
Solution I Resultant orce ( ) ( ) R = 36.8N + 58.8N = 69N rom vector addition, direction angle θ is 58.8 tan 1 N θ = 36.8N = 67.9 7
Solution II Cartesian Vector Notation 1 = { 600cos30 i + 600sin30 j } N = { -400sin45 i + 400cos45 j } N Thus, R = 1 + = (600cos30ºN - 400sin45ºN)i + (600sin30ºN + 400cos45ºN)j = {36.8i + 58.8j}N The magnitude and direction of R are determined in the same manner as before. 8
Solution II Resultant orce ( ) ( ) R = 36.8N + 58.8N = 69N rom vector addition, direction angle θ is 58.8 tan 1 N θ = 36.8N = 67.9 9
.5 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. 30
.5 Cartesian Vectors 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 application 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 31
.5 Cartesian Vectors 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 3
.5 Cartesian Vectors 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. 33
.5 Cartesian Vectors Magnitude of a Cartesian Vector rom the colored triangle, A = A ' + A z rom the shaded triangle, A ' = A + x A y Combining the equations gives magnitude of A A = A + A + x y A z 34
.5 Cartesian Vectors 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 is cosα = A x A cosγ = A z A cos β = A y A 35
.5 Cartesian Vectors 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 + A + x y A z 36
.5 Cartesian Vectors 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 y α + cos β + cos γ = 1 A as expressed in Cartesian vector form is A = Au A = Acosαi + Acosβj + Acosγk = A x i + A y j + A Z k z 37
.6 Addition and Subtraction of Cartesian Vectors Concurrent orce Systems orce resultant is the vector sum of all the forces in the system R = = x i + y j + z k 38
Example.8 Express the force as Cartesian vector. 39
Solution Since two angles are specified, the third angle is found by cos cos cos α = α + cos α + cos 1 - Two possibilities exist, namely ( ) 60 - α = cos 1 0. 5 = ( 0.5) 10 α = cos 1 = β + cos ( ) - ( ) 0. 5 + cos 60 γ = 1 45 0. 707 = 1 = ± 0. 5 40
Solution By inspection, α = 60º since x is in the +x direction Given = 00N = cosαi + cosβj + cosγk = (00cos60ºN)i + (00cos60ºN)j + (00cos45ºN)k = {100.0i + 100.0j + 141.4k}N Checking: = x + y + z = ( ) ( ) 100.0 100.0 ( 141.4) + + = 00N 41
.7 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. 4
.7 Position Vectors 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 43
.7 Position Vectors 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 44
.7 Position Vectors 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 represent the length of cable Angles, α, β and γ represent the direction of the cable Unit vector, u = r/r 45
Example.1 An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B. 46
Solution Position vector r = [-m 1m]i + [m 0]j + [3m (-3m)]k = {-3i + j + 6k}m 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 47
Solution α = cos -1 (-3/7) = 115 β = cos -1 (/7) = 73.4 γ = cos -1 (6/7) = 31.0 48
.8 orce Vector Directed along a Line In 3D problems, direction of is specified by points, through which its line of action lies can be formulated as a Cartesian vector = u = (r/r) Note that has units of forces (N) unlike r, with units of length (m) 49
.8 orce Vector Directed along a Line orce acting along the chain can be presented as a Cartesian vector by - Establish x, y, z axes - orm 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 = u 50
Example.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction. 51
Solution End points of the cord are A (0m, 0m, 7.5m) and B (3m, -m, 1.5m) 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 5
Solution orce has a magnitude of 350N, direction specified by u. = u = 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 53
.9 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 54
.9 Dot Product 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. Distribution law A (B + D) = (A B) + (A D) 55
.9 Dot Product Cartesian Vector ormulation - 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 = 0 j k = 0 56
.9 Dot Product Cartesian Vector ormulation Dot product of vectors A and B 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 57
Example.17 The frame is subjected to a horizontal force = {300j} N. Determine the components of this force parallel and perpendicular to the member AB. 58
Solution Since r i j k u B + 6 + 3 B = = rb = 0.86i + 0.857 j + 0.49k Thus AB = =. u B = = 57.1N ( ) + ( 6) + ( 3) cosθ ( 300 j ) ( 0.86i + 0.857 j + 0.49k ) = (0)(0.86) + (300)(0.857) + (0)(0.49) 59
Solution Since result is a positive scalar, AB has the same sense of direction as u B. Express in Cartesian form ( 57.1N )( 0.86i + 0.857 j + 0.49k ) = = {73.5i Perpendicular component AB = = AB u + 0 j + 110k} N AB AB = 300 j (73.5i + 0 j + 110k ) = { 73.5i + 80 j 110k} N 60
Solution Magnitude can be determined from or from Pythagorean Theorem, = = ( 300N ) ( 57.1N ) = 155N AB 61