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Question Number 217447 by Tawa11 last updated on 14/Mar/25

Commented by mr W last updated on 14/Mar/25

$$questioniswrong.thefriction$$$$coefficientbetweencylinderand$$$$stickmaynotbethesameasthe$$$$frictioncoefficientbetweencylinder$$$$andplane.theformermustbe$$$$largerthanthelatter!ithinkthis$$$$canbesaidevenjustbyusingyour$$$$feelingwithoutphysicscalculations.$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 14/Mar/25

$$wheredidyouhavethisquestion?$$$$dotheyhaveananswer?$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Sir},\mathrm{I}\mathrm{will}\mathrm{confirm}\mathrm{if}\mathrm{there}\mathrm{is}\mathrm{answer}\mathrm{now}.$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Sir},\mathrm{they}\mathrm{got}\frac{3\sin \theta }{1+\cos \theta }$$

Commented by mr W last updated on 14/Mar/25

$$thenthequestionanddiagramare$$$$wrong.$$$$ifthequestionandthediagramwere$$$$right,thenwehad:$$

Commented by mr W last updated on 14/Mar/25

Commented by mr W last updated on 14/Mar/25

$$thethreeforcesmustmeetatthe$$$$samepointasshown.buttheyare$$$$notinequilibrium!$$$$(with\tan \varphi =\mu )$$

Commented by mr W last updated on 14/Mar/25

$$inthiscase,suchthatequilibrium$$$$ispossible,thefrictioncoefficients$$$$mustbedifferent.thenwehave:$$

Commented by mr W last updated on 14/Mar/25

Commented by mr W last updated on 14/Mar/25

$$\tan {\varphi }_1={\mu }_1$$$$\tan {\varphi }_2={\mu }_2$$$$\theta =2{\varphi }_2\Rightarrow {\varphi }_2=\frac{\theta }{2}$$$${\mu }_2={\tan }^{-1}{\varphi }_2={\tan }^{-1}\frac{\theta }{2}$$$${\varphi }_1>{\varphi }_2\Rightarrow {\mu }_1>{\mu }_2$$$$\frac{R_1}{\sin {\varphi }_2}=\frac{R_2}{\sin {\varphi }_1}=\frac{Mg}{\sin ({\varphi }_1-{\varphi }_2)}$$$$\Rightarrow R_1=\frac{\sin {\varphi }_{2}Mg}{\sin ({\varphi }_1-{\varphi }_2)}$$$$\Rightarrow R_2=\frac{\sin {\varphi }_{1}Mg}{\sin ({\varphi }_1-{\varphi }_2)}$$$$N_1=R_1\cos {\varphi }_1=\frac{mg}{2}$$$$\frac{\cos {\varphi }_1\sin {\varphi }_{2}Mg}{\sin ({\varphi }_1-{\varphi }_2)}=\frac{mg}{2}$$$$ifM=masgiven$$$$\frac{\cos {\varphi }_1\sin {\varphi }_2}{\sin ({\varphi }_1-{\varphi }_2)}=\frac{1}{2}$$$$2\cos {\varphi }_1\sin {\varphi }_2=\sin {\varphi }_1\cos {\varphi }_2-\cos {\varphi }_1\sin {\varphi }_2$$$$3\cos {\varphi }_1\sin {\varphi }_2=\sin {\varphi }_1\cos {\varphi }_2$$$$3\tan {\varphi }_2=\tan {\varphi }_1$$$$\Rightarrow 3{\mu }_2={\mu }_1=3{\tan }^{-1}\frac{\theta }{2}$$

Commented by mr W last updated on 15/Mar/25

$$whatdidtheysayasyoutoldthem$$$$thatthequestioniswrongand$$$$thereforetheiransweriswrongtoo?$$

Commented by Tawa11 last updated on 15/Mar/25

$$\mathrm{They}\mathrm{admit}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{not}\mathrm{well}$$$$\mathrm{contructed}\mathrm{sir}.$$$$\mathrm{The}\mathrm{lecturer}\mathrm{said}\mathrm{he}\mathrm{will}\mathrm{change}\mathrm{it}.$$$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by mr W last updated on 15/Mar/25

$$so{i}^{\prime }mright.butiwonderhowthey$$$$couldgetthatanswer\frac{3\sin \theta }{1+\cos \theta }$$$$nevertheless.$$

Answered by mr W last updated on 14/Mar/25

$$ifthequestionismodifiedlike$$$$following,thenitcanbesolved.$$

Commented by mr W last updated on 14/Mar/25

Commented by mr W last updated on 15/Mar/25

Commented by mr W last updated on 15/Mar/25

$$\tan \varphi =\mu$$$$with0<\phi <\frac{\pi }{2}$$$$\frac{R_1}{\sin (\theta -\varphi )}=\frac{R_2}{\sin (\phi -\theta +2\varphi )}=\frac{Mg}{\sin (\phi -\varphi )}$$$$R_1=\frac{Mg\sin (\theta -\varphi )}{\sin (\phi -\varphi )}$$$$R_1\cos (\phi +\varphi )=\frac{mg}{2}$$$$\frac{Mg\sin (\theta -\varphi )\cos (\phi +\varphi )}{\sin (\phi -\varphi )}=\frac{mg}{2}$$$$sinceM=m,$$$$\Rightarrow \frac{\sin (\theta -\varphi )\cos (\phi +\varphi )}{\sin (\phi -\varphi )}=\frac{1}{2}$$$$2(\sin \theta \cos \varphi -\cos \theta \sin \varphi )(\cos \phi \cos \varphi -\sin \phi \sin \varphi )=\sin \phi \cos \varphi -\cos \phi \sin \varphi$$$$2(\tan \theta -\tan \varphi )(1-\tan \phi \tan \varphi )=(\tan \phi -\tan \varphi )\sqrt{(1+{\tan }^2\theta )(1+{\tan }^2\varphi )}$$$$\Rightarrow 2(\tan \theta -\mu )(1-\mu \tan \phi )-(\tan \phi -\mu )\sqrt{(1+{\tan }^2\theta )(1+{\mu }^2)}=0$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Have}\mathrm{been}\mathrm{waiting}\mathrm{for}\mathrm{something}\mathrm{to}$$$$\mathrm{drop}\mathrm{here},\mathrm{as}\mathrm{I}\mathrm{saw}\mathrm{empty}\mathrm{space}.$$$$\mathrm{Hahahahahaha}$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 14/Mar/25

$$rightornot?$$

Commented by Tawa11 last updated on 14/Mar/25

$$\mathrm{Right}\mathrm{sir}.$$

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