A-rectangular-wire-frame-ABCD-is-in-vertical-plane-is-moving-with-a-constant-acceleration-a-into-the-plane-Direction-of-gravity-is-shown-in-figure-A-collar-can-move-on-wire-AC-of-length-l-Coefficie

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Question Number 23489 by Tinkutara last updated on 31/Oct/17

$$\mathrm{A}\mathrm{rectangular}\mathrm{wire}\mathrm{frame}ABCD\mathrm{is}\mathrm{in}$$$$\mathrm{vertical}\mathrm{plane}\mathrm{is}\mathrm{moving}\mathrm{with}\mathrm{a}\mathrm{constant}$$$$\mathrm{acceleration}a\mathrm{into}\mathrm{the}\mathrm{plane}.\mathrm{Direction}$$$$\mathrm{of}\mathrm{gravity}\mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{figure}.\mathrm{A}\mathrm{collar}$$$$\mathrm{can}\mathrm{move}\mathrm{on}\mathrm{wire}AC\mathrm{of}\mathrm{length}l.$$$$\mathrm{Coefficient}\mathrm{of}\mathrm{friction}\mathrm{between}\mathrm{wire}$$$$\mathrm{and}\mathrm{collar}\mathrm{is}\mu .\mathrm{Find}$$$$\left(\mathrm{i}\right)\mathrm{The}\mathrm{minimum}\mathrm{acceleration}a\mathrm{so}\mathrm{that}$$$$\mathrm{collar}\mathrm{does}\mathrm{not}\mathrm{slip}\mathrm{on}\mathrm{wire}.$$$$\left(\mathrm{ii}\right)\mathrm{The}\mathrm{time}\mathrm{taken}\mathrm{by}\mathrm{collar}\mathrm{to}\mathrm{reach}C$$$$\mathrm{if}\mathrm{acceleration}\mathrm{is}\mathrm{half}\mathrm{the}\mathrm{value}$$$$\mathrm{calculated}\mathrm{in}\mathrm{part}\left(\mathrm{i}\right)$$

Commented by Tinkutara last updated on 31/Oct/17

Commented by mrW1 last updated on 31/Oct/17

$$\left(\mathrm{i}\right)$$$$\mu \mathrm{m}\left(\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2{\cos }^2\theta }\right)⩾\mathrm{mg}\sin \theta$$$${\mathrm{a}}^2⩾(\frac{{\sin }^2\theta }{{\mu }^2}-{\cos }^2\theta ){\mathrm{g}}^2$$$$\mathrm{a}⩾\frac{\mathrm{g}}{\mu }\sqrt{{\sin }^2\theta -{\mu }^2{\cos }^2\theta }$$$$\left(\mathrm{ii}\right)$$$$\mathrm{with}\mathrm{a}=\frac{\mathrm{g}}{2\mu }\sqrt{{\sin }^2\theta -{\mu }^2{\cos }^2\theta }$$$${\mathrm{ma}}_{\mathrm{s}}=\mathrm{mg}\sin \theta -\mu \mathrm{mg}\left(\sqrt{\frac{{\sin }^2\theta -{\mu }^2{\cos }^2\theta }{4{\mu }^2}+{\cos }^2\theta }\right)$$$${\mathrm{a}}_{\mathrm{s}}=\frac{\mathrm{g}}{2}(2\sin \theta -\sqrt{{\sin }^2\theta +3{\mu }^2{\cos }^2\theta })$$$$\mathrm{t}=\sqrt{\frac{2\mathrm{L}}{{\mathrm{a}}_{\mathrm{s}}}}=2\sqrt{\frac{\mathrm{L}}{(2\sin \theta -\sqrt{{\sin }^2\theta +3{\mu }^2{\cos }^2\theta })\mathrm{g}}}$$

Answered by ajfour last updated on 31/Oct/17

Commented by ajfour last updated on 31/Oct/17

$$N_z=ma$$$$N_x=mg\cos \theta$$$$N=m\sqrt{a^2+g^2\cos {}^2\theta }$$$$\left(i\right)f-mg\sin \theta =0$$$$\Rightarrow \mu N⩾mg\sin \theta$$$$or{a}^2+g^2\cos {}^2\theta =\frac{g^2\sin {}^2\theta }{{\mu }^2}$$$$a⩾\frac{g}{\mu }\sqrt{\sin {}^2\theta -{\mu }^2\cos {}^2\theta }$$$$\left(ii\right)l=\frac{1}{2}{A}_y{t}^2$$$$mg\sin \theta -\mu N={mA}_y$$$$N=m\sqrt{a^2/4+g^2\cos {}^2\theta }$$$$A_y=g\sin \theta -\mu \sqrt{\frac{g^2}{4{\mu }^2}(s\mathrm{in}{}^2\theta -{\mu }^2\cos {}^2\theta )+g^2\cos {}^2\theta }$$$$t=\sqrt{\frac{2l}{A_y}}$$$$=\sqrt{\frac{2l}{g\sin \theta -\frac{g}{2}\sqrt{\sin {}^2\theta +3{\mu }^2\cos {}^2\theta }}}.$$$$$$

Commented by Tinkutara last updated on 01/Nov/17

$$Igotthat\mu N⩾mg\sin \theta .Butonlydoubt$$$$isthatNisoftheform\stackrel{\wedge }{i}+\stackrel{\wedge }{k}.Then$$$$whyitshouldbeequatedwithmg\sin \theta \stackrel{\wedge }{y}?$$$$Also,howyouwrite{3}^{rd}lineof\left(ii\right)part?$$

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