1. TOPIC 10 Moment of a Force

2. So far we have mainly considered particles. With larger bodies forces may act in many different positions and we have to consider the possibility of rotation. The MOMENT of a force F about a point P is found by multiplying the magnitude of the force by the perpendicular distance from P to the line of action of the force. The unit used for MOMENTS is Nm (newton metres) eg Moment = F x d (Nm) P F d

3. Moment of a Force If the point P lies on the line of action of the force the moment is zero because d = 0 eg Moment = F x 0 = 0 It is also necessary to specify the direction of a moment i.e. clockwise or anticlockwise. For a body in equilibrium the sum of the moments of all forces acting must be zero about any point i.e. sum of anticlockwise moments = sum of clockwise moments P F

4. Moment of a Force UNIFORM means that the weight of the body acts at its ‘centre of mass’. This is the mid-point of a rod Example A uniform beam 6m long and of mass 40kg is supported on 2 trestles P and Q at points 1m and 1.5m from the ends of the beam. (a) Find the reactions at the supports when an 80kg man stands at a point 1m in from Q (b) How far past Q may the man walk before the beam overturns? Answer (a) RS 1m1.5m PQ 40g 80g 2m 0.5m 1m

5. Moment of a Force Resolving vertically R + S = 40g + 80g R + S = 120g R + S = 120 x 10 R + S = 1200N Take moments about P 40g x 2 + 80g x 2.5 = S x 3.5 80g + 200g = 3.5S 280g = 3.5S 280 x 10 = 3.5S 2800 = 3.5S 2800 = S 3.5 S = 800N SoR = 1200 – 800 = 400N

6. Moment of a Force (b) When the beam is about to overturn the reaction at P is equal to 0 i.e. R 2 = 0 Taking moments about Q 80g x d = 40g x 1.5 80g x d = 60g d =60g 80g d = 0.75m R2R2 S2S2 1m1.5m P Q 40g 80g 2md