11-30-07 - Vector Addition
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This document discusses concurrent and equilibrant forces. It defines concurrent forces as acting at the same time and place, with the resultant being the sum of two or more vectors. The equilibrant force is a single additional force with the same magnitude but opposite direction as the resultant, producing equilibrium when combined with the resultant. An example problem calculates the resultant and equilibrant forces of several concurrent forces acting on an object.
11-30-07 - Vector Addition
- 1. Vector Addition Concurrent and Equilibrant Forces
- 2. Definitions Concurrent Forces – Acting at the same time and same place Resultant – Sum of 2 or more vectors Equilibrant Force – A single, additional force that is exerted on an object Same magnitude, but opposite direction of the Resultant When combined with the Resultant, it produces equilibrium Net force = 0
- 3. Example Problem Question: Find the Equilibrant force of these Concurrent forces analytically 12 N south, 31 N west, 29 N north, 56 N west 2 ways to determine resultant Simplify to 2 vectors (use Parallelogram method to find resultant) OR Draw all 4 Head-to-Tail (to find resultant – Start at the beginning and end at the end)
- 4. Example Problem 12 N south, 31 N west, 29 N north, 56 N west Simplify to 2 vectors (use Parallelogram method to find resultant) 12 N south + 29 N north = ? (-12 N north) + 29 N north = 17 N north 31 N west + 56 N west = 87 N west
- 5. Example Problem – Not Drawn to Scale 12 N south, 31 N west, 29 N north, 56 N west Simplify to 17 N north and 87 N west Use Parallelogram method to find resultant 17 N 87 N F R
- 6. Example Problem – Not Drawn to Scale 17 N north and 87 N west Solve for F R using Pythagorean Theorem a 2 + b 2 = c 2 (17 2 ) + (87 2 ) = F R 2 √ ( 7858) = 89 N 17 N 87 N F R = 89 N
- 7. Example Problem – Not Drawn to Scale Magnitude of F R = 89 N To find direction ( Θ ) we will use the Tangent function TOA: T angent Θ = O pposite/ A djacent tan Θ = (17 N) / (87 N) Θ = tan -1 (0.2) Θ = 11° 17 N 87 N (Adjacent) F R = 89 N Θ = 11 ° 17 N (Opposite)
- 8. Example Problem – Not Drawn to Scale Magnitude of F R = 89 N Direction = ? 180° - 11° = 169° Therefore F R = 89 N @ 169° Θ = 11 ° 0 ° 180 ° 169 °
- 9. Example Problem – Not Drawn to Scale F R = 89 N @ 169° Original Question - Find the Equilibrant force of these concurrent forces analytically Equilibrant is same magnitude, opposite direction of Resultant 0 ° 180 ° 169 ° F R F E
- 10. Example Problem – Not Drawn to Scale F R = 89 N @ 169° F E = 89 N @ ?? Because we know that it is the exact opposite direction – we can add 180 ° to the direction of F R 169 ° + 180 ° = 349 ° 0 ° 180 ° 169 ° F R F E 180 ° 349 °
- 11. Example Problem – Not Drawn to Scale F R = 89 N @ 169° F E = 89 N @ 349 ° 0 ° 180 ° 169 ° F R F E 180 ° 349 °
- 12. Solve Analytically – Using equations 30 N @ 0 ° ; 40 N @ 90 ° 20 N @ 180 ° ; 15 N @ 270 ° 18 N @ 360 ° ; 22 N @ 270 ° 44 N @ 270 ° ; 12 N @ 360 ° 10 N @ 0 ° ; 20 N @ 180 ° ; 14 N @ 90 ° ; 20 N @ 270 °