Question Detail
Question
A 1 m long uniform beam of 2 kg mass is hinged at the 0 cm mark. What minimum vertical force applied at the 100 cm mark is required to lift the beam? Acceleration due to gravity, g = 10 m/s².
Correct Answer
Option C is the correct answer.
Detailed Explanation
The weight of the beam (2 kg × 10 m/s² = 20 N) acts at the center of mass, which is at the 50 cm mark. For the beam to just lift off, the torque due to the applied force at 100 cm must equal the torque due to weight about the hinge at 0 cm. So F × 100 cm = 20 N × 50 cm. F = (20 × 50)/100 = 1000/100 = 10 N.
Hint
Apply the principle of moments about the hinge: clockwise torque = counterclockwise torque.