the ruler is exactly 1 m long and has mass 542 g the mass

- the ruler is exactly 1 m long and has mass 542 g
- the mass hangers have negligible mass.

1-a: Suppose the ruler in procedure 2 is asymmetrical,(center of mass is not at 50 cm) and it is balanced on a fulcrum at the 45.6 cm mark (at the center of mass). Now, mass M is hung on the ruler at the 90 cm mark.
- Where must you hang mass 2.2M so the system remains in equilibrium?
At the _________ cm mark

b: The method of procedure 2 can be used to accurately determine the mass of a light coin. Suppose the ruler is now perfectly symmetrical, and it balances at the center. The coin of unknown mass is placed at the 0 cm mark, it is balanced by a mass m = 39.5 g placed at the 75.9 cm mark. Find the mass of the coin.

mcoin = _______ g

Solution

Since the ruler itself is balanced, we can leave it out of the calculations.

The moment for the first mass, M, is the distance from the pivot, 90–45.6 = 44.4 cm multiplied by the mass, or 44.4M

The second mass, 2.2M is at distance x from the pivot. Set them equal and solve for x
x = 44.4M / 2.2M = 20.18 cm.
That is from the pivot point
The distance from the 0 cm point, the left, that is 42.4–33.1 = 9.3 cm

M (coin) . 50 = m ( 39.5 ) . (75.9 – 50 )

M (coin) = 39.5 . 25.9 / 50 = 20.46 g