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- #1

- Mar 10, 2012

- 835

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Write $\underbrace{\mathbb R^3\times \cdots \times \mathbb R^3}_{m\text{ times}}$ as $(\mathbb R^3)^m$.

Let $p=(p_1,\ldots,p_m)\in (\mathbb R^3)^m$.

Let the set of all translations in $\mathbb R^3$ be written as $\mathcal T$ and the set of all special orthogonal transformations in $\mathbb R^3$ be wriiten as $SO(3)$.

Define $M(p)=\{(\tau\circ R(q_1),\ldots,\tau\circ R(p_m)) \in (\mathbb R^3)^m : \tau \in \mathcal T \text{ and } R \in SO(3)\}$.

$M(p)$ happens to be a manifold. (I am not sure about the details of the proof but it's true. See rigid transformation - To Prove That a Certain Set is a Manifold - Mathematics Stack Exchange)

Wrtie $T(p)$ as the tangent space at point $p$ of the manifold $M(p)$.

Let $F=(F_1,\ldots,F_m) \in (\mathbb R^3)^m$.

Imagine a rigid body $\Omega$ in $\mathbb R^3$ which is large enough so that each $p_i$ is on it.

Apply a force $F_i$ at the point $p_i$ of $\Omega$ and write $F=(F_1,\ldots,F_m)$.

Then $\Omega$ is in equilibrium if $F\in T(p)^{\perp}$.

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Can somebody give some insight as to why this should be true? Normally equilibrium is talked about in terms of torques about points etc. How is this characterization equivalent to the usual one?