My question is regarding Torque, and specifically the definition of the lever arm.

As far as I've understood, the lever arm is the line perpendicular to the line of action to your reference point. The line of action is stretched from your force vector.

Now I have the magnitude of torque in two equivalant ways

$\tau = \underbrace{|\vec{r}-\vec{r_0}|\sin{\phi}}_{a}|\vec{F}| = aF.$

$\tau = \underbrace{|\vec{r}-\vec{r_0}|}_{r} \ |\vec{F}|\sin{\phi} = r F_{\perp}.$

The first I think is most common in 2-d examples with levers, and the second in examples with rotating bodies in 3-d, where often $F_{\perp}$ is written as $F_{\tan}$

So my hang up is as follows; How I think it is: $a$ is defined as the lever arm, while $r$ is not. I have a book/compendium here that defines both $r$ and $a$ as the lever arm. As if the lever arm can both be defined in respect to the line of action and the line of the action drawn from the perpendicular component of force.

Is there a difference of what you call the lever arm in the vectorial vs magnitude sense? Is the book's assertment correct? I'm confused.

  • 1
    $\begingroup$ The two interpretations are identical (or more accurately, dual to each other). $\endgroup$ Commented Feb 10, 2016 at 20:55
  • $\begingroup$ Look into Frege's distinction between Sinn and Bedeutung (which I dare not attempt to translate) oxfordreference.com/display/10.1093/oi/… I was about to ask if I am the only one who has two definitions of what "lever arm" means. $\endgroup$ Commented Feb 15, 2023 at 10:29

2 Answers 2


In 2D, torque is $\tau = (r \sin \phi) F = r ( F \sin \phi)$. These are equivalent statements. You either consider the perpendicular distance to the line of action of $F$, or the perpendicular component of $F$ along $r$. Either way the result is the same. It really doesn't matter how you interpret this expression, as both ways are valid.

In 3D, torque is ${\bf \tau} = {\bf r}\times {\bf F}=(r_y F_z - r_z F_y, \; r_z F_x - r_x F_z, \; r_x F_y - r_y F_x)$. There is direct combination of the vector components so there is not question of interpretation. The $x$ component of ${\bf r}$ combines with the $z$ component of ${\bf F}$ in the y direction and the $y$ component of ${\bf F}$ in the z direction.

Another way to think about it is to look an work done by a force. The definition is $\Delta W = \vec{F} \cdot \Delta \vec{s}$ which can be interpreted as

  • Multiply the component of the force along the displacement with the displacement.
  • Multiply the force with the component of the displacement along the force.

A way to understand this is as follows . According to a definition, moment arm is the perpendicular distance from the axis to the line of action of force and the lever arm is the length of the line that connects the axis to the point of action of force. Hence, when the force is perpendicular to the lever arm, the lever arm and the moment arm coincide. For clarity, refer http://www.aaronswansonpt.com/wp-content/uploads/2011/07/Moment-Arm.png


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.