1: - the ruler is exactly 1 m long and has mass 664 g - the mass hangers have ne

Question

1:

- the ruler is exactly 1 m long and has mass 664 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 48.7 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 2.92M so the system remains in equilibrium?
At the _________ cm mark

1-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 = 27.2 g placed at the 82 cm mark. Find the mass of the coin.

mcoin = _______ g

Explanation / Answer

a) 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 – 48.7 = 41.3 cm multiplied by the mass, or 41.3M
The second mass, 2.92M is at distance x from the pivot. Set them equal and solve for x
x = 41.3M/2.92M = 14.14 cm.
That is from the pivot point
The distance from the 0 cm point, the left, that is 48.7 - 14.14 = 34.56 cm

b) M (coin) *45 = (27.2)*(82 – 45)

M (coin) = 27.2 * 37/45 = 22.36 g

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.