A meter rule AB of mass 0.2kg was balanced horizontally on a single support when loads of 0.1kg and 0.08kg were placed at a distance of 60cm and 90cm respectively from A. find the position of the support and the magnitude of the downward force

Answer 1

Fdown = ( .2 +.1 + .08 ) g Newtons = 0.308 g Newtons = Fup
Clockwise turning moment = (.2*50 +.1*60 + .08 *90) g = 23.2 g Newton cm
counterclockwise moment = 0.308 g * x
so if it does not spin
0.308 g x = 23.2 g
x = 75.3 cm