If the coefficient of friction between `M` and the inclined surfaces is `mu = 1//sqrt(3)` find the minimum mass `m` of the rod so that the of mass `M = 10 kg` remain stationary on the inclined plane

Explanation

Angle of repose ` phi = tan^(-1)((1)/(sqrt(3)))= 30^(@)`
Here ` theta gt phi` Hence , the block has a tendency to slide down

`Mg cos theta + N' , N = mg cos theta`
`N' = Mg cos theta + mg cos theta`….(i)
For the block to remain stationary .
`f = Mg sin theta`...(ii)
If f is static as nature `f lt f_(max)`
`Mg sin theta lt mu N'`
`Mg sin theta lt mu(Mg cos theta + mg cos theta)`
`Mg sin theta - mu mg cos theta lt mu mg cos thetan`
`m gt M((tan theta)/(mu) -1) rArr m gt 10((sqrt(3))/(1//sqrt(3)) - 1) rArr m gt 20 kg`