Question

A uniform wooden stick of mass is 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 degree with the wall and bottom of the stick is on a rough floor. the reaction force 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 frictional force f at the bottom of the stick are (g=10m/s square) (take the normal reaction at the wall equal to the normal reaction at the floor) ​

Answer 1

The forces acting to the stick are its weight mg at the center of mass C, normal reaction NA at the contact point A due to the

Floor, frictional force f at point A due to the uneven floor, and the normal for the reaction NB at the contact point B due to the wall. Since the stick is uniform, its center of mass C lies at the middle point, i.e.

The vertical directions are zero, i.e.,

so we assume that NA + NB sin 30◦ − mg = 0 ---------> (1)

NB cos 30◦ − f = 0.

mg(l/2) cos 60◦ − (NB cos 30◦)h − (NB sin 30◦)h tan 30◦ = 0.

Substitute NB = NA and solve equations to get as shown in the figure.