2. 1© Manhattan Press (H.K.) Ltd. Centre of gravity Centre of mass Centre of mass 1.6 Centre of gravity Determine the centre of gravity of a lamina Determine the centre of gravity of a lamina

3. 2 © Manhattan Press (H.K.) Ltd. Centre of gravity 1.6 Centre of gravity (SB p. 68) The centre of gravity (CG) of an object is the point where the line of action of the weight of the object passes. move fingers towards each other centre of gravity = fingers meet (in equilibrium)

4. 3 © Manhattan Press (H.K.) Ltd. Centre of gravity 1.6 Centre of gravity (SB p. 68) Bend over with your legs against wall without toppling? You cannot Line of action of your weight acts outside the base of your feet and you topple over.

5. 4 © Manhattan Press (H.K.) Ltd. Centre of gravity 1.6 Centre of gravity (SB p. 68) You must not lean against the wall. Go to Example 12 Example 12

6. 5 © Manhattan Press (H.K.) Ltd. Centre of mass 1.6 Centre of gravity (SB p. 69) The centre of mass of an object is the point where the whole mass of the object is assumed to be concentrated. Centre of mass = Centre of gravity if the object is small or it is in a uniform gravitational field

7. 6 © Manhattan Press (H.K.) Ltd. Determine the centre of gravity of a lamina 1.6 Centre of gravity (SB p. 70) 1. Experimental method hung freely at A at rest, draw vertical line hung freely at B at rest, draw vertical line G is centre of gravity Line of action of its weight must pass through the centre of gravity.

8. 7 © Manhattan Press (H.K.) Ltd. Determine the centre of gravity of a lamina 1.6 Centre of gravity (SB p. 70) 2. Mathematical method Assume: Lamina is made up of n particles coordinate of each particle coordinate of CG

9. 8 © Manhattan Press (H.K.) Ltd. Determine the centre of gravity of a lamina 1.6 Centre of gravity (SB p. 71) 2. Mathematical method Taking moments about the y-axis, Mg( ) = (m 1 g 1 )x 1 + (m 2 g 2 )x 2 + (m 3 g 3 )x 3 +... + (m n g n )x n

10. 9 © Manhattan Press (H.K.) Ltd. Determine the centre of gravity of a lamina 1.6 Centre of gravity (SB p. 71) 2. Mathematical method If g = constant (in a uniform gravitational field), g 1 = g 2 = g 3 =... = g n = g, then Similarly, by taking moments about the x-axis,

11. 10 © Manhattan Press (H.K.) Ltd. Determine the centre of gravity of a lamina 1.6 Centre of gravity (SB p. 71) 2. Mathematical method coordinates of centre of gravity: Note: These are in fact the coordinates of the centre of mass. Go to Example 13 Example 13 Go to Example 14 Example 14 Go to Example 15 Example 15

12. 11 © Manhattan Press (H.K.) Ltd. End

13. 12 © Manhattan Press (H.K.) Ltd. Q: Q: Fig. (a) shows a person bending over with his back horizontal to lift a load of 60 N. The spine is considered as a rod pivoted at its base. The weight of the upper part of the body is 250 N acting at the centre of gravity G as shown in Fig. (b). Find (a) the tension T in his back muscle, (b) the horizontal compressional force H acted on the spine, and (c) the vertical reaction V at the pivot. Solution 1.6 Centre of gravity (SB p. 68) Fig. (a) Fig. (b)

14. 13 © Manhattan Press (H.K.) Ltd. Solution: (a) Since the spine is in equilibrium, sum of the moments about the pivot = 0. (b) Resultant of the forces along horizontal = 0. T cos12° − H = 0 H = 1 335 cos12° = 1 306 N (c) Resultant of forces vertically = 0. T sin12° + V – 60 – 250 = 0 V = 310 – 1 335 sin12° = 32.4 N 1.6 Centre of gravity (SB p. 69) It is noted that to lift heavy objects by the above way can strain the muscles in the back. Thus, we should lift objects with our “legs” instead, i.e. bend our keens while keeping the back straight and then pick up the object slowly. Return to Text

15. 14 © Manhattan Press (H.K.) Ltd. Q: Q: An object consists of two masses m 1 and m 2 attached by a light rod of length d as shown in the figure. Find the centre of gravity of the object in terms of m 1, m 2 and d. Solution 1.6 Centre of gravity (SB p. 71)

16. 15 © Manhattan Press (H.K.) Ltd. Solution: Let G be the centre of gravity of the object, 1.6 Centre of gravity (SB p. 71) Return to Text

17. 16 © Manhattan Press (H.K.) Ltd. Q: Q: Find the CG of the system of particles as shown in the figure. Solution 1.6 Centre of gravity (SB p. 72)

18. 17 © Manhattan Press (H.K.) Ltd. Solution: 1.6 Centre of gravity (SB p. 72) Return to Text

19. 18 © Manhattan Press (H.K.) Ltd. Q: Q: The figure shows a metal sheet in the shape of a square ABCE and an isosceles triangle DEC. If the length of the square is 12 cm and the height of the triangle is 9 cm, find the centre of mass of the metal sheet. Note: Centre of mass of a triangle = 2/3length of median from the apex Solution 1.6 Centre of gravity (SB p. 73)

20. 19 © Manhattan Press (H.K.) Ltd. Solution: Since the metal sheet is symmetric about OD, the centre of mass should lie on OD, i.e. x = 6 cm from AE. Centre of mass of the square is at G 1, 6 cm from the point O. Centre of mass of the triangle is at G 2, (2/3) × 9 = 6 cm from D. ∴ The distance OG 2 = 12 + 3 = 15 cm The mass of each portion is directly proportional to its area. ∴ Mass of the square ABCE = 144 units Mass of the triangle DEC = 54 units Total mass of the metal sheet = 198 units If the distance of the centre of mass from O along OD is y, then 198 y = 144 × 6 + 54 × 15 ∴ y = 8.45 cm from O 1.6 Centre of gravity (SB p. 73) Return to Text