Moment (torque)

Every time you open a door, tighten a bolt with a spanner, or adjust a bike's handlebars, you're creating a turning effect. The further from the pivot you apply a force, and the more directly you push, the stronger that effect. This turning effect is called a moment or torque, and this resource shows you how to calculate and apply it. You will also explore how the physics of moments apples to Aboriginal digging sticks.

Moment \(M\) in 2D

When a net force is applied through a point that is not the centre of mass, a moment (or torque) \(M\) is applied and rotation occurs. Just as a net force causes linear motion (an object moving in a straight line), a moment causes rotational motion – it is the rotational equivalent of a pushing or pulling force, also called the rotational analogue of force.

The size of a moment depends on:

• the size of the force
• the distance from the pivot to where the force is applied
• the angle of the force relative to the lever arm.

The turning force is most effective when it is acting at right angles to the rotating object. For example, the turning force applied to the beam on the left (approximately \(45^{\circ}\) does not produce as great a moment compared to the beam on the right, where the force is at right angles to the beam.

Moment is found using the equation:

\[M=F\sin\theta\times r\quad\text{or}\quad M=F_{\perp}\times r\]

where:

• \(M\) is the moment in Newton metres \(\text{N m}\)
• \(F\) is the force applied in Newtons \(\text{N}\)
• \(\theta\) is the angle between the force and object in degrees
• \(r\) is the distance from the axis of rotation (or moment arm) in metres \(\text{m}\).

Here, \(F\sin\theta\) is the component of \(F\) acting perpendicular to the moment arm – the part of the force that actually produces rotation.

Moment is a vector quantity. In 2D problems, we describe it as either clockwise or anticlockwise. In 3D problems, it has a direction in space.

Example – calculating moment using force components

A force \(\mathbf{F}\) is applied to the handle of a hammer being used to extract a nail. The force must produce a clockwise moment of \(\mathbf{75}\,\textbf{N m}\) about point A (the nail head). Determine the magnitude of \(\mathbf{F}\).

Since moment is a vector, we take clockwise moments as negative. The force \(F\) acts \(30^{\circ}\) to the handle, so we resolve it into two components:

• \(F\cos30^{\circ}\) acts along the handle, with a perpendicular moment arm of \(150\,\text{mm}=0.15\,\text{m}\) from point A
• \(F\sin30^{\circ}\) acts perpendicular to the handle, with a moment arm of \(450\,\text{mm}=0.45\,\text{m}\) from point A.

Both components produce a clockwise (negative) moment about A. Setting up the moment equation:

\[\begin{align*} \sum M_{A} & = \sum \left( F_{\perp}\times r\right)_{A} \\[6pt]
-75 & = -\left(F\cos30^{\circ}\times0.15\right) - \left(F\sin30^{\circ}\times0.45\right) \end{align*}\]

Factoring out \(F\):

\[\begin{align*} -75 & = -F\left(0.15\cos30^{\circ}+0.45\sin30^{\circ}\right) \\[6pt]
& = -F\left( (0.15\times0.866)_+(0.45\times0.500) \right) \\[6pt]
& = -F\left(0.1299+0.225\right) \\[6pt]
& = -0.3549F \\[6pt]
F & = \frac{75}{0.3549} \\[6pt]
& = 211.3\,\text{N} \end{align*}\]

The magnitude of the force applied to the handle of the hammer is \(211.3\,\text{N}\).

Moment \(M\) in 3D

The moment of a force can also be written as a vector cross product in the 3D space.

\[\vec{M}=\vec{r}\times\vec{F}\]

where:

• \(\vec{M}\) is the moment vector in Newton metres \(\text{N m}\)
• \(\vec{r}\) is the moment arm vector in metres \(\text{m}\)
• \(\vec{F}\) is the force vector in Newtons \(\text{N}\).

Vector \(\vec{M}\) is perpendicular to the plane containing \(\vec{r}\) and \(\vec{F}\).

To find the direction of \(\vec{M}\) using the right-hand rule: point your right-hand fingers in the direction of \(\vec{r}\) (the moment arm, from the point O towards where the force is applied), then curl them towards \(\vec{F}\). Your thumb points in the direction of \(\vec{M}\) – perpendicular to the plane containing \(\vec{r}\) and \(\vec{F}\). The order of the cross product matters: \(\vec{r}\times\vec{F}\) gives a different direction to \(\vec{F}\times\vec{r}\), so always write the moment arm first.

Since \(\vec{r}\) and \(\vec{F}\) are written as \(x, y, z\) coordinates, then \(\vec{M}\) can be found by finding the determinant.

\[\mathbf{M} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\
r_{x} & r_{y} & r_{z} \\
F_{x} & F_{y} & F_{z} \end{vmatrix} \]

where:

• \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) are the unit vectors in the \(x\), \(y\) and \(z\) directions, respectively
• \(r_{x}\), \(r_{y}\) and \(r_{z}\) are \(x\), \(y\) and \(z\) components of the moment arm vector \(\vec{r}\), respectively
• \(F_{x}\), \(F_{y}\) and \(F_{z}\) are the \(x\), \(y\) and \(z\) components of the force vector \(\vec{F}\), respectively.

If you need a refresher on vector or cross product, visit this page.

Example – calculating moment using vector or cross product

Two forces act at point A on a 3D frame. If \(\mathbf{\vec{F}_{1}=80\hat{i}-100\hat{j}+60\hat{k}}\,\textbf{N}\) and \(\mathbf{\vec{F}_{2}=-150\hat{i}+200\hat{j}+90\hat{k}}\,\mathbf{N}\), determine the resultant moment produced by these forces about point O. Express your answer as a Cartesian vector.

First, we need to find the resultant force \(\vec{F}_{r}\). Since both forces act at the same point A, we can combine them into a single resultant force before computing the moment. This simplifies the cross product to a single calculation.

\[\begin{align*} \vec{F}_{R} & = \vec{F}_{1}+\vec{F}_{2} \\[6pt]
& = \left( 80\hat{i}-100\hat{j}+60\hat{k} \right) + \left( -150\hat{i}+200\hat{j}+90\hat{k} \right) \\[6pt]
& = \left( -70\hat{i}+100\hat{j}+150\hat{k}\right) \,\text{N} \end{align*}\]

Now, we identify the position vector \(\vec{r}_{A}\). From the diagram, point A is located at coordinates \((1,0.6,0.8)\,\text{m}\) from the origin O. The position vector is therefore:

\[\vec{r}_{A} = \{ 1.0\hat{i}+0.6\hat{j}+0.8\hat{k} \} \,\text{m}\]

We are ready to calculate the moment using the cross product.

\[\begin{align*} \vec{M} & = \vec{r}_{A} \times \vec{F}_{R} \\[6pt]
& = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\
1.0 & 0.6 & 0.8 \\
-70 & 100 & 150 \end{vmatrix} \end{align*}\]

Expanding the determinant:

\[\begin{align*} M_{i} & = \left( r_{y}\times F_{z} \right) - \left( r_{z}\times F_{y} \right) \\[6pt]
& = (0.6\times150)-(0.8\times100)\\[6pt]
& = 90-80\\[6pt]
& = 10 \end{align*}\]
\[\begin{align*} M_{j} & = -[ \left( r_{x}\times F_{z} \right) - \left(r_{z}\times F_{x} \right) ] \\[6pt]
& = -[ (1.0\times150) - (0.8\times-70) ] \\[6pt]
& = -(150+56) \\[6pt]
& = -206 \end{align*}\]
\[\begin{align*} M_{k} & = \left(r_{x}\times F_{y}\right) - \left(r_{y}\times F_{x}\right) \\[6pt]
& = (1.0\times100)-(0.6\times-70) \\[6pt]
& = 100+42 \\[6pt]
& = 142 \end{align*}\]

The resultant moment about O is \(\vec{M} = \{ 10\hat{i}-206\hat{j}+142\hat{k} \}\,\text{N m}\).

Exercise – calculating moment

1. Two workers are using ropes to prevent a vertical pole from rotating about its pin at point A. The worker at point B exerts a force of \(P=180\,\text{N}\) on their rope at \(45^{\circ}\) to the horizontal, attached \(5.4\,\text{m}\) above point A. The worker at point C is holding a rope at \(3.6\,\text{m}\) above point A and pulls in a direction described by a 3–4–5 ratio (3 vertical, 4 horizontal). Determine the magnitude of the force \(F\) the worker at point C must exert to keep the resultant moment about A equal to zero.
   

   

2. A force of \(F=480\,\text{N}\) acts from point B towards point A along the pipe. Point B is located at \(\{0.3\hat{i}+1.2\hat{j}+0.6\hat{k}\}\,\text{m}\) from point O, and point A is at \(\{0\hat{i}+1.2\hat{j}+0\hat{k}\}\,\text{m}\). Determine the moment of force \(F\) about point O. Express the result as a Cartesian vector.

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