Question 17: Static Equilibrium of Bent Rod
Method 1: Forces and Torques
Let the bead be at the origin.
- Let $\theta$ be the angle of the left arm with the horizontal.
- Horizontal distance to left wall: $x_1 = L_1 \cos \theta$.
- Horizontal distance to right wall: $x_2 = L_2 \sin \theta$ (since the right arm is at $90-\theta$).
Horizontal Force Balance: $N_1 \sin \theta = N_2 \cos \theta \implies N_1/N_2 = \cot \theta$.
Combining these: $$ \frac{L_2}{L_1} = \frac{\cos \theta}{\sin \theta} \implies L_1 \cos \theta = L_2 \sin \theta \implies x_1 = x_2 $$
Aliter: Minimum Potential Energy
The system seeks the configuration of minimum Potential Energy, which corresponds to the lowest possible vertical position of the massive bead.
Geometric Constraint:
The bead is the vertex of a right angle ($90^\circ$) whose sides pass through two fixed points (the corners of the platforms).
The locus of a point $P$ such that the angle subtended by two fixed points $C_1$ and $C_2$ is $90^\circ$ is a circle with the segment $C_1 C_2$ as the diameter.
Finding the Minimum:
The bead moves along this circular path. The potential energy is minimum at the lowest point of this circle.
For a circle, the lowest point lies on the vertical axis passing through the center.
The center of the circle is the midpoint of the horizontal distance between the walls.
$$ x_{center} = \frac{x_{left\_wall} + x_{right\_wall}}{2} $$
Therefore, at equilibrium, the bead must be located at this horizontal midpoint, making it equidistant from both vertical faces.