Force Vector (cont’)

1. Force Vector (cont’)

2. 2.8 Force Vector Directed along a Line • Force F acting along the chain can be presented as a Cartesian vector by - Establish x, y, z axes - Form a position vector r along length of chain • Unit vector, u = r/r that defines the direction of both the chain and the force • We get F = Fu

3. Example 2.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction.

4. Solution End 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}m Magnitude = length of cord AB Unit vector, u = r /r = 3/7i - 2/7j - 6/7k

5. Solution Force F has a magnitude of 350N, direction specified 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°

6. 2.9 Dot Product • Dot product of vectors A and B is written as A·B (Read A dot B) • Define the magnitudes of A and B and the angle between their tails A·B = AB cosθ where 0°≤θ≤180° • Referred to as scalar product of vectors as result is a scalar

7. 2.9 Dot Product • Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D)

8. 2.9 Dot Product • Cartesian Vector Formulation - Dot product of Cartesian unit vectors i·i = (1)(1)cos0° = 1 i·j = (1)(1)cos90° = 0 - Similarly i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1

9. 2.9 Dot Product • Cartesian Vector Formulation • Dot product of 2 vectors A and B A·B = AxBx + AyBy + AzBz • Applications • The angle formed between two vectors or intersecting lines. θ = cos-1 [(A·B)/(AB)] 0°≤θ≤180° • The components of a vector parallel and perpendicular to a line. Aa = A cos θ = A·u

10. Example 2.17 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.

11. Solution Since Thus

12. Solution Since result is a positive scalar, FAB has the same sense of direction as uB. Express in Cartesian form Perpendicular component

13. Solution Magnitude can be determined from F┴ or from Pythagorean Theorem,

14. Quiz

15. y x 30° F = 80 N 1. Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity 2. For vector addition, you have to use ______ law. A) Newton’s Second B) the arithmetic C) Pascal’s D) the parallelogram 3. Can you resolve a 2-D vector along two directions, which are not at 90° to each other? A) Yes, but not uniquely. B) No. C) Yes, uniquely. 4. Can you resolve a 2-D vector along three directions (say at 0, 60, and 120°)? A) Yes, but not uniquely. B) No. C) Yes, uniquely. 5. Resolve F along x and y axes and write it in vector form. A) 80 cos (30°) i – 80 sin (30°) j B) 80 sin (30°) i + 80 cos (30°) j C) 80 sin (30°) i – 80 cos (30°) j D) 80 cos (30°) i + 80 sin (30°) j

16. 6. Determine the magnitude of the resultant (F1 + F2) force in N when F1={ 10i + 20j }N and F2={ 20i + 20j } N . A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N 7. Vector algebra, as we are going to use it, is based on a ___________ coordinate system. A) Euclidean B) Left-handed C) Greek D) Right-handed E) Egyptian 8. The symbols , , and  designate the __ of a 3-D Cartesian vector. A) Unit vectors B) Coordinate direction angles C) Greek societies D) X, Y and Z components 9. What is not true about an unit vector, uA ? A) It is dimensionless. B) Its magnitude is one. C) It always points in the direction of positive X- axis. D) It always points in the direction of vector A.

17. 10. If F = {10 i+ 10 j+ 10 k} N and G = {20 i+ 20j + 20 k } N, then F +G = { ____ } N A) 10 i + 10 j + 10 k B) 30 i + 20 j + 30 k C) – 10 i – 10 j – 10 k D) 30 i + 30 j + 30 k 12. A force of magnitude F, directed along a unit vector U, is given by F = A) F (U)B) U / FC) F / UD) F + U(E)F – U 13. P and Q are two points in a 3-D space. How are the position vectors rPQ and rQP related? A) rPQ = rQP B) rPQ = - rQP C) rPQ = 1/rQP D) rPQ = 2 rQP 14. If F and r are force vector and position vectors, respectively, in SI units, what are the units of the expression (r * (F / F)) ? A) Newton B) Dimensionless C) Meter D) Newton – Meter E) The expression is algebraically illegal.

18. 15. Two points in 3 – D space have coordinates of P (1, 2, 3) and Q (4, 5, 6) meters. The position vector rQP is given by A) {3 i + 3 j + 3 k}m B) {– 3 i – 3 j – 3 k}m C) {5 i + 7 j + 9 k} mD) {– 3 i + 3j + 3 k} m E) {4 i + 5 j + 6 k} m 16. Force vector, F, directed along a line PQ is given by A) (F/ F) rPQ B) rPQ/rPQ C) F(rPQ/rPQ) D) F(rPQ/rPQ) 17. The dot product of two vectors P and Q is defined as A) P Q cos  B) P Q sin  C) P Q tan  D) P Q sec  18. The dot product of two vectors results in a _________ quantity. A) Scalar B) Vector C) Complex D) Zero

19. 19. If a dot product of two non-zero vectors is 0, then the two vectors must be _____________ to each other. A) Parallel (pointing in the same direction) B) Parallel (pointing in the opposite direction) C) Perpendicular D) Cannot be determined. 20. If a dot product of two non-zero vectors equals -1, then the vectors must be ________ to each other. A) Parallel (pointing in the same direction) B) Parallel (pointing in the opposite direction) C) Perpendicular D) Cannot be determined. 21. The dot product can be used to find all of the following except ____ . A) sum of two vectors B) angle between two vectors C) component of a vector parallel to another line D) component of a vector perpendicular to another line

20. 22. Find the dot product of the two vectors P and Q. P = {5 i + 2 j + 3 k} m Q = {-2 i + 5 j + 4 k} m A) -12 m B) 12 m C) 12 m2 D) -12 m2 E) 10 m2

21. Next Chapter...

22. Equilibrium of a particle

23. Chapter Objectives • Concept of the free-body diagram for a particle • Solve particle equilibrium problems using the equations of equilibrium

24. Chapter Outline • Condition for the Equilibrium of a Particle • The Free-Body Diagram • Coplanar Systems • Three-Dimensional Force Systems

25. 3.1 Condition for the Equilibrium of a Particle • Particle at equilibrium if - At rest - Moving at constant a constant velocity • Newton’s first law of motion ∑F = 0 where ∑F is the vector sum of all the forces acting on the particle

26. 3.1 Condition for the Equilibrium of a Particle • Newton’s second law of motion ∑F = ma • When the force fulfill Newton's first law of motion, ma = 0 a = 0 therefore, the particle is moving in constant velocity or at rest

27. 3.2 The Free-Body Diagram • Best representation of all the unknown forces (∑F) which acts on a body • A sketch showing the particle “free” from the surroundings with all the forces acting on it • Consider two common connections in this subject – • Spring • Cables and Pulleys

28. 3.2 The Free-Body Diagram • Spring • Linear elastic spring: change in length is directly proportional to the force acting on it • spring constant or stiffness k: defines the elasticity of the spring • Magnitude of force when spring is elongated or compressed F = ks

29. 3.2 The Free-Body Diagram • Cables and Pulley • Cables (or cords) are assumed negligible weight and cannot stretch • Tension always acts in the direction of the cable • Tension force must have a constant magnitude for equilibrium • For any angle θ, the cableis subjected to a constant tension T

30. Procedure for Drawing a FBD 1. Draw outlined shape 2. Show all the forces - Active forces: particle in motion - Reactive forces: constraints that prevent motion 3. Identify each forces - Known forces with proper magnitude and direction - Letters used to represent magnitude and directions

31. Example 3.1 The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C.

32. Solution FBD at Sphere Two forces acting, weight and the force on cord CE. Weight of 6kg (9.81m/s2) = 58.9N Cord CE Two forces acting: sphere and knot Newton’s 3rd Law: FCE is equal but opposite FCE and FEC pull the cord in tension For equilibrium, FCE = FEC

33. Solution FBD at Knot 3 forces acting: cord CBA, cord CE and spring CD Important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord CE

34. 3.3 Coplanar Systems • A particle is subjected to coplanar forces in the x-y plane • Resolve into i and j components for equilibrium ∑Fx= 0 ∑Fy= 0 • Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero

35. Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes - Label all the unknown and known forces 2. Equations of Equilibrium - Apply F = ks to find spring force - When negative result force- indicates its sense is reverse of that shown on the free body diagram - Apply the equations of equilibrium ∑Fx= 0 ∑Fy= 0

36. Example 3.4 Determine the required length of the cord AC so that the 8kg lamp is suspended. The undeformed length of the spring AB is l’AB = 0.4m, and the spring has a stiffness of kAB = 300N/m.

37. Solution 1. Draw FBD at Point A Three forces acting, force by cable AC, force in spring AB and weight of the lamp. If force on cable AB is known, stretch of the spring is found by F = ks. +→ ∑Fx= 0; TAB–TAC cos30º = 0 +↑ ∑Fy= 0; TABsin30º –78.5N= 0 Solving, TAC = 157.0kN TAB = 136.0kN

38. Solution TAB = kABsAB; 136.0N = 300N/m(sAB) sAB = 0.453N For stretched length, lAB =l’AB+ sAB lAB =0.4m + 0.453m = 0.853m For horizontal distance BC, 2m = lACcos30° + 0.853m lAC = 1.32m

39. 3.4 Three-Dimensional Force Systems • For particle equilibrium ∑F = 0 • Resolving into i, j, k components ∑Fxi + ∑Fyj + ∑Fzk = 0 • Three scalar equations representing algebraic sums of the x, y, z forces ∑Fxi = 0 ∑Fyj = 0 ∑Fzk = 0

40. Procedure for Analysis Free-body Diagram • Establish the z, y, z axes • Label all known and unknown force Equations of Equilibrium • Apply ∑Fx = 0, ∑Fy= 0 and ∑Fz= 0 • Substitute vectors into ∑F = 0 and set i, j, k components = 0 • Negative results indicate that the sense of the force is opposite to that shown in the FBD.

41. Example 3.7 Determine the force developed in each cable used to support the 40kN crate.

42. Solution 1. Draw FBD at Point A To expose all three unknown forces in the cables. 2. Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB(rB / rB) = -0.318FBi – 0.424FBj + 0.848FBk FC = FC(rC / rC) = -0.318FCi – 0.424FCj + 0.848FCk FD = FDi W = -40k

43. Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 -0.318FBi – 0.424FBj + 0.848FBk - 0.318FCi – 0.424FCj + 0.848FCk + FDi - 40k = 0 ∑Fx = 0; -0.318FB - 0.318FC+ FD = 0 ∑Fy= 0; – 0.424FB– 0.424FC= 0 ∑Fz= 0; 0.848FB + 0.848FC- 40 = 0 Solving, FB = FC = 23.6kN FD = 15.0kN

44. QUIZ

45. 1. When a particle is in equilibrium, the sum of forces acting on it equals ___ . (Choose the most appropriate answer) • A constant • A positive number • Zero • A negative number • An integer

46. T1 T2 2. For a frictionless pulley and cable, tensions in the cables are related as A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin 

47. 100 N 100 N 100 N ( A ) ( B ) ( C ) 3. Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above? 4. Why? A) The weight is too heavy. B) The cables are too thin. C) There are more unknowns than equations. D) There are too few cables for a 100 kg weight.

48. A 30 40 F1 F2 A A) B) 100 kg 30 40° 100 kg A F2 F F1 D) C) 30° 40° 30° A A 100 kg 100 kg 5. Select the correct FBD of particle A.

49. 6. Using this FBD of Point C, the sum of forces in the x-direction ( FX)is ___ . Use a sign convention of +  . A) F2 sin 50° – 20 = 0 B) F2 cos 50° – 20 = 0 C) F2 sin 50° – F1 = 0 D) F2 cos 50° + 20 = 0 F2 50o C 20kN

50. 7. In 3-D, when a particle is in equilibrium, which of the following equations apply? A) ( Fx) i + ( Fy)j + ( Fz) k = 0 B)  F = 0 C)  Fx =  Fy =  Fz = 0 D) All of the above. E) None of the above.