TOPIC 10

Moment of a Force

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

PF

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

ExampleA 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)

R S

1m 1.5m

P Q40g 80g

2m 0.5m 1m

Moment of a Force

Resolving verticallyR + S = 40g + 80g

R + S = 120gR + S = 120 x 10R + S = 1200N

Take moments about P40g x 2 + 80g x 2.5 = S x 3.5

80g + 200g = 3.5S280g = 3.5S

280 x 10 = 3.5S2800 = 3.5S

2800 = S 3.5

S = 800NSo R = 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. R2 = 0

Taking moments about Q80g x d = 40g x 1.5

80g x d = 60gd = 60g

80gd = 0.75m

R2 S2

1m 1.5m

PQ40g 80g

2m d