Question Detail
Question
A 1 m long uniform beam of 2 kg mass is hinged at the 0 cm mark. What minimum vertical force at the 100 cm mark is required to lift the beam? g=10 m/s².
Correct Answer
Option C is the correct answer.
Detailed Explanation
Beam's weight (20 N) acts at center (50 cm). To lift, apply force F at 100 cm. Moment about hinge: 20 N × 50 cm = F × 100 cm → F = 10 N? Wait, that's 10 N, option C. But minimum force to just lift? Possibly when beam is horizontal, F_min = weight/2 = 10 N. So C. But options include 1.0 N, 2.0 N, etc. Recheck: weight = 2 kg * 10 = 20 N. Center at 0.5 m from hinge. Moment due to weight = 20 * 0.5 = 10 N·m. Force F at 1 m gives moment F*1. So F=10 N. So C is correct.
Hint
Balance moments about the hinge.