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Question Number 23666 by Tinkutara last updated on 03/Nov/17

A uniform rope of length L and mass per unit λ having one end fixed with the ceiling is released from rest. Find the tension in the fixed end as a function of the distance travelled by the movable end.

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{rope}\:\mathrm{of}\:\mathrm{length}\:{L}\:\mathrm{and}\:\mathrm{mass} \\ $$$$\mathrm{per}\:\mathrm{unit}\:\lambda\:\mathrm{having}\:\mathrm{one}\:\mathrm{end}\:\mathrm{fixed}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{ceiling}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{fixed}\:\mathrm{end}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{movable} \\ $$$$\mathrm{end}. \\ $$

Commented by Tinkutara last updated on 03/Nov/17

Answered by ajfour last updated on 03/Nov/17

F = ((𝛌gl)/2)(1+((3x)/l)) .

$$\:\:\:\:{F}\:=\:\frac{\boldsymbol{\lambda{gl}}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{3}\boldsymbol{{x}}}{\boldsymbol{{l}}}\right)\:. \\ $$

Answered by ajfour last updated on 03/Nov/17

Commented by ajfour last updated on 04/Nov/17

mg−F=(dp/dt) =(d/dt)[(m/l)(((l−x)/2))v] =(m/(2l))(d/dt)[(l−x)v] =(m/(2l))[−v^2 +(l−x)g]=(m/(2l))[−2gx−gx+gl] ⇒ F= mg+(m/(2l))(3gx−gl) =((mg)/2)(1+((3x)/l)) F =((𝛌lg)/2)(1+((3x)/l)) .

$${mg}−{F}=\frac{{dp}}{{dt}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{d}}{{dt}}\left[\frac{{m}}{{l}}\left(\frac{{l}−{x}}{\mathrm{2}}\right){v}\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{{m}}{\mathrm{2}{l}}\frac{{d}}{{dt}}\left[\left({l}−{x}\right){v}\right] \\ $$$$\:\:=\frac{{m}}{\mathrm{2}{l}}\left[−{v}^{\mathrm{2}} +\left({l}−{x}\right){g}\right]=\frac{{m}}{\mathrm{2}{l}}\left[−\mathrm{2}{gx}−{gx}+{gl}\right] \\ $$$$\:\:\:\:\Rightarrow\:\:\:{F}=\:{mg}+\frac{{m}}{\mathrm{2}{l}}\left(\mathrm{3}{gx}−{gl}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{mg}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{3}{x}}{{l}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{F}}\:=\frac{\boldsymbol{\lambda{lg}}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{3}\boldsymbol{{x}}}{\boldsymbol{{l}}}\right)\:. \\ $$

Commented by Tinkutara last updated on 04/Nov/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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