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Vector Moment Analysis (4.2 - 4.4)

Vector Moment Analysis (4.2 - 4.4). M O = F d. M O ┴ F ┴ d. d. Magnitude of moment is still given by product of force and perpendicular distance. Direction is perpendicular to line of F and d. Can be difficult to determine in 3D using scalars. Vector (Cross) Product. C = A x B.

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Vector Moment Analysis (4.2 - 4.4)

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  1. Vector Moment Analysis (4.2 - 4.4) MO = F d MO┴ F ┴d d • Magnitude of moment is still given by product of force and perpendicular distance. • Direction is perpendicular to line of F and d. • Can be difficult to determine in 3D using scalars.

  2. Vector (Cross) Product C = A x B A x B = − (B x A) • Vector product of two vectors Aand Bis vector C: • Line of action of Cis perpendicular to plane of Aand B. • Magnitude of Cis • Direction of Cis obtained from the right-hand rule.

  3. Cartesian Unit Vectors

  4. MO = r x F Magnitude of MO MO = r F sinq MO = F d Any position vector r that connects O to line of F gives same result.

  5. Example in 2D: Find MO N 2.89 m F = 100 N 5 m 30° O

  6. Evaluating Cross Products

  7. Example 2a: Find moment of F about point O

  8. Example 2a’: Find moment of F about point O

  9. Example 2b: Find perpendicular distance from O to line of F

  10. Example 2c: Find moment of F about OA axis(Projection of MO along line of OA)

  11. EXAMPLE # 3 Given: a = 3 in, b = 6 in and c = 2 in. Find: Moment of F about point O. Plan: 1) Find rOA. 2) Determine MO = rOA F. o SolutionrOA= {3 i + 6 j– 0 k} in i j k MO = = [{6(-1) – 0(2)} i– {3(-1) – 0(3)} j + {3(2) – 6(3)} k] lb·in = {-6 i + 3 j – 12 k} lb·in 3 6 0 3 2 -1

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