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The formula for moment about a point is:

$$M = Fd$$

After looking at other answers on stackexchange, I'm still not convinced with the 'intuitive' explanations that are given. I understand the cross product relationship between F and d and how to compute the moment, but I'm not searching for a mathematical explanation. I would prefer an explanation based purely on the explanation of concepts intuitively.

I'm not sure how the formula was decided to express the 'turning effect intensity' of a certain force. Why specifically this formula and not some other form? Maybe there's an explanation using rigid bodies?

XXb8
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    In this context, do you mean the same thing as the torque about a point, $\mathbf{\tau}=\mathbf{r}\times\mathbf{F}$? – DanDan0101 Oct 18 '21 at 19:10
  • If it is defined that way then it gives the change in angular momentum –  Oct 18 '21 at 20:25
  • Did you come across this similar question "Why is moment dependent on the distance from the point of rotation to the force?" https://physics.stackexchange.com/questions/80538/why-is-moment-dependent-on-the-distance-from-the-point-of-rotation-to-the-force/665835#665835 – Not_Einstein Oct 19 '21 at 00:58
  • @DanDan0101 yes, that's what I meant – XXb8 Oct 19 '21 at 06:18
  • @Not_Einstein yes. The most convincing answer seemed to be the energy tradeoff, but how would energy relate to the 'turning effect' in that answer? – XXb8 Oct 19 '21 at 06:20
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    @XXb8 Because it takes a certain amount of energy to turn the bolt one full revolution. So E there is a given and the last equation shows that the force to turn the bolt has to be larger if it is applied at a shorter distance, and can be smaller if applied at a larger distance. – Not_Einstein Oct 19 '21 at 14:14
  • @Not_Einstein This makes sense. However, shouldn't the distance be considered the circumference of the rotation of the force in this case of $Fd$? – XXb8 Oct 21 '21 at 06:57
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    @XXb8 As d (circumference) and r (radius) just differ by a factor of 2, you could say that. A larger r requires a smaller force, but that smaller force has to be applied over a larger distance d. – Not_Einstein Oct 21 '21 at 13:16
  • @Not_Einstein Why is the $2pi$ not present in the formula for a moment then, given that the actual distance covered is the circumference? – XXb8 Oct 21 '21 at 18:19
  • @XXb8 The moment of force is defined as the vector product r x F and it is this quantity that is equal to the rate of change of angular momentum. The distance covered is used in the definition of work. – Not_Einstein Oct 21 '21 at 18:31
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    @XXb8 Also the 2pi is just for the special case of the force being applied along a circle, so it's not in the general formula for moment of force. – Not_Einstein Oct 21 '21 at 22:14
  • @Not_Einstein Even then, why isn't the 2pi included in the moments for problems involving circles then? – XXb8 Oct 22 '21 at 06:17
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    @XXb8 It's a matter of definition. The moment of force equals the rate of change of angular momentum. If you say the moment of force is 2(pi)rxF then that would be equal to 2(pi) times the rate of change of angular momentum. It's useful to have one general definition, not different ones for different situations. – Not_Einstein Oct 22 '21 at 21:51

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Consider a level of the second type with load $m$ at distance $r$ and a force perpendicular to a lever acting at a distance $d$. The system is in zero gravity.

If the lever rotated by a small angle $\Delta\phi$, then the force has done the work $W=F\Delta s=Fd\Delta\phi$. On the other hand this energy was spent to increase the kinetic energy of the load: $$ \Delta T=mv\Delta v = m(r\omega)(r\Delta\omega) = m\left(r\frac{\Delta\phi}{\Delta t}\right)(r\Delta\omega) = mr^2\frac{\Delta\omega}{\Delta t}\Delta\phi $$ Since $W=\Delta T$, we can conclude that: $$ (mr^2)\frac{\Delta\omega}{\Delta t}=Fd. $$ Compare this to the Newton's law: $m\frac{\Delta v}{\Delta t}=F$. You can see that in “rotational world”, $I=mr^2$ plays role of inertia (mass), $\Delta\omega/\Delta t$ is angular acceleration, so “turning effect” $Fd$ is an analogue of force.

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Vasily Mitch
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