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JEE Advanced 2016 · Paper 1, Physics question 2

A uniform wooden stick of mass 1.6 kg and length ll rests in an inclined manner on a smooth, vertical wall of height h (<l)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 h/lh/l and the frictional force ff at the bottom of the stick are (g=10 m s−2)(g = 10\ \mathrm{m\,s^{-2}})

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Answer and worked solution

Official answer: D

Forces. The stick makes 30° with the vertical. The wall's reaction NN is perpendicular to the stick, so it has a horizontal part Ncos⁡30°N\cos30° and an upward part Nsin⁡30°N\sin30°. At the floor there is a normal force RR and friction ff.

Balance, with N=RN=R.
R+Nsin⁡30°=mg=16  ⟹  32R=16  ⟹  R=N=323 N.R+N\sin30°=mg=16\implies\tfrac32R=16\implies R=N=\tfrac{32}3\text{ N}.
f=Ncos⁡30°=323⋅32=1633 N.f=N\cos30°=\frac{32}3\cdot\frac{\sqrt3}2=\frac{16\sqrt3}3\text{ N}.

Torque about the bottom. Let the contact with the wall edge be at distance dd along the stick. The weight acts at l/2l/2 with horizontal lever arm l2sin⁡30°\frac l2\sin30°:
Nd=mg⋅l4  ⟹  d=4l32/3=3l8.Nd=mg\cdot\frac l4\implies d=\frac{4l}{32/3}=\frac{3l}8.

Height. h=dcos⁡30°=3l8⋅32h=d\cos30°=\dfrac{3l}8\cdot\dfrac{\sqrt3}2, so hl=3316\dfrac hl=\dfrac{3\sqrt3}{16}.

(D)

Topics:Equilibrium of rigid bodiesFriction
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