TOPIC 10 Moment of a Force

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

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

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

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 g x 2.5 = S x g + 200g = 3.5S 280g = 3.5S 280 x 10 = 3.5S 2800 = 3.5S 2800 = S 3.5 S = 800N SoR = 1200 – 800 = 400N

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 g x d = 60g d =60g 80g d = 0.75m R2R2 S2S2 1m1.5m P Q 40g 80g 2md