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Consider a rigid body $\mathcal{B}$ modeled by a system of $n$ point masses $B_1,B_2,\dots, B_n$ constrained to keep constant distance from each other. I wonder how it is possible to mathematically explain why, for any set of external forces applied on this system, there is always some configuration of the internal forces which maintains its rigidity.

More explicitly, suppose that on each mass point $B_i$, an external force $\mathbf{F_i}$ is applied. For simplicity, suppose that the net force is $0$ and that the net torque is $0$: $$\sum_{i=1}^{n} \mathbf{F_i} =0, \\\ \sum_{i=1}^{n} \mathbf{r_i}\times\mathbf{F_i} =0. $$

In this case, with the suitable initial conditions, the rigid body should remain motionless. This means that the internal forces between the point masses perfectly cancel the external forces. This is the mathematical statement that the following system of equations has a solution. The variables are the vectors $\mathbf{F_{ij}}$ for $i\neq j$, and the equations are:

  • $ \mathbf{F_{ji}} + \mathbf{F_{ij}}=0$ (third law).
  • $ (\mathbf{r_i}-\mathbf{r_j})\times\mathbf{F_{ij}}=0$ (forces are central, required for angluar momentum conservation).
  • $ \sum_{j:j\neq i} \mathbf{F_{ij}} =- \mathbf{F_i} $ (forces cancel out).

I cannot see mathematically why this system of equations must have a solution.

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    $\begingroup$ If only distances are kept constant, then generally this is not the case - a body can be deformed by changing angles between the bonds. $\endgroup$
    – Roger V.
    May 7, 2023 at 8:32
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    $\begingroup$ Mote that the equations are linear. If you have enough bodies ($n \ge 3$ ?) then you should find after eliminating redundant equations that you have more variables than equations, so the equations are under constrained. $\endgroup$
    – gandalf61
    May 7, 2023 at 8:53
  • $\begingroup$ How is directly showing that the scheme is impossible as stated in your title somehow not helpful? I also forgot to mention that the actual forces often are non-central. Otherwise, you cannot even keep a cubic system cubic. $\endgroup$ May 7, 2023 at 10:08
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    $\begingroup$ @gandalf61 It sounds like the right approach, but I am stuck with trying to identify the redundant equations and eliminate them in the right way. If you take the third set of equations, which contains $n$ equations, then they all sum up to $0$ (conservation of momentum), so the last one can be eliminated. But I still need to use conservation of angular momentum somehow. $\endgroup$
    – 35T41
    May 10, 2023 at 7:51
  • $\begingroup$ @35T41 Half of your first set of equations are redundant - you only need the ones where $i<j$. And once you know that $F_{ji}+F_{ij}=0$ then half of your angular momentum equations are also redundant - once again, you only need to consider the ones where $i<j$. $\endgroup$
    – gandalf61
    May 10, 2023 at 8:36

3 Answers 3

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The system is a consistent, underconstrained system of linear equations, so it must have a solution.

First note that we can remove the $i\neq j$ condition and just write $\mathbf F_{ii} = \mathbf 0$.

From the second equation, it follows that $\mathbf F_{ij} = k_{ij}(\mathbf r_i - \mathbf r_j)$. From the first equation, it then follows that $k_{ij}=k_{ji}$. The third equation is then

$$\sum_j k_{ij}(\mathbf r_i - \mathbf r_j) = -\mathbf F_i.$$

Let's show this is consistent with the constraints on the forces. First, sum over $i$ to get

$$\sum_{ij} k_{ij}(\mathbf r_i - \mathbf r_j) = \mathbf 0.$$

This is identically true due to the symmetry of $k_{ij}$, so the first constraint is always satisifed. Good.

Secondly, cross the equation with $\mathbf r_i$ and sum over $i$ to get

$$\sum_{ij} k_{ij}\mathbf r_i \times (\mathbf r_i - \mathbf r_j) = \sum_{ij} -k_{ij}\mathbf r_i \times \mathbf r_j = \mathbf 0.$$

You can again see that due to the symmetry of $k_{ij}$ (and anti-symmetry of the cross product), the second constraint is always satisfied as well.

Thus, the constraints are always satisifed -- the equations are consistent.

Then, we need to determine the $n^2$ values of $k_{ij}$. Due to symmetry and the condition $k_{ii}=0$, there are only $\frac {n(n-1)} 2$ independent components.

The equations is actually $3n$ equations -- one for each of the 3 components of the $\mathbf F_i$. However, these are not all independent due to the constraints - there are only $3n-6$ independent components.

Since $\frac {n(n-1)} 2 \geq 3n-6$, the equations are underconstrained (or have a unique solution if $n=3$).

You might notice a problem with $n=2$ -- however, there are $3n-5$ components then, since the system is linear and one of the torque conditions is then vacuous. This results in a system with as many equations as variables, and thus a unique solution (as you can check by hand!).

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  • $\begingroup$ Thank you for the answer! I understand everything that you write, but the part I am struggling with is formalizing your claim that "there are $3n-6$ independent components". I tried to simply eliminate 6 redundant equations; e.g. you can eliminate the last 3 equations (for the $n$-th body) because they are simply minus the sum of all the previous ones. But the next $3$ ones are trickier. Maybe I'm missing something and there is a more elegant way to formally justify this? $\endgroup$
    – 35T41
    May 11, 2023 at 14:52
  • $\begingroup$ The next 3 ones are due to the condition $\sum_i \mathbf r_i \times \mathbf F_i = \mathbf 0$. I'm not sure whether there is a simple interpretation in terms of eliminating it from the other equations. $\endgroup$
    – FusRoDah
    May 11, 2023 at 20:47
  • $\begingroup$ But it is a linear condition (or rather 3, since we're dealing with 3D vectors), so you should in principle be able to do so (by linear algebra). $\endgroup$
    – FusRoDah
    May 11, 2023 at 20:48
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In general it's false. Any set of coplanar masses with external forces that don't lie in the plane but satisfy the constraints is a counterexample (for example, three equally spaced collinear masses on the $x$ axis with forces proportional to $\hat y, -2 \hat y, \hat y$).

If the masses are in general position, then a solution exists by the following argument:

If $n>d$, then the $n{-}1\ge d$ struts connected to mass $n$ span the space, so some linear combination of them cancels $\mathbf F_n$. This alters the net force on the other masses in a way that maintains zero total force and torque. By induction on $n$, you can cancel those forces using the struts connecting the other masses.

If $n\le d$, the masses lie in a $d{-}1$ dimensional plane. Let $N_i$ be the components of the external forces normal to the plane. The zero net force and torque constraints imply $\sum_i N_i=0$ and $\sum_i N_i \mathbf r_i=0$. If any $N_k\ne 0$, it follows that

$$\mathbf r_k = \frac{\sum_{i\ne k} N_i \mathbf r_i}{\sum_{i\ne k} N_i},$$

i.e., one of the points is an affine combination of the others, violating the assumption that they're in general position. Therefore the external forces all lie in the plane, and can be canceled by induction on $d$.

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$\def \b {\mathbf}$

if we put the inertial coordinate system at the center of mass you obtain:

Translation

\begin{align*} &\b r_{\rm cm}=\frac{\sum m_i\,\b r_i}{\sum m_i} \quad\Rightarrow\quad \b{\ddot{r}}_{\rm cm}\,\sum m_i=\sum m_i\,\b {\ddot{r}}_i=\sum\b F_i=\b 0 \end{align*}

Rotation

\begin{align*} &\sum \b r_i\times \,m_i\b {\ddot{r}}_i=\tau= \sum \b r_i\times\b F_i+\sum_{j\ne i} (\b r_i-\b r_j)\times \b F_{ij} \end{align*}

if $~\b F_{ij}~$ is central force , or for rigid body $~\b r_i-\b r_j=$ constant ($~ \b F_{i,j}=\b 0~$) , then \begin{align*} &\sum_{j\ne i} \underbrace{(\b r_i-\b r_j)}_{\b r_{ij}}\times \b F_{ij}=\b 0 \end{align*}

the conservation of the angular momentum is achieve if

$~\b r_i=\b r~\quad\Rightarrow \sum \b r_i\times\b F_i=\b 0~ $

or

$~\sum \b r_i\times\b F_i=\b 0~$


I don't see this equation ?

$ \sum_{j:j\neq i} \mathbf{F_{ij}} =- \mathbf{F_i} $

enter image description here

\begin{align*} & \tau=\b r_1\times(\b F_1+\b F_{12}+\b F_{13})+ \b r_2\times(\b F_2+\b F_{23}-\b F_{12})+ \b r_3\times(\b F_3-\b F_{13}-\b F_{23}) \end{align*}

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