I understand the torque vector to be the cross product of the radius (moment arm) and force vectors, but that means the torque would be perpendicular to the radius and force vectors, which makes no sense to me, e.g. a force applied tangent to the surface of a car tire creates a torque along the line of the axle.

I'm pretty sure I am just misunderstanding a simple formula, so I wanted to make sure.

And, when you use the formula for torque, is torque defined as a vector or just a scalar? I would think it would be a vector.


3 Answers 3


We need a quantity that conveys information about the direction of the radius vector and the direction of the force vector.

Two vectors determine a plane. So we need a quantity that specifies a plane. One way to do that is to specify the vector normal to the plane. That's what the cross product does for us. There is an ambiguity as to direction: there are two normal vectors to a plane. We solve this by choosing one arbitrarily: we decide by convention to use the right-hand rule.

There are other ways to represent a torque that some would argue are more natural, for example, the bivector. These other ways are usually extensible to dimensions higher than three, whereas the cross product works only in three dimensions. Well, we live in a world having three spatial dimensions. That fact, and years and years of usage and tradition, has cemented the cross product into our toolbox.

The cross product has a few oddities associated with it, but it does the job. Some people think we should do away with the cross product. It might be nice to do that and use a more natural mathematical construct, but trying to make a change like that is like rolling a very large boulder up a hill.


You are right in saying that the torque points along the line of axle. This doesn't make sense intuitively, but if you look at the formalism of angular momentum vector, this would be obvious.

So angular momentum is defined as $L = r \times p$. And torque is defined as $\tau = r \times F$ . It is clear that $$\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} = \frac{\vec{p}}{m} \times \vec{p} + \vec{r} \times \vec{F} = \vec{r} \times \vec{F} = \vec{\tau}$$

When you look at the car tire case. The friction creates a torque along the axle, which increases the angular momentum of tire in that axle direction. What does that mean? Since $$L=r \times p$$ and $r$ does not change, $p$ must increase. Which makes sense, since friction makes the wheel spin faster!

In summary, the direction of torque vector is determined by the definition of angular momentum and angular velocity vectors which capture more measurable kinematic quantities. This formalism is a bit counterintuitive at first. But you will eventually get used to it...

  • $\begingroup$ I think torque is a rotational vector like angular velocity. We know surface vector of any surface is perpendicular to that surface. It does't mean that the surface is along that perpendicular. Similarly in case of angular velocity, $\omega$ define a rotation in xy plane (say) but the direction of $\omega$ vector along the perpendicular to xy plane, like along +z or -z axis. Same thing occurs in your case. The torque vector is along the axle of the wheel means that the rotational torque's acting around the wheel, which helps in moving the wheel. $\endgroup$ Oct 12, 2015 at 4:38
  • $\begingroup$ Yeah Rajesh, you are right. I am just saying that when students are first introduced to torque, the direction torque points in doesn't make immediate sense since it is not the direction that the wheel rotates. $\endgroup$ Oct 12, 2015 at 5:37
  • $\begingroup$ I'm agree with you Zhengyan. Student who first come to know about torque usually mixed up the concept of linear and angular vector. So it's very important to clear this concept first. $\endgroup$ Oct 12, 2015 at 5:53
  • $\begingroup$ it is important to state the location where vectors such as $\vec{L} $, $\vec{\tau}$ and $\vec{v}$ are measured at. Otherwise you might add the confusion of the OP. $\endgroup$ Dec 19, 2017 at 4:11

A force acts upon a line of action in 3D space. The force vector can be anywhere along this line and it won't change the situation.

Torque is the moment of force because it conveys the (perpendicular) distance where this force acts upon. Any component of location along the line of action needs to be ignored and this is achieved with the vector cross product.

Other common moments in mechanics are:

  • Moment of Rotation - Linear velocity is the moment of rotation because the velocity of a point A depends on the perpendicular distance to the axis of rotation $$\mathbf{v}_A = \mathbf{r} \times {\boldsymbol \omega}$$ where $\mathbf{r}$ is the location of the axis relative to A.
  • Moment of Force - Torque is the moment of force because the equipollent torque at a point A depends on the perpendicular distance to the line of action $${\boldsymbol \tau}_A = \mathbf{r} \times \mathbf{F}$$ where $\mathbf{r}$ is the location of the line of action relative to A.
  • Moment of Momentum - Angular momentum is the moment of momentum because the angular momentum about a point A depends on the perpendicular distance to the axis of momentum $$\mathbf{L}_A = \mathbf{r} \times \mathbf{p}$$ where $\mathbf{r}$ is the location of the axis of momentum relative to A.

All of the above are similar because they are manifestations of the same law. The law described by Julius Plücker when he used the moment of a line to describe the location of a 3D line in space. Rotation, Momentum and Forces all contain lines in space. Their moments are a set of (homogeneous) coordinates describing the closest point on the line to the point of measurement.


Your Answer

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

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