Question 14: Toppling of U-Shaped Cardboard
Goal: Find the angle $\theta$ at which the cardboard strip begins to topple (limiting case). This occurs when the line of action of the total weight passes through the pivot edge.
Step 1: Determine COM Coordinates
We use the coordinates relative to the corner of the base (the pivot point). Total mass $M_{total} = m(1 + 2\eta)$.
The horizontal position (x) of the center of mass is found by summing moments and dividing by $M_{total}$:
$$ X_{cm} = x = \frac{2 \cdot (\eta m) \cdot (\frac{\eta l}{2})}{m(1 + 2\eta)} = \frac{\eta^2 l}{2\eta + 1} $$The vertical position (y) of the center of mass is half the length of the vertical section:
$$ Y_{cm} = \frac{l}{2} $$Step 2: Apply Toppling Condition
The object begins to topple when the angle of inclination $\theta$ is equal to the angle the vector from the pivot to the COM makes with the normal to the base. Since $\frac{r_x}{r_y} = \tan \theta$, we use the geometry:
$$ \tan \theta = \frac{X_{cm}}{Y_{cm}} $$ $$ \tan \theta = \frac{\left( \frac{\eta^2 l}{2\eta + 1} \right)}{l/2} $$Step 3: Final Simplification
$$ \tan \theta = \frac{2\eta^2}{2\eta + 1} $$
Answer: $\theta = \tan^{-1} \left( \frac{2\eta^2}{2\eta + 1} \right)$