Question

A physics experiment is done in which three masses are used to balance a bar on a fulcrum. Two masses (21 kg and 12 kg) are placed on one side of the fulcrum, 12 cm and 21 cm away, respectively. Where must the third mass (18 kg) be placed in order to make the bar balance?

Answer 1

You have to balance the torque in order to make the bar balance
torque on the 21 kg mass is equal to = 21*10*0.12 = 25.2 NM
torque on the 12 kg mass is equal to = 12*10*0.21=
25.2 NM
THUS total comes to 50.4 NM
to to balance the torque
50.4/180 = 0.28 that's 28 cm from the centre to the opposite of the both masses