1. **Problem Statement:** We have a uniform rod AB with weight 8 N acting at its center of gravity, which is at the midpoint of the rod. A 10 N weight is fixed at point B. We want to find the weight $W$ that should be hung at point A to keep the rod in horizontal equilibrium. 2. **Given Data:** - Weight of rod, $W_{rod} = 8$ N - Weight at B, $W_B = 10$ N - Length segments: $BM = 20$ cm (since 30 cm left of M to A and 10 cm right of M to B, total length 40 cm) - The rod is supported at points M and A. 3. **Key Concept:** For equilibrium, the sum of moments about any point must be zero. 4. **Step 1: Locate the center of gravity of the rod** Since the rod is uniform and length $AB = 40$ cm, its center of gravity is at midpoint, 20 cm from B (point M). 5. **Step 2: Take moments about point M** - Moment due to weight at B: $10 \times 20 = 200$ Ncm (clockwise) - Moment due to rod weight: $8 \times 0 = 0$ Ncm (since rod weight acts at M) - Moment due to weight at A: $W \times 30$ Ncm (counterclockwise) 6. **Step 3: Set sum of moments to zero for equilibrium** $$10 \times 20 = W \times 30$$ 7. **Step 4: Solve for $W$** $$W = \frac{10 \times 20}{30} = \frac{200}{30} = \frac{20}{3} \approx 6.67$$ 8. **Answer:** The weight to be hung at A is approximately $6.67$ N. Among the options given, the closest is option d) [1,62], but none exactly matches 6.67 N. "slug":"rod equilibrium","subject":"statics","desmos":{"latex":"y=0","features":{"intercepts":true,"extrema":true}},"q_count":1