Recomendados
Más contenido relacionado
La actualidad más candente
La actualidad más candente (20)
Destacado
Destacado (12)
Similar a 6161103 2.3 vector addition of forces
Similar a 6161103 2.3 vector addition of forces (20)
Más de etcenterrbru
Más de etcenterrbru (20)
Último
Último (20)
6161103 2.3 vector addition of forces
- 1. 2.3 Vector Addition of Forces When two or more forces are added, successive applications of the parallelogram law is carried out to find the resultant Eg: Forces F1, F2 and F3 acts at a point O - First, find resultant of F1 + F2 - Resultant, FR = ( F1 + F2 ) + F3
- 2. 2.3 Vector Addition of Forces Example Fa and Fb are forces exerting on the hook. Resultant, Fc can be found using the parallelogram law Lines parallel to a and b from the heads of Fa and Fb are drawn to form a parallelogram Similarly, given Fc, Fa and Fb can be found
- 3. 2.3 Vector Addition of Forces Procedure for Analysis Parallelogram Law - Make a sketch using the parallelogram law - Two components forces add to form the resultant force - Resultant force is shown by the diagonal of the parallelogram - The components is shown by the sides of the parallelogram
- 4. 2.3 Vector Addition of Forces Procedure for Analysis Parallelogram Law To resolve a force into components along two axes directed from the tail of the force - Start at the head, constructing lines parallel to the axes - Label all the known and unknown force magnitudes and angles - Identify the two unknown components
- 5. 2.3 Vector Addition of Forces Procedure for Analysis Trigonometry - Redraw half portion of the parallelogram - Magnitude of the resultant force can be determined by the law of cosines - Direction if the resultant force can be determined by the law of sines
- 6. 2.3 Vector Addition of Forces Procedure for Analysis Trigonometry - Magnitude of the two components can be determined by the law of sines
- 7. 2.3 Vector Addition of Forces Example 2.1 The screw eye is subjected to two forces F1 and F2. Determine the magnitude and direction of the resultant force.
- 8. 2.3 Vector Addition of Forces Solution Parallelogram Law Unknown: magnitude of FR and angle θ
- 9. 2.3 Vector Addition of Forces Solution Trigonometry Law of Cosines FR = (100 N )2 + (150 N )2 − 2(100 N )(150 N ) cos115o = 10000 + 22500 − 30000(− 0.4226 ) = 212.6 N = 213 N
- 10. 2.3 Vector Addition of Forces Solution Trigonometry Law of Sines 150 N 212.6 N = sin θ sin 115o 150 N sin θ = (0.9063) 212.6 N sin θ = 39.8o
- 11. 2.3 Vector Addition of Forces Solution Trigonometry Direction Φ of FR measured from the horizontal φ = 39.8 + 15 o o = 54.8o ∠φ
- 12. 2.3 Vector Addition of Forces Example 2.2 Resolve the 1000 N ( ≈ 100kg) force acting on the pipe into the components in the (a) x and y directions, (b) and (b) x’ and y directions.
- 13. 2.3 Vector Addition of Forces Solution (a) Parallelogram Law F = Fx + Fy From the vector diagram, Fx = 1000 cos 40o = 766 N Fy = 1000 sin 40o = 643 N
- 14. 2.3 Vector Addition of Forces Solution (b) Parallelogram Law F = Fx + Fy '
- 15. 2.3 Vector Addition of Forces Solution (b) Law of Sines Fx ' 1000 N = sin 50o sin 60o sin 50o Fx ' = 1000 N sin 60o = 884.6 N Fy 1000 N o = sin 70 sin 60o sin 70o Fy = 1000 N sin 60o = 1085 N NOTE: A rough sketch drawn to scale will give some idea of the relative magnitude of the components, as calculated here.
- 16. 2.3 Vector Addition of Forces Example 2.3 The force F acting on the frame has a magnitude of 500N and is to be resolved into two components acting along the members AB and AC. Determine the angle θ, measured below the horizontal, so that components FAC is directed from A towards C and has a magnitude of 400N.
- 17. 2.3 Vector Addition of Forces Solution Parallelogram Law 500 N = FAB + FAC
- 18. 2.3 Vector Addition of Forces Solution Law of Sines 400 N 500 N = sin φ sin 60o 400 N sin φ = sin 60 o 500 N sin φ = 0.6928 φ = 43.9o
- 19. 2.3 Vector Addition of Forces Solution Hence, θ θ = 180 − 60 − 43.9 = 76.1 ∠ o o o o By Law of Cosines or Law of Sines Hence, show that FAB has a magnitude of 561N
- 20. 2.3 Vector Addition of Forces Solution F can be directed at an angle θ above the horizontal to produce the component FAC. Hence, show that θ = 16.1° and FAB = 161N
- 21. 2.3 Vector Addition of Forces Example 2.4 The ring is subjected to two forces F1 and F2. If it is required that the resultant force have a magnitude of 1kN and be directed vertically downward, determine (a) magnitude of F1 and F2 provided θ = 30°, and (b) the magnitudes of F1 and F2 if F2 is to be a minimum.
- 22. 2.3 Vector Addition of Forces Solution (a) Parallelogram Law Unknown: Forces F1 and F2
- 23. 2.3 Vector Addition of Forces Solution Law of Sines F1 1000 N o = sin 30 sin 130o F1 = 643N F2 1000 N o = sin 20 sin 130o F2 = 446 N
- 24. 2.3 Vector Addition of Forces Solution (b) Minimum length of F2 occur when its line of action is perpendicular to F1. Hence when θ = 90o − 20o = 70o F2 is a minimum
- 25. 2.3 Vector Addition of Forces Solution (b) From the vector diagram F1 = 1000 sin 70o N = 940 N F2 = 1000 cos 70o N = 342 N