WJEC Physics for AS Level Student Book 2nd Edition

20 WJEC Physics for AS Level Fig. 1.1.21 Centres of gravity Study point Note that the dotted lines in Fig. 1.1.21 are not planes of symmetry but they do cross at the intersection of the planes of symmetry. With the aid of the PoM, we are in a position to solve a real-life problem! Example In Fig. 1.1.20 where must the fat cat sit to balance the other two on the see-saw? Answer The weights of the cats are (using mg ), 19.62 N , 29.43 N and 53.96 N respectively. Using PoM, the resultant moment about the pivot must be zero. \ Taking clockwise as positive: 53.96 d – 29.43 × 2.0 – 19.62 × 3.0 = 0 \ Solving this equation d = 2.18 m \ The fat cat must sit 2.18 m from the pivot. (c) The centre of gravity (C of G) In the last example, we treated the cats (of all sizes) as though they were point masses. This is obviously not true – they are spread out. However, for any object we can identify a point at which we can consider all its weight to act. This is called its centre of gravity . In a uniform gravitational field (which will always be the case in AS Physics) the C of G of a symmetric body, of uniform density, will lie on any plane of symmetry. Fig. 1.1.21 has the examples you are likely to meet: Fig. 1.1.20 Calculate d d 3.0 m 2.0 m 2.0 kg 3.0 kg 5.5 kg Study point Notice that the see-saw in the Example is pivoted at the centre. We’ll deal with more difficult situations in Section 1.1.7. Study point An alternative working is 5.5 g × d − 3.0 g × 2.0 − 2.0 g × 3.0 = 0 Cancelling the gs and rearranging 5.5 × d = 12 , etc. For a standing object, such as a bus or a racing car, the lower the centre of gravity and the wider the base, the more stable it is. That means that objects with low centres of gravity can be tipped more before they topple over. Knowledge check If the mass of the see-saw plank in the example is 10 kg , show that the pivot must exert an upward force of ~ 200 N on the plank. 1.1.10