Question

An 80g meter stick is supported at its 30 cm mark by a string attached to the ceiling. A 20 g mass is hung from the 80 cm mark what mass should be hung at the 5 cm mark on the meter stick to keep it horizontal and in equilibrium?

Answer 1

`104\; {\rm g}`, assuming that the meter stick is uniform with the center of mass precisely at the
`50\; {\rm cm}` mark.

Step-by-step explanation:

Refer to the diagram attached. The meter stick could be considered as a lever. The string at the
`30\; {\rm cm}` mark would then act as the fulcrum of this lever.

The
`m_(A) = 20\; {\rm g}` mass at the
`80\; {\rm cm}` mark is at a distance of
`r_(A) = 50\; {\rm cm}` to the right of the fulcrum at
`30\; {\rm cm}`.

The weight of the
`80\; {\rm g}` meter stick acts like a weight of
`m_(B) = 80\; {\rm g}` attached to the center of mass of this meter stick. Under the assumptions, this center of mass of this meter stick would be at the
`50\; {\rm cm}` mark, which is
`r_(B) = 20\; {\rm cm}` to the right of the fulcrum at
`30\; {\rm cm}`.

Let
`m_(C)` be the mass attached to the meter stick at the
`5\; {\rm cm}` mark. This mass would be at a distance of
`r_(C) = 25\; {\rm cm}` to the left of the fulcrum at
`30\; {\rm cm}`.

At equilibrium:

`\begin{aligned} & m_(C)\, r_(C) && (\text{mass on the left of fulcrum})\\ &= m_(A)\, r_(A) + m_(B) \, r_(B) && (\text{mass on the right of fulcrum})\end{aligned}`.

Solve for
`m_(C)`, the unknown mass attached to the meter stick at the
`5\; {\rm cm}` mark:

`\begin{aligned}m_(C) &= (m_(A)\, r_(A) + m_(B)\, r_(B))/(r_(C)) \\ &= \frac{20\; {\rm g} * 50\; {\rm cm} + 80\; {\rm g} * 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}`.