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Question Number 15534 by ajfour last updated on 11/Jun/17

Commented by ajfour last updated on 11/Jun/17

$$Findtheexistingstaticfriction$$$$coefficientbetweenropeand$$$$pulleyifthetensioninropeat$$$$thehooktowhich{it}^{\prime }soneend$$$$istiedisequalto\frac{Mg}{2}.$$$$Assumemassofthreadis$$$$negligiblecomparedtothe$$$$massoftheblock\left(M\right).$$

Answered by mrW1 last updated on 11/Jun/17

$$\mathrm{T}=\mathrm{tension}\mathrm{in}\mathrm{rope}$$$$\mathrm{r}=\mathrm{radius}\mathrm{of}\mathrm{pulley}$$$$\mathrm{dT}=-\mu \frac{\mathrm{T}}{\mathrm{r}}\mathrm{rd}\theta =-\mu \mathrm{Td}\theta$$$$\frac{\mathrm{dT}}{\mathrm{T}}=-\mu \mathrm{d}\theta$$$$\ln \mathrm{T}=-\mu \theta +\mathrm{C}$$$$$$$$\mathrm{at}\theta =0:\mathrm{T}=\mathrm{Mg}$$$$\Rightarrow \mathrm{C}=\ln \mathrm{Mg}$$$$\Rightarrow \ln \frac{\mathrm{T}}{\mathrm{Mg}}=-\mu \theta$$$$\Rightarrow \mathrm{T}=\mathrm{Mg}\times {\mathrm{e}}^{-\mu \theta }=\frac{\mathrm{Mg}}{{\mathrm{e}}^{\mu \theta }}$$$$$$$$\mathrm{at}\theta =\frac{3\pi }{2}:\mathrm{T}=\frac{\mathrm{Mg}}{2}$$$$\frac{\mathrm{Mg}}{2}=\frac{\mathrm{Mg}}{{\mathrm{e}}^{\frac{3\pi \mu }{2}}}$$$$\Rightarrow {\mathrm{e}}^{\frac{3\pi \mu }{2}}=2$$$$\Rightarrow \mu =\frac{2\ln 2}{3\pi }\approx 0.147$$

Commented by ajfour last updated on 11/Jun/17

$$thankyousir.answeriscorrect.$$

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