Question Number 203726 by mr W last updated on 27/Jan/24
Commented by mr W last updated on 26/Jan/24
$${solution}\:{to}\:{Q}#\mathrm{203636} \\ $$
Answered by mr W last updated on 27/Jan/24
Commented by mr W last updated on 26/Jan/24
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 26/Jan/24
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 27/Jan/24
![the center of mass of the three charges is G, which in current case must be the centroid of ΔABC, since they have the same mass. G must lie vertically under the hanging point S on the ceiling. say the sides of ΔABC are p, q, r respectively. m_A =AP=((√(2(q^2 +r^2 )−p^2 ))/2) m_B =BQ=((√(2(r^2 +p^2 )−q^2 ))/2) m_C =CR=((√(2(p^2 +q^2 )−r^2 ))/2) d_A =(2/3)m_A =distance from G to A d_B =(2/3)m_B =distance from G to B d_C =(2/3)m_C =distance from G to C say SG=h, SG is vertical. F_1 =((kQ^2 )/p^2 ) with Q=charge, k=Coulomb constant F_2 =((kQ^2 )/q^2 ) F_3 =((kQ^2 )/r^2 ) say F_A is the resultant from F_2 and F_3 F_B is the resultant from F_3 and F_1 F_C is the resultant from F_1 and F_2 T_A , F_A and mg must be in the same plane and in equilibrium. (F_A /d_A )=(T_A /a)=((mg)/h) ⇒F_A =((mgd_A )/h), T_A =((mga)/h) similarly F_B =((mgd_B )/h), T_B =((mgb)/h) F_C =((mgd_C )/h), T_C =((mgc)/h) (F_2 /(0.5q))=(F_3 /(0.5r))=(F_A /m_A )=((mgd_A )/(hm_A ))=((2mg)/(3h)) F_2 =((mgq)/(3h))=((kQ^2 )/q^2 ) ⇒q=(((3kQ^2 h)/(mg)))^(1/3) F_3 =((mgr)/h)=((kQ^2 )/r^2 ) ⇒r=(((3kQ^2 h)/(mg)))^(1/3) similarly F_1 =((mgp)/h)=((kQ^2 )/p^2 ) ⇒p=(((3kQ^2 h)/(mg)))^(1/3) that means p=q=r=(((3kQ^2 h)/(mg)))^(1/3) =((3μ^2 h))^(1/3) with μ=(√((kQ^2 )/(mg))) SP=h_P =((√(2(b^2 +c^2 )−p^2 ))/2) m_A (h^2 +(1/3)m_A ×(2/3)m_A )=(m_A /3)×a^2 +((2m_A )/3)×h_P ^2 h^2 +(2/9)m_A ^2 =(a^2 /3)+((2h_P ^2 )/3) h^2 +(2/9)×((2(q^2 +r^2 )−p^2 )/4)=(a^2 /3)+((2(b^2 +c^2 )−p^2 )/6) h^2 +((p^2 +q^2 +r^2 )/9)=((a^2 +b^2 +c^2 )/3) 9h^2 +p^2 +q^2 +r^2 =3(a^2 +b^2 +c^2 ) 9h^2 +3((9μ^4 h^2 ))^(1/3) −3(a^2 +b^2 +c^2 )=0 let δ=(√(a^2 +b^2 +c^2 )) ((h^2 )^(1/3) )^3 +(1/3)((9μ^4 ))^(1/3) ((h^2 )^(1/3) )−(δ^2 /3)=0 ⇒(h^2 )^(1/3) =(((1/3)(√((δ^4 /4)+(μ^4 /9)))+(δ^2 /6)))^(1/3) −(((1/3)(√((δ^4 /4)+(μ^4 /9)))−(δ^2 /6)))^(1/3) ⇒h=[(((1/3)(√((δ^4 /4)+(μ^4 /9)))+(δ^2 /6)))^(1/3) −(((1/3)(√((δ^4 /4)+(μ^4 /9)))−(δ^2 /6)))^(1/3) ]^(3/2) (T_A /(mg))=(a/h)=(a/([(((1/3)(√((δ^4 /4)+(μ^4 /9)))+(δ^2 /6)))^(1/3) −(((1/3)(√((δ^4 /4)+(μ^4 /9)))−(δ^2 /6)))^(1/3) ]^(3/2) )) (T_B /(mg))=(b/h)=(b/([(((1/3)(√((δ^4 /4)+(μ^4 /9)))+(δ^2 /6)))^(1/3) −(((1/3)(√((δ^4 /4)+(μ^4 /9)))−(δ^2 /6)))^(1/3) ]^(3/2) )) (T_C /(mg))=(c/h)=(c/([(((1/3)(√((δ^4 /4)+(μ^4 /9)))+(δ^2 /6)))^(1/3) −(((1/3)(√((δ^4 /4)+(μ^4 /9)))−(δ^2 /6)))^(1/3) ]^(3/2) )) special case: a=b=c=s T_A =T_B =T_C =T (T/(mg))=(s/([(((√((s^4 /4)+(μ^4 /(81))))+(s^2 /2)))^(1/3) −(((√((s^4 /4)+(μ^4 /(81))))−(s^2 /2)))^(1/3) ]^(3/2) ))](https://www.tinkutara.com/question/Q203733.png)
$${the}\:{center}\:{of}\:{mass}\:{of}\:{the}\:{three}\: \\ $$$${charges}\:{is}\:{G},\:{which}\:{in}\:{current}\:{case} \\ $$$${must}\:{be}\:{the}\:{centroid}\:{of}\:\Delta{ABC}, \\ $$$${since}\:{they}\:{have}\:{the}\:{same}\:{mass}. \\ $$$${G}\:{must}\:{lie}\:{vertically}\:{under}\:{the} \\ $$$${hanging}\:{point}\:{S}\:{on}\:{the}\:{ceiling}. \\ $$$${say}\:{the}\:{sides}\:{of}\:\Delta{ABC}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively}. \\ $$$${m}_{{A}} ={AP}=\frac{\sqrt{\mathrm{2}\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)−{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{B}} ={BQ}=\frac{\sqrt{\mathrm{2}\left({r}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)−{q}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{C}} ={CR}=\frac{\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−{r}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${d}_{{A}} =\frac{\mathrm{2}}{\mathrm{3}}{m}_{{A}} ={distance}\:{from}\:{G}\:{to}\:{A} \\ $$$${d}_{{B}} =\frac{\mathrm{2}}{\mathrm{3}}{m}_{{B}} ={distance}\:{from}\:{G}\:{to}\:{B} \\ $$$${d}_{{C}} =\frac{\mathrm{2}}{\mathrm{3}}{m}_{{C}} ={distance}\:{from}\:{G}\:{to}\:{C} \\ $$$${say}\:{SG}={h},\:{SG}\:{is}\:{vertical}. \\ $$$${F}_{\mathrm{1}} =\frac{{kQ}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\:{with}\:{Q}={charge},\:{k}={Coulomb}\:{constant} \\ $$$${F}_{\mathrm{2}} =\frac{{kQ}^{\mathrm{2}} }{{q}^{\mathrm{2}} } \\ $$$${F}_{\mathrm{3}} =\frac{{kQ}^{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$${say} \\ $$$${F}_{{A}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{2}} \:{and}\:{F}_{\mathrm{3}} \\ $$$${F}_{{B}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{3}} \:{and}\:{F}_{\mathrm{1}} \\ $$$${F}_{{C}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} \\ $$$$ \\ $$$${T}_{{A}} ,\:{F}_{{A}} \:{and}\:{mg}\:{must}\:{be}\:{in}\:{the}\:{same} \\ $$$${plane}\:{and}\:{in}\:{equilibrium}. \\ $$$$\frac{{F}_{{A}} }{{d}_{{A}} }=\frac{{T}_{{A}} }{{a}}=\frac{{mg}}{{h}} \\ $$$$\Rightarrow{F}_{{A}} =\frac{{mgd}_{{A}} }{{h}},\:{T}_{{A}} =\frac{{mga}}{{h}} \\ $$$${similarly} \\ $$$${F}_{{B}} =\frac{{mgd}_{{B}} }{{h}},\:{T}_{{B}} =\frac{{mgb}}{{h}} \\ $$$${F}_{{C}} =\frac{{mgd}_{{C}} }{{h}},\:{T}_{{C}} =\frac{{mgc}}{{h}} \\ $$$$ \\ $$$$\frac{{F}_{\mathrm{2}} }{\mathrm{0}.\mathrm{5}{q}}=\frac{{F}_{\mathrm{3}} }{\mathrm{0}.\mathrm{5}{r}}=\frac{{F}_{{A}} }{{m}_{{A}} }=\frac{{mgd}_{{A}} }{{hm}_{{A}} }=\frac{\mathrm{2}{mg}}{\mathrm{3}{h}} \\ $$$${F}_{\mathrm{2}} =\frac{{mgq}}{\mathrm{3}{h}}=\frac{{kQ}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:\Rightarrow{q}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{kQ}^{\mathrm{2}} {h}}{{mg}}} \\ $$$${F}_{\mathrm{3}} =\frac{{mgr}}{{h}}=\frac{{kQ}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\:\Rightarrow{r}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{kQ}^{\mathrm{2}} {h}}{{mg}}} \\ $$$${similarly} \\ $$$${F}_{\mathrm{1}} =\frac{{mgp}}{{h}}=\frac{{kQ}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\:\Rightarrow{p}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{kQ}^{\mathrm{2}} {h}}{{mg}}} \\ $$$${that}\:{means}\:{p}={q}={r}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{kQ}^{\mathrm{2}} {h}}{{mg}}}=\sqrt[{\mathrm{3}}]{\mathrm{3}\mu^{\mathrm{2}} {h}} \\ $$$${with}\:\mu=\sqrt{\frac{{kQ}^{\mathrm{2}} }{{mg}}} \\ $$$$ \\ $$$${SP}={h}_{{P}} =\frac{\sqrt{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${m}_{{A}} \left({h}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{m}_{{A}} ×\frac{\mathrm{2}}{\mathrm{3}}{m}_{{A}} \right)=\frac{{m}_{{A}} }{\mathrm{3}}×{a}^{\mathrm{2}} +\frac{\mathrm{2}{m}_{{A}} }{\mathrm{3}}×{h}_{{P}} ^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}{m}_{{A}} ^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}{h}_{{P}} ^{\mathrm{2}} }{\mathrm{3}} \\ $$$${h}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}×\frac{\mathrm{2}\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)−{p}^{\mathrm{2}} }{\mathrm{4}}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{p}^{\mathrm{2}} }{\mathrm{6}} \\ $$$${h}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{9}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\mathrm{9}{h}^{\mathrm{2}} +{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\mathrm{9}{h}^{\mathrm{2}} +\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{9}\mu^{\mathrm{4}} {h}^{\mathrm{2}} }−\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\delta=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\left(\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\mathrm{9}\mu^{\mathrm{4}} }\left(\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }\right)−\frac{\delta^{\mathrm{2}} }{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{\delta^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$\Rightarrow{h}=\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\frac{{T}_{{A}} }{{mg}}=\frac{{a}}{{h}}=\frac{{a}}{\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{{T}_{{B}} }{{mg}}=\frac{{b}}{{h}}=\frac{{b}}{\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{{T}_{{C}} }{{mg}}=\frac{{c}}{{h}}=\frac{{c}}{\left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}+\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\delta^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{9}}}−\frac{\delta^{\mathrm{2}} }{\mathrm{6}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$ \\ $$$${special}\:{case}:\:{a}={b}={c}={s} \\ $$$${T}_{{A}} ={T}_{{B}} ={T}_{{C}} ={T} \\ $$$$\frac{{T}}{{mg}}=\frac{{s}}{\left[\sqrt[{\mathrm{3}}]{\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Commented by ajfour last updated on 26/Jan/24
$${Sir}\:{please}\:{take}\:{a}={b}={c}={s}\:\:{and} \\ $$$${N}=\mathrm{3},\:{see}\:{if}\:{it}\:{matches}\:{with}\:{result} \\ $$$${of}\:{q}.\mathrm{1}\:\:\:\:\:{Amazing}\:{Excellent}\:{solution}, \\ $$$${m}\:{not}\:{through}\:{yet}. \\ $$
Commented by mr W last updated on 26/Jan/24
$${thank}\:{you}\:{for}\:{viewing}\:{sir}! \\ $$$${i}'{ll}\:{check}\:{later}. \\ $$
Commented by mr W last updated on 27/Jan/24
$${i}\:{have}\:{fixed}\:{a}\:{big}\:{error}. \\ $$$${for}\:{the}\:{special}\:{case}\:{with}\:{a}={b}={c}={s}, \\ $$$${the}\:{result}\:{is}\:{the}\:{same}\:{as}\:{in}\:{question}\:\mathrm{1} \\ $$$${with}\:{n}=\mathrm{3}. \\ $$$${it}'{s}\:{interesting}\:{to}\:{notice}\:{that}\:\Delta{ABC} \\ $$$${is}\:{always}\:{equilateral},\:{i}.{e}.\:{p}={q}={r}. \\ $$
Commented by ajfour last updated on 27/Jan/24
At least i understood now that p=q=r, very remarkable!
Answered by mr W last updated on 27/Jan/24
$${General}\:{case}: \\ $$$${the}\:{particles}\:{have}\:{different}\:{masses} \\ $$$${and}\:{charges}. \\ $$
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 28/Jan/24
Commented by mr W last updated on 28/Jan/24
Commented by mr W last updated on 27/Jan/24
Commented by mr W last updated on 28/Jan/24
![say the center of mass of the three charges is G, which must lie vertically under the hanging point S on the ceiling. say the sides of ΔABC are p, q, r respectively. (p_1 /p)=(M_C /(M_B +M_C )) ⇒p_1 =((M_C p)/(M_B +M_C )) (p_2 /p)=(M_B /(M_B +M_C )) ⇒p_2 =((M_B p)/(M_B +M_C )) m_A =AP p(m_A ^2 +p_1 p_2 )=p_1 q^2 +p_2 r^2 m_A ^2 +((M_B M_C p^2 )/((M_B +M_C )^2 ))=((M_C q^2 +M_B r^2 )/(M_B +M_C )) ⇒m_A ^2 =((M_C q^2 +M_B r^2 )/(M_B +M_C ))−((M_B M_C p^2 )/((M_B +M_C )^2 )) (d_A /m_A )=((M_B +M_C )/(M_A +M_B +M_C )) (e_A /m_A )=(M_A /(M_A +M_B +M_C )) say SG=h, SG is vertical. F_1 =((kQ_B Q_C )/p^2 ) with k=Coulomb constant F_2 =((kQ_C Q_A )/q^2 ) F_3 =((kQ_A Q_B )/r^2 ) say F_A is the resultant from F_2 and F_3 F_B is the resultant from F_3 and F_1 F_C is the resultant from F_1 and F_2 T_A , F_A and M_A g must be in the same plane and in equilibrium. (F_A /d_A )=(T_A /a)=((M_A g)/h) ⇒F_A =((M_A gd_A )/h), T_A =((M_A ga)/h) similarly F_B =((M_B gd_B )/h), T_B =((M_B gb)/h) F_C =((M_C gd_C )/h), T_C =((M_C gc)/h) (F_2 /((p_1 /p)×q))=(F_3 /((p_2 /p)×r))=(F_A /m_A )=((M_A gd_A )/(hm_A ))=((M_A (M_B +M_C )g)/((M_A +M_B +M_C )h)) ⇒F_2 =((M_C M_A gq)/((M_A +M_B +M_C )h))=((kQ_C Q_A )/q^2 ) ⇒q^3 =((kQ_C Q_A (M_A +M_B +M_C )h)/(gM_C M_A )) ⇒F_3 =((M_A M_B gr)/((M_A +M_B +M_C )h))=((kQ_A Q_B )/r^2 ) ⇒r^3 =((kQ_A Q_B (M_A +M_B +M_C )h)/(gM_A M_B )) similarly ⇒p^3 =((kQ_B Q_C (M_A +M_B +M_C )h)/(gM_B M_C )) let μ_1 =(√((kQ_B Q_C (M_A +M_B +M_C ))/(gM_B M_C ))) μ_2 =(√((kQ_C Q_A (M_A +M_B +M_C ))/(gM_C M_A ))) μ_3 =(√((kQ_A Q_B (M_A +M_B +M_C ))/(gM_A M_B ))) ⇒p=((μ_1 ^2 h))^(1/3) , q=((μ_2 ^2 h))^(1/3) , r=((μ_3 ^2 h))^(1/3) SP=h_P p(h_P ^2 +p_1 p_2 )=p_1 c^2 +p_2 b^2 h_P ^2 +((M_B M_C p^2 )/((M_B +M_C )^2 ))=((M_C c^2 +M_B b^2 )/(M_B +M_C )) h_P ^2 =((M_B b^2 +M_C c^2 )/(M_B +M_C ))−((M_B M_C p^2 )/((M_B +M_C )^2 )) m_A (h^2 +e_A d_A )=e_A a^2 +d_A ×h_P ^2 h^2 +e_A d_A =((M_A a^2 )/(M_A +M_B +M_C ))+(((M_B +M_C )h_P ^2 )/(M_A +M_B +M_C )) h^2 +((M_A (M_B +M_C )m_A ^2 )/((M_A +M_B +M_C )^2 ))=((M_A a^2 )/(M_A +M_B +M_C ))+(((M_B +M_C )h_P ^2 )/(M_A +M_B +M_C )) h^2 +((M_A (M_B +M_C ))/((M_A +M_B +M_C )^2 ))×[((M_C q^2 +M_B r^2 )/(M_B +M_C ))−((M_B M_C p^2 )/((M_B +M_C )^2 ))]=((M_A a^2 )/(M_A +M_B +M_C ))+(((M_B +M_C ))/(M_A +M_B +M_C ))×[((M_B b^2 +M_C c^2 )/(M_B +M_C ))−((M_B M_C p^2 )/((M_B +M_C )^2 ))] h^2 +((M_B M_C p^2 +M_C M_A q^2 +M_A M_B r^2 )/((M_A +M_B +M_C )^2 ))=((M_A a^2 +M_B b^2 +M_C c^2 )/(M_A +M_B +M_C )) h^2 +(((M_B M_C (μ_1 ^4 )^(1/3) +M_C M_A (μ_2 ^4 )^(1/3) +M_A M_B (μ_3 ^4 )^(1/3) )(h^2 )^(1/3) )/((M_A +M_B +M_C )^2 ))−((M_A a^2 +M_B b^2 +M_C c^2 )/(M_A +M_B +M_C ))=0 let ξ=((M_B M_C (μ_1 ^4 )^(1/3) +M_C M_A (μ_2 ^4 )^(1/3) +M_A M_B (μ_3 ^4 )^(1/3) )/((M_A +M_B +M_C )^2 )) or ξ=((((M_A M_B M_C )/((M_A +M_B +M_C )^4 ))×(((kQ_A Q_B Q_C )/g))^2 ))^(1/3) [(1/( ((M_A Q_A ^2 ))^(1/3) ))+(1/( ((M_B Q_B ^2 ))^(1/3) ))+(1/( ((M_C Q_C ^2 ))^(1/3) ))] δ=((M_A a^2 +M_B b^2 +M_C c^2 )/(M_A +M_B +M_C )) ((h^2 )^(1/3) )^3 +ξ((h^2 )^(1/3) )−δ=0 ⇒(h^2 )^(1/3) =(((√((δ^2 /4)+(ξ^3 /(27))))+(δ/2)))^(1/3) −(((√((δ^2 /4)+(ξ^3 /(27))))−(δ/2)))^(1/3) ⇒h=[(((√((δ^2 /4)+(ξ^3 /(27))))+(δ/2)))^(1/3) −(((√((δ^2 /4)+(ξ^3 /(27))))−(δ/2)))^(1/3) ]^(3/2) (T_A /(M_A g))=(a/h) (T_B /(M_B g))=(b/h) (T_C /(M_C g))=(c/h)](https://www.tinkutara.com/question/Q203766.png)
$${say}\:{the}\:{center}\:{of}\:{mass}\:{of}\:{the}\:{three}\: \\ $$$${charges}\:{is}\:{G},\:{which}\:{must}\:{lie}\: \\ $$$${vertically}\:{under}\:{the}\:{hanging}\:{point}\:{S} \\ $$$${on}\:{the}\:{ceiling}. \\ $$$${say}\:{the}\:{sides}\:{of}\:\Delta{ABC}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively}. \\ $$$$\frac{{p}_{\mathrm{1}} }{{p}}=\frac{{M}_{{C}} }{{M}_{{B}} +{M}_{{C}} }\:\Rightarrow{p}_{\mathrm{1}} =\frac{{M}_{{C}} \:{p}}{{M}_{{B}} +{M}_{{C}} } \\ $$$$\frac{{p}_{\mathrm{2}} }{{p}}=\frac{{M}_{{B}} }{{M}_{{B}} +{M}_{{C}} }\:\Rightarrow{p}_{\mathrm{2}} =\frac{{M}_{{B}} \:{p}}{{M}_{{B}} +{M}_{{C}} } \\ $$$${m}_{{A}} ={AP} \\ $$$${p}\left({m}_{{A}} ^{\mathrm{2}} +{p}_{\mathrm{1}} {p}_{\mathrm{2}} \right)={p}_{\mathrm{1}} {q}^{\mathrm{2}} +{p}_{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${m}_{{A}} ^{\mathrm{2}} +\frac{{M}_{{B}} {M}_{{C}} \:{p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }=\frac{{M}_{{C}} \:{q}^{\mathrm{2}} +{M}_{{B}} \:{r}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} } \\ $$$$\Rightarrow{m}_{{A}} ^{\mathrm{2}} =\frac{{M}_{{C}} \:{q}^{\mathrm{2}} +{M}_{{B}} \:{r}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} }−\frac{{M}_{{B}} {M}_{{C}} \:{p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} } \\ $$$$\frac{{d}_{{A}} }{{m}_{{A}} }=\frac{{M}_{{B}} +{M}_{{C}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$$\frac{{e}_{{A}} }{{m}_{{A}} }=\frac{{M}_{{A}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$$ \\ $$$${say}\:{SG}={h},\:{SG}\:{is}\:{vertical}. \\ $$$${F}_{\mathrm{1}} =\frac{{kQ}_{{B}} {Q}_{{C}} }{{p}^{\mathrm{2}} }\:{with}\:{k}={Coulomb}\:{constant} \\ $$$${F}_{\mathrm{2}} =\frac{{kQ}_{{C}} {Q}_{{A}} }{{q}^{\mathrm{2}} } \\ $$$${F}_{\mathrm{3}} =\frac{{kQ}_{{A}} {Q}_{{B}} }{{r}^{\mathrm{2}} } \\ $$$${say} \\ $$$${F}_{{A}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{2}} \:{and}\:{F}_{\mathrm{3}} \\ $$$${F}_{{B}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{3}} \:{and}\:{F}_{\mathrm{1}} \\ $$$${F}_{{C}} \:{is}\:{the}\:{resultant}\:{from}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} \\ $$$$ \\ $$$${T}_{{A}} ,\:{F}_{{A}} \:{and}\:{M}_{{A}} {g}\:{must}\:{be}\:{in}\:{the}\:{same} \\ $$$${plane}\:{and}\:{in}\:{equilibrium}. \\ $$$$\frac{{F}_{{A}} }{{d}_{{A}} }=\frac{{T}_{{A}} }{{a}}=\frac{{M}_{{A}} {g}}{{h}} \\ $$$$\Rightarrow{F}_{{A}} =\frac{{M}_{{A}} {gd}_{{A}} }{{h}},\:{T}_{{A}} =\frac{{M}_{{A}} {ga}}{{h}} \\ $$$${similarly} \\ $$$${F}_{{B}} =\frac{{M}_{{B}} {gd}_{{B}} }{{h}},\:{T}_{{B}} =\frac{{M}_{{B}} {gb}}{{h}} \\ $$$${F}_{{C}} =\frac{{M}_{{C}} {gd}_{{C}} }{{h}},\:{T}_{{C}} =\frac{{M}_{{C}} {gc}}{{h}} \\ $$$$ \\ $$$$\frac{{F}_{\mathrm{2}} }{\frac{{p}_{\mathrm{1}} }{{p}}×{q}}=\frac{{F}_{\mathrm{3}} }{\frac{{p}_{\mathrm{2}} }{{p}}×{r}}=\frac{{F}_{{A}} }{{m}_{{A}} }=\frac{{M}_{{A}} {gd}_{{A}} }{{hm}_{{A}} }=\frac{{M}_{{A}} \left({M}_{{B}} +{M}_{{C}} \right){g}}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}} \\ $$$$\Rightarrow{F}_{\mathrm{2}} =\frac{{M}_{{C}} {M}_{{A}} {gq}}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}}=\frac{{kQ}_{{C}} {Q}_{{A}} }{{q}^{\mathrm{2}} } \\ $$$$\Rightarrow{q}^{\mathrm{3}} =\frac{{kQ}_{{C}} {Q}_{{A}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}}{{gM}_{{C}} {M}_{{A}} } \\ $$$$\Rightarrow{F}_{\mathrm{3}} =\frac{{M}_{{A}} {M}_{{B}} {gr}}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}}=\frac{{kQ}_{{A}} {Q}_{{B}} }{{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}^{\mathrm{3}} =\frac{{kQ}_{{A}} {Q}_{{B}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}}{{gM}_{{A}} {M}_{{B}} } \\ $$$${similarly} \\ $$$$\Rightarrow{p}^{\mathrm{3}} =\frac{{kQ}_{{B}} {Q}_{{C}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right){h}}{{gM}_{{B}} {M}_{{C}} } \\ $$$${let}\: \\ $$$$\mu_{\mathrm{1}} =\sqrt{\frac{{kQ}_{{B}} {Q}_{{C}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)}{{gM}_{{B}} {M}_{{C}} }} \\ $$$$\mu_{\mathrm{2}} =\sqrt{\frac{{kQ}_{{C}} {Q}_{{A}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)}{{gM}_{{C}} {M}_{{A}} }} \\ $$$$\mu_{\mathrm{3}} =\sqrt{\frac{{kQ}_{{A}} {Q}_{{B}} \left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)}{{gM}_{{A}} {M}_{{B}} }} \\ $$$$\Rightarrow{p}=\sqrt[{\mathrm{3}}]{\mu_{\mathrm{1}} ^{\mathrm{2}} {h}},\:{q}=\sqrt[{\mathrm{3}}]{\mu_{\mathrm{2}} ^{\mathrm{2}} {h}},\:{r}=\sqrt[{\mathrm{3}}]{\mu_{\mathrm{3}} ^{\mathrm{2}} {h}} \\ $$$$ \\ $$$${SP}={h}_{{P}} \\ $$$${p}\left({h}_{{P}} ^{\mathrm{2}} +{p}_{\mathrm{1}} {p}_{\mathrm{2}} \right)={p}_{\mathrm{1}} {c}^{\mathrm{2}} +{p}_{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${h}_{{P}} ^{\mathrm{2}} +\frac{{M}_{{B}} {M}_{{C}} {p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }=\frac{{M}_{{C}} {c}^{\mathrm{2}} +{M}_{{B}} {b}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} } \\ $$$${h}_{{P}} ^{\mathrm{2}} =\frac{{M}_{{B}} {b}^{\mathrm{2}} +{M}_{{C}} {c}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} }−\frac{{M}_{{B}} {M}_{{C}} {p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$$${m}_{{A}} \left({h}^{\mathrm{2}} +{e}_{{A}} {d}_{{A}} \right)={e}_{{A}} {a}^{\mathrm{2}} +{d}_{{A}} ×{h}_{{P}} ^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} +{e}_{{A}} {d}_{{A}} =\frac{{M}_{{A}} {a}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} }+\frac{\left({M}_{{B}} +{M}_{{C}} \right){h}_{{P}} ^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$${h}^{\mathrm{2}} +\frac{{M}_{{A}} \left({M}_{{B}} +{M}_{{C}} \right){m}_{{A}} ^{\mathrm{2}} }{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }=\frac{{M}_{{A}} {a}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} }+\frac{\left({M}_{{B}} +{M}_{{C}} \right){h}_{{P}} ^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$${h}^{\mathrm{2}} +\frac{{M}_{{A}} \left({M}_{{B}} +{M}_{{C}} \right)}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }×\left[\frac{{M}_{{C}} \:{q}^{\mathrm{2}} +{M}_{{B}} \:{r}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} }−\frac{{M}_{{B}} {M}_{{C}} \:{p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }\right]=\frac{{M}_{{A}} {a}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} }+\frac{\left({M}_{{B}} +{M}_{{C}} \right)}{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} }×\left[\frac{{M}_{{B}} {b}^{\mathrm{2}} +{M}_{{C}} {c}^{\mathrm{2}} }{{M}_{{B}} +{M}_{{C}} }−\frac{{M}_{{B}} {M}_{{C}} {p}^{\mathrm{2}} }{\left({M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }\right] \\ $$$${h}^{\mathrm{2}} +\frac{{M}_{{B}} {M}_{{C}} {p}^{\mathrm{2}} +{M}_{{C}} {M}_{{A}} \:{q}^{\mathrm{2}} +{M}_{{A}} {M}_{{B}} \:{r}^{\mathrm{2}} }{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }=\frac{{M}_{{A}} {a}^{\mathrm{2}} +{M}_{{B}} {b}^{\mathrm{2}} +{M}_{{C}} {c}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$${h}^{\mathrm{2}} +\frac{\left({M}_{{B}} {M}_{{C}} \sqrt[{\mathrm{3}}]{\mu_{\mathrm{1}} ^{\mathrm{4}} }+{M}_{{C}} {M}_{{A}} \:\sqrt[{\mathrm{3}}]{\mu_{\mathrm{2}} ^{\mathrm{4}} }+{M}_{{A}} {M}_{{B}} \:\sqrt[{\mathrm{3}}]{\mu_{\mathrm{3}} ^{\mathrm{4}} }\right)\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} }−\frac{{M}_{{A}} {a}^{\mathrm{2}} +{M}_{{B}} {b}^{\mathrm{2}} +{M}_{{C}} {c}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} }=\mathrm{0} \\ $$$${let}\: \\ $$$$\xi=\frac{{M}_{{B}} {M}_{{C}} \sqrt[{\mathrm{3}}]{\mu_{\mathrm{1}} ^{\mathrm{4}} }+{M}_{{C}} {M}_{{A}} \:\sqrt[{\mathrm{3}}]{\mu_{\mathrm{2}} ^{\mathrm{4}} }+{M}_{{A}} {M}_{{B}} \:\sqrt[{\mathrm{3}}]{\mu_{\mathrm{3}} ^{\mathrm{4}} }}{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{2}} } \\ $$$${or} \\ $$$$\xi=\sqrt[{\mathrm{3}}]{\frac{{M}_{{A}} {M}_{{B}} {M}_{{C}} }{\left({M}_{{A}} +{M}_{{B}} +{M}_{{C}} \right)^{\mathrm{4}} }×\left(\frac{{kQ}_{{A}} {Q}_{{B}} {Q}_{{C}} }{{g}}\right)^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{M}_{{A}} {Q}_{{A}} ^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{M}_{{B}} {Q}_{{B}} ^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{M}_{{C}} {Q}_{{C}} ^{\mathrm{2}} }}\right] \\ $$$$\delta=\frac{{M}_{{A}} {a}^{\mathrm{2}} +{M}_{{B}} {b}^{\mathrm{2}} +{M}_{{C}} {c}^{\mathrm{2}} }{{M}_{{A}} +{M}_{{B}} +{M}_{{C}} } \\ $$$$\left(\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }\right)^{\mathrm{3}} +\xi\left(\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }\right)−\delta=\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{{h}^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\delta^{\mathrm{2}} }{\mathrm{4}}+\frac{\xi^{\mathrm{3}} }{\mathrm{27}}}+\frac{\delta}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\delta^{\mathrm{2}} }{\mathrm{4}}+\frac{\xi^{\mathrm{3}} }{\mathrm{27}}}−\frac{\delta}{\mathrm{2}}} \\ $$$$\Rightarrow{h}=\left[\sqrt[{\mathrm{3}}]{\sqrt{\frac{\delta^{\mathrm{2}} }{\mathrm{4}}+\frac{\xi^{\mathrm{3}} }{\mathrm{27}}}+\frac{\delta}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\delta^{\mathrm{2}} }{\mathrm{4}}+\frac{\xi^{\mathrm{3}} }{\mathrm{27}}}−\frac{\delta}{\mathrm{2}}}\right]^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\frac{{T}_{{A}} }{{M}_{{A}} {g}}=\frac{{a}}{{h}} \\ $$$$\frac{{T}_{{B}} }{{M}_{{B}} {g}}=\frac{{b}}{{h}} \\ $$$$\frac{{T}_{{C}} }{{M}_{{C}} {g}}=\frac{{c}}{{h}} \\ $$
Commented by ajfour last updated on 28/Jan/24
$${This}\:{is}\:{beyond}\:{praise}!\:{Sir}.\:{The}\:{idea} \\ $$$${effort}\:{n}\:{presentation}\:{all}.\:{Thanks}. \\ $$