1. Chapter 2 Statics of Particles

2. Addition of Forces Parallelogram Rule: The addition of two forces P and Q : A →P→P →Q→Q →P→P →Q→Q += →R→R Draw the diagonal vector represent the resultant force Draw two parallel lines of vectors to form a parallelogram

3. Addition of Forces Triangle Rule: A →P→P →Q→Q Alternative method to Parallelogram Rule Draw the vector from starting point to the tip of second vector Arrange two vectors in tip-to-tail fashion →P→P →Q→Q += →R→R

4. Addition of Forces Triangle Rule: A →P→P →Q→Q The order of the vectors does not matter →P→P →Q→Q + = →Q→Q →P→P + Addition of two forces is commutative →P→P →Q→Q += →R→R

5. Addition of Forces More than two forces: A →P→P →Q→Q Arrange the given vectors in tip-to-tail fashion →S→S Connect the tail of first vector to the tip of last vector →P→P →Q→Q += →R→R →S→S +

6. Addition of Forces More than two forces: A →P→P →Q→Q →S→S →P→P →Q→Q += →R→R →S→S + A →P→P →Q→Q →S→S The order of the vectors does not matter →S→S →P→P + = →Q→Q + →Q→Q →S→S + →P→P + = →P→P →Q→Q + →S→S +

7. Rectangular Components of a Force →F→F →Fx→Fx = →Fy→Fy + →Fx→Fx,Rectangular components of →F→F →Fy→Fy θ: Angle between x-axis and F measured from positive side of x-axis

8. Rectangular Components of a Force Unit Vectors → i, Unit vectors →j→j

9. Rectangular Components of a Force Unit Vectors → i, Unit vectors →j→j

10. Rectangular Components of a Force Example A Force of 800 N is applied on a bolt A. Determine the horizontal and vertical components of the force

11. Rectangular Components of a Force Example A man pulls with a force of 300 N on a rope attached to a building. Determine the horizontal and vertical components of the force applied by the rope at point A

12. Addition of Forces by Summing Their components A →P→P →Q→Q →S→S →P→P →Q→Q += →R→R →S→S +

13. Addition of Forces by Summing Their components - Example Problem 2.22 on page 33 7 kN 9 kN 5 kN Determine the resultant force applied on the boltForceMagnitude x Component y Component F 1 F 1 5 kN 5* cos (40) = 3.83 kN 5* sin (40) = 3.21 kN F 2 F 2 7 kN 7*cos(110)= -2.39 kN 7* sin (110)= 6.57 kN F 3 F 3 9 kN 9*cos(160)= -8.46 kN 9* sin(160)= 3.08 kN R -7.01 kN -7.01 kN 12.86 kN 12.86 kN

14. Equilibrium of Particle When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium

15. Equilibrium of Particle - Example Determine the resultant force applied on point AForceMagnitude x Component y Component F1F1F1F13003000 F2F2F2F2173.20-173.2 F3F3F3F3200-100-173.2 F4F4F4F4400-200346.4 R 0 0 N N N N 200*cos(240) = -100 200*sin(240) = -173.2

16. Equilibrium of Particle – Newton’s First Law If the resultant force acting on a particle is zero, the particle remains at rest (if originally at rest) the particle moves with a constant speed in a straight line ( if originally in motion) Equilibrium State

17. Equilibrium of Particle - Example Load with mass of 75 kg Determine the tensions in each ropes of AB and AC N N N 200*cos(240) = -100 200*sin(240) = -173.2 Since the load is in equilibrium state, The resultant force at A is zero.

18. Equilibrium of Particle – Example (continued) N N N 200*cos(240) = -100 200*sin(240) = -173.2 (1) (2)

19. Equilibrium of Particle – Example-2 Two cables are tied together at C and they are loaded as shown Determine the tensions in cable AC and BC Problem 2.44 / page 41

20. Equilibrium of Particle – Example-2 C Draw “Free Body” Diagram

21. Equilibrium of Particle – Example-2 C For simplicity

22. Equilibrium of Particle – Example-2 C Since the system is in equilibrium The resultant force at C is zero.

23. Equilibrium of Particle – Example-2 Since the system is in equilibrium The resultant force at C is zero. C

24. Equilibrium of Particle – Example-2 (1) (2) C

25. Equilibrium of Particle – Example-2 (1) (2) C

26. Forces in Space ( 3 D )

28. The three angle define the direction of the force F They are measured from the positive side of the axis to the force F They are always between 0 and 180º

29. Forces in Space ( 3 D )

30. The vector is the unit vector along the line of action of F

31. Forces in Space ( 3 D ) The vector is the unit vector along the line of action of F

32. Forces in Space ( 3 D )

33. Direction of the force is defined by the location of two points,

34. Forces in Space ( 3 D )

35. Sample Problem 2.7 The tension in the guy wire is 2500 N. Determine: a) components F x, F y, F z of the force acting on the bolt at A, b) the angles  x,  y,  z  defining the direction of the force

36. Determine the components of the force. Sample Problem 2.7

37. Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. or Sample Problem 2.7

38. Addition of Concurrent Forces in Space The resultant R of two or more vectors in space

39. Sample Problem 2.8 A concrete wall is temporarily held by the cables shown. The tension is 840 N in cable AB and 1200 N in cable AC. Determine the magnitude and direction of the resultant vector on stake A m m m m

40. Sample Problem 2.8

41. Equilibrium of a Particle in Space When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium

42. Problem 2.103 on page 60