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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?

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    $\begingroup$ In this context, do you mean the same thing as the torque about a point, $\mathbf{\tau}=\mathbf{r}\times\mathbf{F}$? $\endgroup$
    – DanDan0101
    Oct 18 at 19:10
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    $\begingroup$ @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. $\endgroup$ Oct 19 at 14:14
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    $\begingroup$ @XXb8 As d (circumference) and r (radius) just differ by a factor of 2đťś‹, you could say that. A larger r requires a smaller force, but that smaller force has to be applied over a larger distance d. $\endgroup$ Oct 21 at 13:16
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    $\begingroup$ @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. $\endgroup$ Oct 21 at 22:14
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    $\begingroup$ @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. $\endgroup$ Oct 22 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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