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Engineering Statics: Open and Interactive

Section 4.3 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
The 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 moments of the individual components, and finally summing them to find the net moment about the point.
This may sound like more work than just finding the moment of the original force, but in practice, it is often easier.
When applying Varignon’s Theorem, you are free to select components in any direction, but the two most convenient choices are discussed below.

Subsection 4.3.1 Parallel and Perpendicular Components

The first common way to apply Verignon’s Theorem is to break the force creating the moment into two components, one parallel with the line segment connecting the moment center with the point of application of the force, and the other perpendicular to it.
Consider the wrench in the interactive diagram below. Assume that the length of the handle is \(d\text{,}\) the magnitude of the force is \(F\) and the angle between the force and the handle is \(\theta\text{.}\) The components parallel to the wrench handle and perpendicular to it are
\begin{align*} F_\| \amp= F \cos \theta \amp F_\perp \amp= F \sin \theta\text{.} \end{align*}
By Verignon’s theorem and the definition of the moment, (4.1.1), the sum of the moments of the components simplifies to:
\begin{equation} M = F_\perp d\text{,}\tag{4.3.1} \end{equation}
because the contribution of the parallel component 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 4.3.1. Varignon’s Theorem: Perpendicular Component
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 (4.1.1) and (4.3.1) not only produce the same result, but they are also 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 \(d_\perp = d \sin \theta\text{,}\) and \(F_\perp = F \sin \theta\text{.}\) Using either equation to calculate the moment gives
\begin{equation} M = F\ d_\perp = F_\|\ d = F\ d\ \sin \theta\text{.}\tag{4.3.2} \end{equation}

Example 4.3.2. Verignon’s Theorem: Parallel and Perpendicular Components.

A gardener is moving soil with a wheelbarrow. The length of the handle from \(A\) to \(B\) is \(\inch{45}\text{,}\) and in the position shown, \(\theta\) is \(\ang{25}\text{.}\)
Find the moment created about axle \(A\) when she applies a vertical \(\lb{30}\) force at \(B\) by resolving force \(F\) into components parallel and perpendicular to the handle.
\begin{equation*} M_A = \inlb{1224} \end{equation*}
When the handle makes an angle of \(\theta\) with the horizontal axis, the perpendicular component of \(F\) makes the same angle with the vertical, so resolving \(F\) into perpendicular and parallel components gives:
\begin{align*} F_\perp \amp = F \cos \theta = \lb{27.19} \amp F_\| \amp = F \sin \theta = \lb{12.68} \end{align*}
Summing the moments of the components we find
\begin{equation*} M_A = F_\perp (\inch{45}) + F_\| (0) = \inlb{1224} \end{equation*}
Note that the parallel component does not contribute to the moment.

Subsection 4.3.2 Horizontal and Vertical Components

Varignon’s theorem is particularly convenient in situations where horizontal and vertical dimensions are provided, as is often the case. If you decompose the forces into horizontal and vertical components you can find the moments of the components without difficulty. The resultant moment is the sum of the moments of the components.
Verignon’s Equation is (4.3.3) and Figure 4.3.3 is the corresponding diagram for the moment of force \(F\) about point \(A\text{.}\) Notice that the horizontal component is multiplied by the vertical distance, and the vertical component is multiplied by the horizontal distance since these are the perpendiculars, and that the signs on the terms can be either positive or negative, depending on the direction of the component moments.
\begin{equation} M = \pm (F_x\ d_y) \pm (F_y\ d_x)\tag{4.3.3} \end{equation}
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 4.3.3. Verignon’s Theorem Horizontal and Vertical Components
When applying Verignon’s Theorem (4.3.3), take care to assign the correct sign to the individual moment terms; counterclockwise moments are positive, and clockwise negative are negative according to the right-hand rule, regardless of whether \(F_x\) acts left or right, or whether \(F_y\) acts up or down.

Example 4.3.4. Varignon’s Theorem: Horizontal and Vertical Components.

A \(\lb{750}\) force is applied to the frame as shown. Determine the moment this force makes about point \(A\) by resolving the force into horizontal and vertical components and applying Varignon’s Theorem.
\begin{equation*} \vec{M}_A = \ftlb{174} \text{ Clockwise}\text{.} \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}\text{.} \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}\text{.} \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}\text{.} \end{equation*}

Subsection 4.3.3 Moment Cross Products

The most robust and general method to a moment of a force is to use the vector cross product
\begin{equation} \vec{M} = \vec{r} \times \vec{F}\text{,}\tag{4.3.4} \end{equation}
where \(\vec{F}\) is the force creating the moment, and \(\vec{r}\) is a position vector from the moment center to any point on the line of action of the force, as shown in Figure 4.3.5. The cross product is a vector multiplication operation and the result is a vector perpendicular to the two vectors being multiplied.


This interactive shows the force \(\vec{F}\) and position vector \(\vec{r}\) for use in the moment cross product. The position vector is a vector from the moment center to any point on the line of action of the force.
You can move the tip of \(\vec{r}\) anywhere along the line of action of \(\vec{F}\) without changing the moment, since the moment arm \(d_\perp = |\vec{r}| \sin \theta\) remains unchanged as you do.
Figure 4.3.5. Moment cross product. \(\vec{M} = \vec{r} \times \vec{F}\)
The cross product method is most often used to find moments in three dimensions, as will be discussed in Section 4.4, but it can also be applied in two-dimensional problems.
The next example demonstrates the technique and compares it to the other methods we have already discussed. You will see that the cross product method is mathematically identical to applying Verignon’s theorem.

Example 4.3.6. 2D Moments - Four Ways.

A 500-pound-force vector pulls from point D located at (0,2) feet and we want to find the moment of this force around point A at (-4,-3) feet. A position vector r connects point A with point D
Force \(\vec{F}\) has a magnitude of \(\lbf{500}\) and acts on point \(B\) in the direction shown.
Find the moment caused by force \(\vec{F}\) around point \(A=\ft{(-4,-3)}\) using different methods and verify that they give the same result.
This problem demonstrates four different ways you can solve the problem. The first two methods use vector algebra; the second two take a scalar approach that uses geometry and right-triangle trigonometry. All four methods are mathematically identical.
Find the moment of \(\vec{F}\) about point \(A\) using Varignon’s Theorem,
\begin{equation*} \vec{M}_A = \left(\vec{r}_x \times \vec{F}_y\right) + \left(\vec{r}_y \times \vec{F}_x\right)\text{.} \end{equation*}
Varignon’s Theorem states that the moment of a force is the sum of the moments of its components. In this example we will determine the vertical and horizontal components of \(\vec{r}\) and \(\vec{F}\text{,}\) then add the cross products of the two perpendicular pairs.
The 3:2 slope of \(\vec{F}\) can be expressed as an angle.
\begin{equation*} \theta = \tan^{-1}\ \frac{3}{2} = \ang{56.3} \end{equation*}
Find the components of \(\vec{r}\) and \(\vec{F}\text{.}\)
\begin{align*} \vec{F} \amp = \lbf{500} \langle \cos \ang{56.3}, \sin \ang{56.3} \rangle\\ \amp = \lbf{\langle 277.35, 416.025 \rangle}\\ \vec{r} \amp = \ft{\langle 6,3 \rangle} \end{align*}
Finally, following Varignon’s Theorem, add the cross products of the perpendicular component pairs.
\begin{align*} \vec{M}_A \amp = \left(\vec{r}_x \times \vec{F}_y\right) + \left(\vec{r}_y \times \vec{F}_x\right)\\ \amp = \ft{6} \cdot \lbf{416.025} \left(\khat\right) + \ft{3}\cdot \lbf{277.35}\left(-\khat\right)\\ \amp = \ftlbf{1664.1}\left(+\khat\right) \end{align*}
  • When finding the moment of two-dimensional vectors in component form, this is often the preferred method, as it is quick and most find the process intuitive.
  • The first cross product, \(\vec{r}_x \times \vec{F}_y\text{,}\) has a positive value because \(\ihat\times\jhat=+\khat\text{,}\) not because you are simply multiplying two positive components.
  • The second cross product, \(\vec{r_y} \times \vec{F_x}\text{,}\) results in a negative value because \(\jhat\times\ihat=-\khat\text{.}\)
  • All moments have units of force times distance, in this case [ft \(\cdot\) lbf].
  • The overall sign of \(\vec{M_A}\) determines the final direction. A positive value corresponds to a counterclockwise moment – right thumb out of the page – and a negative value indicates a clockwise moment. See Figure 4.1.6 for the hand diagram.
Find the moment of \(\vec{F}\) about point \(A\) using a vector cross product,
\begin{equation*} \vec{M}_A = \vec{r} \times \vec{F}\text{.} \end{equation*}
A diagram with both position vector r and force vector F broken into x and y components.
We can also solve for the moment \(\vec{M}_A\) using the vector determinant method of Subsection 2.8.3. We can use the values of \(\theta\text{,}\) \(\vec{r}\text{,}\) and \(\vec{F}\) computed in part (a) above. Jumping straight into the vector determinant, we find:
\begin{align*} \vec{M}_A \amp = \vec{r} \times \vec{F}\\ \amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ r_x \amp r_y \amp 0 \\ F_x \amp F_y \amp 0 \end{vmatrix}\\ \amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ 6 \amp 3 \amp 0 \\ 277.35 \amp 416.025 \amp 0 \end{vmatrix}\\ \amp = \ft{6}\left(\lbf{416.025}\right)\left(\khat\right)+\ft{3}\left(\lbf{277.35}\right)\left(-\khat\right)\\ \amp = \ftlbf{1664.1}\left(+\khat\right) \end{align*}
  • Determinants are a robust way to compute two-dimensional cross products but take a bit more effort than Varignon’s Theorem. The math is exactly the same either way, which means that Varignon’s Theorem is just a shortcut to working through a two-dimensional vector determinant.
  • The signs on the cross-product terms \(\vec{r}_x \times \vec{F}_y\) and \(\vec{r}_y \times \vec{F}_x\) still come from the right-hand rule, and conveniently the process of multiplying diagonals in the determinant takes care of the signs.
  • Recognize that the reason we multiply diagonals in a determinant is that we only want to multiply the perpendicular components.
Find the moment of \(\vec{F}\) about point \(A\) by finding the perpendicular distance \(d_\perp\text{,}\)
\begin{equation*} M_A=F\ d_{\perp}\text{.} \end{equation*}
A diagram where the line-of-action of F is extended down past A and a perpendicular distance is drawn between point A and this line-of-action.
This solution requires you to find the perpendicular distance \(d_{\perp}\) between the point \(A\) and line-of-action of \(\vec{F}\text{.}\) One way to find this distance is shown below.
  1. Draw a moderately large and accurate diagram. Too much confusion has been created by small, inaccurately-drawn diagrams. Include a right triangle whose hypotenuse connects the moment center \(A\) with a convenient point on the line of action of \(\vec{F}\text{.}\) We have chosen point \(C = (0,-3)\text{,}\) although point \(B\) would work as well. Can you prove that \(C\) is on the line of action?
  2. Since the angle \(\theta\) between the line of action of the force and any horizontal line is the same everywhere, \(\theta\) appears three more times on this diagram, and gives us an angle in the right triangle.
  3. Use right triangle trigonometry to find \(d_{\perp}\text{.}\)
    \begin{align*} \sin \theta \amp = \frac{d_{\perp}}{AC}\\ d_{\perp} \amp = AC\left(\sin \theta\right)\\ \amp = 4\left(\sin \ang{56.3}\right)\\ \amp = \ft{3.328} \end{align*}
  4. Finally, compute the moment about \(A\text{.}\)
    \begin{align*} M_A \amp=F\ d_{\perp}\\ \amp =\ft{3.328}\left(\lbf{500}\right)\\ \amp = \ftlbf{1664.10}\\ \vec{M}_A \amp =\ftlbf{1664.10}\left(+\khat\right) \end{align*}
    The \(\left(+\khat\right)\) direction of \(\vec{M}_A\) comes from the observation of the right-hand rule, as scalar moment computations are not directional.
Find the moment of \(\vec{F}\) about point \(A\) by finding the perpendicular component of \(\vec{F}\text{,}\)
\begin{equation*} M_A =F_{\perp}\ d\text{.} \end{equation*}
A diagram where the portion of force F perpendicular to the segment connecting A to point B where the force is apples is shown.
This solution requires you to find the rectangular component of force \(\vec{F}\) perpendicular to segment \(d\)connecting the moment center \(A\) with a point on the line of action of \(\mathbf{F}\text{.}\)
One approach to finding \(F_{\perp}\) is shown below.
  1. Draw a large and accurate diagram to assist in finding the distances and angles in this problem.
    Draw a line through point \(A\) and the point where \(\vec{F}\) acts, or to any other convenient point on its line of action. Then draw a rectangle aligned with this line indicating the parallel and perpendicular components of \(\vec{F}\text{.}\) This rectangle consists of two congruent right triangles, having the force as their common hypotenuse. You will need to find an angle inside in the triangle to find the components.
  2. Use triangle \(ABD\) to compute angle \(\beta\) and the length of the hypotenuse \(d\text{.}\)
    \begin{equation*} \beta=\tan^{-1} \left( \frac{3}{6} \right) = \ang{26.565} \end{equation*}
    \begin{equation*} d=\sqrt{6^2+3^2}=\ft{6.708} \end{equation*}
  3. Recalling that we have previously found \(\theta = \ang{56.31}\text{,}\) find \(\alpha\text{.}\)
    \begin{equation*} \alpha=\theta-\beta = \ang{29.74} \end{equation*}
  4. Find \(\vec{F_{\perp}}\) using the force triangle.
    \begin{align*} F_\perp \amp= F \sin \alpha \\ \amp= \lbf{500} \sin\ang{29.74}\\ \amp = \lbf{248.07} \end{align*}
  5. Finally, compute the magnitude \(M_A\) and the vector \(\vec{M_A}\text{.}\)
    \begin{align*} M_A \amp = F_{\perp} d \\ \amp =\lbf{248.07} \cdot \ft{6.708} \\ \vec{M_A} \amp =\ftlbf{1664.10}\left(+\khat\right) \end{align*}
    The counterclockwise direction \(\left(+\khat\right)\) comes from the right-hand rule, since scalar moment computations are not directional.
We have discussed several different approaches for finding the moment of a force about a point in the preceding sections. While all of these methods produce the same result, frequently you will find that one method is computationally simpler to apply than another. The simplest method generally depends on what is know and unknown in a given situation, so we recommend that you become familiar with all these methods.
The interactive diagram below will help you visualize the different approaches for finding moments that were covered.


This interactive demonstrates four methods to visualize and compute the moment of a force about point in two dimensions.
Figure 4.3.7. Four equivalent approaches to finding a moment about a point.