WJEC Physics for AS Level Student Book 2nd Edition

21 1.1 Basic physics Fig. 1.1.23 shows the principle of this using a tall rectangular block. The block is just on the edge of tipping – it could go either way because the C of G is vertically above the balance point. From the geometry, tan θ = w h , where K and w are as shown in the diagram. Racing cars have a very wide wheel base and are very low on the ground, so their centres of gravity are also low. Fig. 1.1.24 shows a tilt test on an F1 car. 1.1.7 Conditions for equilibrium A body is said to be in equilibrium if it is moving and rotating at a constant rate . In many cases, especially when applied to engineering objects, such as bridges and buildings, this means it is not moving at all. In order for an object to be in equilibrium: 1. The resultant force on the object must be zero, and 2. The resultant moment (about any point) must be zero (the principle of moments). The object in Fig. 1.1.25 is clearly not in equilibrium: the resultant force is downwards and to the right and the resultant moment is clockwise (about the centre of gravity). What about the metre rule arrangement in Fig. 1.1.26? The rule weighs 1.0 N and we can Stretch & challenge For a rigid body, the position of the C of G is fixed within the body; but not for flexible bodies. the position moves. Where is the high jumper's C of G? Fig. 1.1.22 Fig. 1.1.23 Stability of a block Fig. 1.1.24 Stability testing w h θ θ Fig. 1.1.26 Balanced rule B F 2.0 N 3.0 N 1.0 N A 10 40 50 d distance fromA ( cm ) consider this weight to act at the 50 cm mark, the centre of gravity. Assuming it balances, what are the values of d and F ? Applying condition 1: The resultant force = 0 \ F = 3.0 N + 1.0 N + 2.0 N = 6.0 N , \ The pivot exerts an upward force of 6.0 N for the ruler to be in equilibrium. Fig. 1.1.25 This object is not in equilibrium W F 2 F 1 Study point In Fig. 1.1.26, there are two unknown quantities ( F and d ), so we need two equations to calculate them. Knowledge check Find d in Fig. 1.1.26 by assuming F = 6.0 N , (from applying Ȉ F = 0 ) and taking moments about B . 1.1.11