Question-203726

Tinku Tara January 26, 2024 Electrostatics 0 Comments

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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

solution to Q#203636

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

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) ))

t h e c e n t e r o f m a s s o f t h e t h r e e

c h a r g e s i s G, w h i c h i n c u r r e n t c a s e

m u s t b e t h e c e n t r o i d o f Δ A B C,

s i n c e t h e y h a v e t h e s a m e m a s s .

G m u s t l i e v e r t i c a l l y u n d e r t h e

h a n g i n g p o i n t S o n t h e c e i l i n g .

s a y t h e s i d e s o f Δ A B C a r e p, q, r

r e s p e c t i v e l y .

m_{A} = A P = \frac{\sqrt{2 (q^{2} + r^{2}) - p^{2}}}{2}

m_{B} = B Q = \frac{\sqrt{2 (r^{2} + p^{2}) - q^{2}}}{2}

m_{C} = C R = \frac{\sqrt{2 (p^{2} + q^{2}) - r^{2}}}{2}

d_{A} = \frac{2}{3} m_{A} = d i s t a n c e f r o m G t o A

d_{B} = \frac{2}{3} m_{B} = d i s t a n c e f r o m G t o B

d_{C} = \frac{2}{3} m_{C} = d i s t a n c e f r o m G t o C

s a y S G = h, S G i s v e r t i c a l .

F_{1} = \frac{{k Q}^{2}}{p^{2}} w i t h Q = c h a r g e, k = C o u l o m b c o n s t a n t

F_{2} = \frac{{k Q}^{2}}{q^{2}}

F_{3} = \frac{{k Q}^{2}}{r^{2}}

s a y

F_{A} i s t h e r e s u l t a n t f r o m F_{2} a n d F_{3}

F_{B} i s t h e r e s u l t a n t f r o m F_{3} a n d F_{1}

F_{C} i s t h e r e s u l t a n t f r o m F_{1} a n d F_{2}

T_{A}, F_{A} a n d m g m u s t b e i n t h e s a m e

p l a n e a n d i n e q u i l i b r i u m .

\frac{F_{A}}{d_{A}} = \frac{T_{A}}{a} = \frac{m g}{h}

\Rightarrow F_{A} = \frac{{m g d}_{A}}{h}, T_{A} = \frac{m g a}{h}

s i m i l a r l y

F_{B} = \frac{{m g d}_{B}}{h}, T_{B} = \frac{m g b}{h}

F_{C} = \frac{{m g d}_{C}}{h}, T_{C} = \frac{m g c}{h}

\frac{F_{2}}{0.5 q} = \frac{F_{3}}{0.5 r} = \frac{F_{A}}{m_{A}} = \frac{{m g d}_{A}}{{h m}_{A}} = \frac{2 m g}{3 h}

F_{2} = \frac{m g q}{3 h} = \frac{{k Q}^{2}}{q^{2}} \Rightarrow q = \sqrt[3]{\frac{3 {k Q}^{2} h}{m g}}

F_{3} = \frac{m g r}{h} = \frac{{k Q}^{2}}{r^{2}} \Rightarrow r = \sqrt[3]{\frac{3 {k Q}^{2} h}{m g}}

s i m i l a r l y

F_{1} = \frac{m g p}{h} = \frac{{k Q}^{2}}{p^{2}} \Rightarrow p = \sqrt[3]{\frac{3 {k Q}^{2} h}{m g}}

t h a t m e a n s p = q = r = \sqrt[3]{\frac{3 {k Q}^{2} h}{m g}} = \sqrt[3]{3 μ^{2} h}

w i t h μ = \sqrt{\frac{{k Q}^{2}}{m g}}

S P = 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{2 m_{A}}{3} \times h_{P}^{2}

h^{2} + \frac{2}{9} m_{A}^{2} = \frac{a^{2}}{3} + \frac{2 h_{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}

9 h^{2} + p^{2} + q^{2} + r^{2} = 3 (a^{2} + b^{2} + c^{2})

9 h^{2} + 3 \sqrt[3]{9 μ^{4} h^{2}} - 3 (a^{2} + b^{2} + c^{2}) = 0

l e t δ = \sqrt{a^{2} + b^{2} + c^{2}}

{(\sqrt[3]{h^{2}})}^{3} + \frac{1}{3} \sqrt[3]{9 μ^{4}} (\sqrt[3]{h^{2}}) - \frac{δ^{2}}{3} = 0

\Rightarrow \sqrt[3]{h^{2}} = \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} + \frac{δ^{2}}{6}} - \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} - \frac{δ^{2}}{6}}

\Rightarrow h = {[\sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} + \frac{δ^{2}}{6}} - \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} - \frac{δ^{2}}{6}}]}^{\frac{3}{2}}

\frac{T_{A}}{m g} = \frac{a}{h} = \frac{a}{{[\sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} + \frac{δ^{2}}{6}} - \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} - \frac{δ^{2}}{6}}]}^{\frac{3}{2}}}

\frac{T_{B}}{m g} = \frac{b}{h} = \frac{b}{{[\sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} + \frac{δ^{2}}{6}} - \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} - \frac{δ^{2}}{6}}]}^{\frac{3}{2}}}

\frac{T_{C}}{m g} = \frac{c}{h} = \frac{c}{{[\sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} + \frac{δ^{2}}{6}} - \sqrt[3]{\frac{1}{3} \sqrt{\frac{δ^{4}}{4} + \frac{μ^{4}}{9}} - \frac{δ^{2}}{6}}]}^{\frac{3}{2}}}

s p e c i a l c a s e : a = b = c = s

T_{A} = T_{B} = T_{C} = T

\frac{T}{m g} = \frac{s}{{[\sqrt[3]{\sqrt{\frac{s^{4}}{4} + \frac{μ^{4}}{81}} + \frac{s^{2}}{2}} - \sqrt[3]{\sqrt{\frac{s^{4}}{4} + \frac{μ^{4}}{81}} - \frac{s^{2}}{2}}]}^{\frac{3}{2}}}

Commented by ajfour last updated on 26/Jan/24

Sir please take a=b=c=s and N=3, see if it matches with result of q.1 Amazing Excellent solution, m not through yet.

S i r p l e a s e t a k e a = b = c = s a n d

N = 3, s e e i f i t m a t c h e s w i t h r e s u l t

o f q . 1 A m a z i n g E x c e l l e n t s o l u t i o n,

m n o t t h r o u g h y e t .

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

thank you for viewing sir! i′ll check later.

t h a n k y o u f o r v i e w i n g s i r!

i^{'} l l c h e c k l a t e r .

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 1 with n=3. it′s interesting to notice that ΔABC is always equilateral, i.e. p=q=r.

i h a v e f i x e d a b i g e r r o r .

f o r t h e s p e c i a l c a s e w i t h a = b = c = s,

t h e r e s u l t i s t h e s a m e a s i n q u e s t i o n 1

w i t h n = 3 .

{i t}^{'} s i n t e r e s t i n g t o n o t i c e t h a t Δ A B C

i s a l w a y s e q u i l a t e r a l, 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.

G e n e r a l c a s e :

t h e p a r t i c l e s h a v e d i f f e r e n t m a s s e s

a n d c h a r g e s .

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)

s a y t h e c e n t e r o f m a s s o f t h e t h r e e

c h a r g e s i s G, w h i c h m u s t l i e

v e r t i c a l l y u n d e r t h e h a n g i n g p o i n t S

o n t h e c e i l i n g .

s a y t h e s i d e s o f Δ A B C a r e p, q, r

r e s p e c t i v e l y .

\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} = A P

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}}

s a y S G = h, S G i s v e r t i c a l .

F_{1} = \frac{{k Q}_{B} Q_{C}}{p^{2}} w i t h k = C o u l o m b c o n s t a n t

F_{2} = \frac{{k Q}_{C} Q_{A}}{q^{2}}

F_{3} = \frac{{k Q}_{A} Q_{B}}{r^{2}}

s a y

F_{A} i s t h e r e s u l t a n t f r o m F_{2} a n d F_{3}

F_{B} i s t h e r e s u l t a n t f r o m F_{3} a n d F_{1}

F_{C} i s t h e r e s u l t a n t f r o m F_{1} a n d F_{2}

T_{A}, F_{A} a n d M_{A} g m u s t b e i n t h e s a m e

p l a n e a n d i n e q u i l i b r i u m .

\frac{F_{A}}{d_{A}} = \frac{T_{A}}{a} = \frac{M_{A} g}{h}

\Rightarrow F_{A} = \frac{M_{A} {g d}_{A}}{h}, T_{A} = \frac{M_{A} g a}{h}

s i m i l a r l y

F_{B} = \frac{M_{B} {g d}_{B}}{h}, T_{B} = \frac{M_{B} g b}{h}

F_{C} = \frac{M_{C} {g d}_{C}}{h}, T_{C} = \frac{M_{C} g c}{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} {g d}_{A}}{{h m}_{A}} = \frac{M_{A} (M_{B} + M_{C}) g}{(M_{A} + M_{B} + M_{C}) h}

\Rightarrow F_{2} = \frac{M_{C} M_{A} g q}{(M_{A} + M_{B} + M_{C}) h} = \frac{{k Q}_{C} Q_{A}}{q^{2}}

\Rightarrow q^{3} = \frac{{k Q}_{C} Q_{A} (M_{A} + M_{B} + M_{C}) h}{{g M}_{C} M_{A}}

\Rightarrow F_{3} = \frac{M_{A} M_{B} g r}{(M_{A} + M_{B} + M_{C}) h} = \frac{{k Q}_{A} Q_{B}}{r^{2}}

\Rightarrow r^{3} = \frac{{k Q}_{A} Q_{B} (M_{A} + M_{B} + M_{C}) h}{{g M}_{A} M_{B}}

s i m i l a r l y

\Rightarrow p^{3} = \frac{{k Q}_{B} Q_{C} (M_{A} + M_{B} + M_{C}) h}{{g M}_{B} M_{C}}

l e t

μ_{1} = \sqrt{\frac{{k Q}_{B} Q_{C} (M_{A} + M_{B} + M_{C})}{{g M}_{B} M_{C}}}

μ_{2} = \sqrt{\frac{{k Q}_{C} Q_{A} (M_{A} + M_{B} + M_{C})}{{g M}_{C} M_{A}}}

μ_{3} = \sqrt{\frac{{k Q}_{A} Q_{B} (M_{A} + M_{B} + M_{C})}{{g M}_{A} M_{B}}}

\Rightarrow p = \sqrt[3]{μ_{1}^{2} h}, q = \sqrt[3]{μ_{2}^{2} h}, r = \sqrt[3]{μ_{3}^{2} h}

S P = 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]{μ_{1}^{4}} + M_{C} M_{A} \sqrt[3]{μ_{2}^{4}} + M_{A} M_{B} \sqrt[3]{μ_{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

l e t

ξ = \frac{M_{B} M_{C} \sqrt[3]{μ_{1}^{4}} + M_{C} M_{A} \sqrt[3]{μ_{2}^{4}} + M_{A} M_{B} \sqrt[3]{μ_{3}^{4}}}{{(M_{A} + M_{B} + M_{C})}^{2}}

o r

ξ = \sqrt[3]{\frac{M_{A} M_{B} M_{C}}{{(M_{A} + M_{B} + M_{C})}^{4}} \times {(\frac{{k Q}_{A} Q_{B} Q_{C}}{g})}^{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}}}]

δ = \frac{M_{A} a^{2} + M_{B} b^{2} + M_{C} c^{2}}{M_{A} + M_{B} + M_{C}}

{(\sqrt[3]{h^{2}})}^{3} + ξ (\sqrt[3]{h^{2}}) - δ = 0

\Rightarrow \sqrt[3]{h^{2}} = \sqrt[3]{\sqrt{\frac{δ^{2}}{4} + \frac{ξ^{3}}{27}} + \frac{δ}{2}} - \sqrt[3]{\sqrt{\frac{δ^{2}}{4} + \frac{ξ^{3}}{27}} - \frac{δ}{2}}

\Rightarrow h = {[\sqrt[3]{\sqrt{\frac{δ^{2}}{4} + \frac{ξ^{3}}{27}} + \frac{δ}{2}} - \sqrt[3]{\sqrt{\frac{δ^{2}}{4} + \frac{ξ^{3}}{27}} - \frac{δ}{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

This is beyond praise! Sir. The idea effort n presentation all. Thanks.

T h i s i s b e y o n d p r a i s e! S i r . T h e i d e a

e f f o r t n p r e s e n t a t i o n a l l . T h a n k s .

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