A uniform rod AB of length 3a and mass m rests in a horizontal position, supported by a light rod CD which connects a point C vertically below A to a point D on AB, where AD = a/2. The angle ACD is 30

Here's the solution for part (a) of the problem.

Part (a) (i): Show that $F = 2\sqrt{3}mg$.

Step 1: Identify the forces and geometry.
The uniform rod AB has length $3a$ and mass $m$. Its weight $mg$ acts downwards at its midpoint, which is at a distance of $1.5a$ from the hinge A. The rod AB is horizontal.
The light rod CD connects point C (vertically below A) to point D on AB. The distance $AD = \frac{a}{2}$. The angle $ACD = 30^\circ$.
Since C is vertically below A and D