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Question-226785

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Question Number 226785 by fantastic2 last updated on 14/Dec/25
Commented by fantastic2 last updated on 15/Dec/25
there are three small balls of mass m_(1,) m_2 and m_3 all are hanging from a point O by a string length l . what charge should be given to them so that they make angle α with the vertical going through point O?
$${there}\:{are}\:{three}\:{small}\:{balls}\:{of}\:{mass}\:{m}_{\mathrm{1},} {m}_{\mathrm{2}} \:{and}\:{m}_{\mathrm{3}} \\ $$$${all}\:{are}\:{hanging}\:{from}\:{a}\:{point}\:{O}\:{by}\:{a}\:{string} \\ $$$${length}\:{l}\:. \\ $$$${what}\:{charge}\:{should}\:{be}\:{given}\:{to}\:{them}\:{so} \\ $$$${that}\:{they}\:{make}\:{angle}\:\alpha\:{with}\:{the}\:{vertical} \\ $$$${going}\:{through}\:{point}\:{O}? \\ $$
Commented by fantastic2 last updated on 14/Dec/25
learnt some new things today :)
$$\left.{learnt}\:{some}\:{new}\:{things}\:{today}\::\right) \\ $$$$ \\ $$
Commented by fantastic2 last updated on 14/Dec/25
in the vector representation of Columbs law F_(12) ^→ =((q_1 q_2 )/(4πκε_0 r_(21) ^2 ))r_(21) ^� =((q_1 q_2 )/(4πκε_0 r_(21) ^3 ))r_(21) ^→ and F_(21) ^→ =((q_1 q_2 )/(4πκε_0 r_(12) ^2 ))r_(12) ^� =((q_1 q_2 )/(4πκε_0 r_(12) ^2 ))r_(12) ^→ what does F_(12) ,F_(21) ,r_(12,) r_(21) mean and how r_(12) ^→ =r_(12) r_(12) ^� vise versa pleas explain especially the last one
$${in}\:{the}\:{vector}\:{representation}\:{of}\:{Columbs} \\ $$$${law} \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{12}} =\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{\mathrm{4}\pi\kappa\epsilon_{\mathrm{0}} {r}_{\mathrm{21}} ^{\mathrm{2}} }\hat {{r}}_{\mathrm{21}} =\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{\mathrm{4}\pi\kappa\epsilon_{\mathrm{0}} {r}_{\mathrm{21}} ^{\mathrm{3}} }\overset{\rightarrow} {{r}}_{\mathrm{21}} \\ $$$${and} \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{21}} =\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{\mathrm{4}\pi\kappa\epsilon_{\mathrm{0}} {r}_{\mathrm{12}} ^{\mathrm{2}} }\hat {{r}}_{\mathrm{12}} =\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{\mathrm{4}\pi\kappa\epsilon_{\mathrm{0}} {r}_{\mathrm{12}} ^{\mathrm{2}} }\overset{\rightarrow} {{r}}_{\mathrm{12}} \\ $$$${what}\:{does}\:{F}_{\mathrm{12}} ,{F}_{\mathrm{21}} ,{r}_{\mathrm{12},} {r}_{\mathrm{21}} \:{mean} \\ $$$${and}\:{how}\:\overset{\rightarrow} {{r}}_{\mathrm{12}} ={r}_{\mathrm{12}} \hat {{r}}_{\mathrm{12}} \:{vise}\:{versa} \\ $$$${pleas}\:{explain}\:{especially}\:{the}\:{last}\:{one} \\ $$
Commented by mr W last updated on 15/Dec/25
r_(12) =distance between q_1 to q_2 (scalar) r_(12) ^→ =vector from q_1 to q_2 , ∣r_(12) ^(→) ∣=∣r_(21) ^(→) ∣=r_(12) r_(12) ^� =unit vector from q_1 to q_2 , ∣r_(12) ^� ∣=∣r_(21) ^� ∣=1 r_(12) ^→ =r_(12) r_(12) ^� , r_(21) ^(→) =r_(12) r_(21) ^�
$${r}_{\mathrm{12}} ={distance}\:{between}\:{q}_{\mathrm{1}} \:{to}\:{q}_{\mathrm{2}} \:\left({scalar}\right) \\ $$$$\overset{\rightarrow} {{r}}_{\mathrm{12}} ={vector}\:{from}\:{q}_{\mathrm{1}} \:{to}\:{q}_{\mathrm{2}} ,\:\mid\overset{\rightarrow} {{r}_{\mathrm{12}} }\mid=\mid\overset{\rightarrow} {{r}_{\mathrm{21}} }\mid={r}_{\mathrm{12}} \\ $$$$\hat {{r}}_{\mathrm{12}} ={unit}\:{vector}\:{from}\:{q}_{\mathrm{1}} \:{to}\:{q}_{\mathrm{2}} ,\:\mid\hat {{r}}_{\mathrm{12}} \mid=\mid\hat {{r}}_{\mathrm{21}} \mid=\mathrm{1} \\ $$$$\overset{\rightarrow} {{r}}_{\mathrm{12}} ={r}_{\mathrm{12}} \hat {{r}}_{\mathrm{12}} ,\:\overset{\rightarrow} {{r}_{\mathrm{21}} }={r}_{\mathrm{12}} \hat {{r}}_{\mathrm{21}} \\ $$
Commented by fantastic2 last updated on 15/Dec/25
thanks!
$${thanks}! \\ $$
Commented by fantastic2 last updated on 15/Dec/25
(given: it makes an equilateral triangle when the point masses are connected)
$$\left({given}:\:{it}\:{makes}\:{an}\:{equilateral}\:{triangle}\right. \\ $$$$\left.{when}\:{the}\:{point}\:{masses}\:{are}\:{connected}\right) \\ $$
Commented by mr W last updated on 15/Dec/25
Commented by mr W last updated on 15/Dec/25
see a more interesting general case in Q203726
$${see}\:{a}\:{more}\:{interesting}\:{general}\:{case} \\ $$$${in}\:{Q}\mathrm{203726} \\ $$
Commented by mr W last updated on 15/Dec/25
we got for m_1 =m_2 =m_3 =m, a=b=c=l: ⇒q=l sin α(√(((√3)mg tan α)/k))
$${we}\:{got}\:{for}\:{m}_{\mathrm{1}} ={m}_{\mathrm{2}} ={m}_{\mathrm{3}} ={m},\:{a}={b}={c}={l}: \\ $$$$\Rightarrow{q}={l}\:\mathrm{sin}\:\alpha\sqrt{\frac{\sqrt{\mathrm{3}}{mg}\:\mathrm{tan}\:\alpha}{{k}}} \\ $$
Commented by fantastic2 last updated on 15/Dec/25
wonderful !have no words
$${wonderful}\:!{have}\:{no}\:{words} \\ $$
Commented by fantastic2 last updated on 15/Dec/25
i got the same. wow
$${i}\:{got}\:{the}\:{same}.\:{wow} \\ $$
Commented by fantastic2 last updated on 15/Dec/25
Commented by fantastic2 last updated on 15/Dec/25
the PM s(point mass) have made an equi. triangle. all the three were at the centroid of the equi. triangle. they have made an equi. triangle after repelling themselves. so all have the same charge q(let) if length of median is M (2/3)M=lsin α M=((3lsin α)/2) if side length of equi. triangle is r ((√3)/2)r=M r=(√3)lsin α repelling force does not depend on mass ∴ F=k(q^2 /r^2 ) F_(net) =(√(F^2 +F^2 +2F×Fcos 60^0 ))=F(√3) ∴F_(net) =k(q^2 /r^2 )(√3) m_1 g=T_1 cos α F_(net) =T_1 sin α ⇒tan α=(F_(net) /(m_1 g)) similarly tan α=(F_(net) /(m_1 g))=(F_(net) /(m_3 g)) ∴(F_(net) /(m_1 g))=(F_(net) /(m_3 g))=(F_(net) /(m_1 g)) ⇒m_1 =m_2 =m_3 =m F_(met) =tan αmg ⇒k(q^2 /r^2 )(√3)=tan αmg ⇒q=(√((r^2 mgtan α)/(k(√3))))=(√((3(lsin α)^2 mgtan α)/(k(√3)))) ∴q=lsin α(√((((√3)mgtan α)/k) ))✓
$${the}\:{PM}\:{s}\left({point}\:{mass}\right)\:{have}\:{made}\:{an}\:{equi}. \\ $$$${triangle}.\:{all}\:{the}\:{three}\:{were}\:{at}\:{the}\:{centroid} \\ $$$${of}\:{the}\:{equi}.\:{triangle}.\:{they}\:{have}\:{made}\:{an}\:{equi}. \\ $$$${triangle}\:{after}\:{repelling}\:{themselves}. \\ $$$${so}\:{all}\:{have}\:{the}\:{same}\:{charge}\:{q}\left({let}\right) \\ $$$$\:{if}\:{length}\:{of}\:{median}\:{is}\:{M} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{M}={l}\mathrm{sin}\:\alpha \\ $$$${M}=\frac{\mathrm{3}{l}\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$${if}\:{side}\:{length}\:{of}\:{equi}.\:{triangle}\:{is}\:{r} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{r}={M} \\ $$$${r}=\sqrt{\mathrm{3}}{l}\mathrm{sin}\:\alpha \\ $$$${repelling}\:{force}\:{does}\:{not}\:{depend}\:{on}\:{mass} \\ $$$$\therefore\:{F}={k}\frac{{q}^{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$${F}_{{net}} =\sqrt{{F}^{\mathrm{2}} +{F}^{\mathrm{2}} +\mathrm{2}{F}×{F}\mathrm{cos}\:\mathrm{60}^{\mathrm{0}} }={F}\sqrt{\mathrm{3}} \\ $$$$\therefore{F}_{{net}} ={k}\frac{{q}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\sqrt{\mathrm{3}} \\ $$$${m}_{\mathrm{1}} {g}={T}_{\mathrm{1}} \mathrm{cos}\:\alpha \\ $$$${F}_{{net}} ={T}_{\mathrm{1}} \mathrm{sin}\:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{{F}_{{net}} }{{m}_{\mathrm{1}} {g}} \\ $$$${similarly}\: \\ $$$$\mathrm{tan}\:\alpha=\frac{{F}_{{net}} }{{m}_{\mathrm{1}} {g}}=\frac{{F}_{{net}} }{{m}_{\mathrm{3}} {g}} \\ $$$$\therefore\frac{{F}_{{net}} }{{m}_{\mathrm{1}} {g}}=\frac{{F}_{{net}} }{{m}_{\mathrm{3}} {g}}=\frac{{F}_{{net}} }{{m}_{\mathrm{1}} {g}} \\ $$$$\Rightarrow{m}_{\mathrm{1}} ={m}_{\mathrm{2}} ={m}_{\mathrm{3}} ={m} \\ $$$${F}_{{met}} =\mathrm{tan}\:\alpha{mg} \\ $$$$\Rightarrow{k}\frac{{q}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\sqrt{\mathrm{3}}=\mathrm{tan}\:\alpha{mg} \\ $$$$\Rightarrow{q}=\sqrt{\frac{{r}^{\mathrm{2}} {mg}\mathrm{tan}\:\alpha}{{k}\sqrt{\mathrm{3}}}}=\sqrt{\frac{\mathrm{3}\left({l}\mathrm{sin}\:\alpha\right)^{\mathrm{2}} {mg}\mathrm{tan}\:\alpha}{{k}\sqrt{\mathrm{3}}}} \\ $$$$\therefore{q}={l}\mathrm{sin}\:\alpha\sqrt{\frac{\sqrt{\mathrm{3}}{mg}\mathrm{tan}\:\alpha}{{k}}\:}\checkmark \\ $$

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