A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h (<l) such that a small portion of the stick extends beyond the wall. The reaction force of the 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 of h/l and the frictional force f at the bottom of the stick are (g = 10 ms⁻²):
hl=3√3/16,f=16√3/3N
hl=3/16,f=16√3/3N
hl=√3/16,f=16√3/3N
hl=3√3/16,f=8√3/3N
Solution:
Balancing torque about the lowest point, we get Nhsin60°=mgl/2cos60° Also, N+N₂=mg N=2mg/3 4mg/3√3 h=mgl/4 hl=3√3/16 f=Nsin60°=mg√3/2=16√3/3 Answer is option A.