The product of a force and the perpendicular distance of the line of actions of the force from a point is known as moment of the force about that point.

Let F = A force acting on a body as shown in Fig.2.28

r = Perpendicular distance from the point.

O on the line of action of force F.

Then moment (M) of the force F about O is given by, M = F × r

The tendency of this moment is to rotate the body in the clockwise direction about O. Hence this moment is called cloakwise moment If the tendency of a moment is to rotate the body in anti-clockwise direction, then that moment is known as anti-cloakwise moment. If cloakwise moment is taken –ve then anti-cloakwise moment will be +ve.

In S.I.system, moment is expressed in N m (Newton meter).

Fig. 2.29 shows a body on which three forces F_{1}, F_{2} and F_{3} are acting. Suppose it is required to find the resultant moments if these forces about point O.

Let r_{1} = Perpendicular distance from O on the line of action of force F_{1}.

r_{2} and r_{3} = Perpendicular distance from O on the line of action of force F_{2} & F_{3} respectively.

Moment of F_{1} about O = F_{1} × r_{1} (cloakwise)(-)

Moment of F_{2} about O = F_{2} × r_{2} (cloakwise)(-)

Moment of F_{3} about O = F_{3} × r_{3} (anti-cloakwise)(+)

The resultant moment will be algebraic sum of all the moments.

(so) The resultant moment of F_{1, }F_{2 }andF_{3} about O = – F_{1 }× r_{1} – F_{2} × r_{2 }+ F_{3} × r_{3}

**Problem 2.15.** *Four forces of magnitude 10 N, 20 N, 30 N and 40 N are acting respectively along the four sides of a square ABCD as shown in Fig.2.30. Determine the resultant moment about the point A. Each slide of the square is given 2 m.*

**Sol.** Given :

Length AB = BC = CD

= DA = 2m

Force at B = 10 N

Force at C = 20 N

Force at D = 30 N

Force at A = 40 N

The resultant moment about A is to be determined. Perpendicular distance from A on the lines of action of these forces will be zero.

Hence their moments about A will be zero. The moment of the force at C about point A

= Force at C × distance from A on the line of action of force at C

= (20 N) × (Length AB)

= 20 × 2 N m = 40 N m (anti-cloakwise)

The moment of force at D about point A

= Force at D × distance from A on the line of action of force at D.

= (30 N) × (Length AD)

= 30 × 2 N m = 60 N m (anti-cloakwise)

Resultant moment of all forces about A

= 40 + 60 = 100 N m (anti-cloakwise). **Ans.**

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