I have an intuition problem calculating torque using the cross product formula. As for example let the magnitude of the force be 50 lbs and length of the wrench be one foot and you are exerting force in a clockwise motion and the angle you apply the force 60 degrees. This is an example so I can ask my question. Using the right hand rule the torque points perpendicular to the force you are applying to the bolt. In this case since the sine of 60 degrees is about .86 it would be (.86)(50) foot lbs. How can the bolt turn clockwise if the force is concentrated perpendicular to where it needs to turn? The cross product formula demands the torque be perpendicular. Obviously my mistake but I don't see where.

  • $\begingroup$ The cross product yields the perpendicular distance to the line of action of a force. $\endgroup$ – ja72 Mar 26 '17 at 22:24
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    $\begingroup$ Related: physics.stackexchange.com/q/82874/2451 and links therein. $\endgroup$ – Qmechanic Mar 27 '17 at 6:56
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    $\begingroup$ It's purely a convention -- a way to express torque in one vector value. There is no "logical" reason for using the right hand vs left hand vs big toe, but making the vector be the axis of rotation (vs, say, making it a tangent to some circle around the axis, or simply representing it as a scalar) allows the axis to be identified as a part of the vector, vs needing a separate quantity. $\endgroup$ – Hot Licks Mar 27 '17 at 22:59

How can the bolt turn clockwise if the force is concentrated perpendicular to where it needs to turn?

Because that force is perpendicular to the direction towards the rotation-centre. Not to the turning direction. The bolt does indeed turn in the same way as the force pulls it.

When you define a torque vector direction, you have a problem. You can't define a vector direction as something that turns around. The direction must be along a straight line. So instead of choosing the torque "turn", we could choose the torque axis as the vector direction.

Have a look at this picture:

enter image description here

The axis is vertical through the bolt along the two upwards/downwards arrows. If you choose to define the torque vector direction along this axis, all fits. We just have to remember that choice.

Torque is: $$\vec \tau = \vec F \times \vec r$$

The force vector $\vec F$ times the vector towards the rotation-centre $\vec r$ gives the torque vector. The result of a cross-product is mathematically a vector pointing vertically upwards, so this fits perfectly to that choice. The torque vector $\vec \tau$ that you get from this calculation has the torque magnitude but the torque-axis direction.

As long as you remember this choice - this definition - all is good. Everytime you hear "the direction of the torque is horizontal", you know that this is only the axis of the torque; the torque (the turn) is then upright.

  • $\begingroup$ @Steeven....can you also explain how the angle plays into this ...in my sample problem I believe it was 60 degrees.... perhaps you are assuming a 90 degree angle with sine ( theta ) = 1 ? $\endgroup$ – Sedumjoy Mar 26 '17 at 22:35
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    $\begingroup$ @Sedumjoy The vector equation $$\vec \tau=\vec F\times \vec r$$ is the full, general formula. A cross-product already takes the angle into account (the cross-product of two parallel vectors is 0, for example). Only if you want to use the non-vector-version (magnitude-version), you will see the angle being a part of the equation. It is usually written like this: $$\tau=F_\perp\;r\quad\text{ or }\quad \tau=F\;r_\perp$$ where the $~_\perp$ is the perpendicular component. Multiplying sine onto this takes care of the angle. So the usual magnitude formula becomes: $$\tau=F\;r\;\sin(\theta)$$ $\endgroup$ – Steeven Mar 26 '17 at 23:28
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    $\begingroup$ @Sedumjoy On the picture, the force is directly away ($90^\circ$) from the wrench, yes. If you pull a bit sideways and not straight perpendicularly away, you would have an angle. Only the perpendicular component of the force has an influence (anything that pulls parallel to the wrench, at $0^\circ$, causes no turning at all). So if there is a non-$90^\circ$-angle, you get the perpendicular force component by multiplying the sine of the angle. $\endgroup$ – Steeven Mar 26 '17 at 23:33
  • $\begingroup$ In the USA at least, the radial vector points outward towards the point where the force is applied, not inwards towards the center of rotation, so the convention I was taught would have the torque vector vertically downwards in the pictured example. See: en.wikipedia.org/wiki/Torque $\endgroup$ – Todd Wilcox Mar 28 '17 at 5:50

To add to Steeven's answer and in particular his very pertinent statement:

You can't define a vector direction as something that turns around.

It may help you to understand that torque as a vector is actually cheating a little bit: it's a "simplification" that we can only get away with in two and three dimensions, which is why the "direction" seems a little abstract. The torque "vector" direction defines the axis of the motion that it tends to induce, and for the same reason that torque as a vector is a bit of a trick, even the notion of axis only works in two and three dimensions.

Torque is about rotation, and rotations primarily are about transformations that are confined to planes. For example, a rotation about the $z$-axis is a transformation that churns up the $x-y$ plane - it transforms the $x$ and $y$ co-ordinates of things - but leaves the $z$ co-ordinates unchanged.

When we do higher dimensional geometry, rotations change planes and leave more than one dimension invariant. In a four dimensional rotation, it's incomplete to speak of a rotation about an axis, because, for example, you can have a rotation that transforms the $x$ and $y$ co-ordinates of points invariant, but it leaves the $z$ and $w$ co-ordinate invariant.

So, in general, the easiest way to specify a rotation is by specifying the plane that it changes, rather than specifying the subspace that it leaves invariant.

It just so happens that in three dimensions, the subspace left invariant is a line or an "axis"- so the two approaches amount to the same thing. We can define a plane in three dimensions by specifying a vector normal to it, which is why we can get away with a torque or angular velocity as a vector. In general these quantities are directed planes, not lines with direction.

  • $\begingroup$ This is interesting. You mind me asking....Cross product magnitude is an area of the two tail vectors. We can prove this. Do you agree? In the diagrams of torque problems the vector is moved to head , tail vs. tail tail. It's not the case that the magnitude of the diagonal of the parallelogram is the same as the area is it ? What would that length correspond to in relation to the cross product? $\endgroup$ – Sedumjoy Mar 27 '17 at 5:57
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    $\begingroup$ The bivector (plane of rotation) and the vector torque are hodge duals of one another. Vector calculus is usually presented to students using only vectors (to "simplify" things) rather than using "exterior algebra" which happily handles planes, volumes and so on. The cross product is another example. Whenever something that is not a vector or scalar turns up it is converted to its hodge dual. Once you deal with more dimensions you have to give this up. $\endgroup$ – Francis Davey Mar 27 '17 at 8:02
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    $\begingroup$ @FrancisDavey Indeed, although I didn't think the language of exterior algebra was appropriate for the question. And in that language, the cross product is an excellent vehicle of intuition to introduce the Hodge dual. $\endgroup$ – WetSavannaAnimal Mar 27 '17 at 9:33
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    $\begingroup$ Absolutely fair enough, it is just that sometimes discussions of this imply that torque is "really" a vector and that there is nothing problematic about it. My guess is that some of the unnaturalness new students find about it (well some students) is because something unnatural is being done. $\endgroup$ – Francis Davey Mar 27 '17 at 9:59
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    $\begingroup$ To amplify a little: length x force looks like it should have some area in it on dimensional or geometric grounds, so the idea that it is really some kind of directed area (a bivector) comes out quite naturally. But I appreciate the vast majority of students do not learn this way. $\endgroup$ – Francis Davey Mar 27 '17 at 10:02

Consider the definition of torque $\vec{\tau}$ due to a force $\vec{F}$ passing through a point $\vec{r}$ $$\vec{\tau} = \vec{r} \times \vec{F}$$

Using the cross product identity $\| \vec{A} \times \vec{B} \| = \| A \| \|B \| \sin \theta$ where $\theta$ is angle formed by the two vectors we can write the following

$$ \| \vec{\tau} \| = \| \vec{r} \| \| \vec{F} \| \sin \theta $$ $$ \tau = F (r \cos \varphi) = F \, d$$ since $\theta = \frac{\pi}{2}+\varphi$ and $d = r \cos\varphi$ is the perpendicular distance to the force line of action.

In Summary, the cross product removes any influcence of the location of the force along the line of action and only considers the perpendicular distance for measuring torque.



Torque is the moment of the line of action of a force. It is defined as $\vec{\tau} = \vec{r} \times \vec{F}$

Velocity is the moment of the line of rotation of a rigid body. It is defined as $\vec{v} = \vec{r} \times \vec{\omega}$

Both quantities ($\vec{\tau}$ and $\vec{v}$) contain the information about the distance (position) to a line in space. This can be recovered by

$$ \begin{align} \vec{r}_{\perp} &= \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2} & \vec{r}_{\perp} &= \frac{\vec{F} \times \vec{\tau}}{\| \vec{F} \|^2} \end{align} $$

The direction of the torque vector is similar to the direction of the velocity vector on a rotating rigid body. It is a circumferential vector perpendicular to both the line of action and the location of the line. It is best explained motion motion as the tangential velocity of an extended rotating body under the coordinate origin.

See this answer for a more detailed explanation of the geometry in mechanics.

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    $\begingroup$ This is very interesting....I did not see this before I commented.... let me digest this fine piece of art before proceeding. $\endgroup$ – Sedumjoy Mar 27 '17 at 16:15
  • $\begingroup$ See my edit. I am providing a little more detail in this subject that has a lot of nuances and it dear to my heart. $\endgroup$ – ja72 Mar 27 '17 at 19:56

I believe that your question is best answered by the gyroscope experiments. First, the gyroscope not spinning, is supported on both ends. One support is then removed, and the gyro "falls." However, when this experiment is repeated with the gyro spinning, the gyro, instead of falling, it spins around the supporting end! This motion is perpendicular to both, the gravity force vector and the torque vector. This proves that torque generates a vector that is perpendicular to the plane of rotation.


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