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6161103 4.3 moment of force vector formulation

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1) Moment of force is calculated using the cross product of the position vector r and force vector F. 2) The magnitude of the moment is equal to the force F multiplied by the perpendicular distance d between the line of action of F and the point of reference. 3) The direction of the moment is determined by the right-hand rule applied to r and F.

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4.3 Moment of Force - Vector Formulation Moment of force F about point O can be expressed using cross product MO = r X F where r represents position vector from O to any point lying on the line of action of F
4.3 Moment of Force - Vector Formulation Magnitude For magnitude of cross product, MO = rF sinθ where θ is the angle measured between tails of r and F Treat r as a sliding vector. Since d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd
4.3 Moment of Force - Vector Formulation Direction Direction and sense of MO are determined by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F - Direction of MO is the same as the direction of the thumb
4.3 Moment of Force - Vector Formulation Direction *Note: - “curl” of the fingers indicates the sense of rotation - Maintain proper order of r and F since cross product is not commutative
4.3 Moment of Force - Vector Formulation Principle of Transmissibility For force F applied at any point A, moment created about O is MO = rA x F F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O
4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation For force expressed in Cartesian form, r r r i j k r r r M O = r XF = rx ry rz Fx Fy Fz where rx, ry, rz represent the x, y, z components of the position vector and Fx, Fy, Fz represent that of the force vector
4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation With the determinant expended, MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k MO is always perpendicular to the plane containing r and F Computation of moment by cross product is better than scalar for 3D problems
4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation Resultant moment of forces about point O can be determined by vector addition MRo = ∑(r x F)
4.3 Moment of Force - Vector Formulation Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant Given the perpendicular distance from A to cable is rd MA = rdF In 3D problems, MA = rBC x F
4.3 Moment of Force - Vector Formulation Example 4.4 The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.
4.3 Moment of Force - Vector Formulation Solution Either one of the two position vectors can be used for the solution, since MA = rB x F or MA = rC x F Position vectors are represented as rB = {1i + 3j + 2k} m and rC = {3i + 4j} m Force F has magnitude 60N and is directed from C to B
4.3 Moment of Force - Vector Formulation Solution r r F = (60 N )u F r  (1 − 3)i + 93 − 4) r + 92 − 0) k  r j = (60 N )     (−2) + (−1) + ( 2) 2 2 2   { } r r r = − 40i − 20 j + 40k N Substitute into determinant formulation r r r i j k r r r M A = rB XF = 1 3 2 − 40 − 20 40 { } r r r = [3(40) − 2(−20)]i − [1(40) − 2(−40)] j + [1(−20) − 3( 40)]k
4.3 Moment of Force - Vector Formulation Solution r r r i j k Or r r r M A = rC XF = 3 4 0 − 40 − 20 40 { } r r r = [4(40) − 0(−20)]i − [3(40) − 0(−40)] j + [3(−20) − 4(40)]k Substitute into determinant formulation { } r r r r M A = 160i − 120 j + 100k N .m For magnitude, r M A = (160) 2 + (120) 2 + (100) 2 = 224 N .m
4.3 Moment of Force - Vector Formulation Example 4.5 Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.
4.3 Moment of Force - Vector Formulation Solution Position vectors are directed from point O to each force rA = {5j} m and rB = {4i + 5j - 2k} m For resultant moment about O, r r r r r r r r M Ro = Σ( r XF ) = rA XF1 + rB XF2 + rC XF3 r r r r r r r r r i j k i j k i j k r r r = 0 5 0 + 0 5 0 + 4 5 − 2 = {30i − 40 j + 60k }N .m − 60 40 20 0 50 0 80 40 − 30
4.3 Moment of Force - Vector Formulation Solution For magnitude r M Ro = (30) 2 + (−40) 2 + (60) 2 = 78.10 N .m For unit vector defining the direction of moment axis, r r r r r M Ro 30i − 40 j + 60k u= r = M Ro 78.10 r r r = 0.3941i − 0.5121 j + 0.76852k
4.3 Moment of Force - Vector Formulation Solution For the coordinate angles of the moment axis, cos α = 0.3841; α = 67.4o cos β = −0.5121; β = 121o cos γ = 0.7682; γ = 39.8o

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6161103 4.3 moment of force vector formulation

  • 1. 4.3 Moment of Force - Vector Formulation Moment of force F about point O can be expressed using cross product MO = r X F where r represents position vector from O to any point lying on the line of action of F
  • 2. 4.3 Moment of Force - Vector Formulation Magnitude For magnitude of cross product, MO = rF sinθ where θ is the angle measured between tails of r and F Treat r as a sliding vector. Since d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd
  • 3. 4.3 Moment of Force - Vector Formulation Direction Direction and sense of MO are determined by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F - Direction of MO is the same as the direction of the thumb
  • 4. 4.3 Moment of Force - Vector Formulation Direction *Note: - “curl” of the fingers indicates the sense of rotation - Maintain proper order of r and F since cross product is not commutative
  • 5. 4.3 Moment of Force - Vector Formulation Principle of Transmissibility For force F applied at any point A, moment created about O is MO = rA x F F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O
  • 6. 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation For force expressed in Cartesian form, r r r i j k r r r M O = r XF = rx ry rz Fx Fy Fz where rx, ry, rz represent the x, y, z components of the position vector and Fx, Fy, Fz represent that of the force vector
  • 7. 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation With the determinant expended, MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k MO is always perpendicular to the plane containing r and F Computation of moment by cross product is better than scalar for 3D problems
  • 8. 4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation Resultant moment of forces about point O can be determined by vector addition MRo = ∑(r x F)
  • 9. 4.3 Moment of Force - Vector Formulation Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant Given the perpendicular distance from A to cable is rd MA = rdF In 3D problems, MA = rBC x F
  • 10. 4.3 Moment of Force - Vector Formulation Example 4.4 The pole is subjected to a 60N force that is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.
  • 11. 4.3 Moment of Force - Vector Formulation Solution Either one of the two position vectors can be used for the solution, since MA = rB x F or MA = rC x F Position vectors are represented as rB = {1i + 3j + 2k} m and rC = {3i + 4j} m Force F has magnitude 60N and is directed from C to B
  • 12. 4.3 Moment of Force - Vector Formulation Solution r r F = (60 N )u F r  (1 − 3)i + 93 − 4) r + 92 − 0) k  r j = (60 N )     (−2) + (−1) + ( 2) 2 2 2   { } r r r = − 40i − 20 j + 40k N Substitute into determinant formulation r r r i j k r r r M A = rB XF = 1 3 2 − 40 − 20 40 { } r r r = [3(40) − 2(−20)]i − [1(40) − 2(−40)] j + [1(−20) − 3( 40)]k
  • 13. 4.3 Moment of Force - Vector Formulation Solution r r r i j k Or r r r M A = rC XF = 3 4 0 − 40 − 20 40 { } r r r = [4(40) − 0(−20)]i − [3(40) − 0(−40)] j + [3(−20) − 4(40)]k Substitute into determinant formulation { } r r r r M A = 160i − 120 j + 100k N .m For magnitude, r M A = (160) 2 + (120) 2 + (100) 2 = 224 N .m
  • 14. 4.3 Moment of Force - Vector Formulation Example 4.5 Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.
  • 15. 4.3 Moment of Force - Vector Formulation Solution Position vectors are directed from point O to each force rA = {5j} m and rB = {4i + 5j - 2k} m For resultant moment about O, r r r r r r r r M Ro = Σ( r XF ) = rA XF1 + rB XF2 + rC XF3 r r r r r r r r r i j k i j k i j k r r r = 0 5 0 + 0 5 0 + 4 5 − 2 = {30i − 40 j + 60k }N .m − 60 40 20 0 50 0 80 40 − 30
  • 16. 4.3 Moment of Force - Vector Formulation Solution For magnitude r M Ro = (30) 2 + (−40) 2 + (60) 2 = 78.10 N .m For unit vector defining the direction of moment axis, r r r r r M Ro 30i − 40 j + 60k u= r = M Ro 78.10 r r r = 0.3941i − 0.5121 j + 0.76852k
  • 17. 4.3 Moment of Force - Vector Formulation Solution For the coordinate angles of the moment axis, cos α = 0.3841; α = 67.4o cos β = −0.5121; β = 121o cos γ = 0.7682; γ = 39.8o