1. **Problem Statement:** A triangular structure with a horizontal base of 6 m and a side of 5 m inclined upwards supports a vertical weight of 500 N at the top vertex. A small horizontal segment of 0.50 m is marked from the left base corner. We need to find the tension in the string holding the structure in equilibrium under a 150 N force range. 2. **Relevant Physics Concept:** For equilibrium, the sum of forces and the sum of moments (torques) about any point must be zero. 3. **Assumptions and Setup:** - Let the tension in the string be $T$. - The 150 N force acts horizontally. - The weight $W = 500$ N acts vertically downward. - The base length is 6 m, and the small segment is 0.50 m from the left. 4. **Calculate the moment about the left base corner:** Taking moments about the left base corner to eliminate the tension force's moment (assuming it acts at the top vertex): - Moment due to weight $W$ at the top vertex (horizontal distance = 0.50 m): $$ M_W = W \times 0.50 = 500 \times 0.50 = 250 \text{ Nm} $$ - Moment due to horizontal force 150 N at the base (distance 0.50 m): $$ M_F = 150 \times 0.50 = 75 \text{ Nm} $$ 5. **Equilibrium condition for moments:** Sum of clockwise moments = Sum of anticlockwise moments Assuming tension $T$ creates a moment balancing these: $$ T \times 5 = 250 + 75 = 325 $$ 6. **Solve for tension $T$:** $$ T = \frac{325}{5} = 65 \text{ N} $$ 7. **Conclusion:** The tension in the string is $\boxed{65}$ N. This solution follows the Sri Lankan A/L physics syllabus principles of equilibrium and moments.