1. **Stating the problem:** A 150 N weight is placed on a wire stretched along a triangular frame. The triangle has a base of 6 m and a left side of 5 m. A 500 N weight hangs from the top vertex. We need to find the tension in the wire. 2. **Understanding the setup:** The wire is along the base of the triangle (6 m). The 150 N weight is on the wire, and the 500 N weight hangs from the top vertex, creating tension in the wire. 3. **Key formulas and rules:** - The tension in the wire can be found by analyzing the forces and moments (torques) acting on the system. - The sum of moments about any point must be zero for equilibrium. 4. **Assign variables:** Let $T$ be the tension in the wire. 5. **Calculate the height of the triangle:** Using Pythagoras theorem for the triangle with sides 5 m and 6 m base: $$h = \sqrt{5^2 - (6/2)^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ m}$$ 6. **Calculate moments about the left corner of the base:** - The 150 N weight acts at 0.5 m from the left corner. - The 500 N weight acts at the top vertex, which is horizontally 3 m from the left corner (midpoint of base). Sum of moments about left corner: $$\text{Clockwise moments} = 150 \times 0.5 + 500 \times 3 = 75 + 1500 = 1575 \text{ Nm}$$ 7. **Tension force acts along the base (6 m), so its moment arm is 6 m:** Let the tension force $T$ act at the right corner, producing a counterclockwise moment: $$T \times 6 = 1575$$ 8. **Solve for tension $T$:** $$T = \frac{1575}{6} = 262.5 \text{ N}$$ **Final answer:** The tension in the wire is $262.5$ N.