# Varignon’s theorem

## Varignon’s theorem varignon’s theorem examples

**What is Varignon’s theorem?**

Varignon’s theorem, in Mechanics, states that the sum of the moments produced by a system of concurrent forces with respect to a certain point is equal to the moment of the resultant force with respect to the same point. varignon’s theorem examples

For this reason this theorem is also known asÂ *the principle of moments*Â .varignon’s theorem examples

Although the first to enunciate it was the Dutch Simon Stevin (1548-1620), the creator of the hydrostatic paradox, the French mathematician Pierre Varignon (1654-1722) was the one who later gave it its final form.

An example of how Varignon’s theorem works in Mechanics is the following: suppose that a simple system of two coplanar and concurrent forcesÂ **FÂ **_{1}Â andÂ **FÂ **_{2}Â acts on a pointÂ (denoted in bold because of their vector character).Â These forces result in a net or resultant force, calledÂ **FÂ **_{R}Â .

Each force exerts a torque or moment about a point O, which is calculated by the vector product between the position vectorÂ **rÂ **_{OP}Â and the forceÂ **F**Â , whereÂ **rÂ **_{OP} Â is directed from O to the point of concurrency P:

**MÂ **_{O1}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{1}**Â **

**MÂ **_{O2}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{2}

SinceÂ **FÂ **_{R}Â =Â **FÂ **_{1}Â +Â **FÂ **_{2}Â , then:

**MÂ **_{O}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{1}Â +Â **rÂ **_{OP}Â Ã—Â **FÂ **_{2}Â =Â **MÂ **_{O1}Â +Â **MÂ **_{O2}

But sinceÂ **rÂ **_{OP}Â is a common factor, then, applying distributive property to the cross product:

**MÂ **_{O}Â =Â **rÂ **_{OP}Â Ã— (Â **FÂ **_{1}Â +Â **FÂ **_{2}Â ) =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{RÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â }

Therefore, the sum of the moments or torques of each force with respect to point O is equivalent to the moment of the resultant force with respect to the same point. varignon’s theorem examples

__Statement and proof__

__Statement and proof__

Let a system of N concurrent forces, formed byÂ **FÂ **_{1}Â ,Â **FÂ **_{2}Â ,Â **FÂ **_{3}Â â€¦Â **FÂ **_{N}Â , whose lines of action intersect at point P (see figure 1), the moment of this force systemÂ **MÂ **_{O}Â , with respect to at a point O is given by:

**MÂ **_{O}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{1}Â +Â **rÂ **_{OP}Â Ã—Â **FÂ **_{2}Â +Â **rÂ **_{OP}Â Ã—Â **FÂ **_{3}Â +â€¦Â **rÂ **_{OP}Â Ã—Â **FÂ **_{N}Â =Â **rÂ **_{OP}Â Ã— (Â **FÂ **_{1}Â +Â **FÂ **_{2}Â +Â **FÂ **_{3}Â +â€¦Â **FÂ **_{N}Â )

**Demonstration**

To prove the theorem, use is made of the distributive property of the vector product between vectors.

Let the forcesÂ **FÂ **_{1}Â ,Â **FÂ **_{2}Â ,Â **FÂ **_{3}Â â€¦Â **FÂ **_{N be}Â Â applied to points AÂ _{1}Â , AÂ _{2}Â , AÂ _{3}Â â€¦ AÂ _{N}Â and concurrent at point P. The moment resulting from this system, with respect to a point O, calledÂ **MÂ **_{O}Â is the sum of the moments of each force with respect to that point:

**MÂ **_{O}Â = âˆ‘Â **rÂ **_{OAi}Â Ã—Â **FÂ **_{i}

Where the sum goes from i = 1 to i = N, since there are N forces.Â Since we are dealing with concurrent forces and since the vector product between parallel vectors is zero, it happens that:

**rÂ **_{PAi}Â Ã—Â **FÂ **_{i}Â =Â **0**

With the null vector denoted asÂ **0**Â .

The moment of one of the forces with respect to O, for example that of the forceÂ **FÂ **_{i}Â applied on AÂ _{i}Â , is written like this:

**MÂ **_{Oi}Â =Â **rÂ **_{OAi}Â Ã—Â **FÂ **_{i}

The position vectorÂ **rÂ **_{OAi}Â Â can be expressed as the sum of two position vectors:

**rÂ **_{OAi}Â =Â **rÂ **_{OP}Â +Â **rÂ **_{PAi}

In this way, the moment about O of the forceÂ **FÂ **_{i}Â is:

**MÂ **_{Oi}Â = (Â **rÂ **_{OP}Â +Â **rÂ **_{PAi}Â ) Ã—Â **FÂ **_{i}Â = (Â **rÂ **_{OP}Â Ã—Â **FÂ **_{i}Â ) + (Â **rÂ **_{PAi}Â Ã—Â **FÂ **_{i}Â )

But the last term is null, as explained above, becauseÂ **rÂ **_{PAi}Â lies on the line of action ofÂ **FÂ **_{i}Â , therefore:

**MÂ **_{Oi}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{i}

Knowing that the moment of the system with respect to point O is the sum of all the individual moments of each force with respect to said point, then:

**MÂ **_{O}Â = âˆ‘Â **MÂ **_{Oi}Â = âˆ‘Â **rÂ **_{OP}Â Ã—Â **FÂ **_{i}

SinceÂ **rÂ **_{OP}Â is constant it comes out of the sum:

**MÂ **_{O}Â =Â **rÂ **_{OP}Â Ã— (âˆ‘Â **FÂ **_{i}Â )

But âˆ‘Â **FÂ **_{i}Â is simply the net force or resultant forceÂ **FÂ **_{R}Â , therefore it is immediately concluded that:

**MÂ **_{O}Â =Â **rÂ **_{OP}Â Ã—Â **FÂ **_{R}

__Example__

__Example__

Varignon’s theorem facilitates the calculation of the moment of the forceÂ **F**Â with respect to the point O in the structure shown in the figure, if the force is decomposed into its rectangular components and the moment of each of them is calculated:

__Applications of Varignon’s theorem__

__Applications of Varignon’s theorem__

When the resultant force of a system is known, Varignon’s theorem can be applied to replace the sum of each of the moments produced by the forces that compose it by the moment of the resultant. varignon’s theorem examples

If the system consists of forces on the same plane and the point with respect to which the moment is to be calculated belongs to that plane, the resulting moment is perpendicular.

For example, if all the forces are in the xy plane, the moment is directed in the z axis and it only remains to find its magnitude and its sense, such is the case of the example described above.

In this case, Varignon’s theorem allows us to calculate the resulting moment of the system through the summation. It is very useful in the case of a three-dimensional force system, for which the direction of the resulting moment is not known a priori. varignon’s theorem examples

To solve these exercises, it is convenient to decompose forces and position vectors into their rectangular components, and from the sum of the moments, determine the components of the net moment.

__Exercise resolved__

__Exercise resolved__

Using Varignon’s theorem, calculate the moment of the force F around the point O shown in the figure if the magnitude of F is 725 N.

To apply Varignon’s theorem, the force **F** is decomposedÂ into two components, whose respective moments around O are calculated and added to obtain the resulting moment. varignon’s theorem examples

FÂ _{x}Â = 725 N âˆ™ cos 37 Âº = 579.0 N

FÂ _{y}Â = – 725 NN âˆ™ sin 37 Âº = âˆ’436.3 N

Similarly, the position vectorÂ **r**Â directed from O to A has the components:

rÂ _{x}Â = 2.5 m

rÂ _{y}Â = 5.0 m

The moment of each component of the force about O is found by multiplying the force and the perpendicular distance.

Both forces tend to rotate the structure in the same direction, which in this case is clockwise, to which a positive sign is arbitrarily assigned: varignon’s theorem examples

MÂ _{Ox}Â = FÂ _{x}Â âˆ™ rÂ _{y}Â âˆ™ sin 90Âº = 579.0 N âˆ™ 5.0 m = 2895 N âˆ™ m

MÂ _{Oy}Â = FÂ _{y}Â âˆ™ rÂ _{x}Â âˆ™ sin (âˆ’90Âº) = âˆ’436.3 N âˆ™ 2.5 m âˆ™ (âˆ’1) = 1090.8 N âˆ™ m

The resulting moment about O is:

**MÂ **_{O}Â =Â **MÂ **_{Ox}Â +Â **MÂ **_{Oy} = 3985.8 N âˆ™ m perpendicular to the plane and in a clockwise direction. varignon’s theorem examples