1. Teach A Level Maths Moment of a Force

2. Volume 4: Mechanics 1 Moment of a Force Volume 4: Mechanics 1 Moment of a Force

3. We can balance a ruler on an outstretched finger if the middle of the ruler is on our finger. A small movement of the ruler results in it starting to tip ( and then it will probably slide ). It’s impossible to balance the ruler if we place our finger away from the centre. We can model the ruler as a thin, uniform rod; its mass is evenly distributed.

4. However the forces are the same in the 2 situations. but this is not. In the 2 nd situation, the ruler is not in equilibrium because there is a turning effect. R W R W Drawing the forces on the ruler, this is possible... When we draw a force diagram, because the body is uniform, we can show the weight acting at the centre. The turning effect of a force is called the moment of the force.

5. W A R 2 The moment of the weight in this diagram has an anti-clockwise moment about A. By also supporting the ruler to the left of W, we can again make it rest in equilibrium. R W A B R 2 The normal reaction at B has a clockwise moment about A.

6. W A R 2 B R 2 The force at B is half the size of the weight, yet its moment balances that of the weight. Distance as well as magnitude effects the size of the moment of a force.

7. W A x The moment of a force, F, about a point A is defined as F  d where d is the perpendicular distance from A to the line of action of F. d The line of action of W An anti-clockwise moment is taken as positive, so the moment of W about A is...  Wd The moment of W about P is... P x  Wx. x e.g.

8. W P A The moment of a force, F, about a point A is defined as F  d where d is the perpendicular distance from A to the line of action of F. If supports are placed at A and P, there will be reaction forces at these points. RN What do you notice about the moment of R about P. Which force has the same moment about A ? The line of action of R passes through P so the distance from P to the line of action is zero. Therefore the moment of R about P is zero. Similarly, the moment of N about A is zero.

9. Watch out for perpendicular distances. A x This isn’t perpendicular to the line of action. We need this length.

10. Watch out for perpendicular distances. A x Tip: If you are not sure whether the moment of a force is clockwise or anti-clockwise, imagine the perpendicular line to be fixed at the point you are taking moments about... fix pull and pull the other end of the perpendicular in the direction of the force. In your mind you will see it start to turn.

11. For a rigid body to rest in equilibrium, the resultant force is zero, and the sum of the moments about any point is zero. When we are solving problems involving a body such as a ruler, plank or ladder, we model the body as a rod. The rod is an example of a rigid body, one which cannot sag or bend.

12. A B 1 e.g.1.A plank of mass 18 kg and length 4 m rests in equilibrium on smooth supports at A and B. A is at one end of the plank and B is 1 m from the other end. Modelling the plank as a uniform rod, find the magnitudes of the reactions on the plank at A and B, giving the answers in terms of g.

13. Resolving: R  N  18g  0 - - - - - - (1) A : This means “anti-clockwise moments about A ”.  18g(2) length: 4 m - - - - - - (2) From (2), N  36g 3  N  12g newtons Subs. in (1) : R  12g  18g  0  R  6g newtons N(3)  0 Solution: The rod is uniform, so the weight acts at the centre. A B 1 R 18g N 2 mass: 18 kg

14. e.g.2.A package of mass 6 kg is placed at B on a uniform plank AB of length 2 m and mass 9 kg. The plank rests on supports at A and P as shown. A B P Modelling the plank as a uniform rod and the package as a particle at B, find how far P is from B if the plank is on the point of tipping.

15. B P A R 9g9g 6g6g x 1 Uniform plank AB: length 2 m, mass 9 kg Package: mass 6 kg Solution: Ans:As the plank tips, it turns about P and the force at A disappears. We can take the force at A as zero to get the point when the plank is about to tip, since even the smallest movement of P will cause it to tip. N = 0 How far is P from B if the plank is about to tip ? 1 What does the wording “the plank is about to tip” tell us about the force at A ? We are not interested in the forces on the package, so we ignore the reaction between the plank and package and draw the weight as if it is part of the plank.

16. We can take moments about A, B, P or even the centre of the rod. Choosing to take moments about P means we only need one equation but it’s a bit more awkward than using B, so I’ve chosen B. Uniform plank AB: length 2 m, mass 9 kg Package: mass 6 kg Solution: How far is P from B if the plank is about to tip ? B P A R 9g9g 6g6g x 1 N = 0 1

17. Resolving: R  9g  6g  0 - - - - - - (2) B : 9g(1)  R x  0 - - - - - - (1)  x   R  15g Subst. in (1) : 9g  15g x  0  x  0·6 9 15 P is 0·6 m from B. Uniform plank AB: length 2 m, mass 9 kg Package: mass 6 kg Solution: How far is P from B if the plank is about to tip ? B P A R 9g9g 6g6g x 1 N = 0 1

18. A B A non-uniform rod does not have its mass spread evenly along its length. The point at which we can balance a non-uniform rod is called the centre of mass. On a diagram we show the weight acting through this point. mg centre of mass e.g.

19. e.g.3.A non-uniform rod AB of weight 70 newtons and length 5 m hangs horizontally in equilibrium held by ropes at A and B. B G The centre of mass is at a point G, 2 m from A. Modelling the ropes as light inextensible strings, find the tension in each rope. A 2 x

20. B G T2T2 70 A T1T1 2 weight: 70 newtons length: 5 m centre of mass at G x Resolving: T 1  T 2  70  0 - - - - - - (2) A : T 2 (5)  70(2)  0 - - - - - - (1)  T 1  70  T 2 Solution: T 2  28   T 1  42 The tension at A is 42 newtons and at B is 28 newtons. 3

21. SUMMARY  The moment of a force, F, about a point A is defined as F  d where d is the perpendicular distance from A to the line of action of F.  For a rigid body to be in equilibrium, the resultant force is zero and the sum of the moments about any point is zero.  Moments can be taken about any convenient point.  The centre of mass is the point where we can assume all the weight acts.  We usually solve problems by resolving and taking moments about one point.

22. EXERCISE 1.A uniform rod AB of weight 20 newtons and length 6 m rests horizontally in equilibrium on supports at A and P, where P is 4 m from A. Find the magnitudes of the forces on the rod at A and P. A B 4 P

23. A B 4 EXERCISE Rod:weight 20 newtons length 6 m Find R and N. P RN 20 Solution: Resolving: R  N  20  0 - - - - - - (2) N(4)  20(3)  0 - - - - - - (1)  R  20  15 The force at A has magnitude 5 newtons and at P magnitude 1 5 newtons. A : 3  4N  60  N  15  R  5

24. EXERCISE 2.A non-uniform rod AB of weight 20 newtons and length 5 m rests horizontally in equilibrium on supports at A and B. The reaction on the rod at A is 12 newtons. A B 5 (a)the reaction at B, and (b)the distance of G from A. The centre of mass is at G. Find, G x

25. A B 5 G x EXERCISE non-uniform rod, AB centre of mass: G weight: 20 newtons length: 5 m reaction at A : 12 newtons (a)Find the reaction at B Solution: 12R 20 Resolving: 12  R  20  0 - - - - - - (1)  R  8 The reaction at B is 8 newtons.

26. A B 5 G x EXERCISE non-uniform rod, AB centre of mass: G weight: 20 newtons length: 5 m reaction at A : 12 newtons Solution: 12R 20 (b)Find the distance of G from A. x R  8 R(5)  20(x)  0 - - - - - - (2) A :  40  20x  0  x  2 Subs. R  8 G is 2 m from A.

28. The following page contains the summary in a form suitable for photocopying.

29. Summary MOMENT OF A FORCE TEACH A LEVEL MATHS – MECHANICS 1  The moment of a force, F, about a point A is defined as F  d where d is the perpendicular distance from A to the line of action of F.  For a rigid body to be in equilibrium, the resultant force is zero and the sum of the moments about any point is zero.  Moments can be taken about any convenient point.  The centre of mass is the point where we can assume all the weight acts.  We usually solve problems by resolving and taking moments about one point.  A rigid body cannot sag or bend.