1. **Problem statement:** A wire DE made of A-36 steel with diameter 5 mm supports a beam with a man of mass 80 kg sitting at seat C. We need to find the elongation of wire DE due to the man's weight. 2. **Given data:** - Diameter of wire, $d = 5$ mm - Mass of man, $m = 80$ kg - Lengths: $AB = 800$ mm, $BC = 600$ mm, vertical distance $AE = 600$ mm - Material: A-36 steel (Young's modulus $E \approx 200 \times 10^3$ MPa) 3. **Step 1: Calculate the force due to the man’s weight** $$W = mg = 80 \times 9.81 = 784.8\ \text{N}$$ 4. **Step 2: Determine the geometry to find the length and angle of wire DE** Point D is at $B + BC/2$ horizontally (since wire attaches between B and C, assume at C for simplicity), so horizontal distance from wall to D is $AB + BC = 800 + 600 = 1400$ mm. Vertical distance from D to E is $AE = 600$ mm. Length of wire DE: $$L = \sqrt{1400^2 + 600^2} = \sqrt{1,960,000 + 360,000} = \sqrt{2,320,000} = 1523.16\ \text{mm}$$ 5. **Step 3: Calculate the tension in wire DE** The man’s weight acts vertically downward at C, so the tension in wire DE must balance the vertical component of the force. Angle $\theta$ of wire DE with horizontal: $$\theta = \tan^{-1}\left(\frac{600}{1400}\right) = \tan^{-1}(0.4286) = 23.2^\circ$$ Vertical component of tension $T$: $$T \sin \theta = W \Rightarrow T = \frac{W}{\sin \theta} = \frac{784.8}{\sin 23.2^\circ} = \frac{784.8}{0.393} = 1997.4\ \text{N}$$ 6. **Step 4: Calculate the cross-sectional area of the wire** $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5}{2}\right)^2 = \pi \times 2.5^2 = 19.635\ \text{mm}^2$$ 7. **Step 5: Calculate elongation $\delta$ using Hooke’s Law** $$\delta = \frac{FL}{AE}$$ Where: - $F = T = 1997.4$ N - $L = 1523.16$ mm - $A = 19.635$ mm$^2$ - $E = 200 \times 10^3$ MPa = $200 \times 10^3$ N/mm$^2$ Calculate elongation: $$\delta = \frac{1997.4 \times 1523.16}{200 \times 10^3 \times 19.635} = \frac{3,042,000}{3,927,000} = 0.774\ \text{mm}$$ **Final answer:** The elongation of wire DE is approximately **0.77 mm**.