__What is the Principle of Moment?__

Principle of moment states that *–*

*If a number of co-planer forces are in static equilibrium, then the algebraic sum of their moments about any point in their plane is zero.*

This principle of moment is very useful in solving numerical problems related to –

- A system of
*parallel forces.* *Shear force and bending moment diagrams.*- Structural members,
*cantilevers*etc. *Deflection in beams*, columns and struts etc.

Principle of moment is based upon *Verignon’s theorem.*

__Varignon’s Theorem__

Varignon’s theorem can be stated as –

*The algebraic sum of moments of two forces about any point in their plane of action is equal to the moment of their resultant force about the same point.*

__PROOF__

Let, two co-planer forces ( P ) and ( Q ) are acting along the lines of action EX and EY respectively. Point E is the point of intersection of line of action of these forces. Space diagram of above two force system is shown in figure.

Consider about a point O in the plane of forces ( P ) and ( Q ) .

From point O a line OC is drawn parallel to EX which intersects EY at B . Let the length of EB is representing the magnitude of force ( Q ) .

Therefore, scale of the drawing is –

= \frac {\text {Magnitude of force Q}}{\text {Length of EB}}

A line segment EA is now cut from EX by using the above scale. Thus EA will represent the magnitude of force ( P ) .

Now parallelogram EBCA is completed by drawing line ( AC \parallel EB ) .

According to *parallelogram law of forces,* the diagonal EC will represent the resultant of forces ( P ) and ( Q ) .

Point O is joined to E and A .

Then, as per theory for geometry of a moment, we find that –

- 2 \times \triangle OAE will represent the moment of force ( P ) about point O .
- 2 \times \triangle OEB will represent the moment of force ( Q ) about point O .
- 2 \times \triangle OEC will represent the moment of resultant force ( R ) about point O .

From geometry of the figure, we have –

( \triangle OEC ) = ( \triangle OEB + \triangle BEC ) .

Also ( \triangle BEC ) = ( \triangle EAC )

Because each triangle is half of area of parallelogram ( EBCA )

Therefore, \quad ( \triangle OEC ) = ( \triangle OEB + \triangle EAC )

Again ( \triangle EAC ) = ( \triangle OEA ) ( Both triangles are of same base and having same altitude due to laying between two parallel lines. )

Therefore, \quad ( \triangle OEC ) = ( \triangle OEB + \triangle OEA )

Multiplying each term by 2 we get –

2 \times \triangle OEC = 2 \times \triangle OEB + 2 \times \triangle OEA

Or, Moment of ( EC ) = Moment of ( EB ) + Moment of ( EA )

Therefore, * Moment of resultant force ( R ) about O = Moment of force ( P ) about O \ + Moment of force ( Q ) about O *.

*This relationship is applicable for any number of forces.*

__Geometry of a Moment__

*Representation of a moment of force in a graphical way is called geometry of a moment.*

Geometry of moment is obtained from the principle of moment as follows –

Consider about a force represented by ( \vec {AB} ) as shown in figure.

( \vec {AB} ) is representing the force in magnitude, direction and line of action. Let O is any point outside the line of action of this force.

A perpendicular OM is drawn to AB . Point O is joined to point A and point B .

Then, moment of force ( \vec {AB} ) about the point O will be –

( AB \times OM ) = 2 \times \text {Area of triangle} \ \triangle {OAB}

*Therefore, a moment can be expressed as a triangle. This is called geometrical representation of a moment.*

## System of Parallel Forces

*Varignon’s theorem can be used in finding the solution for resultant force of a parallel force system.*

A system of two parallel forces can be of two types –

- Like parallel force system.
- Unlike parallel force system.

**LIKE PARALLEL FORCES** – Two parallel forces are said to be like when they act in the same sense i.e. they act in the same direction on same body.

**UNLIKE PARALLEL FORCES – **Two parallel forces are said to be unlike when they act in opposite sense i.e. when they act in opposite direction on same body.

### Resultant of Parallel Forces

Let, two parallel forces ( P ) and ( Q ) are acting at points A and B of a body. Assume ( P > Q ) and the resultant force ( R ) cuts the line AB at point C as shown in figure (A) and (B).

Figure (A) shows two like forces system and figure (B) shows two unlike forces system.

Since, both forces have parallel line of action, so their resultant will be as following –

- Resultant of the like parallel forces in figure (A) will be [ R = ( P + Q ) ] .
- Resultant of the unlike parallel forces in figure (B) will be [ R = ( P - Q ) ]

And, the direction of resultant ( R ) will also be parallel to the direction of forces ( P ) and ( Q ) .

Since, the resultant force ( R ) is passing through the point C , so the moment of resultant force ( R ) about point C must be zero.

Therefore, according to *Varignon’s theorem,* the algebraic sum of the moments of forces ( P ) and ( Q ) about point C must also be zero.

So, the position of point C will be such that, the moment of force ( P ) is just equal in magnitude but opposite in direction to the moment of force ( Q ) . This will happen –

- For like forces – if the point C will lie internal to the line AB .
- For unlike forces – if the point C will lie external to the line AB as shown in figures.

Now, through point C a line MN is drawn perpendicular to the direction of forces ( P ) and ( Q ) . Let it makes an angle ( \theta ) with the line AB as shown in figure.

Then, from geometry of the figure, we have ( CM = AC \cos \theta ) \ \text {and} \ ( CN = BC \cos \theta )

Now, taking the moments of forces about point C , we will get –

P \times CM = Q \times CN

Or, \quad P \times AC \cos \theta = Q \times BC \cos \theta

Dividing both sides by ( \cos \theta ) , we will get –

P \times AC = Q \times BC

Or, \quad \left ( \frac {P}{BC} \right ) = \left ( \frac {Q}{AC} \right )

Therefore, we have got the following conclusions –

- For like parallel forces – the point C divides the line AB internally in the inverse ratio of magnitude of forces.
- For unlike parallel forces – the point C divides the line AB externally in the inverse ratio of the magnitude of forces.
- It is clear that, the position of point C is independent of the inclination of the direction of forces to the line AB .
- The point C is called the centre of the parallel forces.