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Question Number 203726 by mr W last updated on 27/Jan/24

Commented by mr W last updated on 26/Jan/24

$$\text{You can't use 'macro parameter character \#' in math mode}$$

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

$$thecenterofmassofthethree$$$$chargesisG,whichincurrentcase$$$$mustbethecentroidof\mathrm{\Delta }ABC,$$$$sincetheyhavethesamemass.$$$$Gmustlieverticallyunderthe$$$$hangingpointSontheceiling.$$$$saythesidesof\mathrm{\Delta }ABCarep,q,r$$$$respectively.$$$$m_A=AP=\frac{\sqrt{2(q^2+r^2)-p^2}}{2}$$$$m_B=BQ=\frac{\sqrt{2(r^2+p^2)-q^2}}{2}$$$$m_C=CR=\frac{\sqrt{2(p^2+q^2)-r^2}}{2}$$$$d_A=\frac{2}{3}{m}_A=distancefromGtoA$$$$d_B=\frac{2}{3}{m}_B=distancefromGtoB$$$$d_C=\frac{2}{3}{m}_C=distancefromGtoC$$$$saySG=h,SGisvertical.$$$$F_1=\frac{{kQ}^2}{p^2}withQ=charge,k=Coulombconstant$$$$F_2=\frac{{kQ}^2}{q^2}$$$$F_3=\frac{{kQ}^2}{r^2}$$$$say$$$$F_{A}istheresultantfrom{F}_{2}and{F}_3$$$$F_{B}istheresultantfrom{F}_{3}and{F}_1$$$$F_{C}istheresultantfrom{F}_{1}and{F}_2$$$$$$$$T_A,F_{A}andmgmustbeinthesame$$$$planeandinequilibrium.$$$$\frac{F_A}{d_A}=\frac{T_A}{a}=\frac{mg}{h}$$$$\Rightarrow F_A=\frac{{mgd}_A}{h},T_A=\frac{mga}{h}$$$$similarly$$$$F_B=\frac{{mgd}_B}{h},T_B=\frac{mgb}{h}$$$$F_C=\frac{{mgd}_C}{h},T_C=\frac{mgc}{h}$$$$$$$$\frac{F_2}{0.5q}=\frac{F_3}{0.5r}=\frac{F_A}{m_A}=\frac{{mgd}_A}{{hm}_A}=\frac{2mg}{3h}$$$$F_2=\frac{mgq}{3h}=\frac{{kQ}^2}{q^2}\Rightarrow q=\sqrt[3]{\frac{3{kQ}^{2}h}{mg}}$$$$F_3=\frac{mgr}{h}=\frac{{kQ}^2}{r^2}\Rightarrow r=\sqrt[3]{\frac{3{kQ}^{2}h}{mg}}$$$$similarly$$$$F_1=\frac{mgp}{h}=\frac{{kQ}^2}{p^2}\Rightarrow p=\sqrt[3]{\frac{3{kQ}^{2}h}{mg}}$$$$thatmeansp=q=r=\sqrt[3]{\frac{3{kQ}^{2}h}{mg}}=\sqrt[3]{3{\mu }^{2}h}$$$$with\mu =\sqrt{\frac{{kQ}^2}{mg}}$$$$$$$$SP=h_P=\frac{\sqrt{2(b^2+c^2)-p^2}}{2}$$$$m_A(h^2+\frac{1}{3}{m}_A\times \frac{2}{3}{m}_A)=\frac{m_A}{3}\times a^2+\frac{2m_A}{3}\times h_P^2$$$$h^2+\frac{2}{9}{m}_A^2=\frac{a^2}{3}+\frac{2h_P^2}{3}$$$$h^2+\frac{2}{9}\times \frac{2(q^2+r^2)-p^2}{4}=\frac{a^2}{3}+\frac{2(b^2+c^2)-p^2}{6}$$$$h^2+\frac{p^2+q^2+r^2}{9}=\frac{a^2+b^2+c^2}{3}$$$$9h^2+p^2+q^2+r^2=3(a^2+b^2+c^2)$$$$9h^2+3\sqrt[3]{9{\mu }^4{h}^2}-3(a^2+b^2+c^2)=0$$$$let\delta =\sqrt{a^2+b^2+c^2}$$$${\left(\sqrt[3]{h^2}\right)}^3+\frac{1}{3}\sqrt[3]{9{\mu }^4}\left(\sqrt[3]{h^2}\right)-\frac{{\delta }^2}{3}=0$$$$\Rightarrow \sqrt[3]{h^2}=\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}+\frac{{\delta }^2}{6}}-\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}-\frac{{\delta }^2}{6}}$$$$\Rightarrow h={[\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}+\frac{{\delta }^2}{6}}-\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}-\frac{{\delta }^2}{6}}]}^{\frac{3}{2}}$$$$\frac{T_A}{mg}=\frac{a}{h}=\frac{a}{{[\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}+\frac{{\delta }^2}{6}}-\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}-\frac{{\delta }^2}{6}}]}^{\frac{3}{2}}}$$$$\frac{T_B}{mg}=\frac{b}{h}=\frac{b}{{[\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}+\frac{{\delta }^2}{6}}-\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}-\frac{{\delta }^2}{6}}]}^{\frac{3}{2}}}$$$$\frac{T_C}{mg}=\frac{c}{h}=\frac{c}{{[\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}+\frac{{\delta }^2}{6}}-\sqrt[3]{\frac{1}{3}\sqrt{\frac{{\delta }^4}{4}+\frac{{\mu }^4}{9}}-\frac{{\delta }^2}{6}}]}^{\frac{3}{2}}}$$$$$$$$specialcase:a=b=c=s$$$$T_A=T_B=T_C=T$$$$\frac{T}{mg}=\frac{s}{{[\sqrt[3]{\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}+\frac{s^2}{2}}-\sqrt[3]{\sqrt{\frac{s^4}{4}+\frac{{\mu }^4}{81}}-\frac{s^2}{2}}]}^{\frac{3}{2}}}$$

Commented by ajfour last updated on 26/Jan/24

$$Sirpleasetakea=b=c=sand$$$$N=3,seeifitmatcheswithresult$$$$ofq.1AmazingExcellentsolution,$$$$mnotthroughyet.$$

Commented by mr W last updated on 26/Jan/24

$$thankyouforviewingsir!$$$$i^{\prime }llchecklater.$$

Commented by mr W last updated on 27/Jan/24

$$ihavefixedabigerror.$$$$forthespecialcasewitha=b=c=s,$$$$theresultisthesameasinquestion1$$$$withn=3.$$$${it}^{\prime }sinterestingtonoticethat\mathrm{\Delta }ABC$$$$isalwaysequilateral,i.e.p=q=r.$$

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

$$Generalcase:$$$$theparticleshavedifferentmasses$$$$andcharges.$$

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

$$saythecenterofmassofthethree$$$$chargesisG,whichmustlie$$$$verticallyunderthehangingpointS$$$$ontheceiling.$$$$saythesidesof\mathrm{\Delta }ABCarep,q,r$$$$respectively.$$$$\frac{p_1}{p}=\frac{M_C}{M_B+M_C}\Rightarrow p_1=\frac{M_{C}p}{M_B+M_C}$$$$\frac{p_2}{p}=\frac{M_B}{M_B+M_C}\Rightarrow p_2=\frac{M_{B}p}{M_B+M_C}$$$$m_A=AP$$$$p(m_A^2+p_1{p}_2)=p_1{q}^2+p_2{r}^2$$$$m_A^2+\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}=\frac{M_C{q}^2+M_B{r}^2}{M_B+M_C}$$$$\Rightarrow m_A^2=\frac{M_C{q}^2+M_B{r}^2}{M_B+M_C}-\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}$$$$\frac{d_A}{m_A}=\frac{M_B+M_C}{M_A+M_B+M_C}$$$$\frac{e_A}{m_A}=\frac{M_A}{M_A+M_B+M_C}$$$$$$$$saySG=h,SGisvertical.$$$$F_1=\frac{{kQ}_B{Q}_C}{p^2}withk=Coulombconstant$$$$F_2=\frac{{kQ}_C{Q}_A}{q^2}$$$$F_3=\frac{{kQ}_A{Q}_B}{r^2}$$$$say$$$$F_{A}istheresultantfrom{F}_{2}and{F}_3$$$$F_{B}istheresultantfrom{F}_{3}and{F}_1$$$$F_{C}istheresultantfrom{F}_{1}and{F}_2$$$$$$$$T_A,F_{A}and{M}_{A}gmustbeinthesame$$$$planeandinequilibrium.$$$$\frac{F_A}{d_A}=\frac{T_A}{a}=\frac{M_{A}g}{h}$$$$\Rightarrow F_A=\frac{M_A{gd}_A}{h},T_A=\frac{M_{A}ga}{h}$$$$similarly$$$$F_B=\frac{M_B{gd}_B}{h},T_B=\frac{M_{B}gb}{h}$$$$F_C=\frac{M_C{gd}_C}{h},T_C=\frac{M_{C}gc}{h}$$$$$$$$\frac{F_2}{\frac{p_1}{p}\times q}=\frac{F_3}{\frac{p_2}{p}\times r}=\frac{F_A}{m_A}=\frac{M_A{gd}_A}{{hm}_A}=\frac{M_A(M_B+M_C)g}{(M_A+M_B+M_C)h}$$$$\Rightarrow F_2=\frac{M_C{M}_{A}gq}{(M_A+M_B+M_C)h}=\frac{{kQ}_C{Q}_A}{q^2}$$$$\Rightarrow q^3=\frac{{kQ}_C{Q}_A(M_A+M_B+M_C)h}{{gM}_C{M}_A}$$$$\Rightarrow F_3=\frac{M_A{M}_{B}gr}{(M_A+M_B+M_C)h}=\frac{{kQ}_A{Q}_B}{r^2}$$$$\Rightarrow r^3=\frac{{kQ}_A{Q}_B(M_A+M_B+M_C)h}{{gM}_A{M}_B}$$$$similarly$$$$\Rightarrow p^3=\frac{{kQ}_B{Q}_C(M_A+M_B+M_C)h}{{gM}_B{M}_C}$$$$let$$$${\mu }_1=\sqrt{\frac{{kQ}_B{Q}_C(M_A+M_B+M_C)}{{gM}_B{M}_C}}$$$${\mu }_2=\sqrt{\frac{{kQ}_C{Q}_A(M_A+M_B+M_C)}{{gM}_C{M}_A}}$$$${\mu }_3=\sqrt{\frac{{kQ}_A{Q}_B(M_A+M_B+M_C)}{{gM}_A{M}_B}}$$$$\Rightarrow p=\sqrt[3]{{\mu }_1^{2}h},q=\sqrt[3]{{\mu }_2^{2}h},r=\sqrt[3]{{\mu }_3^{2}h}$$$$$$$$SP=h_P$$$$p(h_P^2+p_1{p}_2)=p_1{c}^2+p_2{b}^2$$$$h_P^2+\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}=\frac{M_C{c}^2+M_B{b}^2}{M_B+M_C}$$$$h_P^2=\frac{M_B{b}^2+M_C{c}^2}{M_B+M_C}-\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}$$$$$$$$m_A(h^2+e_A{d}_A)=e_A{a}^2+d_A\times h_P^2$$$$h^2+e_A{d}_A=\frac{M_A{a}^2}{M_A+M_B+M_C}+\frac{(M_B+M_C)h_P^2}{M_A+M_B+M_C}$$$$h^2+\frac{M_A(M_B+M_C)m_A^2}{{(M_A+M_B+M_C)}^2}=\frac{M_A{a}^2}{M_A+M_B+M_C}+\frac{(M_B+M_C)h_P^2}{M_A+M_B+M_C}$$$$h^2+\frac{M_A(M_B+M_C)}{{(M_A+M_B+M_C)}^2}\times [\frac{M_C{q}^2+M_B{r}^2}{M_B+M_C}-\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}]=\frac{M_A{a}^2}{M_A+M_B+M_C}+\frac{(M_B+M_C)}{M_A+M_B+M_C}\times [\frac{M_B{b}^2+M_C{c}^2}{M_B+M_C}-\frac{M_B{M}_C{p}^2}{{(M_B+M_C)}^2}]$$$$h^2+\frac{M_B{M}_C{p}^2+M_C{M}_A{q}^2+M_A{M}_B{r}^2}{{(M_A+M_B+M_C)}^2}=\frac{M_A{a}^2+M_B{b}^2+M_C{c}^2}{M_A+M_B+M_C}$$$$h^2+\frac{(M_B{M}_C\sqrt[3]{{\mu }_1^4}+M_C{M}_A\sqrt[3]{{\mu }_2^4}+M_A{M}_B\sqrt[3]{{\mu }_3^4})\sqrt[3]{h^2}}{{(M_A+M_B+M_C)}^2}-\frac{M_A{a}^2+M_B{b}^2+M_C{c}^2}{M_A+M_B+M_C}=0$$$$let$$$$\xi =\frac{M_B{M}_C\sqrt[3]{{\mu }_1^4}+M_C{M}_A\sqrt[3]{{\mu }_2^4}+M_A{M}_B\sqrt[3]{{\mu }_3^4}}{{(M_A+M_B+M_C)}^2}$$$$or$$$$\xi =\sqrt[3]{\frac{M_A{M}_B{M}_C}{{(M_A+M_B+M_C)}^4}\times {\left(\frac{{kQ}_A{Q}_B{Q}_C}{g}\right)}^2}[\frac{1}{\sqrt[3]{M_A{Q}_A^2}}+\frac{1}{\sqrt[3]{M_B{Q}_B^2}}+\frac{1}{\sqrt[3]{M_C{Q}_C^2}}]$$$$\delta =\frac{M_A{a}^2+M_B{b}^2+M_C{c}^2}{M_A+M_B+M_C}$$$${\left(\sqrt[3]{h^2}\right)}^3+\xi \left(\sqrt[3]{h^2}\right)-\delta =0$$$$\Rightarrow \sqrt[3]{h^2}=\sqrt[3]{\sqrt{\frac{{\delta }^2}{4}+\frac{{\xi }^3}{27}}+\frac{\delta }{2}}-\sqrt[3]{\sqrt{\frac{{\delta }^2}{4}+\frac{{\xi }^3}{27}}-\frac{\delta }{2}}$$$$\Rightarrow h={[\sqrt[3]{\sqrt{\frac{{\delta }^2}{4}+\frac{{\xi }^3}{27}}+\frac{\delta }{2}}-\sqrt[3]{\sqrt{\frac{{\delta }^2}{4}+\frac{{\xi }^3}{27}}-\frac{\delta }{2}}]}^{\frac{3}{2}}$$$$\frac{T_A}{M_{A}g}=\frac{a}{h}$$$$\frac{T_B}{M_{B}g}=\frac{b}{h}$$$$\frac{T_C}{M_{C}g}=\frac{c}{h}$$

Commented by ajfour last updated on 28/Jan/24

$$Thisisbeyondpraise!Sir.Theidea$$$$effortnpresentationall.Thanks.$$

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