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In defining the moment of a force about a point as "the tendency of one or more applied forces to rotate an object about an axis [going through a point, hence also about a point]", I see it logical to infer that it is somehow related to the angle between the force and the moment arm, but why put it directly proportional to the sine of that angle? Why not $\ln^n(1+\theta)$ for example? This will become zero as the angle tends to zero.

Furthermore, why have the magnitude of the moment arm and the force in the definition? Maybe so that to underline that for two forces of different magnitudes this tendency to rotate is different? And I guess the same for the moment arm's length?

Perhaps there is no aim to describe how this tendency "changes" as a function of the angle but we just want some formula that would tell us that for a given two forces and two moment arms, one is bigger than the other?

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  • $\begingroup$ Torque is a (pseudo)vector. In 3D space it has three components. The easiest way to get them is by taking the cross product of the two relevant vectors, $\mathbf{r}$ and $\mathbf{F}$. This makes computations very straightforward. What exactly is your counterproposal for writing the torque vector? $\endgroup$ – G. Smith Feb 21 at 4:22
  • $\begingroup$ I am asking about its definition in the "philosophical" sense not its mathematical definition through its current definition. As I said in the question, why not just consider the sine of the angle between the force and the moment arm. $\endgroup$ – Luyw Feb 21 at 5:23
  • $\begingroup$ Actually, you asked about “practicality”, which has little to do with philosophy. Do you want to edit your question to be more clear about what kind of answer you seek? $\endgroup$ – G. Smith Feb 21 at 5:32
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Torque is defined as

$$ \boldsymbol {\tau} = \mathbf r \times \mathbf F$$

But the reason we define it that way is because we have found a conserved quantity called angular momentum, $\boldsymbol {\ell} = \mathbf r \times \mathbf p$, which is related to torque as:

$$\boldsymbol {\tau} = \frac {d\boldsymbol {\ell}}{dt}$$

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