1. 2.6 Addition and Subtraction of Cartesian VectorsExampleGiven: A = Axi + Ayj + AZkand B = Bxi + Byj + BZkVector AdditionResultant R = A + B = (Ax + Bx)i + (Ay + By )j + (AZ + BZ) kVector SubstractionResultant R = A - B = (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k
2. 2.6 Addition and Subtraction of Cartesian Vectors Concurrent Force Systems - Force resultant is the vector sum of all the forces in the system FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzkwhere ∑Fx , ∑Fy and ∑Fz represent the algebraic sums of the x, y and z or i, j or k components of each force in the system
3. 2.6 Addition and Subtraction of Cartesian VectorsForce, F that the tie down rope exerts on theground support at O is directed along the ropeAngles α, β and γ can be solved with axes x, yand z
4. 2.6 Addition and Subtraction of Cartesian VectorsCosines of their values forms a unit vector u thatacts in the direction of the ropeForce F has a magnitude of F F = Fu = Fcosαi + Fcosβj + Fcosγk
5. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.8Express the force F as Cartesian vector
6. 2.6 Addition and Subtraction of Cartesian VectorsSolutionSince two angles are specified, the thirdangle is found bycos 2 α + cos 2 β + cos 2 γ = 1cos 2 α + cos 2 60o + cos 2 45o = 1cos α = 1 − (0.5) − (0.707 ) = ±0.5 2 2Two possibilities exit, namelyα = cos −1 (0.5) = 60o or α = cos −1 (− 0.5) = 120o
7. 2.6 Addition and Subtraction of Cartesian VectorsSolutionBy inspection, α = 60° since Fx is in the +x directionGiven F = 200N F = Fcosαi + Fcosβj + Fcosγk = (200cos60°N)i + (200cos60°N)j + (200cos45°N)k = {100.0i + 100.0j + 141.4k}NChecking: F = Fx2 + Fy2 + Fz2 = (100.0) + (100.0) + (141.4) 2 2 2 = 200 N
8. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.9Determine the magnitude and coordinatedirection angles of resultant force acting onthe ring
9. 2.6 Addition and Subtraction of Cartesian VectorsSolutionResultant force FR = ∑F = F1 + F2 = {60j + 80k}kN + {50i - 100j + 100k}kN = {50j -40k + 180k}kNMagnitude of FR is found by FR = (50)2 + (− 40)2 + (180)2 = 191.0 = 191kN
10. 2.6 Addition and Subtraction of Cartesian VectorsSolutionUnit vector acting in the direction of FR uFR = FR /FR = (50/191.0)i + (40/191.0)j + (180/191.0)k = 0.1617i - 0.2094j + 0.9422kSo that cosα = 0.2617 α = 74.8° cos β = -0.2094 β = 102° cosγ = 0.9422 γ = 19.6°*Note β > 90° since j component of uFR is negative
11. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.10Express the force F1 as a Cartesian vector.
12. 2.6 Addition and Subtraction of Cartesian VectorsSolutionThe angles of 60° and 45° are not coordinatedirection angles.By two successive applications ofparallelogram law,
13. 2.6 Addition and Subtraction of Cartesian VectorsSolutionBy trigonometry, F1z = 100sin60 °kN = 86.6kN F’ = 100cos60 °kN = 50kN F1x = 50cos45 °kN = 35.4kN kN F1y = 50sin45 °kN = 35.4kNF1y has a direction defined by –j,Therefore F1 = {35.4i – 35.4j + 86.6k}kN
14. 2.6 Addition and Subtraction of Cartesian VectorsSolutionChecking:F1 = F12 + F12 + F12 x y z= (35.4)2 + (− 35.4)2 + (86.6)2 = 100 NUnit vector acting in the direction of F1 u1 = F1 /F1 = (35.4/100)i - (35.4/100)j + (86.6/100)k = 0.354i - 0.354j + 0.866k
15. 2.6 Addition and Subtraction of Cartesian VectorsSolution α1 = cos-1(0.354) = 69.3° β1 = cos-1(-0.354) = 111° γ1 = cos-1(0.866) = 30.0°Using the same method, F2 = {106i + 184j - 212k}kN
16. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.11Two forces act on the hook. Specify thecoordinate direction angles of F2, so that theresultant force FR acts along the positive y axisand has a magnitude of 800N.
18. 2.6 Addition and Subtraction of Cartesian VectorsSolutionSince FR has a magnitude of 800N and actsin the +j directionFR = F1 + F2800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)kTo satisfy the equation, the correspondingcomponents on left and right sides must be equal
20. 2.7 Position Vectorsx,y,z Coordinates- Right-handed coordinate system- Positive z axis points upwards, measuringthe height of an object or the altitude of apoint- Points are measured relative to theorigin, O.
21. 2.7 Position Vectorsx,y,z CoordinatesEg: For Point A, xA = +4m along the x axis,yA = -6m along the y axis and zA = -6malong the z axis. Thus, A (4, 2, -6)Similarly, B (0, 2, 0) and C (6, -1, 4)
22. 2.7 Position VectorsPosition Vector- Position vector r is defined as a fixed vectorwhich locates a point in space relative to anotherpoint.Eg: If r extends from theorigin, O to point P (x, y, z)then, in Cartesian vectorform r = xi + yj + zk
23. 2.7 Position Vectors Position VectorNote the head to tail vector addition of thethree componentsStart at origin O, one travels x in the +i direction,y in the +j direction and z in the +k direction,arriving at point P (x, y, z)
24. 2.7 Position Vectors Position Vector - Position vector maybe directed from point A to point B - Designated by r or rABVector addition gives rA + r = rBSolving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)kor r = (xB – xA)i + (yB – yA)j + (zB –zA)k
25. 2.7 Position Vectors Position Vector - The i, j, k components of the positive vector r may be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB)Note the head to tail vector addition of thethree components
26. 2.7 Position VectorsLength and direction ofcable AB can be found bymeasuring A and B usingthe x, y, z axesPosition vector r can beestablishedMagnitude r representthe length of cable
27. 2.7 Position VectorsAngles, α, β and γrepresent the directionof the cableUnit vector, u = r/r
28. 2.7 Position VectorsExample 2.12An elastic rubber band isattached to points A and B.Determine its length and itsdirection measured from Atowards B.
29. 2.7 Position VectorsSolutionPosition vector r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}mMagnitude = length of the rubber band r= (− 3)2 + (2)2 + (6)2 = 7mUnit vector in the director of r u = r /r = -3/7i + 2/7j + 6/7k
31. 2.8 Force Vector Directed along a Line In 3D problems, direction of F is specified by 2 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 units offorces (N) unlike r, withunits of length (m)
32. 2.8 Force Vector Directed along a LineForce F acting along the chain can bepresented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length ofchain
33. 2.8 Force Vector Directed along a LineUnit vector, u = r/r that defines the directionof both the chain and the forceWe get F = Fu
34. 2.8 Force Vector Directed along a LineExample 2.13The man pulls on the cordwith a force of 350N.Represent this force actingon the support A, as aCartesian vector anddetermine its direction.
35. 2.8 Force Vector Directed along a LineSolutionEnd points of the cord are A (0m, 0m, 7.5m)and B (3m, -2m, 1.5m)r = (3m – 0m)i + (-2m – 0m)j + (1.5m – 7.5m)k = {3i – 2j – 6k}mMagnitude = length of cord AB r = (3m ) + (− 2m ) + (− 6m ) = 7m 2 2 2Unit vector, u = r /r = 3/7i - 2/7j - 6/7k
36. 2.8 Force Vector Directed along a LineSolutionForce F has a magnitude of 350N, directionspecified by u F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149°
37. 2.8 Force Vector Directed along a LineExample 2.14The circular plate ispartially supported bythe cable AB. If theforce of the cable on thehook at A is F = 500N,express F as aCartesian vector.
38. 2.8 Force Vector Directed along a LineSolutionEnd points of the cable are (0m, 0m, 2m) and B(1.707m, 0.707m, 0m) r = (1.707m – 0m)i + (0.707m – 0m)j + (0m – 2m)k = {1.707i + 0.707j - 2k}mMagnitude = length of cable ABr= (1.707m )2 + (0.707m )2 + (− 2m )2 = 2.723m
39. 2.8 Force Vector Directed along a LineSolutionUnit vector, u = r /r = (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k = 0.6269i + 0.2597j – 0.7345kFor force F, F = Fu = 500N(0.6269i + 0.2597j – 0.7345k) = {313i - 130j - 367k} N
40. 2.8 Force Vector Directed along a LineSolutionCheckingF= (313) + (130) + (− 367 ) 2 2 2= 500 NShow that γ = 137° andindicate this angle on thediagram
41. 2.8 Force Vector Directed along a LineExample 2.15The roof is supported bycables. If the cables exertFAB = 100N and FAC = 120Non the wall hook at A,determine the magnitude ofthe resultant force acting atA.
44. 2.8 Force Vector Directed along a LineSolutionFR = FAB + FAC = {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} NMagnitude of FR FR = (150.7 )2 + (40)2 + (− 150.7 )2 = 217 N
45. 2.9 Dot ProductDot product of vectors A and B is writtenas A·B (Read A dot B)Define the magnitudes of A and B and theangle between their tails A·B = AB cosθ where 0°≤ θ ≤180°Referred to as scalarproduct of vectors asresult is a scalar
46. 2.9 Dot Product Laws of Operation1. Commutative law A·B = B·A2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a3. Distribution law A·(B + D) = (A·B) + (A·D)
48. 2.9 Dot Product Cartesian Vector Formulation - Dot product of 2 vectors A and B A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk) = AxBx(i·i) + AxBy(i·j) + AxBz(i·k) + AyBx(j·i) + AyBy(j·j) + AyBz(j·k) + AzBx(k·i) + AzBy(k·j) + AzBz(k·k) = AxBx + AyBy + AzBzNote: since result is a scalar, be careful of including any unit vectors in the result
49. 2.9 Dot Product Applications - The angle formed between two vectors or intersecting lines θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°Note: if A·B = 0, cos-10= 90°, A isperpendicular to B
50. 2.9 Dot ProductApplications- The components of a vector parallel andperpendicular to a line- Component of A parallel or collinear with line aa’ isdefined by A║ (projection of A onto the line) A║ = A cos θ- If direction of line is specified by unit vector u (u =1), A║ = A cos θ = A·u
51. 2.9 Dot ProductApplications- If A║ is positive, A║ has a directionalsense same as u- If A║ is negative, A║ has a directionalsense opposite to u- A║ expressed as a vector A║ = A cos θ u = (A·u)u
52. 2.9 Dot ProductApplicationsFor component of A perpendicular to line aa’1. Since A = A║ + A┴, then A┴ = A - A║2. θ = cos-1 [(A·u)/(A)] then A┴ = Asinθ3. If A║ is known, by Pythagorean Theorem A⊥ = A2 + A||2
53. 2.9 Dot ProductFor angle θ between therope and the beam A,- Unit vectors along thebeams, uA = rA/rA- Unit vectors along theropes, ur=rr/rr- Angle θ = cos-1(rA.rr/rArr) = cos-1 (uA· ur)
54. 2.9 Dot ProductFor projection of the forcealong the beam A- Define direction of the beam uA = rA/rA- Force as a Cartesian vector F = F(rr/rr) = Fur- Dot product F║ = F║·uA
55. 2.9 Dot ProductExample 2.16The frame is subjected to a horizontal forceF = {300j} N. Determine the components ofthis force parallel and perpendicular to themember AB.
56. 2.9 Dot ProductSolutionSince r r r rr r 2i + 6 j + 3ku B = rB = rB (2)2 + (6)2 + (3)2 r r r= 0.286i + 0.857 j + 0.429kThen r rFAB = F cosθ rr r r r r= F .u B = (300 j ) ⋅ (0.286i + 0.857 j + 0.429k )= (0)(0.286) + (300)(0.857) + (0)(0.429)= 257.1N
57. 2.9 Dot ProductSolutionSince result is a positive scalar,FAB has the same sense ofdirection as uB. Express inCartesian formr r rFAB = FAB u AB r r r= (257.1N )(0.286i + 0.857 j + 0.429k ) r r r= {73.5i + 220 j + 110k }NPerpendicular componentr r r r r r r r r rF⊥ = F − FAB = 300 j − (73.5i + 220 j + 110k ) = {−73.5i + 80 j − 110k }N
58. 2.9 Dot ProductSolutionMagnitude can be determinedFrom F┴ or from PythagoreanTheoremr r2 r 2F⊥ = F − FAB= (300 N )2 − (257.1N )2= 155 N
59. 2.9 Dot ProductExample 2.17The pipe is subjected to F = 800N. Determine theangle θ between F and pipe segment BA, and themagnitudes of the components of F, which areparallel and perpendicular to BA.
60. 2.9 Dot ProductSolutionFor angle θrBA = {-2i - 2j + 1k}mrBC = {- 3j + 1k}mThus, r r rBA ⋅ rBC (− 2 )(0 ) + (− 2 )(− 3) + (1)(1) cos θ = r r = rBA rBC 3 10 = 0.7379 θ = 42.5o
61. 2.9 Dot ProductSolutionComponents ofr F r r rr r (−2i − 2 j + 1k )u AB = rAB = rAB 3 r  − 2 i +  − 2  r +  1  k r=   j    3   3   3 r rrFAB = F .u B r r  2 r  2  r  1  r= (− 758.9 j + 253.0k ) ⋅  − i +  −  j +  k  3   3   3= 0 + 506.0 + 84.3= 590 N
62. 2.9 Dot ProductSolutionChecking from trigonometry,r rFAB = F cos θ= 800 cos 42.5o N= 540 NMagnitude can be determinedFrom F┴r rF⊥ = F sin θ = 800 sin 42.5o = 540 N
63. 2.9 Dot ProductSolutionMagnitude can be determined from F┴ or fromPythagorean Theoremr r2 r 2F⊥ = F − FAB= (800)2 − (590)2= 540 N
64. Chapter SummaryParallelogram Law Addition of two vectors Components form the side and resultant form the diagonal of the parallelogram To obtain resultant, use tip to tail addition by triangle rule To obtain magnitudes and directions, use Law of Cosines and Law of Sines
65. Chapter SummaryCartesian Vectors Vector F resolved into Cartesian vector form F = Fxi + Fyj + Fzk Magnitude of F 2 2 2 F = Fx + Fy + Fz Coordinate direction angles α, β and γ are determined by the formulation of the unit vector in the direction of F u = (Fx/F)i + (Fy/F)j + (Fz/F)k
66. Chapter SummaryCartesian Vectors Components of u represent cosα, cosβ and cosγ These angles are related by cos2α + cos2β + cos2γ = 1Force and Position Vectors Position Vector is directed between 2 points Formulated by distance and direction moved along the x, y and z axes from tail to tip
67. Chapter SummaryForce and Position Vectors For line of action through the two points, it acts in the same direction of u as the position vector Force expressed as a Cartesian vector F = Fu = F(r/r)Dot Product Dot product between two vectors A and B A·B = AB cosθ
68. Chapter SummaryDot Product Dot product between two vectors A and B (vectors expressed as Cartesian form) A·B = AxBx + AyBy + AzBz For angle between the tails of two vectors θ = cos-1 [(A·B)/(AB)] For projected component of A onto an axis defined by its unit vector u A = A cos θ = A·u
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