por What is Varignon’s theorem?
Varignon’s theorem, in Mechanics, affirms 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 state it was the Dutchman 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 and F2, (denoted in bold because of its vector character). These forces give rise to a net or resultant force, called FR.
Each force exerts a torque or moment about a point O, which is calculated by the cross product between the position vector rOP and the force Fwhere rOP is directed from O to the point of concurrency P:
mO1 = rOP × F1
mO2 = rOP × F2
Given the FR = F1 + F2, then:
mOR = rOP × F1 + rOP × F2 = mO1 + mo2
But how rOP is a common factor, so, applying the distributive property to the cross product:
mOR = rOP × (F1 + F2) = rOP × FR.
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, F2, F3… FN, whose lines of action intersect at point P (see figure 1), the moment of this system of forces mO, with respect to a point O is given by:
mOR = rOP × F1 + rOP × F2 + rOP × F3 + … rOP × FNo = rOP × (F1 + F2 + F3 +… FN)
Demonstration
To prove the theorem, use is made of the distributive property of the vector product between vectors.
be the forces F1, F2, F3… FN applied to points A1, A2, A3… AN and concurrent at point P. The resulting moment of this system, with respect to a point O, called mOr, is the sum of the moments of each force, with respect to said point:
mOR = ∑ rOAi × FYo
where the sum goes from i = 1 to i = N, since there are N forces. Since we are dealing with concurrent forces and since the vector product between parallel vectors is zero, it happens that:
rPAi × Fi = 0
With the null vector denoted as 0.
The moment of one of the forces about O, for example that of the force Fi applied to Ai, is written like this:
mhey = rOAi × FYo
The position vector rOAi can be expressed as the sum of two position vectors:
rOAi = rOP+ rPAI
Thus, the moment about O of the force Fi is:
moh = (rOP+ rPAi) × Fi = (rOP × Fi) + (rPAi × FYo)
But the last term is zero, as explained above, because rPAi is located on the line of action of Fand therefore:
mhey = rOP × FYo
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:
mOR = ∑ mOi = ∑ rOP × FYo
As rOP is constant comes out of the sum:
mOR = rOP × (∑ FYo)
But ∑ Fi is simply the net force or resultant force FR, therefore it follows immediately that:
mOR = rOP × FR.
Example
Varignon’s theorem makes it easy to calculate the moment of force F With respect to point O in the structure shown in the figure, if the force is broken down 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 about 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 along the z axis and all that remains is to find its magnitude and direction, 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 resultant moment is not known a priori.
To solve these exercises, it is convenient to break down forces and position vectors into their rectangular components, and from the sum of the moments, determine the components of the net moment.
solved exercise
Using Varignon’s theorem, calculate the moment of the force F about the point O shown in the figure if the magnitude of F is 725 N.
Solution
To apply Varignon’s theorem, we break down the force F into two components, whose respective moments about O are calculated and added to obtain the resulting moment.
Fx = 725 N ∙ cos 37º = 579.0 N
Fy = − 725 NN ∙ sin 37 º = −436.3 N
Similarly, the position vector r directed from O to A has the components:
rx = 2.5m
ry = 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:
MOx = Fx∙ ry ∙ sin 90º = 579.0 N ∙ 5.0 m = 2895 N∙m
MOy = Fy∙ rx ∙ sin (−90º) = −436.3 N ∙ 2.5 m ∙ (−1) = 1090.8 N∙m
The resultant moment about O is:
mOR = moh + mOy = 3985.8 N∙m perpendicular to the plane and clockwise.
References
Bedford, 2000. A. Engineering Mechanics: Statics. Addison Wesley.
Beer, F. 2010. Statics. McGraw Hill. 9na. Edition.
Hibbeler, R. 1992. Mechanics for Engineers. 6th. Edition. CECSA. HK Engineering. Varignon’s theorem. Recovered from: youtube.com. Wikipedia. Varignon’s theorem (Mechanics). Retrieved from: en.wikipedia.org.