WJEC Physics for AS Level Student Book 2nd Edition

Example Fig. 1.1.28 shows a load on a bridge. Calculate the forces, F 1 and F 2 , provided by the supports. Answer Apply the conditions for equilibrium. 1. Resultant force = 0 \ F 1 + F 2 = 10 000 + 5000 = 15 000 N [1] 2. Resultant moment about any point = 0 Take moments about the left-hand support with clockwise being positive. \ 10 000 N × 2.5 m + 5000 N × 4 m – F 2 × 5 m = 0 \ 5 F 2 = 45 000 N \ F 2 = 9000 N [2] Substitute the value for F 2 in equation [1] to calculate F 1 l F 1 = 6000 N 1.1 Basic physics Study point In finding two unknown forces, you need two equations. You could: 1. resolve once and take moments once, or 2. take moments about two different point. Hint: choose the point so that the moment of one of the unknown forces is zero. This simplifies the algebra. In the example (Fig. 1.1.28), this means taking moments about either the left- hand or the right-hand support. Fig. 1.1.28 Forces on a bridge F 2 F 1 5 m 1 m 5000 N 10000 N Example A pub sign is supported on a vertical wall by a hinge, H and a wire, W , as shown in Fig. 1.1.29. Calculate the tension, T , in W . Answer Take moments about H . ACM positive. Using the principle of moments: T sin 40° × 0.90 m – 80.0 N × 0.55 m = 0 \ T = 76.1 N Following on from this example, we can also use the equilibrium conditions to find the force exerted by the hinge on the pub sign bar – F in Fig. 1.1.29. We could do this using the triangle of forces: The three forces on the sign have a 0 resultant. Using the result for the tension, the triangle looks like this (Fig. 1.1.30). To find F and θ we use components and the fact that the resultant horizontal and vertical components are zero: \ Horizontally: F sin θ = T cos 40° \ F sin θ = 58.3 N [1] and vertically F cos θ + T cos 50° = 80 \ F cos θ = 31.1 N [2] If we then divide equation [1] by equation [2] we get tan θ = 58.3 31.1 = 1.875 . So we can calculate θ and hence F . Fig. 1.1.29 Pub sign 0.90 m 40° F H 0.55 m W T 80 N 23 Knowledge check Calculate the magnitude and direction of the force F in Fig. 1.1.29. 1.1.14 θ F 7 = 76.1 N 80 N 50° θ Fig 1.1.30 Triangle of forces