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Engineering Statics: Open and Interactive

Section 4.4 Varignon’s Theorem

Varignon’s Theorem is a method to calculate moments developed in 1687 by French mathematician Pierre Varignon (1654 – 1722). It states that sum of the moments of several concurrent forces about a point is equal to the moment of the resultant of those forces, or alternately, the moment of a force about a point equals the sum of the moments of its components.
This means you can find the moment of a force by first breaking it into components, evaluating the scalar moments of the individual components, and finally summing them to find the net moment about the point. The scalar moment of a component is the magnitude of the component times the perpendicular distance to the moment center by the definition of a moment, with a positive or negative sign assigned to indicate its direction.
This may sound like more work than just finding the moment of the original force, but in practice it is often easier. Consider the interactive to the right. If we break the force into components along the wrench handle and perpendicular to it, the sum of the moments of these component is
\begin{equation} M = F_\perp d \tag{4.4.1} \end{equation}
, where \(d\) is the length of the handle, and \(F_\perp\) is the component of \(F\) perpendicular to the handle. Here, the contribution of the parallel component to the sum is zero, since its line of action passes through the moment center \(A\text{.}\)


This interactive shows the force applied to the wrench broken into components perpendicular and parallel to the handle. The moment is the perpendicular component times the length of the handle.
Figure 4.4.1. Varignon's Theorem: \(M = F_\perp d\)
This result agrees with our intuitive understanding of how a wrench works; the greatest torque is developed when the force is applied at a right angle to the handle.
Equations (4.2.1) and (4.4.1) not only produce the same result, they are completely identical. If the length of the handle is \(d\) and the angle between the force \(\vec{F}\) and the handle is \(\theta\text{,}\) then \(d_\perp = d \sin \theta\) and \(F_\perp = F \sin \theta\text{.}\) Using either equation to calculate the moment gives
\begin{equation} M = F\ d\ \sin \theta\tag{4.4.2} \end{equation}

Subsection 4.4.1 Rectangular Components

Varignon's theorem is particularly convenient to use the diagram provides horizontal and vertical dimensions, which is often the case. If you decompose forces into horizontal and vertical components you can find the scalar moments of the components without difficulty.
The moment of a force is the sum of the moments of the components, so
\begin{equation} M = \pm F_x d_y \pm F_y d_x\tag{4.4.3} \end{equation}
. Take care to assign the correct sign to the individual moment terms to indicate direction; positive moment tend to rotate the object counter-clockwise and negative moment tend to rotate it clockwise according to the standard right hand rule convention.
Point A is shown as an origin, with a distance dx indicated in the horizontal direction and distance dy indicated in the vertical direction. At coordinates (dx, dy), a Force F originates acting up and to the right with components of Fx in the horizontal direction and Fy in the vertical direction.
Figure 4.4.2. Sum of moments of components. \(M = \pm F_x d_y \pm F_y d_x\)

Example 4.4.3. Varignon's theorem.

A \(\lb{750}\) force is applied to the frame as shown. Determine the moment this force makes about point \(A\text{.}\)
\begin{equation*} \vec{M}_A = \ftlb{174} \text{ Clockwise}. \end{equation*}
Force \(\vec{F}\) acts \(\ang{60}\) from the vertical with a \(\lb{750}\) magnitude, so its horizontal and vertical components are
\begin{align*} F_x \amp = F \sin \ang{60} = \lb{649.5} \\ F_y \amp = F \cos \ang{60} = \lb{375.0} \end{align*}
For component \(F_x\text{,}\) the perpendicular distance from point \(A\) is \(\ft{2}\) so the moment of this component is
\begin{equation*} M_1 = 2 F_x = \ftlb{1299} \text{ Clockwise} \end{equation*}
For component \(F_x\text{,}\) the perpendicular distance from point \(A\) is \(\ft{3}\) so the moment of this component is
\begin{equation*} M_2 = 3 F_y = \ftlb{1125} \text{ Counter-clockwise} \end{equation*}
Assigning a negative sign to \(M_1\) and a positive sign to \(M_2\) to account for their directions and summing, gives the moment of \(\vec{F}\) about \(A\text{.}\)
\begin{align*} M_A \amp = - M_1 + M_2\\ \amp = - 1299 + 1125\\ \amp = -\ftlb{174} \end{align*}
The negative sign indicates that the resultant moment is clockwise, with a magnitude of \(\ftlb{174}\text{.}\)
\begin{equation*} \vec{M}_A = \ftlb{174} \text{ Clockwise}. \end{equation*}
The interactive diagram below will help you visualize the different tools to find moments that were covered in this section.


This interactive demonstrates three methods to visualize and compute the two-dimensional moment of a force about point.
Figure 4.4.4. Various approaches to find a moment about a point.