Question

Asked May 2, 2019

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.37 g coins stacked over the 30.0 cm mark, the stick is found to balance at the 46.3 cm mark. What is the mass of the meter stick?

Step 1

The meter stick is in equilibrium when the 50.0 cm mark (center of mass) is placed on the knife edge. Two coins of mass 5.37 g are placed at the 30.0 cm mark. Then the new equilibrium position is 46.3 cm.

Step 2

According to principle of moments, the net anticlockwise moment about a pivot point is equal to the net clockwise moment about the same point. The moment of force is given by the product of the force and its perpendicular distance from the pivot point.

Step 3

Equate the moments about the new equilibrium position

Here, *m*1 is the mass of the coins, *m*2 is the mass of the stick, *g* is the acceleration due to gravity, *r*1 is the distance between the new equilibrium...

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