Suppose you are given a line $L$ in space, then if you have a system of particles with forces acting on them the torque about a point on the line, you have to sum $ \sum r_i F_i$ over all particles. However say you wanted to compute the quantity 'torque about the line', then you'd find that this quantity is independent of which point you take. This makes intuitive sense (to me) because we are torque components along this line. It can be said that this torque measures rotation about that line. (Refer here for proof)

However, what does the actual torque which we calculate after we choose an arbitrary point on the line represent? Does it mean rotation perpendicular to that line.... that is something I can't grasp?

I hope for a clear and simple explanation of this quantity and how changes in this quantity and how variations in it change the motion physically.

Illustration of what I mean:

A line in space can be written by the equation $ \vec{r} = \vec{a_o} + \lambda \vec{v}$, now the vector connecting the point $a_o$ (*)to line is $ \vec{b}$. Suppose the location of force is $ \vec{j}$ from origin, the vector connecting the force vector to the line is $\vec{j} - \vec{r}$. We can write torque around $a_o$ as:

$$ \vec{\tau} = ( \vec{j} - \vec{r}) \times \vec{F}$$

Now, it's pretty easy to see that the torque is dependent on the point we take on line as follows:

$$ \vec{ \tau(\lambda)} = ( \vec{j} - (a_o + \lambda \vec{v} ) )\times \vec{F}$$


  • All the above discussion is in cartesian coordinates with an arbitrary origin which is not on the line.
  • (*): Here I have switched between the representation of point using position vector and point $ \vec{a_o}$ is the position vector to the point $a_o$ with some coordinates.
  • $ \vec{v}$ is a vector parallel to the line
  • $ \lambda $ is the parameter for the line.
  • I am looking for further discussion from the answer given by user BMS in this stack


Equation of a line in space

  • $\begingroup$ Can you grasp the the idea of torque causing rotation of a wheel about its axle? The axle is the line. Looking at the wheel face on, the axle is a point. The torque is the same about the line as it is about the point. $\endgroup$
    – Bob D
    Oct 6, 2020 at 17:14
  • $\begingroup$ Right yeah I agree that torque is same along the whole line but we get different values of the torque vector if we actually do evaluate it on different along the line points. However the part along the line is same. So, I'm asking what this extra torque we get from taking different points on the line $\endgroup$
    – Babu
    Oct 6, 2020 at 17:19
  • $\begingroup$ I don't understand you. What "part" along the line is the same. What do you mean by "extra torque"? You need to be clear. Explain it quantitatively. $\endgroup$
    – Bob D
    Oct 6, 2020 at 17:53
  • $\begingroup$ I have edited the question describing the situation using mathematical equations $\endgroup$
    – Babu
    Oct 6, 2020 at 19:11
  • $\begingroup$ Why are you showing your "line in space" as the sum of two vectors? What is the coordinate system for this "line in space"? What is lambda? a constant? In your second to last equation you seem to show it as variable that Tau is a function of. You show $a_o$ as a vector quantity in the equation then you refer to it as a "point". Which is it. What is $v$ in the equation. $\endgroup$
    – Bob D
    Oct 6, 2020 at 20:09

2 Answers 2


Another Answer

I realized I did not answer the question directly. The question of if I know the torque from forces on a system summed up at a point, does this torque represent also the torque about a line passing through the point.

The quick answer is not unless the line is parallel to the net force applied to the system. The proof is in the torque transformation equation from one point to another $$ \vec{\tau}_B = \vec{\tau}_A + \vec{F} \times \vec{r}_{AB} $$

The two torques are equal only if $\vec{F} \times \vec{r}_{AB} = \vec{0}$, or the displacement vector is parallel to the net force.

In addition, the term torque about a line is a bit ambiguous. Does this line represent a kinematic constraint (as rotation about the line) and the torque is the scalar magnitude applied along the line to balance the applied forces? If that is so, there will be other torque components required that are going to be perpendicular to the line which are going to be ignored.

In this case, the component of torque due to a distance force $\vec{F}$ along a line with direction $\hat{z}$ is given by the triple product

$$ \tau_z = \hat{z} \cdot ( \vec{r} \times \vec{F} ) = \vec{F} \cdot ( \hat{z} \times \vec{r} ) = \vec{r} \cdot ( \vec{F} \times \hat{z}) $$

and the above value is the same anywhere along the line. To prove this, consider a different displacement vector with a parallel offset $\vec{r} + \hat{z} d$,

$$ \tau_z = \vec{F} \cdot ( \hat{z} \times (\vec{r} + \hat{z} d)) = \vec{F} \cdot ( \hat{z} \times \vec{r} ) \; \checkmark$$

Original Answer

In some sense torque that is due to a force summed at a point is only an indication of where the line of action of the force is located in space. This line of action is the locus of points where there is no torque (or moment of force), or the torque is only parallel to the line. The direction of the line is described by the direction of the force.


A force of $$\vec{F} = \pmatrix{ 6 \\ -1 \\ 2}$$ has equipollent moment about the origin of $$\vec{\tau} = \pmatrix{7 \\ 16 \\ -13}$$ find the line where the force is applied though.

The location of the line of action is found by

$$ \vec{r} = \tfrac{1}{\| \vec{F} \|^2} (\vec{F} \times \vec{\tau}) $$

$$ \vec{r} = \tfrac{1}{41} \pmatrix{6 \\ -1 \\2 } \times \pmatrix{7 \\ 16 \\ -13} = \tfrac{1}{41} \pmatrix{ -19 \\ 92 \\ 103 } $$

Let us prove this by calculating $\vec{r}\times\vec{F}$

$$ \vec{\tau} \equiv \tfrac{1}{41} \pmatrix{ -19 \\ 92 \\ 103 } \times \pmatrix{6 \\ -1 \\2} = \pmatrix{7 \\ 16 \\-13} \;\; \checkmark$$

In fact, this is the only useful (geometrical) information one can extract from the torque vector. Well, the location of the line of action, and the amount of parallel torque along the line of action.

The parallel torque is calculated by $$ \vec{\tau}_\parallel = \left( \frac{ \vec{F} \cdot \vec{\tau} }{ \| \vec{F} \|^2 } \right) \vec{F}$$ with the part inside the parenthesis called the pitch (it is a scalar value).

In the above example, the pitch was zero since $\vec{F} \cdot \vec{\tau} =0$, but for a general case of 3D torque, it might be 0, positive, negative, or infinite. It is infinite if there is a pure torque with no net force applied.

  • $\begingroup$ Ok, I went on a spree reading your many answers but at the end of it I'm quite confused. These things were never mentioned in any book on classical physics I read and seems like it orginates mostly from robotics. The point which bugs me is that why is there so much mathematics involved in trying to describe a simple physical phenomena... or is it not actually that simple $\endgroup$
    – Babu
    Oct 7, 2020 at 6:20
  • $\begingroup$ What is "equipollent moment "? $\endgroup$
    – Babu
    Oct 7, 2020 at 6:26
  • 1
    $\begingroup$ @Buraian - Equipollent moment is the torque that is a result of a force at a distance. When you transfer a load system from one point to another, you have to add the equipollent torque in order not to change the problem. > EQUIPOLLENT Of two systems of forces, having the same vector sum and the same total torque about an arbitrary point $\endgroup$ Oct 7, 2020 at 11:50
  • $\begingroup$ Also read this almost identical answer to a different but related question. $\endgroup$ Oct 8, 2020 at 20:58
  • 1
    $\begingroup$ I would just like to say a thank you for the extensive explanations that you have done on these screw theory topics throughout the years. It was a great pleasure going over them and I in fact got interested in the topic so much that I went through all the lectures that you had provided. They were difficult at points for me but I still enjoyed it. $\endgroup$
    – Babu
    Oct 9, 2020 at 20:54

However say you wanted to compute the quantity 'torque about the line', then you'd find that this quantity is independent of which point you take.

In a general situation it is not true. Suppose several particles accelerating in the same direction, as a group of racing cars after the start.

If we trace a line perpendicular to the lane, the torque increases as the point is far away from it.


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.