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Question Number 27224 by mrW1 last updated on 03/Jan/18

Commented by mrW1 last updated on 03/Jan/18

$$Thefrictioncoefficientbetweemthe$$$$blocksis\mu .$$$$1)findthetensioninthethread.$$$$2)whatistheinclinationanglesuch$$$$thatthetensioninthreadismaximum?$$

Commented by mrW1 last updated on 03/Jan/18

$$SeealsoQ27202$$

Commented by ajfour last updated on 03/Jan/18

Commented by ajfour last updated on 03/Jan/18

$$Grantedblockslides:$$$$T=N\sin \theta -\mu mg\cos {}^2\theta$$$$=\left(mg\cos \theta \right)\sin \theta -\mu mg\cos {}^2\theta$$$$\frac{dT}{d\theta }=mg(\cos 2\theta +\mu \sin 2\theta )$$$$=0for\tan 2\theta =-\frac{1}{\mu };forthisvalue$$$$\frac{dT}{d{\theta }^2}=mg(-2\sin 2\theta +2\mu \cos 2\theta )$$$$=mg(-\frac{2}{\sqrt{1+{\mu }^2}}-\frac{2{\mu }^2}{\sqrt{1+{\mu }^2}})$$$$=-2mg\sqrt{1+{\mu }^2}<0$$$$hencemaximumtensionfor$$$$\theta =\frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}\left(\frac{1}{\mu }\right)=\alpha \left(say\right).$$$$T_{max}=\frac{mg}{2}(\sin 2\alpha -2\mu \cos {}^2\alpha )$$$$=\frac{mg}{2}[\sin 2\alpha -\mu (1+\cos 2\alpha )]$$$$=\frac{mg}{2}[\frac{1}{\sqrt{1+{\mu }^2}}-\mu (1-\frac{\mu }{\sqrt{1+{\mu }^2}})]$$$$=\frac{mg}{2}[\sqrt{1+{\mu }^2}-\mu ].$$

Answered by mrW1 last updated on 03/Jan/18

$$Q1:$$$$N=mg\cos \theta$$$$f=\mu N=\mu mg\cos \theta$$$$$$$$case1:mg\sin \theta <f=\mu mg\cos \theta$$$$or\tan \theta <\mu or\theta <\alpha ={\tan }^{-1}\mu$$$$blockmrestsonblockM,$$$$\Rightarrow T=0$$$$$$$$case2:\theta ⩾\alpha ={\tan }^{-1}\mu$$$$T=N\sin \theta -f\cos \theta$$$$T=mg\cos \theta \sin \theta -\mu mg{\cos }^2\theta$$$$=\frac{mg}{2}(\sin 2\theta -\mu {2\cos }^2\theta )$$$$=\frac{mg}{2}(\sin 2\theta -\mu \cos 2\theta -\mu )$$$$=\frac{mg}{2}[\sqrt{1+{\mu }^2}(\frac{1}{\sqrt{1+{\mu }^2}}\sin 2\theta -\frac{\mu }{\sqrt{1+{\mu }^2}}\cos 2\theta )-\mu$$$$=\frac{mg}{2}[\sqrt{1+{\mu }^2}(\cos \alpha \sin 2\theta -\sin \alpha \cos 2\theta )-\mu ]$$$$T=\frac{mg}{2}[\sqrt{1+{\mu }^2}\sin (2\theta -\alpha )-\mu ]$$$$with\alpha ={\tan }^{-1}\mu$$$$$$$$Q2:$$$$Tismaximumif2\theta -\alpha =\frac{\pi }{2}$$$$or{\theta }_m=\frac{\pi }{4}+\frac{\alpha }{2}=\frac{\pi }{4}+\frac{1}{2}{\tan }^{-1}\mu =\frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}\frac{1}{\mu }$$$$Example:\mu =0.3,{\theta }_m=53.3°$$

Commented by Tinkutara last updated on 04/Jan/18

Thank you!

Commented by mrW1 last updated on 03/Jan/18

https://play.google.com/store/apps/details?id=com.iudesk.android.photo.editor

Commented by mrW1 last updated on 03/Jan/18

$$Iuse\mathit{P}\mathit{h}\mathit{o}\mathit{t}\mathit{o}\mathit{E}\mathit{d}\mathit{i}\mathit{t}\mathit{o}\mathit{r}.$$

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