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I am studying Statics and saw that:

The moment of a force about a given axis (or Torque) is defined by the equation:

$M_X = (\vec r \times \vec F) \cdot \vec x \ \ \ $ (or $\ \tau_x = (\vec r \times \vec F) \cdot \vec x \ $)

But in my Physics class I saw:

$\vec M = \vec r \times \vec F \ \ \ $ (or $\ \vec \tau = \vec r \times \vec F \ $)

In the first formula, the torque is a triple product vector, that is, a scalar quantity. But in the second, it is a vector. So, torque (or moment of a force) is a scalar or a vector?

Qmechanic
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Vinicius ACP
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    The torque is a vector. In your $M_X$ you simply have the component of the torque along $\vec x$, and this component is a scalar. – ZeroTheHero Aug 02 '17 at 20:16
  • Where have you seen the first version written? – sammy gerbil Aug 02 '17 at 21:29
  • I think that its in Beer's Vectorial Mechanics for Engineers, but i'm not sure. – Vinicius ACP Aug 02 '17 at 21:50
  • Technically, torque is a pseudo (or axial) vector, see https://en.wikipedia.org/wiki/Pseudovector#:~:text=In%20physics%20and%20mathematics%2C%20a%20pseudovector%20%28or%20axial,in%20the%20opposite%20direction%2C%20of%20its%20mirror%20image. – jim Jan 24 '21 at 21:48

3 Answers3

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Torque (Force Moment) is a vector that describes the location of the Force line of action.

  • Lemma: If you give me a force vector ${\vec F}$ and a moment vector about the origin ${\vec M}$ then I can define a line whose points obey the relationship $\vec{M} = {\vec r} \times {\vec F}$. This line has direction parallel to the force ${\vec F}$ and passes through a point (closest to the origin) defined by $${\vec r} = \frac{ {\vec F} \times {\vec M} }{ \| {\vec F} \|^2 } $$

Proof: Use $\vec{M} = {\vec r} \times {\vec F}$ into the equation for the point.

$$ \require{cancel} \frac{ {\vec F} \times {\vec M} }{ \| {\vec F} \|^2 } = \frac{ {\vec F} \times ({\vec r} \times {\vec F}) }{ \| {\vec F} \|^2 } = \frac{ \vec{r} ( \vec{F} \cdot \vec{F}) - \vec{F} (\cancel{\vec{F} \cdot \vec{r}} ) }{ \| {\vec F} \|^2 } = \vec{r} \frac{\| {\vec F} \|^2}{\| {\vec F} \|^2} = \vec{r} $$

This requires that $\vec{F} \cdot \vec{r}=0$ which is true for the point on the line closest to the origin.

It is true for both statics and dynamics that a moment is just a force at a distance. Only when the net force is zero (force couple) the moment is a pure moment and it does not convey any location information.

John Alexiou
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  • @MichaelLevy - thanks. I changed the wording a bit. I hope you agree it is an improvement. – John Alexiou Jan 24 '21 at 20:23
  • @MichaelLevy - check the link. It is not working for me. – John Alexiou Jan 24 '21 at 23:32
  • @MichaelLevy Read also this post and my answer. – John Alexiou Jan 24 '21 at 23:36
  • The requested URL /metric/metric_public/vectors/vector_coordinate_geometry/vector_equation_of_line.html was not found on this server. – John Alexiou Jan 25 '21 at 22:51
  • where is your vector equation of a line? https://en.wikipedia.org/wiki/Line_(geometry)#As_a_vector_equation – Michael Levy Jan 25 '21 at 23:00
  • There are many forms of line equations but in 3D it most common to define a line (or a ray) by a point and direction which is what I have done in the post. You can also write $$ \frac{x-r_x}{F_x} = \frac{y-r_y}{F_y} = \frac{z-r_z}{F_z} $$ or $$ \vec{F} \times \left( \pmatrix{x \ y \ z} - \vec{r} \right) = \vec{0}$$ – John Alexiou Jan 25 '21 at 23:04
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It is obviously a vector, as you can see in the 2nd formula.

What you are doing in the first one is getting the $x$-component of that vector. Rememebr that the scalar product is the projection of one vector over the other one's direction. Actually you should write $\hat{x}$ or $\vec{i}$ or $\hat{i}$ to denote that it is a unit vector. That's because a unit vector satisfies

$\vec{v}\cdot\hat{u}=|v| \cdot |1|\cdot \cos(\alpha)=v \cos(\alpha)$

and so it is the projection of the vector itself.

In conclusion, the moment is a vector, and the first formula is only catching one of its components, as noted by the subindex.

FGSUZ
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There are some application, where we might want to quantify both the torque, which is a vector, and the component of the torque about a particular axis, which is a scalar.

I illustrate an example of this in the figure below, which is from 1 and provided here under fair use for the purpose of scholarship. The door is hinged so that it turns only around the $\mathbf{\widehat{k}}$ axis. Meanwhile, the door knob is located at a position $\mathbf{r}$ relative to the origin. A force $\mathbf{F}$ is applied to the door knob.

By $\boldsymbol{\tau}$, I denote the torque on the door knob, which is $$\boldsymbol{\tau} = \mathbf{r}\times \mathbf{F}.$$ By $\tau_z$, I denote the scalar component of the torque vector about the axis of rotation. So, $$\tau_z = \mathbf{\widehat{k}} \cdot \boldsymbol{\tau} = \left(\mathbf{r}\times \mathbf{F}\right)\cdot \mathbf{\widehat{k}}.$$

enter image description here

Bibliography

1 Mathematical Methods in the Physical Sciences, 3rd Edition, Mary L. Boas, ISBN: 978-0-471-19826-0 July 2005.

Michael Levy
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