2
$\begingroup$

The Problem

There are $3$ positively charged particles fixed in a frictionless horizontal plane, positioned in the vertices of a triangle.

The $i$-th particle has mass $m_i$ and charge $Q_i$.

When they are free to move, their positions always form a triangle that is similar to the first triangle, such as their corresponding sides are always parallel. [So they can't rotate, just spread out]

Determine the largest angle of the triangle, if the charge/mass ratio of the particles is given by:$$\dfrac{Q_1}{m_1}:\dfrac{Q_2}{m_2}:\dfrac{Q_3}{m_3}=1:2:3$$


My Attempt

I tried to approach it with vectors, centering a cartesian referential in the centroid of the triangle. If the position of each particle $i$ is $\vec{r_i}$, then: $$\begin{cases}\displaystyle\vec{r_{21}}+\vec{r_{23}}+\vec{r_{31}}=0\\\vec{a_1}=\frac{KQ_1}{M_1}\left(\frac{Q_2}{r^3_{21}}\vec{r}_{21}+\frac{Q_3}{r^3_{31}}\vec{r}_{31}\right)\\\vec{a_2}=\frac{2KQ_1}{M_1}\left(\frac{Q_1}{r^3_{12}}\vec{r}_{12}+\frac{Q_3}{r^3_{32}}\vec{r}_{32}\right)\\\vec{a_3}=\frac{3KQ_1}{M_1}\left(\frac{Q_1}{r^3_{13}}\vec{r}_{13}+\frac{Q_2}{r^3_{23}}\vec{r}_{23}\right)\\\vec{a_{1}}+\vec{a_{2}}+\vec{a_{3}}=0\end{cases}$$

There must be an elegant solution to this problem... What am I missing here? Is it possible to represent symmetry here without directly operating the position vectors?

$\endgroup$
2
  • $\begingroup$ You probably want the centroid of the triangle. en.wikipedia.org/wiki/Centroid To keep the triangle similar, the vertices will need to move as follows. Each moves along the line joining the centroid to the vertex. And they move at the relative rates such that the distances from the vertices to the centroid keep the same ratio. $\endgroup$
    – Dan
    Commented Feb 25, 2022 at 19:48
  • $\begingroup$ Thank you Dan, I actually already said that in my solution attempt, which translates into the five equations represented above. We can only say that the sum of the position vectors is equal to zero if the origin is the centroid... $\endgroup$
    – nickh
    Commented Feb 25, 2022 at 19:56

2 Answers 2

3
$\begingroup$

Building up from @Dan's comment, the problem seems friendly enough that you only seem to need the equations of motion at time $0$ to find a geometry that follows the desired rules. In reality, things turned out to be trickier.

First, we start by setting the centroid as the origin of coordinates. Then we place charge $1$ at $\mathbf{r_1}\left( t = 0\right) = (1, 0)$, charge $2$ at $\mathbf{r_2}\left( t = 0\right) = (\cos(\phi),\sin(\phi))$ and charge $3$ at $\mathbf{r_3}\left( t = 0\right) = - \mathbf{r_1}\left( t = 0\right) - \mathbf{r_2}\left( t = 0\right) $.

For the condition established in the problem to be satisfied, the net force acting on charge $1$, $\mathbf{f_1}$, must remain horizontal at all times. In fact, all net forces must remain parallel to the corresponding position vectors: $\mathbf{f_i} \parallel \mathbf{r_i}$. In particular this is true at $t=0$. The equation $0 = (0, 1) \cdot \mathbf{f_1}\left( t = 0\right)$ happens to be enough to give you a value for $\phi$, which, in this setup, fully determines the geometry of the system.

After replacing this value and letting the system evolve, it turns out that this is not a general enough ansatz to have a triangle that genuinely remains symmetrical at all times. The initial position of charge $2$ should be less constrained that in my previous ansatz, so we go instead for \begin{eqnarray} \mathbf{r_1}\left( t = 0\right) &=& (1, 0) \, , \\ \mathbf{r_2}\left( t = 0\right) &=& l(\cos(\phi),\sin(\phi)) \, , \\ \mathbf{r_3}\left( t = 0\right) &=& - \mathbf{r_1}\left( t = 0\right) - \mathbf{r_2}\left( t = 0\right) \, , \end{eqnarray} with $l$ some positive number. We exhausted the degree of freedom associated with scaling invariance by setting the norm of $\mathbf{r_1}\left( t = 0\right)$ equal to $1$, and we exhausted rotational invariance by fixing its orientation along the $x$ axis. So setting the norm of $\mathbf{r_2}\left( t = 0\right)$ to $1$ was a mistake in that first ansatz.

With the more general ansatz we are forced to use an additional equation, so that we can solve for both $\phi$ and $l$. This can come, for instance, from $\mathbf{f_2}\left( t = 0\right)$ having no component orthogonal to $\mathbf{r_2}\left( t = 0\right)$. These equations are not particularly nice, but Mathematica is able to provide a solution: \begin{eqnarray} \phi &\approx& 1.802119183 \, \, \mathrm{rad} \approx 103.2538234° \\ &\mathrm{and}& \\ l &\approx& 1.381620315 \, . \end{eqnarray} Notice that $\phi$ is not an angle of the triangle, but the solution allows you to find the largest triangle angle which happens to be \begin{equation} \arccos \left[ \frac{(\mathbf{r_2} - \mathbf{r_1}) \cdot (\mathbf{r_3} - \mathbf{r_1})}{|\mathbf{r_2} - \mathbf{r_1}| |\mathbf{r_3} - \mathbf{r_1}|} \right] \approx 1.470038873 \, \, \mathrm{rad} \approx 84.22702317° \, . \end{equation}

$\endgroup$
4
  • $\begingroup$ First of all, thank you very much for all insight and detail put in your answer! I just did not understand one bit: how did you get a result for phi using only 2 equations, when there are 5 unknowns (the charges, L and phi)? Even using the charge/mass ratios, I got an equation dependent on Q2, Q3, phi and L as a result of dot(f1, r1) = 0. Also, I tried simulating the problem (by brute forcing possible triangles) and got 1.3869 rad = 79.47º as the largest triangle angle, which is close to your 84.23º, but my algorithm may be wrong... $\endgroup$
    – nickh
    Commented Feb 28, 2022 at 9:42
  • 1
    $\begingroup$ Let me point out that the equations are not $\mathbf{f_i} . \mathbf{r_i} = 0$ but instead more like $\mathbf{f_i} \times \mathbf{r_i} = \mathbf{0}$, with $\times$ being the cross product, if you picture them as 3D vectors with a $0$ $z$-component, because we are saying that they must be parallel. $\endgroup$
    – secavara
    Commented Feb 28, 2022 at 12:04
  • 1
    $\begingroup$ It seems I did assume that the ratio $1 : 2 : 3$ was between the charges and not the charge/mass ratios (so as it all the masses were the same), and that allowed me to get to a single numeric answer. Otherwise, you do get something that would depend on two charge ratios, if you only use two equations. Interestingly, Mathematica can still find solutions for $\phi$ and $l$ as a function of $q_1$, $q_2$ and $q_3$, using only two equations, the expressions are just huge. This makes me think that the exercise meant to give information on charge ratios and not mass/charge ratios. $\endgroup$
    – secavara
    Commented Feb 28, 2022 at 12:04
  • $\begingroup$ Ah, indeed, in my first comment I should have said $\mathbf{f_i}\times\mathbf{r_i}=0$, thank you for pointing it out. Still, I am certain that the $1:2:3$ proportion refers to the charge/mass ratios. Perhaps assuming $w=q_1/m_1$ and using all 3 cross-product equations this constant $w$ cancels out. But taking a step back, the problem may not be meant to be solved with direct vector manipulation, I just can't see how one could express the symmetry involved without a cartesian coordinate system... $\endgroup$
    – nickh
    Commented Feb 28, 2022 at 19:07
2
+50
$\begingroup$

A simpler approach is to set the magnitudes of the mutual accelerations proportional to the separations. This is a necessary condition and seems to also be sufficient.

i.e. $\space a_{12}/r_{12} = a_{23}/r_{23} = a_{31}/r_{31}$

taking $a_{ij} = \frac{KQ_iQ_j/r_{ij}^2}{M_{ij}}$ where $M_{ij}=\frac{M_iM_j}{(M_i+M_j)}$ is the reduced mass

giving $a_{ij}/r_{ij} = K(Q_i/M_i)(Q_j/M_j)(M_i+M_j)/r_{ij}^3$

Substituting the $1:2:3$ ratios for $Q_i/M_i$ gives

$1\cdot2\cdot(M_1+M_2)/r_{12}^3 = 2\cdot3\cdot(M_2+M_3)/r_{23}^3 = 3\cdot1\cdot(M_3+M_1)/r_{31}^3$

This immediately gives you the ratio of the three sides and you can use the cosine formula to get the angles. But the answer is not independent of the masses!

Assuming the masses are all equal, we get the sides of the triangle in the ratio $2^{(1/3)}:6^{(1/3)}:3^{(1/3)}$ which gives the same answer of $84.2270^{\circ}$ found by @secavara

$\endgroup$
2
  • 2
    $\begingroup$ That's a great catch! $\endgroup$
    – secavara
    Commented Mar 2, 2022 at 7:04
  • $\begingroup$ That was an excellent approach, thank you so much @Roger Wood! $\endgroup$
    – nickh
    Commented Mar 2, 2022 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.