Question

 If two weights are suspended of masses 25 kg and 50 kg at a distance of 1m and 2m from middle,where should a mass of 74 kg be suspended to balance it?​

Answer 1

Answer:

Explanation:

by applying rotational equilubrium , that is net torque about the axis should be zero

torque due to 25 kg  =  25 * 1 * g = 250 N/m (LET IT BE CLOCKWISE )

torque due to 50 kg = 50 * g * 2 = 1000N/m ( since this is suspended on other side of the scale from that of 25 kg te torque caused will be opposite of that by 25 kg hence let it be anticlockwise )

net torque on system = 750 N/m ( anticlockwise ) and a 74 kg should be used   to balance this on the same side as 25 kg

750 = 74 * g * x

750 = 740 *x

x =  1.01 m