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

Explanation

length of ladder , ` AB , = 4 m `
W= 25 kg at the center C of the ladder
`angleABO = 60^(@) `
`angleBAO = 90^(@) - 60^(@) = 30^(@)`
`R_(1)` is reaction of the wall on the ladder bot OA `R_(2)` is reaction of the ground , normal to ground . Force of friction (f) between the ladder and the ground acts along BO
In equilibrium `R_(2) = W` and `R_(1) = f `
Taking moments about B we get for equilibrium of the ladder
` R_(2) xx 0 - W xx BD + R_(1) xx AO = 0 `
` W xx BD = R_(1) xx AO `
` R_(2) xx BD = f X AO `
` (f)/(R_(2)) = (BD)/(AO) = (BC cos 60^(@))/(AB sin 60^(@)) `
` mu = (AB//2)/(AB) xx (1//2)/(sqrt(3)// 2) `
` mu = (1)/sqrt(3)= 0.29 `.