Question Detail
Question
A girl weighing 20 kg sits at one end of a uniform beam 10 m long. Another girl weighing twice as much as the first girl sits on the other end of the beam. If the beam is pivoted in the middle, at what distance from the 20 kg girl should a 25 kg mass be suspended in order to balance the beam?
Correct Answer
Option A is the correct answer.
Detailed Explanation
Beam weight acts at centre (5 m from ends). Pivot at centre. Let left end be 20 kg girl at 0 m, right end be 40 kg girl at 10 m. Moments about pivot: Left side moment = 20g × 5 = 100g (anticlockwise). Right side moment = 40g × 5 = 200g (clockwise). Net clockwise = 100g. To balance, add 25 kg mass on left side at distance d from left end (so from pivot = d - 5). Moment from 25 kg = 25g × (d - 5) anticlockwise. For equilibrium: 100g + 25g(d-5) = 200g → 100 + 25(d-5) = 200 → 25(d-5) = 100 → d-5 = 4 → d = 9 m from left end. Distance from 20 kg girl = 9 m, not in options. If mass placed on right side: 200g + 25g(5 - d) = 100g → 25(5-d) = -100 → 5-d = -4 → d=9 m same. If we consider the 25 kg at 4 m from left, then at 4 m from left is 1 m left of pivot, moment = 25g, left total = 125g, right = 200g, not balanced. So likely the intended answer is 4 m if they meant distance from pivot?
Hint
Use principle of moments about pivot.