r F F F F MOMENT of FORCE = F x r

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

 r Hence MOMENT = Fr sin  What is the moment of F about O? F d O But d = r sin  Moment = F x d Hence MOMENT = Fr sin 

THE PRINCIPLE OF MOMENTS 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

THE SYSTEM IS IN EQUILIBRIUM Clockwise moment Anticlockwise moment Anticlockwise Moment = 8.0N x 3.0 cm =24.0 N cm Clockwise Moment = 6.0N x 4.0 cm =24.0 N cm THE SYSTEM IS IN EQUILIBRIUM

Why are these systems balanced? 1 2

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

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

This uniform bridge is 20 m long with a mass of 10 tonnes 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. Vertically R1 + R2 = 100 000 + 200 000 = 300 000 6 m R1 10 m R2 200 000 N A B 100 000 N 20 m Taking moments about A eliminates R1 200 000 x 6 + 100 000 x 10 = 20 R2 R1 = 110 000 N Taking moments about B eliminates R2 200 000 x 14 + 100 000 x 10 = 20 R2 R2 = 190 000 N Check 190 000 + 110 000 = 300 000 !