Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

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

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

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

Perpendicular from P to incline  be D.  let time taken to reach P′  be t.  length of wire=Dsec ((π/2)−α−θ)                               =(D/(sin (θ+α)))   acceleration along wire=gsin θ  (D/(sin (θ+α)))=((gt^2 sin θ)/2)  t^2 =((2D)/(gsin θsin (θ+α))) >0     =((4D)/(cos α−cos (2θ+α)))  For t^2  to be minimum,   denominator is a maximum.  that is     cos (2θ−α)=−1                         2θ−α=π                            θ=(π/2)−(α/2) .  wire is then the angular bisector  of ⊥ from P to incline plane and  vertical line from P to incline plane.

$$\mathrm{Perpendicular}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline} \\ $$$$\mathrm{be}\:\mathrm{D}. \\ $$$$\mathrm{let}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{P}'\:\:\mathrm{be}\:\mathrm{t}. \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{wire}=\mathrm{Dsec}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{D}}{\mathrm{sin}\:\left(\theta+\alpha\right)} \\ $$$$\:\mathrm{acceleration}\:\mathrm{along}\:\mathrm{wire}=\mathrm{gsin}\:\theta \\ $$$$\frac{\mathrm{D}}{\mathrm{sin}\:\left(\theta+\alpha\right)}=\frac{\mathrm{gt}^{\mathrm{2}} \mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\mathrm{t}^{\mathrm{2}} =\frac{\mathrm{2D}}{\mathrm{gsin}\:\theta\mathrm{sin}\:\left(\theta+\alpha\right)}\:>\mathrm{0} \\ $$$$\:\:\:=\frac{\mathrm{4D}}{\mathrm{cos}\:\alpha−\mathrm{cos}\:\left(\mathrm{2}\theta+\alpha\right)} \\ $$$$\mathrm{For}\:\mathrm{t}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{be}\:\mathrm{minimum}, \\ $$$$\:\mathrm{denominator}\:\mathrm{is}\:\mathrm{a}\:\mathrm{maximum}. \\ $$$$\mathrm{that}\:\mathrm{is}\:\:\:\:\:\mathrm{cos}\:\left(\mathrm{2}\theta−\alpha\right)=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\theta−\alpha=\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\:. \\ $$$$\mathrm{wire}\:\mathrm{is}\:\mathrm{then}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{bisector} \\ $$$$\mathrm{of}\:\bot\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline}\:\mathrm{plane}\:\mathrm{and} \\ $$$$\mathrm{vertical}\:\mathrm{line}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{incline}\:\mathrm{plane}. \\ $$

Commented by Tinkutara last updated on 19/Jul/17

Thanks Sir!

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com