2. © John Parkinson 1 WHAT WAS THAT?

3. © John Parkinson 2 MOMENTS

4. © John Parkinson 3 F F F F r MOMENT of FORCE = F x r

5. © John Parkinson 4 THE MOMENT OF A FORCE ABOUT A POINT DEPENDS UPON: MOMENT UNITS ? NEWTON METRES (Nm)  THE SIZE OF THE FORCE  THE PERPENDICULAR DISTANCE OF THE FORCE’S LINE OF ACTION FROM THE POINT about a point = FORCE X PERPENDICULAR DISTANCE of the force from the point

6. © John Parkinson 5 r F  O What is the moment of F about O? d Moment =F x d But d = r sin  Hence MOMENT = Fr sin 

7. © John Parkinson 6 FOR A SYSTEM TO BE IN EQUILIBRIUM, THE SUM OF THE CLOCKWISE MOMENTS ABOUT ANY POINT MUST EQUAL THE SUM OF THE ANTICLOCKWISE MOMENTS ABOUT THAT POINT

8. © John Parkinson 7 Clockwise Moment = 6.0N x 4.0 cm =24.0 N cm Anticlockwise Moment = 8.0N x 3.0 cm =24.0 N cm THE SYSTEM IS IN EQUILIBRIUM Clockwise moment Anticlockwise moment

9. © John Parkinson 8 Why are these systems balanced? 1 2

10. © John Parkinson 9 Equilibrium of a Rigid Body Under Coplanar Forces Equilibrium means that….. …there is no rotation. …there is no acceleration. …there is no net force acting on the object.

11. © John Parkinson 10 CONDITIONS FOR THE EQUILIBRIUM OF A BODY * The vector sum of all the forces acting on the body is ZERO * The algebraic sum of all the moments acting about any point is ZERO [Otherwise there would be translational motion] [Otherwise there would be rotational motion]

12. © John Parkinson 11 This uniform bridge is 20 m long with a mass of 10 tonnes. The lorry has a mass of 20 tonnes and its mass centre is situated 6 m from A. Using g = 10 N kg -1, Find the reaction force at each support A and B. A B 200 000 N R1R1 R2R2 100 000 N 10 m 6 m 20 m Vertically R1 R1 + R2R2 = 100 000 + 200 000 = 300 000 Taking moments about A eliminates R 1 200 000 x 6 + 100 000 x 10 = 20 R 2 R 1 = 110 000 N Taking moments about B eliminates R 2 200 000 x 14 + 100 000 x 10 = 20 R 2 R 2 = 190 000 N Check 190 000 + 110 000 = 300 000 !

13. © John Parkinson 12 PAIRS OF EQUAL AND OPPOSITE FORCES THAT PRODUCE ROTATION F F PAIRS OF EQUAL AND OPPOSITE FORCES THAT PRODUCE ROTATION PAIRS OF EQUAL AND OPPOSITE FORCES THAT PRODUCE ROTATION

14. © John Parkinson 13 THE MOMENT OF A COUPLE F F d C G = MOMENT OF THE COUPLE THE MOMENT OF a COUPLE is called a TORQUE HENCE G = F d

15. © John Parkinson 14 B NS I I F F APPLYING FLEMING’S LEFT HAND RULE THE TWO FORCES PRODUCE A COUPLE ON THE COIL CAUSING ROTATION

16. © John Parkinson 15 …of an object is the point at which the entire weight of an object may be considered concentrated. Objects can be balanced with a fulcrum under their centre of gravity. Objects in free fall rotate about their center of gravity. CENTRE OF GRAVITY - CENTRE OF MASS A regular object has its Centre of Mass at its geometrical centre.

17. © John Parkinson 16 WV03 VXP G mg d mg x d is a clockwise moment That’s OK!

18. © John Parkinson 17 WV03 VXP mg x d is now an anticlockwise moment! mg G d WV03 VXP Oh dear! But at least WE know about Moments d mg mg x d kicked it over !!