At the bottom edge of a smooth wall , an inclined plane is kept at an angle of `45^(@)` . A uniform ladder of length `l` and mass `M` rests on the inclined plane against the wall such that it is perpendicular to the inclined.

(a) if the plane is also smooth ,which way will the ladder slide .
(b) What is the minimum coefficient of friction neccessary so that the ladder does not slip on the inclined?

Explanation

a. The ladder has a tendency of slip by rotating clockwise about the point `A`
b. The free body diagram of the ladder is shown in figure.
Balancing torque about point `A`, `fl=Mg(l/2sin45^@)`
`:. F=(Mg)/(2sqrt(2))`…..i


Balancing forces in the vertical direction
`Mg=R_(2)cos45^@+fsin45^@`..............ii
From eqn i and ii we get
`R_(2)=(3Mg)/(2sqrt2)`..........iii
From eqn i and ii `f=muR_(2)`
or `(Mg)/(2sqrt(2))=(3Mg)/(2sqrt(2))mu` or `mu=1/3`