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Question Number 18262 by Tinkutara last updated on 17/Jul/17

$$\mathrm{A}\mathrm{point}P\mathrm{is}\mathrm{located}\mathrm{above}\mathrm{an}\mathrm{inclined}$$$$\mathrm{plane}.\mathrm{It}\mathrm{is}\mathrm{possible}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{plane}$$$$\mathrm{by}\mathrm{sliding}\mathrm{under}\mathrm{gravity}\mathrm{down}\mathrm{a}\mathrm{straight}$$$$\mathrm{frictionless}\mathrm{wire}\mathrm{joining}\mathrm{to}\mathrm{some}\mathrm{point}$$$$P{}^{\prime }\mathrm{on}\mathrm{the}\mathrm{plane}.\mathrm{How}\mathrm{should}P{}^{\prime }\mathrm{be}$$$$\mathrm{chosen}\mathrm{so}\mathrm{as}\mathrm{to}\mathrm{minimize}\mathrm{the}\mathrm{time}$$$$\mathrm{taken}?$$

Commented by Tinkutara last updated on 17/Jul/17

Commented by ajfour last updated on 18/Jul/17

Answered by ajfour last updated on 18/Jul/17

$$\mathrm{Perpendicular}\mathrm{from}\mathrm{P}\mathrm{to}\mathrm{incline}$$$$\mathrm{be}\mathrm{D}.$$$$\mathrm{let}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{reach}{\mathrm{P}}^{\prime }\mathrm{be}\mathrm{t}.$$$$\mathrm{length}\mathrm{of}\mathrm{wire}=\mathrm{Dsec}(\frac{\pi }{2}-\alpha -\theta )$$$$=\frac{\mathrm{D}}{\sin (\theta +\alpha )}$$$$\mathrm{acceleration}\mathrm{along}\mathrm{wire}=\mathrm{gsin}\theta$$$$\frac{\mathrm{D}}{\sin (\theta +\alpha )}=\frac{{\mathrm{gt}}^2\sin \theta }{2}$$$${\mathrm{t}}^2=\frac{2\mathrm{D}}{\mathrm{gsin}\theta \sin (\theta +\alpha )}>0$$$$=\frac{4\mathrm{D}}{\cos \alpha -\cos (2\theta +\alpha )}$$$$\mathrm{For}{\mathrm{t}}^2\mathrm{to}\mathrm{be}\mathrm{minimum},$$$$\mathrm{denominator}\mathrm{is}\mathrm{a}\mathrm{maximum}.$$$$\mathrm{that}\mathrm{is}\cos (2\theta -\alpha )=-1$$$$2\theta -\alpha =\pi$$$$\theta =\frac{\pi }{2}-\frac{\alpha }{2}.$$$$\mathrm{wire}\mathrm{is}\mathrm{then}\mathrm{the}\mathrm{angular}\mathrm{bisector}$$$$\mathrm{of}\mathrm{\perp }\mathrm{from}\mathrm{P}\mathrm{to}\mathrm{incline}\mathrm{plane}\mathrm{and}$$$$\mathrm{vertical}\mathrm{line}\mathrm{from}\mathrm{P}\mathrm{to}\mathrm{incline}\mathrm{plane}.$$

Commented by Tinkutara last updated on 19/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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