A uniform rod of mass $20\text{ kg}$ and length $5\text{ m}$ 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 $g = 10\text{ m/s}^2$):
A
100 N
B
$100\sqrt{3}\text{ N}$
C
200 N
D
$200\sqrt{3}\text{ N}$
The answer is B.
Solution
Torque balance about the ground contact point gives normal force from wall $N_w = \frac{W}{2} \cot 30^\circ = 100\sqrt{3}\text{ N}$. Horizontal equilibrium requires friction $f = N_w = 100\sqrt{3}\text{ N}$.
The rod makes an angle of $60^\circ$ with the vertical wall, so its angle with the horizontal floor is $\theta = 90^\circ - 60^\circ = 30^\circ$. Weight of the rod is $W = mg = 20 \times 10 = 200\text{ N}$, acting at the midpoint. Taking torque about the bottom contact point O: $\sum \tau_O = 0 \implies N_w (l \sin\theta) = W \left(\frac{l}{2}\right) \cos\theta \implies N_w = \frac{W}{2} \cot\theta = \frac{200}{2} \cot 30^\circ = 100\sqrt{3}\text{ N}$. For horizontal translational equilibrium, the friction force exerted by the floor balances $N_w$: $f = N_w = 100\sqrt{3}\text{ N}$.
ConceptLadder / Leaning Rod Equilibrium
System of Particles and Rotational Motion in past papers
100 questions from this chapter have appeared across 27 exam years.
