A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60 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/s2)
- 200 N
- 200 3 N
- 100 N
- 100 3 N
Answer
(4) 100 3 N
Solution
Forces on the rod: normal reaction N1 of the smooth wall (horizontal), weight mg at the centre, normal reaction N2 and friction f at the floor.
[figure]
Horizontal equilibrium: f = N1; vertical equilibrium: N2 = mg
Torque about the foot of the rod (rod makes 60 with the wall):
N1 (L 60 ) = mg ( L2 60 )
N1 = mg 60 2 = 200 32 = 100 3 N
f = N1 = 100 3 N
Answer: (4)