# Force-couple system equivalence: 3 moment equations

In force-couple system equivalence, We can show that two systems are equivalent if the resultant force in both are equal and that it has the same line of action.

So Rx1 = Rx2, Ry1= Ry2, Ma1 = Ma2

Now my book says we can use another form of equation: We can choose 3 point and find the moment about them and use it to find the equivalent force, as long as a, b and c are not on the same line

I fail to see why should this work. Ma = x Ry -y Rx Mb = x Ry - y Rx Mc = x Ry - y Rx

THe x and ys in each equation is different, How do you solve such a system to find Ry and Rx and the line of action? You can also use 2 moments and 1 resultant force equation in x or y direction

• Please use math formatting for readability. Note that the cross product $\times$ is made with \times. Jan 12, 2018 at 1:34
• Can you show an example of how to calculate the force from the three moments. Jan 12, 2018 at 1:45
• This is what I am asking, It said that we can replace the 2 equation of force ( in x and y direction) with 2 moments equation. We also use 3 moments equation to ensure that 1 system is equal to another. The question is I dont know why this would work Jan 12, 2018 at 1:54
• I wonder if this statement works in general in 3D, or only in 2D. I am investigating. Jan 12, 2018 at 15:26
• Please do. I can see we can use it of course if we know two systems are equivalent , we can find some unknowns using moment equation but it is stated in the book that we can also find the resultant force using those 3 moments equation ( instead of the usual 2 force component and 1 moment), we weren't given any example of that. So if you find a solution to this please post it :) Jan 12, 2018 at 16:20

So we have three points on a plane, with coordinates \begin{aligned} \mathbf{r}_A & = \pmatrix{x_A\\y_A\\0} & \mathbf{r}_B & = \pmatrix{x_B\\y_B\\0} & \mathbf{r}_C & = \pmatrix{x_C\\y_C\\0} \end{aligned}

A force is acting with a line of action perpendicular to the plane \begin{aligned} \mathbf{F} & = \pmatrix{0\\0\\F} & \mathbf{r} & = \pmatrix{x \\ y \\ 0} \end{aligned}

But we don't know the force magnitude or the location. But from any two of the three equipollent moments \begin{aligned} \mathbf{M}_A & = \left(\mathbf{r}-\mathbf{r}_A\right) \times \mathbf{F} \\ \mathbf{M}_B & = \left(\mathbf{r}-\mathbf{r}_B\right) \times \mathbf{F} \\ \mathbf{M}_C & = \left(\mathbf{r}-\mathbf{r}_C\right) \times \mathbf{F} \\ \end{aligned} we deduce the force by using the fact that $\mathbf{M}_B - \mathbf{M}_A = \left(\mathbf{r}_A-\mathbf{r}_B \right) \times \mathbf{F}$

$$\mathbf{F} = \frac{ \left( \mathbf{r}_B-\mathbf{r}_A \right) \times \left(\mathbf{M}_B-\mathbf{M}_A\right) }{ \| \mathbf{r}_B-\mathbf{r}_A \|^2 }$$ provided that $\mathbf{F}$ is perpendicular to the plane of A, B and C.

The location of the force is then recovered from any one of the moments by

$$\mathbf{r} = \mathbf{r}_A + \frac{ \mathbf{F} \times \mathbf{M}_A}{\| \mathbf{F} \|^2}$$

Quick Proof

Take $\mathbf{M}_B - \mathbf{M}_A = \left(\mathbf{r}_A-\mathbf{r}_B \right) \times \mathbf{F}$ and plug in the solution for $\mathbf{F}$ as follows:

\require{cancel} \begin{aligned} \mathbf{M}_B - \mathbf{M}_A &= -\left(\mathbf{r}_B-\mathbf{r}_A \right) \times \frac{ \left( \mathbf{r}_B-\mathbf{r}_A \right) \times \left(\mathbf{M}_B-\mathbf{M}_A\right) }{ \| \mathbf{r}_B-\mathbf{r}_A \|^2 } \\ & = -\frac{\left(\mathbf{r}_B-\mathbf{r}_A \right) \cancel{ \left( \left(\mathbf{r}_B-\mathbf{r}_A \right) \cdot\left(\mathbf{M}_B-\mathbf{M}_A\right) \right) } - \left(\mathbf{M}_B-\mathbf{M}_A\right) \| \mathbf{r}_B-\mathbf{r}_A \|^2 }{ \| \mathbf{r}_B-\mathbf{r}_A \|^2 } \\ & \equiv \mathbf{M}_B - \mathbf{M}_A \end{aligned}

Use the vector triple product $a\times(b \times c) = b (a\cdot c) - c (a\cdot b)$

The canceling of $\left(\mathbf{r}_B-\mathbf{r}_A \right) \cdot\left(\mathbf{M}_B-\mathbf{M}_A\right)$ can be proven by the definition of the equipollent moments.

Similarly, take the calculation of $\mathbf{r}$ and expand out the expression $$\mathbf{F} \times \mathbf{r} = \mathbf{F} \times \mathbf{r}_A + \frac{ \mathbf{F} \times \left( \mathbf{F} \times \mathbf{M}_A \right)}{ \| \mathbf{F} \|^2} = \mathbf{F} \times \mathbf{r}_A + \frac{ \mathbf{F} \cancel{ \left( \mathbf{F} \cdot \mathbf{M}_A \right)}-\mathbf{M}_A \| \mathbf{F} \|^2 }{ \| \mathbf{F} \|^2} = \mathbf{F} \times \mathbf{r}_A - \mathbf{M}_A$$

Example

• A force $F=\pmatrix{0\\0\\5}$ acts through the point $\mathbf{r} = \pmatrix{4.2 \\ 2.5 \\ 0}$

• We measure the moment $\mathbf{M}_A = \pmatrix{12.5 \\ 29 \\ 0}$ at $\mathbf{r}_A = \pmatrix{10\\0\\0}$

• We measure the moment $\mathbf{M}_B = \pmatrix{-12.5 \\ -21 \\ 0}$ at $\mathbf{r}_B = \pmatrix{0\\5\\0}$

• Use the following relative quantities \begin{aligned} \mathbf{r}_{AB} &= \mathbf{r}_B - \mathbf{r}_A = \pmatrix{-10 \\ 5 \\ 0} & \mathbf{M}_{AB} &= \mathbf{M}_B - \mathbf{M}_A = \pmatrix{-25 \\ -50 \\ 0} \end{aligned}

• To get the force

$$\mathbf{F} = \frac{ \pmatrix{-1 \\ 5 \\0} \times \pmatrix{-25 \\ -50 \\ 0} }{ \| \pmatrix{-10 \\ 5 \\0} \|^2} = \pmatrix{0 \\ 0 \\ 5} \; \checkmark$$

• and the location

$$\mathbf{r} = \pmatrix{10 \\ 0 \\0} + \frac{ \pmatrix{0\\0\\5} \times \pmatrix{12.5\\29\\0} }{ 5^2 } = \pmatrix{4.2 \\ 2.5 \\ 0} \; \checkmark$$