6161103 2.6 addition and subtraction of cartesian vectors

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6161103 2.6 addition and subtraction of cartesian vectors

  1. 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. 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. 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. 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. 5. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.8Express the force F as Cartesian vector
  6. 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. 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. 8. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.9Determine the magnitude and coordinatedirection angles of resultant force acting onthe ring
  9. 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. 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. 11. 2.6 Addition and Subtraction of Cartesian VectorsExample 2.10Express the force F1 as a Cartesian vector.
  12. 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. 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. 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. 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. 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.
  17. 17. 2.6 Addition and Subtraction of Cartesian VectorsSolutionCartesian vector formFR = F1 + F2F1 = F1cosα1i + F1cosβ1j + F1cosγ1k = (300cos45°N)i + (300cos60°N)j + (300cos120°N)k = {212.1i + 150j - 150k}NF2 = F2xi + F2yj + F2zk
  18. 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
  19. 19. 2.6 Addition and Subtraction of Cartesian VectorsSolutionHence, 0 = 212.1 + F2x F2x = -212.1N 800 = 150 + F2y F2y = 650N 0 = -150 + F2z F2z = 150NSince magnitude of F2 and its componentsare known,α1 = cos-1(-212.1/700) = 108°β1 = cos-1(650/700) = 21.8°γ1 = cos-1(150/700) = 77.6°

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