1.

A massless spool of inner radius r, outer radius R is placed against vertical wall and tilted split floor as shown. A light inextensible thread is tightly wound around the spool through which a mass m is hanging. There exists no friction at point A, while the coefficient of friction between spool and point B is μ . The angle between two surface is theta.

Answer»

The magnitude of force on the spool at B in order to MAINTAIN equilibrium is
`mgsqrt((r/R)^(2)+(1-r/R)^(2)1/(TAN^(2)theta))`
The magnitude of force on the spool at B in order to maintain equilibrium is mg `(1-r/R)1/(tantheta)`
The minimum value of μ for the SYSTEM to remain in equilibrium is `(cottheta)/((R-r)-1)`
The minimum value of μ for the system to remain in equilibrium is `(tantheta)/((R-r)-1)`.

Solution :From the free body diagram we can write that,
mgr= jR ... (i)
`N_(1) sintheta + f= mg` ….(ii)
`N_(1)COSTHETA =N_(2)` …(iii)
From (i), `f=(mgr)/R`
`N_(2)=(mg-f)/(tantheta)=(mg)/(tantheta)[1-r/R]`
Net force at B: `F_(g)=sqrt(f^(2)+N_(2)^(2))=mgsqrt((r/R)^(2)+(1-r/R)1/(tan^(2)theta))`
For minimum value of `mu:flemuN_(2)`
`rArr(mgr)/Rle(MUG)/(tantheta)[1-r/R]rArrmugge(tantheta)/((R//r)-1)`


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