JEE Advanced2016PhysicsLaws of MotionActual
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 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
Options
- Ah l = 3 16 , f = 16 3 3 N
- Bh l = 3 16 , f = 16 3 3 N
- Ch l = 3 3 16 , f = 8 3 3 N
- Dh l = 3 3 16 , f = 16 3 3 N
Correct answer
D. h l = 3 3 16 , f = 16 3 3 N
Step-by-step solution
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