1. MOMENTS

2. If we consider a particle under the action of two equal and opposite forces: The particle will have a zero resultant, and will be in equilibrium. If however the forces do not act through the same point, e.g. on a beam There is still a zero resultant, but clearly the beam would not be in equilibrium. It would rotate. P P P P

3. The moment of a force, about a point is defined as, the product of the magnitude of the force, and the perpendicular distance of the line of action of the force from the point. The clockwise moment of the force of magnitude P, about point A, is given by: G A = A P d Pd The moment of a force is a measure of its ability to produce a turning effect.

4. Example 1: A light rod AD of length 6m has forces acting on it as shown. Find the sum of the clockwise moments about the points A and C. Taking clockwise moments about point A A ( which we can write as:: ). G A = 2 3 1 4 5 7 9 A B C D 9 × 6– 5 × 2 – 7 × 5 = 54 – 10 – 35 = 9 Nm C : G C = + 5 × 39 × 1– 4 × 5 = 9 + 15 – 20= 4Nm

5. Example 2: A light rod AD of length 6m has forces acting on it as shown. Find the sum of the clockwise moments about the points B and D. G B = 2 3 1 4 5 7 8 A B C D 8 × 4– 7 × 3 – 4 × 2 = 32 – 21 – 8 = 3 Nm G D = + 5 × 47 × 1– 4 × 6 = 7 + 20 – 24 = 3 Nm Note in this case, the sum of the moments about any point is the same. B : D : This is because the resultant force on the rod iszero.

6. Example 3: A light rod AD of length 5m has forces acting on it as shown. Find the sum of the clockwise moments about the points A and B. G A = 1 31 3 10 5 2 A B C D 10 × 3– 5 × 4 – 2 × 5 = 30 – 20 – 10 = 0 Nm G B = – 5 × 13 × 3– 2 × 2 = 9 – 5 – 4 = 0 Nm Note in this case, the sum of the moments about any point is the same. A : B : This is because there is a zero resultant force on the rod. In this case, the rod is in equilibrium.

7. Example 4: A uniform steel girder AB, has length 10m and weight 450N. A load of weight 120N is attached to the girder at A. The loaded girder hangs in equilibrium in a horizontal position, held by two vertical steel cables attached at points C and D, where AC = 1m and DB = 3m. Find the tensions in the two cables. 2 1 A B D T2T2 450 The beam is in equilibrium, so resolving: T 1 + 280 = 570 C : G C = 450 × 4– 120 × 1– T 2 × 6 In equilibrium, G C = 00 = 1800 – 120 – 6T 2 120 4 3 T1T1 C T 2 = 280 T 1 = 290

8. Example 5: A uniform beam AB, of length 6m and mass 40kg, is at rest on 2 supports at P and B, where AP = 1m. A mass of 20kg is placed on the beam at a distance x m from B, such that the normal reaction at P is double the reaction at B. Find the distance x. 2 1 A B N 2N 20g 40g P x 3 The beam is in equilibrium, so resolving:3N = 60g B : G B = 2N × 5– 40g × 3– 20g × x In equilibrium, G B = 00 = 200g – 120g – 20g xx = 4 Note: The position of the mass was not known, and it turns out to be to the left of the centre of the rod. N = 20g

9. Summary of key points: This PowerPoint produced by R.Collins ; Updated Apr. 2009 The moment of a force, about a point is defined as, the product of the magnitude of the force, and the perpendicular distance of the line of action of the force from the point. A P d G A =Pd A : If a rod is in equilibrium, the resultant force on the rod will be zero, and the sum of the moments about any point will also be zero. The clockwise moment of a force of magnitude P, about point A, is given by: