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 o…

Question

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 &#60; 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 o 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 frictional force f at the bottom of the stick are: g = 10 &#160; m s - 2

Aicharya · Accepted Answer

Force equation in x-direction, N 1 cos ⁡ 30 o - f = 0 ....(i) Force equation in y-direction, N 1 sin ⁡ 30 o + N 2 - m g = 0 ....(ii) Torque equation about O, m g l 2 cos ⁡ 60 o - N 1 h cos ⁡ 30 o = 0 ...(iii) Also, given N 1 = N 2 .....(iv) [Note taking reaction from floor as normal reaction only] Solving (i), (ii), (iii) and (iv) we have h l = 3 3 16 and f = 16 3 3

A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of…

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