Question

Uniform meter rod is balanced at 70cm mark. A weight of 20gf is suspended at 40cm mark and another 200gf is suspended at 95 cm mark. Find the weight of the meter rod

Answer 1

let the weight of meter rod be x

the weight acts on the center of mass lying on 50cm

the rod is in equilibrium

the clockwise moments = anticlockwise moments

(20x40) + 50x = 200x95

50x = 19000 - 800

50x = 18200

x = 364 gf