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Question Number 52611 by ajfour last updated on 10/Jan/19

Commented by ajfour last updated on 11/Jan/19

$$Findtensionsinbothstringsif$$$$\left(i\right)△ABCbeequilateralofsidea.$$$$\left(ii\right)itbeofsidesa,b,c.$$

Commented by ajfour last updated on 11/Jan/19

$$mrWSir,canyouhelpmefind$$$$yourquestioninwhichihadfound$$$$heightofvertexofpyramidwith$$$$lateraledgesp,q,randsidesofbase$$$$a,b,c?$$

Commented by mr W last updated on 12/Jan/19

$$seeQ47145,Q47113$$

Commented by mr W last updated on 12/Jan/19

$$thanksforcheckingsir!$$

Commented by ajfour last updated on 12/Jan/19

$$youransweriscorrectSir,$$$$for\lambda =2,Tshouldbezero.$$

Answered by mr W last updated on 12/Jan/19

Commented by ajfour last updated on 12/Jan/19

$$ExceptionallygoodSir,\mathcal{SUPERB}!$$

Commented by mr W last updated on 13/Jan/19

$$thegeneralcasewithdifferentside$$$$lengthescanalsobesolvedinthe$$$$samewayashere,butverylengthy.$$$$soonlythecasewithequilateral$$$$triangleistreated.$$$$AB=BC=CA=a$$$$AE=BE=l/2$$$$let\lambda =\frac{l}{a}$$$$CD=\frac{\sqrt{3}a}{2}$$$$DE=\sqrt{{\left(\frac{l}{2}\right)}^2-{\left(\frac{a}{2}\right)}^2}=\frac{\sqrt{l^2-a^2}}{2}$$$$intriangle\mathrm{\Delta }DEC:$$$$\mathrm{\angle }D=\frac{\pi }{2}-\theta$$$$\mathrm{\angle }DEC=\pi -\theta$$$$\mathrm{\angle }DCE=\theta -(\frac{\pi }{2}-\theta )=2\theta -\frac{\pi }{2}$$$$\frac{\sqrt{3}a}{2\sin (\pi -\theta )}=\frac{\sqrt{l^2-a^2}}{2\sin (2\theta -\frac{\pi }{2})}$$$$\Rightarrow \frac{\sqrt{3}}{\sin \theta }=\frac{\sqrt{{\lambda }^2-1}}{-\cos 2\theta }$$$$\Rightarrow \frac{\sqrt{3}}{\sin \theta }=\frac{\sqrt{{\lambda }^2-1}}{2{\sin }^2\theta -1}$$$$\Rightarrow 2\sqrt{3}{\sin }^2\theta -\sqrt{{\lambda }^2-1}\sin \theta -\sqrt{3}=0$$$$\Rightarrow \sin \theta =\frac{\sqrt{{\lambda }^2-1}+\sqrt{{\lambda }^2+23}}{4\sqrt{3}}$$$$\Rightarrow \cos \theta =\frac{\sqrt{2[13-{\lambda }^2-\sqrt{({\lambda }^2-1)({\lambda }^2+23)}]}}{4\sqrt{3}}$$$$T_{DE}=2T\frac{\sqrt{l^2-a^2}}{l}=2T\frac{\sqrt{{\lambda }^2-1}}{\lambda }$$$$T_{DE}=2mg\cos \theta$$$$\Rightarrow T=mg\times \frac{\lambda \cos \theta }{\sqrt{{\lambda }^2-1}}=\frac{mg\lambda \sqrt{2[13-{\lambda }^2-\sqrt{({\lambda }^2-1)({\lambda }^2+23)}]}}{4\sqrt{3({\lambda }^2-1)}}$$$$\Rightarrow T=mg\frac{\lambda }{4}\sqrt{\frac{2[13-{\lambda }^2-\sqrt{({\lambda }^2-1)({\lambda }^2+23)}]}{3({\lambda }^2-1)}}$$$${it}^{\prime }sobviousthata<l⩽2a,i.e.1<\lambda ⩽2.$$

Commented by mr W last updated on 12/Jan/19

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