Question

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.67 g coins stacked over the 36.0 cm mark, the stick is found to balance at the 42.1 cm mark. What is the mass of the meter stick?

Answer 1

The mass of the ruler is uniformely distributed throughout its length therefore we can model it as a point mass acting at its midpoint, the 50.0cm mark. As the pivot is at 42.1 the distance to the mass of the ruler M is 7.9cm. The distance to the coins from the pivot is 42.1-36 = 6.1cm. Taking moments (torque) we have:

11.34g x 6.1 = Mg x 7.9 (the g's cancel)

M = 69.174/7.9 = 8.76 gm

The moments are negative and positve because they are acting on different sides of the pivot (eg the coins are 6.1cm one side and the M is 7.9 cm the other side). By normal force do you mean the reaction force in the pivot? That acts at the 42.1cm mark and is equal to (11.34 + 8.76)g = 20.1g Newtons.