A massles spool of inner radius r and other 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 `mu`. The angle between two surface is `theta`

Options

A. The magnitude of the force on the spool at `B` in order to maintain equilibrium is `mg sqrt((r/R)^(2)+(1-r/R)^(2) 1/(tan^(2)theta))`
B. The magnitude of the force on the spool at `B` in order to maintain equilibrum is `mg (1-r/R)1/(tantheta)`
C. The minimum value of `mu` for the system to remain in equilibrium is `(cottheta)/((R//r)-1)`
D. The minimum values of `mu` for the system to remain in equilibrium is `(tantheta)/((R//r)-1)`

Answer

A, D

Explanation

`mgr=fR`………..i
`N_1 sin theta+f=mg` ………………ii
`N_1costheta=N_2`……………..iii

From eqn i. `f=(mgr)/R`
From eqn ii and iii `N_2=(mg-f)/(tantheta)=(mg)/(tantheta)[1-r/R]`
Net force at `B: f_B=sqrt(f^2+N_2^2)`
For the minimum value of `mu:flemuN_2`
`implies (mgr)/rle(mumg)/(tantheta)[1/r/R]`
`impliesmuge(tantheta)/(R//r-1))`