2. 1© Manhattan Press (H.K.) Ltd. Scalar and vector quantities Manipulation of vectors Manipulation of vectors 1.3 Vectors and resolution of forces Forces in equilibrium Forces in equilibrium

3. 2 © Manhattan Press (H.K.) Ltd. Scalar and vector quantities 1.3 Vectors and resolution of forces (SB p. 47) A scalar quantity is one which can be described fully by just stating its magnitude. 1. Scalar quantities e.g. mass, time, length, temperature, density, speed, energy, volume

4. 3 © Manhattan Press (H.K.) Ltd. Scalar and vector quantities 1.3 Vectors and resolution of forces (SB p. 48) A vector quantity is one which can only be fully described if its magnitude and direction are stated. 2. Vector quantities e.g. displacement, velocity, acceleration, force, momentum, magnetic flux, electric intensity

5. 4 © Manhattan Press (H.K.) Ltd. Scalar and vector quantities 1.3 Vectors and resolution of forces (SB p. 48) 2. Vector quantities Represented by:

6. 5 © Manhattan Press (H.K.) Ltd. Scalar and vector quantities 1.3 Vectors and resolution of forces (SB p. 48) 2. Vector quantities

7. 6 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 48) 1. Addition of vectors (P and Q) parallelogram of vectors Go to Common Error Resultant P + Q

8. 7 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 49) 1. Addition of vectors (P and Q) triangle of vectors

9. 8 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 49) 2. Subtraction of vectors (P and Q)

10. 9 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 49) 3. Resolution of vector vector R = resultant of 2 vectors (A, B or C, D or E, F) A, B or C, D or E, F are components of R

11. 10 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 50) 3. Resolution of vector R to 2 perpendicular directions (resolution of vector)

12. 11 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 50) 3. Resolution of vector R x = R cosθ; R y = R sinθ tanθ = R y /R x R =

13. 12 © Manhattan Press (H.K.) Ltd. Manipulation of vectors 1.3 Vectors and resolution of forces (SB p. 51) 3. Resolution of vector e.g. (a) a horizontal component X = F cosθ (b) a vertical component Y = F sinθ

14. 13 © Manhattan Press (H.K.) Ltd. Forces in equilibrium 1.3 Vectors and resolution of forces (SB p. 52) Note: For simplicity, the vector notation ( → ) is omitted in describing the forces in the following chapters. 1. Addition of forces Go to More to Know 7 More to Know 7 P, Q, R, S, T in equilibrium?

15. 14 © Manhattan Press (H.K.) Ltd. Forces in equilibrium 1.3 Vectors and resolution of forces (SB p. 53) 1. Addition of forces The polygon of forces states that if the forces acting on a point can be represented in magnitude and direction by the sides of a polygon, then the forces are in equilibrium. In equilibrium (can form polygon)

16. 15 © Manhattan Press (H.K.) Ltd. Forces in equilibrium 1.3 Vectors and resolution of forces (SB p. 53) 2. Resultant of a number of forces Not in equilibrium (cannot form polygon) R = resultant of forces Go to Example 3 Example 3 Go to Example 4 Example 4

17. 16 © Manhattan Press (H.K.) Ltd. Forces in equilibrium 1.3 Vectors and resolution of forces (SB p. 55) 3. Forces in equilibrium Go to Example 5 Example 5 In equilibrium, algebraic sum of components = 0 Go to Common Error Go to Example 6 Example 6 Go to Example 7 Example 7 Go to Example 8 Example 8 Go to Example 9 Example 9

18. 17 © Manhattan Press (H.K.) Ltd. End

19. 18 © Manhattan Press (H.K.) Ltd. 1.3 Vectors and resolution of forces (SB p. 48) The magnitude of the resultant force is not necessarily greater than those of its components of forces. Return to Text

20. 19 © Manhattan Press (H.K.) Ltd. 1.3 Vectors and resolution of forces (SB p. 52) Concurrent & coplanar forces The forces shown in Fig. 1.22(a) are concurrent and coplanar forces. When the force vectors all intersect at one point, the forces are known as concurrent. When the problem involves forces in two dimensions only, the forces are said to be coplanar as they all lie on the same plane. Return to Text

21. 20 © Manhattan Press (H.K.) Ltd. Q: Q:The forces acting on a point O are shown in the figure below. Find the magnitude and direction of the resultant. Solution 1.3 Vectors and resolution of forces (SB p. 54)

22. 21 © Manhattan Press (H.K.) Ltd. Solution: Return to Text Algebraic sum of components of forces along Ox, X = A cos15° + B cos30° + C cos120° + D cos225° = 4 cos15° + 3 cos30° + 3 cos120° + 2 cos225° = 3.55 N Algebraic sum of components of forces along Oy, Y = 4 sin15° + 3 sin30° + 3 sin120° + 2 sin225° = 3.72 N The magnitude of the resultant R = = 5.14 N tanθ= ∴ θ= 46.3° The resultant is 5.14 N at an angle 46.3° to the direction Ox. 1.3 Vectors and resolution of forces (SB p. 54)

23. 22 © Manhattan Press (H.K.) Ltd. Q: Q: Sam, Brian and John attempt to push an object in the direction Ox. Sam exerts a force of 200 N at a direction 30° to Ox, and Brian exerts a force of 400 N at 60° to Ox as shown in the figure. What are the magnitude and direction of the smallest force that John should exert such that the resultant of all three forces acts along Ox? Solution 1.3 Vectors and resolution of forces (SB p. 54)

24. 23 © Manhattan Press (H.K.) Ltd. Solution: Return to Text In the figure, PQ represents the force F B = 400 N, QR represents the force F A = 200 N and PS along the Ox direction represents the resultant of the force F A + F B + F C, where F C is the smallest force that John exerts. Therefore, the direction of F C is in the negative Oy direction and the magnitude is: 400 cos30° – 200 cos60° = 246.4 N 1.3 Vectors and resolution of forces (SB p. 55)

25. 24 © Manhattan Press (H.K.) Ltd. Although the forces are in equilibrium, the object is not necessary to be in “static equilibrium”. We will learn the conditions for static equilibrium in Section 1.5. Return to Text 1.3 Vectors and resolution of forces (SB p. 55)

26. 25 © Manhattan Press (H.K.) Ltd. Q: Q: The figure shows a point O in equilibrium under the action of 5 coplanar forces. Calculate the values for θand x. Solution 1.3 Vectors and resolution of forces (SB p. 55)

27. 26 © Manhattan Press (H.K.) Ltd. Solution: Since the forces are in equilibrium, the algebraic sum of the components of the forces in any direction is zero. Algebraic sum of components of forces along OX = 0 3x sinθ + 2x cos90° + 4x cos180° + 13 cos270° + 2x sinθ = 0 5x sinθ − 4x = 0 sinθ = 4/5 θ = 53° 8’ Algebraic sum of components of forces along OY = 0 3x cosθ + 2x − 2x cosθ − 13 = 0 2x + x cosθ = 13 2 x + x (3/5) = 13 (13/5) x = 13 ∴ x = 5 1.3 Vectors and resolution of forces (SB p. 56) Return to Text

28. 27 © Manhattan Press (H.K.) Ltd. Q: Q: The figure shows a body S of weight W hanging vertically by a thread tied at L to the string KLM. Find the tension in the section KL if the system is in equilibrium. Solution 1.3 Vectors and resolution of forces (SB p. 56)

29. 28 © Manhattan Press (H.K.) Ltd. Solution: Since the forces W, T and F are in equilibrium, the resultant of the three forces must be zero. Therefore, the algebraic sum of the components of the forces in any direction must be zero. Algebraic sum of components of forces along LX = 0. F cos30° + (− T) = 0 F cos30° − T = 0 T = F cos30°............... (1) Algebraic sum of components of forces along LY = 0. F sin30° + (− W) = 0 F sin30° − W = 0 W = F sin30°................ (2) (1)/(2) 1.3 Vectors and resolution of forces (SB p. 56) Return to Text

30. 29 © Manhattan Press (H.K.) Ltd. Q: Q: The figure shows a picture of weight 10.0 N hanging freely by a cord EFG. Find the tension T in the cord. Solution 1.3 Vectors and resolution of forces (SB p. 57)

31. 30 © Manhattan Press (H.K.) Ltd. Solution: Since the picture is in equilibrium, algebraic sum of vertical components of the forces = 0. T sin60° + T sin60° − 10.0 = 0 2T sin60° = 10.0 ∴ T = = 5.77 N 1.3 Vectors and resolution of forces (SB p. 58) Return to Text

32. 31 © Manhattan Press (H.K.) Ltd. Q: Q: A mass of 5 kg hangs from two light strings of lengths 3 m and 4 m from two points at the same level and 5 m apart. Find the tension in each of the strings. (Assume g = 9.81 m s −2.) Solution 1.3 Vectors and resolution of forces (SB p. 58)

33. 32 © Manhattan Press (H.K.) Ltd. Solution: Since the forces are in equilibrium, algebraic sum of the forces along CX = 0. T 2 sin  −  T 1 cos  = 0 T 2 =..........................(1) Also, algebraic sum of the forces along CY = 0. T 2 cos  + T 1 sin  −  W = 0 Substituting T 2 from (1), 1.3 Vectors and resolution of forces (SB p. 59) Return to Text

34. 33 © Manhattan Press (H.K.) Ltd. Q: Q: An object of weight W rests in equilibrium on top of a smooth plane inclined at an angle θ to the horizontal as shown in Fig. (a). (a) Find F 1 and R 1 in terms of W and θ. (b) If the forces acted on the object are as shown in Fig. (b) and the object is in equilibrium, find F 2 and R 2 in terms of W and θ. Solution 1.3 Vectors and resolution of forces (SB p. 59) Fig. (a) Fig. (b)

35. 34 © Manhattan Press (H.K.) Ltd. Solution: (a) Since the object is in equilibrium, algebraic sum of components of forces along the inclined plane = 0. F 1 + (− W sinθ) = 0 ∴ F 1 = W sinθ Algebraic sum of components of forces perpendicular to the inclined plane = 0. R 1 + (− W cosθ) = 0 ∴ R 1 = W cosθ 1.3 Vectors and resolution of forces (SB p. 59) Return to Text

36. 35 © Manhattan Press (H.K.) Ltd. Solution (cont’d): (b) Since the object is in equilibrium, algebraic sum of components of forces along the inclined plane = 0. F 2 cosθ + (−W sinθ) = 0 ∴ F 2 = = W tanθ Algebraic sum of components of forces perpendicular to the inclined plane = 0. R 2 + (−F 2 sinθ) + (−W cosθ) = 0 ∴ R 2 = F 2 sinθ + W cosθ Substituting F 2, = (W tanθ) sinθ + W cosθ 1.3 Vectors and resolution of forces (SB p. 60) Return to Text