A 4m 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

In figure, AB is a ladder of weight w which acts at its centre of gravity G.

`angle(ABC)=60^(@) therefore angle (BAC)=30^(@)`
Let `N_(1)` be the reaction of the wall, and `N_(2)` the reaction of the ground.
Force of friction f between the ladder and the ground acts along BC.
For horizontal equilibrium , `f=N_(1)`.........(i)
For vertical equilibrium, `N_(2)=w` ..........(ii)
Taking moments about B, we get for equilibrium,
`N_(1)(4 cos 30^(@))-w( 2 cso 60^(@))=0`.......(iii)
Here, w=250N
Solving these three equations , we get
f=72.17 Nand `N_(2)=250N `
`therefore mu =(f)/(N_(2))=(72.17)/(250)=0.288`