A uniform wooden stick of mass 1.6 kg and length l rests in an inclined mannar on a smooth, vertical wall of height `h(ltl)` such that a small portion of the stick extends beyond the wall. The reaction force of th wall on the stick is perpendicular to the stick. The stick makes an angle of `30^@` with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/l and the friectional force f at the bottom of the stick are `(g = 10 ms^2)`

Options

A. `h/1 = sqrt(3)/16, f = (16sqrt(3))/3` N
B. `h/1 = 3/16, f = (16sqrt(3))/3 N`
C. `h/1 = (3sqrt(3))/16, f = (8sqrt(3))/3` N
D. `h/1 =(3sqrt(3))/16, f = (16sqrt(3))/3` N

Answer

D

Explanation

`Nsin 30^(@) + N =1.6 xx 10, N=32/3`
`f=N cos 30^(@) = 32/3 xx sqrt(3)/2 = (16 sqrt(3))/3`

`tau_(0)` must be zero.
`mg xx L/2 cos 60^(@) =N xx h/(sin 60^(@))`
`16 xx L/2 xx 1/2 = 32/3 xx (2h)/sqrt(3), h/l = (3sqrt(3))/16`