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# 6161103 2.8 force vector directed along a line

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### 6161103 2.8 force vector directed along a line

1. 1. 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)
2. 2. 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
3. 3. 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
4. 4. 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.
5. 5. 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
6. 6. 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°
7. 7. 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.
8. 8. 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
9. 9. 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
10. 10. 2.8 Force Vector Directed along a LineSolutionCheckingF= (313) + (130) + (− 367 ) 2 2 2= 500 NShow that γ = 137° andindicate this angle on thediagram
11. 11. 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.
12. 12. 2.8 Force Vector Directed along a LineSolutionrAB = (4m – 0m)i + (0m – 0m)j + (0m – 4m)k = {4i – 4k}mrAB = (4m )2 + (− 4m )2 = 5.66mFAB = 100N (rAB/r AB) = 100N {(4/5.66)i - (4/5.66)k} = {70.7i - 70.7k} N
13. 13. 2.8 Force Vector Directed along a LineSolutionrAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k = {4i + 2j – 4k}m rAC = (4m )2 + (2m )2 + (− 4m )2 = 6mFAC = 120N (rAB/r AB) = 120N {(4/6)i + (2/6)j - (4/6)k} = {80i + 40j – 80k} N
14. 14. 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