I used to think that torque and force were equally “fundamental”. In other words, my understanding was that we usually use Cartesian coordinates in many common problems because it is a convenient system, so as a result instantaneous forces which act in straight lines seem “easier” mathematically but torques require some extra “baggage”. This baggage includes typically teaching that torque is defined in terms of force.

But if say we happened to choose polar coordinates for the problem the situation would appear the other way around. So it would be arbitrary if we chose to define forces in terms of torques instead.

But later on I learned that angular momentum is conserved independently of regular linear momentum (IIRC). Given the definitions of force & torque as derivatives of momentum, this makes it seem much less certain that one should define either torque in terms of force or vice versa — it gives the impression they are more distinct than it first seemed.

That said, as far as I know a lot of physics is about defining & describing “fundamental forces” — not “fundamental torques”.

So is choosing to use either force or torque as the basis of laws & problems arbitrary? Or is there an actual fundamental rationale to when one or the other should be used?

  • 3
    $\begingroup$ Torque isn't something that arises when we move to polar coordinates. They exist no matter how we choose to represent coordinates. Additionally, you can work with forces in polar coordinates without even referring to torque. Why do you think this, exactly? $\endgroup$ Sep 1, 2019 at 21:54
  • $\begingroup$ @AaronStevens I was only referring to the choice of coordinate system potentially being more convenient depending on whether you chose to express a particular problem in terms of force or torque. $\endgroup$ Sep 2, 2019 at 0:41
  • $\begingroup$ Whether or not forces or torques are useful for the problem doesn't depend on the coordinate system. For example, many introductory statics problems rely on torque to be simplified, yet polar coordinates never need to be mentioned to do this. In fact, nearly all introductory physics problems that involve torque never mention polar coordinates. $\endgroup$ Sep 2, 2019 at 11:17
  • $\begingroup$ whether something is 'Fundamental is a negotiated position, and is usually somewhat circular, with a careful side step at the point that circularity may be observed. In many sense, Torque is about the choice (or lack of) origin. A free force is nice, but a force about an origin is special...! $\endgroup$ Sep 2, 2019 at 19:37
  • 2
    $\begingroup$ translation and rotation, equally fundamental I think. $\endgroup$ Sep 3, 2019 at 21:31

8 Answers 8


In mechanics no. Torque is not a fundamental quantity. it's only job is to describe where in space a force is acting through (the line of action). Torque just describes a force at a distance. Given a force $\boldsymbol{F}$ and a torque $\boldsymbol{\tau}$ you can tell that the force acts along a line in space with direction defined by $\boldsymbol{F}$, but location defined by $\boldsymbol{\tau}$ as follows $$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F} \|^2 } $$

In fact, you can slide the force vector anywhere along its line and it won't change the problem, so the $\boldsymbol{r}$ calculated above happens to be the point on the line closest to the origin.

It might be easier to discuss angular momentum first, since torque is the time derivative of angular momentum, just as force is the time derivative of linear momentum.

For a single particle with linear momentum $\boldsymbol{p} = m\boldsymbol{v}$ located at some instant at a point $\boldsymbol{r}$ the angular momentum is $$ \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$

So where is the momentum line in space? The momentum line is called the axis of percussion. It is located at

$$ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2 } = \frac{\boldsymbol{p} \times ( \boldsymbol{r} \times \boldsymbol{p})}{\| \boldsymbol{p} \|^2} = \frac{ \boldsymbol{r} (\boldsymbol{p} \cdot \boldsymbol{p}) - \boldsymbol{p} ( \boldsymbol{p} \cdot \boldsymbol{r}) }{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \frac{ \| \boldsymbol{p} \|^2}{\| \boldsymbol{p} \|^2} = \boldsymbol{r} \; \checkmark $$

provided that the point $\boldsymbol{r}$ is perpendicular to the momentum $\boldsymbol{p}$. Let me elaborate. Imagine the direction of the line being $\boldsymbol{\hat{e}} = \boldsymbol{p} / \| \boldsymbol{p} \|$, and consider a point $\boldsymbol{r} + t \boldsymbol{\hat{e}}$ for some arbitrary scalar $t$. The angular momentum will be $\boldsymbol{L} = ( \boldsymbol{r} + t \boldsymbol{\hat{e}}) \times \boldsymbol{p} = \boldsymbol{r} \times \boldsymbol{p} $. So where along the line (the value of $t$) doesn't matter. Finally, if $\boldsymbol{r}$ is not perpendicular to $\boldsymbol{p}$ you can always find a value of $t$ that makes the point perpendicular. Set $t = -(\boldsymbol{r} \cdot \boldsymbol{p}) / \| \boldsymbol{p} \|$ and the point will be perpendicular.

Such a point can always be found, and it is the point on the line closest to the origin.

The conservation law for angular momentum (coupled with the conservation law for linear momentum) just states that not only the magnitude and direction of momentum is conserved but also the line in space where moment acts through is also conserved. So not only which direction is momentum point, by where is space it exists.

To visualize this, consider a case where you want to remove the momentum of a freely rotating body that is moving in space. You have a hammer, and you need to find out the following in order to completely stop the body. a) how much momentum to hit it with (the magnitude), b) in which direction to swing (direction) and c) where to hit it (location).

In summary, the common quantities in mechanics are interpreted as follows

$$ \begin{array}{r|l|l} \text{concept} & \text{value} & \text{moment}\\ \hline \text{rotation axis} & \text{rot. velocity}, \boldsymbol{\omega} & \text{velocity}, \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\ \text{line of action} & \text{force}, \boldsymbol{F} & \text{torque}, \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \\ \text{axis of percussion} & \text{momentum}, \boldsymbol{p} & \text{ang. momentum}, \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} \end{array} $$

The stuff under the value column are fundamental quantities that give us the magnitude of something (as well as the direction). The stuff under the moment column are secondary quantities that depend on where they are measured and give use the relative location of the fundamental quantities. Hence the terms torque = moment of force, velocity = moment of rotation and angular momentum = moment of momentum. All that means is that these quantities are $\boldsymbol{r} \times \text{(something fundamental)}$ and they describe the moment arm to this something.

The location of the line in space is always the same formula

$$ \text{(location)} = \frac{ \text{(value)} \times \text{(moment)}}{ \text{(magnitude)}^2} $$

where $\text{(magnitude)}$ is always the magnitude of the $\text{(value)}$ vector.

In statics, for example, we learn to balance forces and moments, which should be interpreted as balancing the force magnitude and the line of action of the force.

  • 3
    $\begingroup$ Your first equation suggests that $\mathbf F$ is always perpendicular to $\mathbf r$. But this certainly is not always true. Am I missing something? $\endgroup$ Sep 2, 2019 at 2:07
  • 1
    $\begingroup$ @AaronStevens - the $\boldsymbol{r}$ can be anywhere along the line of action. The calculation just returns the one closest to the origin that is perpendicular to the direction of the line (and hence the direction of $\boldsymbol{F}$ or $\boldsymbol{p}$ or $\boldsymbol{\omega}$. $\endgroup$ Sep 2, 2019 at 13:18
  • 1
    $\begingroup$ @AaronStevens - but there isn't a single point where a force acts. It is a line in space. It is like saying the point of rotation, as similarly, rotation acts about an axis and not a point. A force acts along an axis and not a point. $\endgroup$ Sep 2, 2019 at 18:56
  • 1
    $\begingroup$ @AaronStevens - I am talking about rigid bodies, so no discussion on contact pressures are relevant here. I know you can apply a force at a point, but for rigid bodies, it does not matter where that point is as you can slide the force vector along the line of action and it will have the same effect. So with torque you cannot extract the location of the force applicaton, just the location of the line of action. $\endgroup$ Sep 3, 2019 at 0:29
  • 1
    $\begingroup$ Yes, I agree that the torque doesn't change if you move the force along that line. But at the end of the day that force is still applied at some specific location, which is not necessarily given by that equation, even if it results in equivalent motion had it been applied there. Anyway, it's a minor thing to discuss at this point. Thanks for the clarification. $\endgroup$ Sep 3, 2019 at 3:45

The original question is tagged "Newtonian mechanics", but the author also talks about "fundamental forces", so I assume it may be of some interest to point out some fundamental phenomena that are observable with classical macroscopic objects, but are, strictly speaking, beyond Newtonian mechanics.

In quantum mechanics, angular momentum of a photon is fundamental. When circularly polarized photons are absorbed or photons are scattered with a change of polarization, transfer of their angular momentum can be detected as macroscopic torque -- for example causing small but macroscopic objects to rotate.

This effect does not depend on where the photon strikes the object with respect to its center of mass.

(Earlier discussion relevant to the OP's question: Is it possible to apply a torque without a moment arm?)

When a photon gets absorbed, its angular momentum causes the change of the angular momentum of the electron(s) in the object. Through some relaxation mechanisms, the angular momentum of electrons gets transferred to the macroscopic rotation of the object. I am not sure what exactly these mechanisms are -- proper treatment of how angular momentum is transferred to the lattice is not a trivial subject: https://arxiv.org/pdf/1802.01638.pdf

Simple classical interpretation of the situation is: the photon induces a dipole moment in the absorbing medium, and this dipole experiences a torque from the rotating electric field of the photon -- the situation resembles what happens in the induction motor.

Early (first?) experiment with the scattering of photons: "Mechanical Detection and Measurement of the Angular Momentum of Light" Richard A. Beth 1936 https://journals.aps.org/pr/abstract/10.1103/PhysRev.50.115

Using photons to spin small but macroscopic objects: "...by utilizing transfer of photon spin angular momentum, it is also possible to set objects into rotational motion simply by targeting them with a beam of circularly polarized light" "Ultrafast Spinning of Gold Nanoparticles in Water Using Circularly Polarized Light" 2013 https://pubs.acs.org/doi/abs/10.1021/nl4010817

(Light can carry angular momentum not only in the spin of photons but also as orbital angular momentum. See Emilio Pisanty's answer here: Photon spin and total angular momentun )

(This answer has been updated to address the questions from the comments.)

  • $\begingroup$ But how do you suppose the light interacts with those objects? $\endgroup$ Sep 2, 2019 at 4:46
  • $\begingroup$ When a photon gets absorbed, its angular momentum causes the change of the angular momentum of the electron, and thus of the whole macroscopic object. This transfer of angular momentum does not depend on where the photon is absorbed with respect to the center of mass of the object. $\endgroup$
    – Gene Ruso
    Sep 2, 2019 at 4:56
  • $\begingroup$ But how does this transfer happen? $\endgroup$ Sep 2, 2019 at 5:12
  • 3
    $\begingroup$ The angular momentum of light is present already in Maxwell's equations and is only inherited by QFT. There is nothing fundamentally new about angular momentum in QFT; it's just quantized classical wave mechanics. See Hans C. Ohanian, "What is spin?", in Am. J. Phys. 54 (6), June 1986 (online here). $\endgroup$
    – benrg
    Sep 3, 2019 at 3:50
  • 1
    $\begingroup$ @AaronStevens I agree that strictly in mechanics, torque is not fundamental. But if we expand the scope to even just classical physics, we may have to admit dipoles as "fundamental" (e.g. compass needle) and then we would have to admit the torques as equally fundamental. $\endgroup$
    – Gene Ruso
    Sep 3, 2019 at 15:17

To the extent that torque (or moment) is derived from force, then force is more "fundamental" than toque.

However, torque is certainly more than just force with additional “baggage”. And it’s more than just about coordinate systems. Torque and force are not a matter of either or. Both are needed for the analysis of motion and equilibrium.

Moment, which is another term for torque, is a fundamental concept in statics. For example in statics both forces and moments are needed to determine static equilibrium . Forces cause straight line motion. Moments cause rotational motion. The requirements for equilibrium are that the sum of both the moments and forces have to be zero. And it goes beyond statics. Bending moments and shear forces are fundamental to the study of mechanics of materials.

You will learn to appreciate the difference if and when you study statics , dynamics, and mechanics of materials.

Hope this helps

  • $\begingroup$ I would argue that when we're talking about fundamental point particles, torque is just a matter of coordinate systems, and only becomes necessary when we consider composite systems. $\endgroup$ Sep 1, 2019 at 21:46
  • $\begingroup$ @Danny what do you mean by saying that torque is just a matter of coordinate systems? $\endgroup$ Sep 1, 2019 at 21:53
  • 3
    $\begingroup$ @BobD Certainly torques make life a whole lot simpler for extended bodies, but I would argue that the concept of force is more fundamental than torque, as torque is defined in terms of force, but not the other way around. Couldn't one argue that a zero net torque is just a simpler way of saying that the net force acting on every part of the body must be $0$? $\endgroup$ Sep 1, 2019 at 21:55
  • $\begingroup$ @AaronStevens I wasn't arguing that torque isn't derived from force or that torque is equally "fundamental" as force. For that matter I'm not sure we can say forces in general are "fundamental" in the same manner as the strong nuclear, weak nuclear, electromagnetic and gravitational forces. Instead my focus was on the OP statement "So is choosing to use either force or torque as the basis of laws & problems arbitrary? Or is there an actual fundamental rationale to when one or the other should be used?" It's the ideal of either or use of force and torque to solve problem I addressed. $\endgroup$
    – Bob D
    Sep 1, 2019 at 22:19
  • 1
    $\begingroup$ @AaronStevens You never know.I've given up trying to figure out why $\endgroup$
    – Bob D
    Sep 2, 2019 at 2:25

From a purely Newtonian mechanics perspective I would argue that force is more fundamental concept than torque. This is mainly because torques is, for lack of a better term, a property of forces. Also, the torque produced by a force depends on your subjective choice of which point you are calculating the torque about. This is all captured in the definition of torque $$\boldsymbol\tau=\mathbf r\times\mathbf F$$ where $\mathbf F$ is the force vector and $\mathbf r$ is the vector pointing from the point about which you are calculating the torque to the point where the force is applied.

Note that this defines torque in terms of a force, but you cannot determine a force from a torque. For a given $\boldsymbol\tau$ and a given $\mathbf r$ there is not a unique force $\mathbf F$. Therefore, this also gives the impression that force is a more fundamental concept.

Also note that the definition of torque does not depend on if we are using polar coordinates or not. You can discuss forces in polar coordinates without referring to torque, and you can talk about torques in Cartesian coordinates.

So torque is not fundamental, however that doesn't mean it's not useful. It is useful in looking at how forces cause extended bodies to move (or not move), and it is useful when motion has rotational symmetry about some point (i.e. when angular momentum is conserved).

Moving beyond Newtonian mechanics, I will say that torque could be viewed as somewhat more fundamental than it is in Newtonian mechanics, but I think even then the focus is just shifted to angular momentum rather than torque.

The reason I say this is because once you get to the level of physics where angular momentum is fundamental you really stop talking about forces and torques anyway, and what is focused on more is momentum and energy. For example, Lagrangian and Hamiltonian mechanics focus more on energy than forces. Schrodinger's equation deals with energies rather than forces.

Additionally, we have spin that has angular momentum associated with it, but there is no classical analog for it. We don't even talk about torques when dealing with spin, yet we do discuss angular momentum a lot. Even then, for non-spin angular momentum, the angular momentum operators are still defined in terms of linear momentum operators.

So, I suppose my point of view in all of this is that torque is not very fundamental. At the classical level, torques are really just properties of forces. Once you get deeper into physics, the focus shifts to energy, momentum, and angular momentum. Torque falls away.

To get to your more practical questions:

So is choosing to use either force or torque as the basis of laws & problems arbitrary? Or is there an actual fundamental rationale to when one or the other should be used?

I wouldn't say it is arbitrary, but I also wouldn't say there are certain rules to follow as to when torque is useful or not. What holds true for using torque in a certain problem really holds true for any type of problem solving strategy. If you realize that thinking about torques will help you solve the problem, then you should use it! For example, if we are interested in the change in angular momentum of a rotating object produced by a force, it would be useful to think about the torque this force produces. However, if you want to analyze the motion of a projectile near the surface of the Earth, it wouldn't be very helpful to think about the torque the gravity force has about some reference point (although it wouldn't necessarily mean you are wrong for attempting to analyze the problem in this way).

  • $\begingroup$ This definitional argument seems to translate almost verbatim to angular momentum vs linear momentum: the former also depends on which point you are calculating the angular momentum about, and it's also captured in the definition of angular momentum: $\vec L=\vec r\times\vec p$. Similarly you can't define linear momentum from angular momentum. Yet both these quantities are usually considered fundamental. $\endgroup$
    – Ruslan
    Sep 2, 2019 at 5:46
  • 1
    $\begingroup$ @Ruslan I suppose you are correct. However, I would argue that angular momentum only becomes fundamental past Newtonian mechanics, which is what this question has been tagged with. $\endgroup$ Sep 2, 2019 at 11:10
  • $\begingroup$ Well, actually, angular momentum is fundamental even in classical mechanics--just as fundamental as linear momentum. The definition of $\vec{L}$ isn't really $\vec{r}\times\vec{p}$--rather, it is a result. The definition of angular momentum is that it is the generator of rotations in space--which is just as fundamental as the definition of linear momentum which is that it is the generator of translations in space. $\endgroup$
    – user87745
    Sep 3, 2019 at 17:29
  • $\begingroup$ To make my point more appealing, I would add that the $\vec{r}\times\vec{p}$ definition is not useful in other than $3$ dimensions. Moreover, in more than $3$ dimensions, angular momentum would have more independent components than linear momentum would. $\endgroup$
    – user87745
    Sep 3, 2019 at 17:33

Yes, the concepts of force and torque are equally fundamental.

Noether's theorem states that each symmetry in a physical system corresponds to a conservation law. The symmetry under translation results in the conservation of momentum, of which force is the derivative (therefore the sum of all forces in a physical system is always 0). The symmetry under rotation results in the conservation of angular momentum, of which torque is the derivative (therefore the sum of all torques in a physical system is always 0).

The symmetries under translation and under rotation are equally fundamental to mechanics, therefore force and torque are equally fundamental concepts.

  • $\begingroup$ I like this argument a lot, but Noether's theorem and its application are above my level of physics. Do the others agree? $\endgroup$ Sep 3, 2019 at 15:27
  • $\begingroup$ Your argument says nothing about what torque is fundamental other than asserting it at the end. Torque can be described in terms of force. It's unnecessary (but useful) to describe mechanical systems: not fundamental. $\endgroup$
    – JimmyJames
    Sep 3, 2019 at 17:33
  • $\begingroup$ @Peter when you see people talking about the "generators of [...]" in some of the other comments they are also playing at this level of abstraction. At the level of group theory and symmetries the angular and translational symmetry are independent of one another. $\endgroup$ Sep 3, 2019 at 17:35

Yes, they are equally fundamental because they are both "forces" in the same sense. The relationship between torque, angular momentum, and angles is identical to that between linear forces, linear momentum, and position.

Take the equation $\mathbf{F} = m \mathbf{a}$. Properly, that equation is $\mathbf{F} = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$. One way to read that equation is "The net force is the time rate of change of linear momentum." We also have $\mathbf{\tau} = \frac{\mathrm{d} \mathbf{L}}{\mathrm{d}t}$ for torques and angular momentum.

Now, the reason why torques might not appear to be as fundamental comes down to the fact that much of classical mechanics is constructed using point-like particles that do not have an orientation to them. It is an interesting fact that you can approximate macroscopic bodies using such a construction, and recover angular momentum as arising from the collective motion of these particles, but that doesn't make the angular quantities non-fundamental.

You could equally do the construction of mechanics in terms of tiny rigid bodies, where each body has a location and orientation. That orientation, defined by angles, will naturally lead to angular momentum, torques, etc, being on equal footing with forces. The reason we don't usually do it that way is because it adds a lot of complication and the approximation that you can ignore the angular momentum of the tiny parts is usually right, so it's a lot of extra work for no benefit.

Back to the equations above, $\mathbf{F} = \frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t}$ and $\mathbf{\tau} = \frac{\mathrm{d} \mathbf{L}}{\mathrm{d}t}$. In more advanced physics courses you learn that linear momentum is the quantity that is conserved because the laws of physics don't depend on your position. You also learn that angular momentum is conserved because those same laws don't depend on how you orient your coordinate axes. Thus, linear and angular momentum are on the same footing, and the above equations can be understood as being about the rate at which the conserved quantities are transferred between two or more bodies.

There are some extra complications that come from the fact that you cannot describe orientation with a vector - you need either three angles, or a rotation matrix. It is these complications that make torque and angular momentum harder to deal with, but that doesn't make them less fundamental.

For example, the arbitrary "right hand rule" that comes in with cross products is an artifact that comes from the fact that you're not really dealing with vectors, but treating something a little more complicated like a vector (in technical terms: rank-2 antisymmetric tensors), which is only possible in 3-d. The right hand rule gets fixed when you decide how to translate the parts of those tensors (matrices) to vector components. For example, we could have the components of angular momentum ($L_x$, $L_y$, $L_z$) be: $$\left[\begin{array}{ccc}0 & L_z & -L_y \\ -L_z & 0 & L_x \\ L_y & -L_x & 0\end{array}\right],$$ but exchanging all of the minus signs is an equally valid choice. It is the decision of where to put the minus signs that produces left vs right hand rules. Equivalently, the choice corresponds to asking, "Which way is a rotation by a positive angle?" The above corresponds to, "Stick the thumb of your right hand in the direction of the axis you're rotating around and grab the axis. Your fingers will wrap around the axis in the direction of positive rotations (looking down at your thumb, counter-clockwise)."


enter image description here

Rigid body will rotate about a point that locate at the center of mass due to a external force $\vec{f}$ that effective at point $p_1$ , the torque is then:

$$\vec{\tau}_{p1}=\vec{r}_c\times \vec{f}$$

if we choose another arbitrary point $p_2$ that lies on the force line then we get for the torque

$$\vec{\tau}_{p2}=\left(\vec{r}_c+\lambda \hat{\vec{f}}\right)\times \vec{f}$$

where $-\infty < \lambda < \infty$

In booth case the magnitude of torque is equal $||\vec{\tau}_{p1}||=||\vec{\tau}_{p2}||$

for $\lambda_\perp=-\frac{\vec{f}^T\,\vec{r}_c}{||\vec{f}||}$ we get the the shortest distance to the center of mass point so:

$$\vec{\tau}_{p\perp}=\underbrace{\left(\vec{r}_c+\lambda_\perp \hat{\vec{f}}\right)}_{\vec{r}_\perp}\times \vec{f}$$

again the magnitude $||\vec{\tau}_{p\perp}||=||\vec{\tau}_{p1}||$

so $p_\perp$ is not a certain point at all!!

If the force $\vec{f}$ is effective an the center of mass, the body can still rotate due to external torque $\vec{\tau}_E$

To answer your question.

If the torque is due to external force then it isn’t fundamental because to calculate the torque you use the force and a point on the force line, but in case that torque is because external torque this is fundamental.

  • $\begingroup$ This is all true, but what is your answer to the OP's question? $\endgroup$ Sep 3, 2019 at 14:21
  • $\begingroup$ @AaronStevens I see, I put some comments to answer the question. I also wanted to show that the perpendicular distance don’t take certain rolls at all $\endgroup$
    – Eli
    Sep 3, 2019 at 15:55

It is possible (and perhaps natural) to turn the question around. As others have mentioned; photon spin, and electron spin in particular, can not be identified with a concept of displacement around a circle in spatial dimension, yet participate in macroscopic angular conservation laws just the same. So arguably, conservation of angular momentum is about as fundamental as it gets. But what about linear momentum?

Well, should you imagine (without too much loss of generality) that we live in (hyper)spherical universe, it is easy to reimagine any statement about linear conservation as a statement of angular conservation around a corresponding pole of the universe instead.

So no, not everything about angular momentum can be derived from linear momentum; but the converse may infact be true.

  • $\begingroup$ Particle spin in QFT can be understood as large-scale circular motion of the field. It is present already in Maxwell's equations. See Hans C. Ohanian, "What is spin?", in Am. J. Phys. 54 (6), June 1986 (online here). $\endgroup$
    – benrg
    Sep 3, 2019 at 3:52
  • $\begingroup$ Interesting reference; thanks for pointing that out; I wasnt aware of that perspective on spin-1/2 particles. Nonetheless, I think its true one may view force as derived from torque as much as the other way around. $\endgroup$ Sep 3, 2019 at 7:56

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.