maintaining balance?

Question

if you hang a meter stick in balance and then hang a 100 g mass on one side and a 200 g mass on the other side at a distance that causes the meter stick to maintain balance
a. will the 200 g mass be farther from the center than the 100 g mass
b. will the 100 g mass be farther away from the center than the 200 g mass
c. will both masses be the same distance from the center
d. would it depend upon the length of the meter stick to determine which mass was farther from the center
and how would you work this out to obtain the answer?

Answer 1

The correct answer would be b.

Answer 2

You work out the answer by making the moments equal on either side of the balance point. The moment is equal to force x distance. You don't have to calculate the actual force so you can just use mass times distance. A 100 g mass 50 cm from the balance point would have a (mass) moment of
100g x 50 cm = 5000 g-cm.
To balance this, the moment of the 200 g mass must also equal 5000 g-cm, so its distance from the pivot point is
5000 g-cm/200 g = 25 cm.

Or you might just intuit that the lighter mass needs to be further out to balance the heavier one. Think about two different-sized kids playing on a seesaw.

Answer 3

This is firstly a very unnecessary question (because everybody knows that) and secondly it does N O T belong to astronomy and space.