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

Here, `m = 1.6kg length =l(gt h)`
`N_(1)` is reaction of wall perpendicular to rod `N_(2)` is reaction of ground normal to floor `f` is force of friction along the floor towards the wall
In equilibrium `Sigma F_(x) =0`
`N_(1) cos30^(@)-f=0`
`Sigma F_(y) = 0` ...(i)
`N_(1) sin 30^(@) + N_(2) - mg =0` ...(ii)
Also tau_(0)=0`
`:.` mgxx(1)/(2)cos60^(@)-N_(1)((h)/(cos30^(@)))=0` ..(iii)
Also `N_(1) N_(2)`
From (i) `f = N_(1) cos 30^(@) = N_(1) sqrt3/(2)`
`N_(1) xx(1)/(2) + N_(1)= mg`
or `(3)/(2)N_(1) =mg`
From (iii) `(1)/(2) xx (1)/(2) = N_(1) (h.2)/(sqrt3) = (2)/(3)mg. (2h)/(sqrt30)`
`(l)/(4) =(4h)/(3sqrt3) :. (h)/(l) =(3sqrt3)/(16)`
We can show that `f =(16sqrt3)/(3)`
.