A`4 m` long ladder weighing `25 kg` rests with its upper and against a smooth wall and lower and on rought ground.What should be the minimum coefficient of friction between the ground and ladder for it to be inclined at `60^(@)` with the horizontal without slipping? `(Take g =10 m//s^(2))`

Explanation

In figure `AB` is a ladder of weight `w` which acts at its center of greavity `G`
`/_ ABC = 60^(@)`
`:. /_ BAC = 30^(@)`
Let `N_(1)` be the reation of the wall , and `N_(2)` reation of the ground .Force of friction `f` between ladder and the ground acts along `BC`. For horizontal equalibrium
`f =N_(1)` ...(i)
For vertical equalibrium
`N_(2) =w` ...(ii)
taking moment about `B` we get for equalibrium
`N_(1) (4 cos 30^(@)) - w(2 cos 60^(@)) = 0` ...(iii)
Here `w = 250 N`
Solving these three equations,we get
`f = 72.17 N`
and `N_(2) = 250 N`
:. `mu = (f)/(N_(2)) = (72.17)/(250)` `= 0.288`