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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:

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

Notes:

  • 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

References:

Equation of a line in space

Clemens Bartholdy
  • 8,309

2 Answers2

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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.

Example:

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.

John Alexiou
  • 40,139
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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.

Claudio Saspinski
  • 17,607