Varignon's theorem

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.

For this reason this theorem is also known as the beginning of moments.

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 forces acts on a point F₁ Y F_two, (denoted in bold because of their vector character). These forces give rise to a net or resultant force, called F_R.

Each force exerts a torque or moment about a point O, which is calculated by the vector product between the position vector r_OP and the strengh F, where r_OP is directed from O to the point of concurrency P:

M_O1 = r_OP × F₁

M_O2 = r_OP × F_two

Given the F_R = F₁ + F_two, then:

M_OR = r_OP × F₁ + r_OP × F_two = M_O1 + M_O2

But how r_OP is a common factor, then, applying distributive property to the cross product:

M_OR = r_OP × (F₁ + F_two) = r_OP × F_R                

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.

Statement and proof

Let be a system of N concurrent forces, formed by F₁, F_two, F₃... F_N, whose lines of action intersect at point P (see figure 1), the moment of this force system M_OR, with respect to a point O is given by:

M_OR = r_OP × F₁ + r_OP × F_two + r_OP × F₃ +... r_OP × F_N = r_OP × (F₁ + F_two + F₃ +... F_N)

Demonstration

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

Be the forces F₁, F_two, F₃... F_N applied to points A₁, TO_two, TO₃… TO_N and concurrent at point P. The resulting moment of this system, with respect to a point O, called M_OR, is the sum of the moments of each force, with respect to said point:

M_OR = ∑ r_OAi × F_i

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

r_PAi × F_i = 0

With the null vector denoted as 0.

The moment of one of the forces with respect to O, for example that of the force F_i applied in A_i, it is written like this:

M_{I heard} = r_OAi × F_i

The position vector r_OAi  can be expressed as the sum of two position vectors:

r_OAi = r_OP + r_PAi

In this way, the moment about O of the force F_i it is:

M_{I heard} = (r_OP + r_PAi) × F_i = (r_OP × F_i) + (r_PAi × F_i)

But the last term is null, as explained above, because r_PAi is on the line of action of F_i, Thus:

M_{I heard} = r_OP × F_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:

M_OR = ∑ M_{I heard} = ∑ r_OP × F_i

What r_OP is constant comes out of the sum:

M_OR = r_OP × (∑ F_i)

But ∑ F_i is simply the net force or resultant force F_R, therefore it is immediately concluded that:

M_OR = r_OP × F_R

Example

Varignon's theorem facilitates the calculation of the moment of force F With respect to 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

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.

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.

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

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.

Solution

To apply Varignon's theorem, the force is decomposed F in two components, whose respective moments around O are calculated and added to obtain the resulting moment.

F_x = 725 N ∙ cos 37 º = 579.0 N

F_Y = - 725 N N ∙ sin 37 º = −436.3 N

Similarly, the position vector r directed from O to A has the components:

r_x = 2.5 m

r_Y = 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:

M_Ox = F_x∙ r_Y ∙ sin 90º = 579.0 N ∙ 5.0 m = 2895 N ∙ m

M_Oy = F_Y∙ r_x ∙ sin (−90º) = −436.3 N ∙ 2.5 m ∙ (−1) = 1090.8 N ∙ m

The resulting moment about O is:

M_OR = M_Ox + M_Oy = 3985.8 N ∙ m perpendicular to the plane and clockwise.

References

1. Bedford, 2000. A. Engineering Mechanics: Statics. Addison wesley. 
2. Beer, F. 2010. Static. McGraw Hill. 9th. Edition.
3. Hibbeler, R. 1992. Mechanics for engineers. 6th. Edition. CECSA.
4. HK Engineering. Varignon's theorem. Recovered from: youtube.com.
5. Wikipedia. Varignon's theorem (Mechanics). Recovered from: en.wikipedia.org.