Question Number 28940 by ajfour last updated on 01/Feb/18
Commented by ajfour last updated on 01/Feb/18
$$\left({i}\right)\:{Find}\:{tension}\:{in}\:{the}\:{three}\: \\ $$$${threads},\:{each}\:{of}\:{length}\:{l}. \\ $$$$\left({ii}\right)\:{Find}\:{tension}\:{for}\:{a}={b}={c}\: \\ $$$$\left({The}\:{blue}\:{triangle}\:{is}\:{a}\:{portion}\:{of}\right. \\ $$$$\left.{the}\:{roof}\:\right)\:. \\ $$
Answered by mrW2 last updated on 02/Feb/18
Commented by mrW2 last updated on 02/Feb/18
![the mass must hang under the circumcenter O of the triangle. let R=radius of circumcircle OA=OB=OC=R angle made by threads and horizontal is θ with cos θ=(R/L) or θ=cos^(−1) (R/L). let′s see the vertical component of tension T_2 : the sum of torque about line BC=0, T_2 sin θ×h_A =mg e_A ⇒T_2 =((mg)/(sin θ×(h_A /e_A ))) h_A =c sin B e_A =R cos ∠BOD=R cos ((∠BOC)/2)=R cos A R=(a/(2 sin A))=(b/(2 sin B))=(c/(2 sin C)) ⇒c=2R sin C ⇒h_A =2R sin C sin B ⇒(h_A /e_A )=((2 sin B sin C)/(cos A)) ⇒=((−cos (B+C)+cos (B−C))/(cos A)) ⇒=((cos A+cos (B−C))/(cos A)) ⇒=1+((cos (B−C))/(cos A)) ⇒T_2 =((mg)/(sin θ (((2 sin B sin C)/(cos A))))) or ⇒T_2 =((mg)/(sin θ [1+((cos (B−C))/(cos A))])) T_1 and T_3 similarly. In case of a=b=c, R=(2/3)×((√3)/2)a=(((√3)a)/3) cos θ=(((√3)a)/(3L)) sin θ=(√(1−(1/3)((a/L))^2 )) A=B=C=60° ⇒T_1 =T_2 =T_3 =((mg)/(3(√(1−(1/3)((a/L))^2 ))))](https://www.tinkutara.com/question/Q28959.png)
$${the}\:{mass}\:{must}\:{hang}\:{under}\:{the} \\ $$$${circumcenter}\:{O}\:{of}\:{the}\:{triangle}. \\ $$$$ \\ $$$${let}\:{R}={radius}\:{of}\:{circumcircle} \\ $$$${OA}={OB}={OC}={R} \\ $$$$ \\ $$$${angle}\:{made}\:{by}\:{threads}\:{and}\:{horizontal} \\ $$$${is}\:\theta\:{with}\:\mathrm{cos}\:\theta=\frac{{R}}{{L}}\:{or}\:\theta=\mathrm{cos}^{−\mathrm{1}} \frac{{R}}{{L}}. \\ $$$$ \\ $$$${let}'{s}\:{see}\:{the}\:{vertical}\:{component}\:{of} \\ $$$${tension}\:{T}_{\mathrm{2}} : \\ $$$${the}\:{sum}\:{of}\:{torque}\:{about}\:{line}\:{BC}=\mathrm{0}, \\ $$$${T}_{\mathrm{2}} \:\mathrm{sin}\:\theta×{h}_{{A}} ={mg}\:{e}_{{A}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mg}}{\mathrm{sin}\:\theta×\frac{{h}_{{A}} }{{e}_{{A}} }} \\ $$$${h}_{{A}} ={c}\:\mathrm{sin}\:{B} \\ $$$${e}_{{A}} ={R}\:\mathrm{cos}\:\angle{BOD}={R}\:\mathrm{cos}\:\frac{\angle{BOC}}{\mathrm{2}}={R}\:\mathrm{cos}\:{A} \\ $$$${R}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{2}\:\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{2}\:\mathrm{sin}\:{C}} \\ $$$$\Rightarrow{c}=\mathrm{2}{R}\:\mathrm{sin}\:{C} \\ $$$$\Rightarrow{h}_{{A}} =\mathrm{2}{R}\:\mathrm{sin}\:{C}\:\mathrm{sin}\:{B} \\ $$$$\Rightarrow\frac{{h}_{{A}} }{{e}_{{A}} }=\frac{\mathrm{2}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{cos}\:{A}} \\ $$$$\Rightarrow=\frac{−\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:\left({B}−{C}\right)}{\mathrm{cos}\:{A}} \\ $$$$\Rightarrow=\frac{\mathrm{cos}\:{A}+\mathrm{cos}\:\left({B}−{C}\right)}{\mathrm{cos}\:{A}} \\ $$$$\Rightarrow=\mathrm{1}+\frac{\mathrm{cos}\:\left({B}−{C}\right)}{\mathrm{cos}\:{A}} \\ $$$$ \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mg}}{\mathrm{sin}\:\theta\:\left(\frac{\mathrm{2}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{cos}\:{A}}\right)} \\ $$$${or} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{{mg}}{\mathrm{sin}\:\theta\:\left[\mathrm{1}+\frac{\mathrm{cos}\:\left({B}−{C}\right)}{\mathrm{cos}\:{A}}\right]} \\ $$$$ \\ $$$${T}_{\mathrm{1}} \:{and}\:{T}_{\mathrm{3}} \:{similarly}. \\ $$$$ \\ $$$${In}\:{case}\:{of}\:{a}={b}={c}, \\ $$$${R}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}{L}} \\ $$$$\mathrm{sin}\:\theta=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{a}}{{L}}\right)^{\mathrm{2}} } \\ $$$${A}={B}={C}=\mathrm{60}° \\ $$$$\Rightarrow{T}_{\mathrm{1}} ={T}_{\mathrm{2}} ={T}_{\mathrm{3}} =\frac{{mg}}{\mathrm{3}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{a}}{{L}}\right)^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 02/Feb/18
$${Understood}\:{Sir},\:{Very}\:{Nice}\:! \\ $$