A uniform metre rule of mass 150g is pivoted freely at 0 cm mark.What force applied vertically upwards at the 60 cm mark is needed to maintain the rule horizontally?

5 answers

this is a class 2 lever

the weight of the rule acts at its center of mass

f * 60 = (.15 g) * 50 ... g is gravitational acceleration ... 9.8 m/s^2
which is the answer
show the workings
i din't get answer
how to find