A uniform rod of length 120 cm and mass 300 g is supported on a knife-edge. A point mass of 150 g is attached to the 0 cm end of the rod. If the system is in horizontal equilibrium, the position of the knife-edge from the 0 cm end is ________ cm .

Step-by-step solution

Let the knife-edge be at a distance x cm from the 0 cm end. The mass of the uniform rod is 300 g , and its weight acts at its geometric centre, which is at the 60 cm mark. For rotational equilibrium about the knife-edge, the net torque must be zero. Torque due to the 150 g mass = Torque due to the mass of the rod 150g x = 300g (60 - x) 150x = 18000 - 300x 450x = 18000 x = 40 The position of the knife-edge is 40 cm from the 0 cm end. Answer: 40