Question: 22. A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 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 \mathrm{~m} / \mathrm{s}^{2})
(1) 100 N
(2) 100 \sqrt{3} \mathrm{~N}
(3) 200 N
(4) 200 \sqrt{3} \mathrm{~N}
Answer: Option (2)
Explanation:
The wall is smooth, so it exerts only a horizontal normal reaction N_{w}.
The floor exerts a vertical normal force N_{f}
and a horizontal friction force f.
Since the rod is in equilibrium, total horizontal force is zero:
f = N_{w}Total vertical force is zero:
N_{f} = mg = 20 \times 10 = 200\ \text{N}Take moments about the point where the rod touches the floor.
Let the rod make angle 30^\circ
with the floor (because it makes 60^\circ with the wall).
The weight acts at the midpoint of the rod (2.5 m from bottom).
Its perpendicular distance from the floor contact is:
2.5 \sin 30^\circ = 2.5 \times \frac{1}{2} = 1.25\ \text{m}The wall’s normal force N_{w} acts horizontally at the top end.
Its perpendicular distance from the floor contact is:
5 \sin 30^\circ = 5 \times \frac{1}{2} = 2.5\ \text{m}Equating moments about the bottom:
mg(1.25) = N_{w}(2.5)Substitute mg = 200:
200 \times 1.25 = 2.5 N_{w} 250 = 2.5 N_{w} N_{w} = 100Since f = N_{w}, the friction force is:
f = 100But this is the horizontal component; the rod makes 60^\circ
with the wall and thus the friction direction resolves into floor coordinates.
The actual friction magnitude required at the floor is:
f = N_{w} \sec 60^\circ = 100 \times \sqrt{3}Thus the friction exerted by the floor is:
100\sqrt{3}\ \text{N}