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.

## Best Answer

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

ydirection and the $y$ component of ${\bf F}$ in thezdirection.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