Let's analyze the problem step by step:

Problem Details:

1. Rod Information:

   • A uniform rod $OC$ is freely hinged at $O$ on a smooth vertical wall.
   • Rod length = $2 \, \text{m}$
   • Mass of the rod = $10 \, \text{kg}$ (hence its weight is $10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N}$).
   • A light inextensible cable $AB$ holds the rod in equilibrium.
   • $B$ is at the midpoint of the rod ($1 \, \text{m}$ from $O$).

2. Angle Information:

   • $\angle OAB = 45^\circ$, making $\triangle OAB$ a 45-45-90 triangle.

3. Forces Acting:

   • Weight of the rod acts at its center of gravity, $1 \, \text{m}$ from $O$.
   • Reaction force $R$ acts at $O$, perpendicular to the wall.
   • Tension $T$ in the cable $AB$ acts at an angle.

Questions to Address:

1. Diagram of Forces:

   • Draw the force diagram, showing:
     • The weight of the rod $W = 98 \, \text{N}$ acting vertically down at the center of the rod.
     • The reaction force $R$ at $O$ (horizontal due to the smooth wall).
     • The tension $T$ in the cable at $45^\circ$ to the wall.

2. Tension in Cable $AB$:

   • Use moments about $O$: $\text{Sum of moments about } O = 0$

3. Reaction at $O$:

   • Use equilibrium of forces in horizontal and vertical directions.

Solutions:

(i) Force Diagram:

Draw forces based on the descriptions:

1. $W = 98 \, \text{N}$, acting downward at the midpoint of the rod.
2. $T$, acting along $AB$ at an angle of $45^\circ$ to the horizontal.
3. $R$, horizontal reaction force at $O$.

(ii) Tension in Cable $T$:

Taking moments about $O$ (clockwise moments = counterclockwise moments): $T \times 2 \cos(45^\circ) = W \times 1$ Substitute: $T \cdot 2 \cdot \frac{\sqrt{2}}{2} = 98 \cdot 1$ $T \cdot \sqrt{2} = 98$ $T = \frac{98}{\sqrt{2}} = 69.3 \, \text{N}$

(iii) Reaction Force at $O$:

Use equilibrium conditions:

1. Horizontal forces: $R = T \cdot \cos(45^\circ) = 69.3 \cdot \frac{\sqrt{2}}{2} = 49 \, \text{N}$

2. Vertical forces: $W = T \cdot \sin(45^\circ)$ Already balanced, since $T \cdot \sin(45^\circ) = 69.3 \cdot \frac{\sqrt{2}}{2} = 49 \, \text{N}$.

Final Results:

1. Tension in cable $AB$: $T = 69.3 \, \text{N}$
2. Reaction force at $O$: $R = 49 \, \text{N}$

Would you like a detailed diagram or further clarification?

Follow-Up Questions:

1. How does the length of the rod affect the tension in the cable?
2. What happens to the reaction at $O$ if the angle changes from $45^\circ$?
3. If the wall were rough, how would it affect the equilibrium equations?
4. How does the position of point $B$ affect the forces?
5. What if the cable $AB$ were not light but had a mass?

Tip: For equilibrium problems, always resolve forces and take moments about a point to simplify calculations.