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Recall the Law of Moments for a one dimensional rod:

"When an object is in equilibrium the sum of the clockwise moments is equal to the sum of the anticlockwise moments."

I understand that we may define a quantity $F \times d$, and call it a moment. Then, experimentally, we can observe that $F_1\times d_1 = F_2\times d_2$ is a necessary condition for equilibrium.

However, is it possible to derive the Law of Moments from some more fundamental principle, e.g. the principle of least action?

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  • $\begingroup$ The principle of least action would be unaltered. Except in the Lagrangian, the rotational/torsional terms would be of much more importance. The derivation proceeds very simply then. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 4 '14 at 17:42
  • $\begingroup$ @TorstenHĕrculĕCärlemän Can you please expand on this? $\endgroup$ – Fly by Night Jan 4 '14 at 17:48
  • $\begingroup$ Let us assume a body which has a degree of freedom about a fixed axis. It might, in actuality be in equilibrium but for the sake of generality add the $1/2*I*\omega^2$ term for the kinetic energy of rotation. Then work out the principle of least action, ending with the Euler Lagrange equations in terms of torques. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 4 '14 at 18:10
  • $\begingroup$ @TorstenHĕrculĕCärlemän I'm sorry, but I don't understand what you mean by "...work out the principle of least action...". I was hoping that someone might be able to show me how to derive it in their answer. $\endgroup$ – Fly by Night Jan 4 '14 at 18:15
  • $\begingroup$ Are you well versed with the principles of variational calculus? That's exactly what you need to do. However, I have realized that my method doesn't cover all the cases, so it is just there for a heuristic check. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 4 '14 at 18:17
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I am not so sure about what you meant by more fundamental principle, but I think you can derive the law of moments by d'Alembert principle.

The equilibrium condition is given by \begin{equation} \sum_i \vec{F}_i \cdot \delta \vec{r}_i = 0 \end{equation} For any virtual displacements on a rigid body, it can be seen as a combination of virtual displacement of center of mass and pure rotation around the center of mass. In the center of mass frame, the virtual distance is $\delta \vec{r}_i = \delta \vec{\theta} \times \vec{r}_i$, where $\delta \vec{\theta}$ is constant for any virtual displacements as a constraint of rigid body. Therefore, \begin{align*} \sum_i \vec{F}_i \cdot (\delta \vec{\theta} \times \vec{r}_i) & = 0\\ \sum_i \delta \vec{\theta} \cdot (\vec{r}_i \times \vec{F}_i) & = 0 \\ \delta \vec{\theta} \cdot \sum_i \vec{r}_i \times \vec{F}_i & = 0 \end{align*} Since $\delta \vec{\theta}$ is arbitrary, $\sum_i \vec{r}_i \times \vec{F}_i = 0$.

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You can derive this law by using the amount of work done by each force when the rod would be rotated by the same angle $\theta$, because when they are in equilibrium than the sum of the two amounts of work done should be zero. $$ W=\int_C{\vec{F}\cdot d\vec{s}}=\int_\theta^0{Frd\theta} $$ So $$ W_1+W_2=0 $$ $$ \int_\theta^0{F_1r_1d\theta}+\int_\theta^0{F_2r_2d\theta}=0 $$ Assuming that $\theta$ is small (but not zero) such that the forces stay approximately perpendicular to the arm, then the sum of these integrals can be solved as: $$ \theta\left(F_1r_1+F_2r_2\right)=0 $$ This would seem to give a wrong answer, since this yields $F_1r_1=-F_2r_2$. But you have to keep in mind what to positive direction of each force would be in this integral, since it is integrated over an angle. This means that the forces on opposite sides of the pivot will have the opposite sign. The side who's forces would change sign will depend on which angular direction you choose as positive.

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  • $\begingroup$ In your paragraph “because when they are in equilibrium than the sum of the two amounts of work done should be zero” are you referring to the principle of virtual work? $\endgroup$ – pppqqq Jan 4 '14 at 19:14
  • $\begingroup$ I derived it from the (static) equilibrium: "A system of particles is in static equilibrium when all the particles of the system are at rest and the total force on each particle is permanently zero." This would mean that a none-rotating rod in equilibrium would remain at rest. So if the total work would be none-zero it would contribute to kinetic energy of the system and therefore induce motion, so no longer meet the definition of static equilibrium. Therefore you can say that the work done has to be zero. $\endgroup$ – fibonatic Jan 4 '14 at 23:30
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I view moments as a vector field around an action axis (force axis), similarly to how linear velocities are a vector field around a motion axis (rotation axis). This all comes from Screw Theory and the law of moments is really a manifestation of screw algebra.

The balance described really refers to a case where the force and motion screws are reciprocal (cancel each other out) to produce a system with no net power in or out. It can be a static or a dynamic equilibrium in general.

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