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)/(l) = (sqrt3)/(16),f = (16sqrt3)/(3)N`
B. `(h)/(l) = (3)/(16),f = (16sqrt3)/(3)N`
C. `(h)/(l) = (3sqrt3)/(16),f (8sqrt3)/(3)N`
D. `(h)/(l) = (3sqrt3)/(16),f (16sqrt3)/(3)N`

Answer

D

Explanation

(d) Considering the normal reaction of the floor and wall
to be N and with reference to the figure.

By vertical equilibrium.
`N + N sin 30^@ = 1.6 g rArr N = (3.2g)/(3) ….(i)`
By horizontal equilibrium
`f = Ncos 30^@ = (sqrt3)/(2) N = (16sqrt3)/(3) From (i)`
Taking torque about A we get
`1.6 g xx AB = Nxx x`
`1.6gxx(l)/(2)cos 60^@ = (3.2g)/(3)xx x :. (3l)/(8) = x ....(ii)`
But `cos 30^@ = (h)/(x) :. x = (h)/(cos 30^@) ...(iii)`
from (ii) and (iii) (h)/(cos 30^@) = (3l)/(8) :. (h)/(l) = (3sqrt3)/(16).`