Question 19: Equilibrium of Cylinder and Rods
Note: The vertical component of friction balances the weight of the cylinder.
From the equilibrium of the rods and the cylinder, we analyze the forces and torques.
1. Torque equation for the rod:
Taking torque about the hinge for one rod (mass $m$, length $l$):
$$ mg \frac{l}{2} \sin\theta = \mathcal{N} x $$
Where $x$ is the distance from the hinge to the point of contact along the rod.
2. Force balance for the cylinder:
Considering the vertical equilibrium:
$$ 2\mathcal{N}(\mu \cos\theta – \sin\theta) = Mg $$
However, using the friction coefficient relation derived in the notes:
$$ \mu = \frac{mg + Mg \frac{l}{x} \sin\theta}{mg \sin\theta \cos\theta \frac{l}{x}} $$
Simplifying based on the key values provided in the problem context:
$$ \mu = \frac{m}{M} \left( \frac{x}{\dots} \right) + \tan\theta $$
Substituting the numerical values from the calculation shown ($m/M$ ratio and angle properties):
$$ \ge \frac{m}{M} (\dots) + \tan\theta $$
$$ = 0.5 + 0.4 $$
$$ = 0.9 $$
Final Answer from Key:
$$ \ge \frac{m}{M}\left(\frac{\eta^2 l^2 + r^2}{rl}\right) + \frac{r}{\eta l} = 0.9 $$