solving system of equations(nonlinear)

Question

I am trying to solve the following system of equations:

$$\frac{kq^2}{d}=mg(L-L\cos(t))+\frac{kq^2}{r}$$ $$\sin(t)=\frac{x}{L}$$ $$r^2=(L-L\cos(t))^2+(x+d)^2$$

The parameters are: $k,L,d,q,m,g$

The variables: $x,r,t$

all parameters and variables $> 0$

I tried using wolfram but it gave me weird answers.

Answer 1

Let $s = \sin(t)$ and $c = \cos(t)$, with the equation $s^2 + c^2 = 1$. Then you have a system of equations in rational functions. One solution is $x=0,r=d,s=0,c=1$. The others depend on solving a quintic equation. If $r$ is a solution of $$ {d}^{2}{g}^{2}{m}^{2}{z}^{5}+{d}^{3}{g}^{2}{m}^{2}{z}^{4}+ \left( -{d} ^{4}{g}^{2}{m}^{2}-4\,Ldgkm{q}^{2} \right) {z}^{3}-{d}^{5}{g}^{2}{m}^{ 2}{z}^{2}+ \left( -4\,L{d}^{3}gkm{q}^{2}+4\,{L}^{2}{k}^{2}{q}^{4}+4\,{ d}^{2}{k}^{2}{q}^{4} \right) z-4\,{L}^{2}d{k}^{2}{q}^{4}-4\,{d}^{3}{k} ^{2}{q}^{4} =0$$ then you have

$$\eqalign{c&=1+{\frac {k{q}^{2}}{gmLr}}-{\frac {k{q}^{2}}{mgLd}}\cr s&=-\dfrac12 \,{\frac {d}{L}}+\dfrac12\,{\frac {{r}^{2}}{Ld}}+{\frac {k{q}^{2}}{mgdr}}-{ \frac {k{q}^{2}}{gm{d}^{2}}}\cr x&=-\dfrac{d}{2}+\dfrac12\,{\frac {{r}^{2}}{d}}+{\frac { Lk{q}^{2}}{mgdr}}-{\frac {Lk{q}^{2}}{gm{d}^{2}}}\cr}$$