1. CH 2: STATICS OF PARTICLES Force componentEquilibrium of a particle Chapter Objectives: To show addition of forces (resultant force) To resolve forces into their components To express force and position in Cartesian vectors. To analyse the equilibrium of forces acted on a particle

2. Introduction to Force Vectors Force » the action of one body on another. » a vector. » characterised by points of application, magnitude and direction. »represented by P, P or 45º 10 N

3. Parallelogram Diagram P + Q = Q + P P + Q ≠ P + Q P – Q = P + (-Q) P Q O R R Vector Addition

4. A B C a c b Law of Sinus a/sin A = b/sin B = c/sin C Law of Cosines a 2 = b 2 + c 2 – 2bc(cosA) b 2 = a 2 + c 2 – 2ac(cosB) c 2 = a 2 + b 2 – 2ab(cosC)

5. VECTOR OPERATION Vector A and its negative counterpart Scalar Multiplication and Division

6. Vector Addition Parallelogram Law Triangle construction

7. Vector Subtraction or R’ = A – B = A + (-B) A - B

8. Resolution of Vector Extend parallel lines from the head of R to form components

9. Sample Problem 2.1 The two forces act on bolt at A Determine their resultant Q = 60N P = 40N

10. Solution Q = 60N P = 40N

11. Solution Q = 60N P = 40N

12. A B C a c b Law of Cosines a 2 = b 2 + c 2 – 2bc(cosA) b 2 = a 2 + b 2 – 2ac(cosB) c 2 = a 2 + b 2 – 2ab(cosC) Solution P = 40N Q = 60N P = 40N Q = 60N

13. Law of Sinus a/sin A = b/sin B = c/sin C Solution Q = 60N P = 40N

14. Solution Sudut A Alternative Trigonometric Solution We construct the right triangle BCD and compute CD = (60N) sin 25˚ = 25.36 N BD = (60N) cos 25˚ = 54.38 N tan A = 25.36 N 94.38 N R = 25.36 sin A A=15.04˚ R=97.73N α = 20º + A = 35.04˚ R = 97.7 N, 35.0˚ P = 40N Q = 60N

15. SP 2.2 A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25 kN force directed along the axis of the barge, determine (a)the tension in each rope given α = 45 , (b) the value of α for which the tension in rope 2 is minimum. Sample Problem No 2.2

16. 25kN Solution Tension for Graphical Solution The parallelogram law is used; the diagonal (resultant) is known to be equal 25kN and to be directed to the right. This sides are drawn paralled to the ropes.If the drawing is done to scale, we measure T 1 = 18.5 kN T 2= 13.0 kN

17. 18.30kN 25kN 12.94kN 25kN

18. 12.5kN 21.65kN

19. 2.10 To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 300 N, determine a) The required angle if the responding R of the two forces applied at A is to be vertical b) The corresponding magnitude of R.

20. Solution (a) Using the triangle rule and the law of sines 360 N 300 N 35˚ R

21. Solution (b) β= ? We need to calculate the value of βto find the value of R Law of Sinus a/sin A = b/sin B = c/sin C R = 513 N 360 N 300 N 35˚ R

22. Rectangular Components of a Force x- y components Perpendicular to each other F = Fx i + Fy j Fx = F cos θ Fy = F sin θ Fy = Fy j Fx = Fx i F O x y θ i j

23. Example 1 A force of 800N is exerted on a bold A as shown in the diagram. Determine the horizontal and vertical components of the force. Fx = -F cos α= -(800 N ) cos 35º = -655N Fy = +F sin α= +(800 N ) sin 35º = +459N The vector components of F are thus Fx = -(655 N) i Fy = +(495 N) j And we may write F in the form F = - (655 N) i + (459 N) j

24. Example 2 A man pulls with a force of 300N on a rope attached to a building as shown in the picture. What are the horizontal and vertical components of the force exerted by the rope at point A? Fx = + (300N) cos αFy = - (300N) sin α Observing that AB = 10m cos α = 8m = 8m = 4 AB 10m 5 sin α = 6m = 6m = 3 AB 10m 5 We thus obtain Fx = + (300N) = +240 N 4545 Fy = - (300N) = -180 N 3535 We write F = + (240N) i - (180 N) j tan θ= Fx Fy (a) (b) F = (2.9)

25. Example 3 A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A. Determine the magnitude of the force and the angle θit forms with the horizontal. 6.7

26. Solution 6.7 First we draw the diagram showing the two rectangular components of the force and the angle θ. From Eq.(2.9), we write Using the formula given before, Fx = F cos θ Fy = F sin θ θ= 65.17˚

27. Addition of Cartesian Vectors R = P + Q Rxi + Ryj = Pxi + Pyj + Qxi + Qyj Rx= Px+ Qx Ry= Py + Qy Rx = Σ Fx Ry = Σ Fy

28. P S Q O R O PyjPyj SyjSyj SxiSxi QyjQyj PxiPxi QxiQxi RxiRxi RyiRyi

29. 2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS How we calculate the force ?

30. 2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

31. SAMPLE PROBLEM 2.3 F 2 = 80 N F 1 = 150 N F 4 = 100 N F 3 = 110 N Four forces act on A as shown. Determine the resultant of the forces on the bolt.

32. Solution F 2 = 80 N F 1 = 150 N F 4 = 100 N F 3 = 110 N - F 3 j - ( F 2 sin 20˚ ) i ( F 2 cos 20˚ ) j - ( F 4 sin 15˚ ) j ( F 4 cos 15˚ ) i ( F 1 sin 30˚ ) j ( F 1 cos 30˚ ) i

33. Solution F 2 = 80 N F 1 = 150 N F 4 = 100 N F 3 = 110 N - F 3 j - ( F 2 sin 20˚ ) i ( F 2 cos 20˚ ) j - ( F 4 sin 15˚ ) j ( F 4 cos 15˚ ) i ( F 1 sin 30˚ ) j ( F 1 cos 30˚ ) i

34. - F 3 j - ( F 2 sin 20˚ ) i ( F 2 cos 20˚ ) j - ( F 4 sin 15˚ ) j ( F 4 cos 15˚ ) i ( F 1 sin 30˚ ) j ( F 1 cos 30˚ ) i Ry = (14.3 N) jRx = (199.1 N) i

35. PROBLEMS

36. 2.21 Determine the x and y component of each of the forces shown.

37. Solution

39. Problems 2.23 1.2 m1.4 m 2.3 m 1.5 m 2.0 m 1800 N 950N 900N Determine the x and y components of each of the forces shown.

40. Solution 2.59m 2.69m 2.5 m

41. Solution 2.59m 2.69m 2.5 m

42. 2.9 EQUILIBRIUM OF A PARTICLE The resultant force acting on a particle is zero R = ΣF = 0 ΣFx = 0 ΣFy = 0 The forces of 20 N acting on the line but in opposite direction, passing through point A having the same magnitude. This produces a resultant of R = 0. A 20 N

43. Equilibrium of Forces F 4 = 400 N F 3 = 200 N F 2 = 173.2 F 1 = 300 O O F 2 = 173.2 N F 3 = 200 F 4 = 400 N 30  An equilibrium system of forces produces a closed force polygon

44. Example Figure 2.27 shows four forces acting on A. In figure 2.28, the resultant of the given forces is determined by the polygon rule. Starting from point O with F1 and arranging the forces in tip-to-tail fashion, we find that the tip of F4 coincides with the starting point O. Thus the resultant R of the given system of forces is zero, and the particle is in equilibrium. The closed polygon drawn in Fig 2.28 provides a graphical expression of the equilibrium of A. To express algebraically the condition for the equilibrium of a particle, we write Resolving each force F into rectangular components, we have We conclude that the necessary and sufficient conditions for the equilibrium of a particle are

45. Returning to the particle shown in figure 2.27, we check that the equilibrium conditions are satisfied. We write

46. Newton ’ s First Law of Motion If the resultant force acting on a particle is zero, the particle will remain at rest or will move with constant speed in a straight line.

47. Sample Problem 2.4 In a ship-unloading operation, a 15.6 kN automobile is supported by a cable. A rope is tied to the cable at A and pulled in order to centre the automobile over its intended position. The angle between the cable and the vertical is 2 , while the angle between the rope and the horizontal is 30 . What is the tension in the rope?

48. 15.6kN T AB T AC T AB

49. Sample problem 2.6 As part of the design of a new sailboat, it is desired to determine the drag force which may be expected at a given speed. To do so, a model of the proposed hull is placed in a test channel and three cables are used to keeps its bow on the centerline of the channel. Dynamometer reading indicate that for a given speed, the tension is 180N in cable AB and 270N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. 2.13m 0.46m 1.22m 180N 270N

50. 2.13m 0.46m 1.22m 180N 270N Solution Determine of the Angles First the angles α and βdefining the direction of cables AB and AC are determined. tanα 1.22m 2.13m = 1.75 = α = 60.26˚ tan β 1.22m 0.46m = 0.375 = β = 20.56˚ T AC T AE = 270 N FDFD T AB = 180 N β= 20.56 ˚ α = 60.26 ˚ Free body diagram Choosing the hull as a free body, we draw the free-body diagram shown. It includes the forces exerted by the three cables an the hull, as well as the drag force F D exerted by the flow.

51. T AC T AE = 270 N FDFD T AB = 180 N β= 20.56 ˚ α = 60.26 ˚ 2.13m 0.46m 1.22m 180N 270N Solution Equilibrium Condition R = T AB + T AC + T AE + F D We express that the hull is in the equilibriumby writing that the resultant of all forces is zero

52. Since more than three forces are involved, we resolved the forces into X and Y components T AC TAE = 270 N FDFD β= 20.56 ˚ α = 60.26 ˚ TAB = 180 N T AB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j T AC = T AC sin 20.56˚ i + T AC cos 20.56˚ j T AE = - (270N) j = -(156.3N) i + (89.3N) j = T AC 0.3512 i + T AC 0.9363 j F D = F D i R = T AB + T AC + T AE + F D from, (-156.3N + T AC 0.3512 + F D ) i + (89.3N + T AC 0.9363 – 270 N) j = 0 (180N) cos 60.26˚ j - (180N) sin 60.26˚ i - (270N) j X Y (180N) cos 60.26˚ j

53. T AB = -(180N) sin 60.26˚ i + (180N) cos 60.26˚ j T AC = T AC sin 20.56˚ i + T AC cos 20.56˚ j T AE = - (270N) j = -(156.3N) i + (89.3N) j = T AC 0.3512 i + T AC 0.9363 j F D = F D i R = T AB + T AC + T AE + F D from, (-156.3N + T AC 0.3512 + F D ) i + (89.3N + T AC 0.9363 – 270 N) j = 0 : - 156.3N + T AC 0.3512 + F D = 0 89.3N + T AC 0.9363 – 270 N = 0 T AC = 193 N F D = 88.5 N

54. THE END