Asked by Criticos

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?
Answered by R_scott
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
Answered by bett
which is the answer
Answered by bett
show the workings
Answered by aron smash
i din't get answer
Answered by rotich
how to find
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