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

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

Commented by Tinkutara last updated on 03/Nov/17

Answered by ajfour last updated on 03/Nov/17

$$F=\frac{\mathit{\lambda }\mathit{g}\mathit{l}}{2}(1+\frac{3\mathit{x}}{\mathit{l}}).$$

Answered by ajfour last updated on 03/Nov/17

Commented by ajfour last updated on 04/Nov/17

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

Commented by Tinkutara last updated on 04/Nov/17

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

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