Question Number 203688 by ajfour last updated on 26/Jan/24
Commented by ajfour last updated on 26/Jan/24
$${Repost}\:\:\:{Q}.\mathrm{203636}\: \\ $$
Commented by ajfour last updated on 26/Jan/24
$${Looks}\:{suuperB}!\:\:{waiting}\:{for}\:{you}\:{to} \\ $$$${elaborate},\:{gracious}. \\ $$
Commented by mr W last updated on 27/Jan/24
![i think i got it, easier than i thought, (T_A /(mg))=(a/( [(((1/3)(√((((a^2 +b^2 +c^2 )^2 )/4)+(μ^4 /9)))+((a^2 +b^2 +c^2 )/6)))^(1/3) −(((1/3)(√((((a^2 +b^2 +c^2 )^2 )/4)+(μ^4 /9)))−((a^2 +b^2 +c^2 )/6)))^(1/3) ]^(3/2) )) with μ=(√((kQ^2 )/(mg))) i hope you may comfirm this.](https://www.tinkutara.com/question/Q203721.png)
$${i}\:{think}\:{i}\:{got}\:{it},\:{easier}\:{than}\:{i}\:{thought}, \\ $$$$\frac{{T}_{{A}} }{{mg}}=\frac{{a}}{\:\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{6}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${with}\:\mu=\sqrt{\frac{{kQ}^{\mathrm{2}} }{{mg}}} \\ $$$${i}\:{hope}\:{you}\:{may}\:{comfirm}\:{this}. \\ $$