3.2: Moment of Force

First, let’s look at a familiar two-dimensional situation. In Figure III.1 I draw a force $\textbf{F}$ and a point O. The moment of the force with respect to O can be defined as

Force times perpendicular distance from O to the line of action of $\textbf{F}$.

Alternatively, (Figure III.2) the moment can be defined equally well by

Transverse component of force times distance from O to the point of application of the force.

Either way, the magnitude of the moment of the force, also known as the torque, is $rF \sin\theta$ We can regard it as a vector, $\boldsymbol\tau$, perpendicular to the plane of the paper:

$$\begin{equation} \ \boldsymbol\tau = \textbf{r} \times \textbf{F}\tag{3.2.1}\label{eq:3.2.1} \end{equation}$$

Now let me ask a question. Is it correct to say the moment of a force with respect to (or “about”) a point or with respect to (or “about”) an axis?

In the above two-dimensional example, it does not matter, but now let me move on to three dimensions, and I shall try to clarify.

In Figure III.3, I draw a set of rectangular axes, and a force $\textbf{F}$, whose position vector with respect to the origin is $\textbf{r}$.

The moment, or torque, of $\textbf{F}$ with respect to the origin is the vector

$$\begin{equation} \ \boldsymbol\tau = \textbf{r}\times \textbf{F}\tag{3.2.2}\label{eq:3.2.2} \end{equation}$$

The $x-, y-$ and $z$ -components of $\boldsymbol\tau$ are the moments of $\textbf{F}$ with respect to the $x-, y-$ and z-axes. You can easily find the components of $\boldsymbol\tau$ by expanding the cross product $\ref{eq:3.2.2}$:

$$\boldsymbol\tau = \hat{\textbf{x}}(yF_{z}-zF_{y})+\hat{\textbf{y}}(yF_{x}-xF_{z})+\hat{\textbf{z}}(xF_{y}-yF_{x}) \tag{3.2.3}\label{eq:3.2.3}$$

where $\bf \hat{x},\hat{y},\hat{z}$ are the unit vectors along the $x,y,z$ axes. In Figure III.4, we are looking down the $x$-axis, and I have drawn the components $F_{y}$ and $F_{z}$, and you can see that, indeed, $\tau_{x} =yF_{z}-zF_{y}$.

The dimensions of moment of a force, or torque, are ML²T^-2, and the SI units are N m. (It is best to leave the units as N m rather than to express torque in joules.)