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 (20 N) acts at its center, which is at the 50 cm mark. To lift the beam, the torque due to the applied force at the 100 cm mark must equal the torque due to weight about the hinge. So F × 1 m = 20 N × 0.5 m => F = 10 N.
Hint
Apply the principle of moments about the hinge.