Let's break this problem down step-by-step:

Step 1: Setup of the System

We are given two smooth uniform rods $OB$ and $OC$, each of length $l$ and weight $W$, jointed at point $O$ to form an inverted "A" shape. They are symmetrically placed on a fixed horizontal rod of length $2a$. There is a light vertical string connected from the joint $O$ to the midpoint of the horizontal rod. The rods are inclined at an angle $\theta$ to the horizontal.

• The weight of each rod $W$ acts vertically downward at their center of mass.
• The string provides tension $T$ in the vertical direction.

Our goal is to find the tension $T$ in the string when the rods are inclined at an angle $\theta$.

Step 2: Symmetry and Force Balance

Since the rods are smooth and symmetrical, the system is in static equilibrium, meaning:

1. Vertical Force Balance: The upward force (from the tension $T$) must balance the downward gravitational force due to the weight of the rods.
2. Horizontal Force Balance: No net horizontal forces act on the system, as the smooth joints do not exert any horizontal reaction forces.

Step 3: Vertical Force Balance

Each rod has a weight $W$, and the total weight of the two rods is $2W$. The string must support this total weight, so the vertical component of the forces acting on the system is crucial.

1. Center of Mass of the Rods:

   • For each rod, the center of mass is at its midpoint, i.e., at $\frac{l}{2}$ from $O$.

2. Vertical Reaction at O:

   • The tension $T$ in the string acts vertically upward at point $O$, and its force must counteract the vertical force due to the weights of both rods.

3. Force Balance: $2W = T$

However, this is not sufficient since the rods are inclined, and there are also moments to consider.

Step 4: Moment Balance (Torque)

To fully solve for the tension $T$, we need to consider the moments (torques) about the point where the rods touch the fixed horizontal rod. Taking moments about the point $O$ (which eliminates the need to deal with the reaction forces at the rod supports), we get:

• The total weight $2W$ of the rods acts at a distance of $\frac{l}{2} \cos \theta$ from $O$.
• The string exerts a force $T$ at point $O$, with no moment arm.

Thus, the equation for the moments about $O$ is: $T a = W \frac{l}{2} \cos \theta$

Step 5: Solve for Tension $T$

Rearranging the above equation for $T$, we get: $T = \frac{W \frac{l}{2} \cos \theta}{a}$ $T = \frac{W l \cos \theta}{2a}$

Hence, the tension $T$ in the string is given by: $T = \frac{W l \cos \theta}{2a}$

This is the required expression for the tension in the string.

Do you need further clarification on any part of this process?

Here are 5 related questions to expand on this problem:

1. How would the tension $T$ change if the rods were not symmetric?
2. What happens to the tension if the rods are inclined at a smaller angle $\theta$?
3. Can you derive the horizontal force components acting at the joint O?
4. How would adding weight to the string affect the overall tension?
5. What would the tension be if the string were not vertical but at an angle?

Tip: Always check for symmetry in physical problems—it often simplifies calculations significantly!