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PhysicsNEET 2025Torque, angular momentum and their conservation

System of Particles and Rotational Motion

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60∘60^\circ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take g=10 m/s2g = 10\ \mathrm{m/s^2}):

  1. A.100 N
  2. B.1003100\sqrt{3} N✓
  3. C.200 N
  4. D.2003200\sqrt{3} N

Solution

Take torques about the foot. The wall pushes horizontally with NN; the rod's weight acts at its midpoint. With the rod at 60∘60^\circ to the wall (i.e. 30∘30^\circ to the floor): N⋅Lsin⁡30∘=mg⋅L2cos⁡30∘⇒N=mgcos⁡30∘2sin⁡30∘=200×3/21=1003N \cdot L\sin30^\circ = mg\cdot\frac{L}{2}\cos30^\circ \Rightarrow N = mg\frac{\cos30^\circ}{2\sin30^\circ} = 200\times\frac{\sqrt3/2}{1} = 100\sqrt3 N. Friction at the floor balances NN: 1003100\sqrt3 N.

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