A sphere is placed rotating with its centre initially at rest in a corner as shown in figure `(a)` and `(b)`. Coefficient of friction between all surfaces and the sphere is `(1)/(3)`. Find the ratio of the frictional force `(f_(a))/(f_(b))` by ground in situations `(a)` and `(b)`.
.

Options

A. 1
B. `9//10`
C. `10//9`
D. none

Answer

B

Explanation

taking equilibrium in vertical direction
`mg = N_(1) + f_(2)` ….(1)
taking equilibrium in horizontal direction
`N_(2) = f_(1)`
therefore `f_(2) = mu N_(2)`

`f_(2) = mu f_(1)` …(2)
and `f_(1) = mu N_(1) rArr N_(1) = (f_(1))/(mu)` ....(3)
From (1), (2) and (3)
`mg = (f_(1))/(mu) + mu f_(1)`
`f_(1) = (mg)/((1)/(mu) + mu)`
`f_(a) = (mg)/(3 + (1)/(3))`
`f_(a) = (3)/(10) mg`
`f_(b) - mu N rArr f_(b) = mu mg rArr f_(b) = (mg)/(3), so (f_(a))/(f_(b)) = (9)/(10)`.
.