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

Answer:

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

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:

$\begin{aligned}m_{C} &= \frac{m_{A}\, r_{A} + m_{B}\, r_{B}}{r_{C}} \\ &= \frac{20\; {\rm g} \times 50\; {\rm cm} + 80\; {\rm g} \times 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}$.