Question

3. In an experiment to prove the concept of the second condition of equilibrium, a boy balanced the meterstick on a peg board exactly at the midpoint. He then put a cylindrical metal at the 10cm mark and moved the meterstick to balance it again. If the meterstick weighs 12g, how much does the cylindrical metal weighs?

Answer 1

In an experiment to prove the concept of the second condition of equilibrium, a boy balanced the meterstick on a peg board exactly at the midpoint. He then put a cylindrical metal at the 10cm mark and moved the meterstick to balance it again. If the meterstick weighs 12g, how much does the cylindrical metal weighs?

Solution:

$L = 1m - \text{length of the meterstick}$

$m_{1} = 12g - \text{weight of the meterstick}$

$m_{2} - \text{weight of the cylindrical metal}$

$d = 10\mathrm{cm} - \text{distance from the end of meterstick to the cylindrical metal}$

$x - \text{displacement of the meterstick}$

Second condition of equilibrium (point A):

A: $M_{1} + M_{2} = 0$ (1)

$M_{1} = m_{1}g\cdot \left(\frac{L}{2} -x - d\right)$ (2)

$M_{2} = -m_{2}g\cdot \left(\frac{L}{2} +x\right)$ (3)

(3) and (2) in (1):

$$m _ {1} g \cdot \left(\frac {L}{2} - x - d\right) - m _ {2} g \cdot \left(\frac {L}{2} + x\right) = 0$$ $$\begin{array}{l} m _ {2} = m _ {1} \cdot \frac {\frac {L}{2} - x - d}{\frac {L}{2} + x} = m _ {1} \cdot \frac {L - 2 x - 2 d}{L + 2 x} = 1 2 g \cdot \frac {1 0 0 c m - 2 x - 2 0 c m}{1 c m + 2 x} \\ = 1 2 g \cdot \frac {8 0 c m - 2 x}{1 0 0 c m + 2 x} \\ \end{array}$$

Hence, to find $m_{2}$ we need to know the displacement of the meterstick $(x)$

Answer: to find $m_{2}$ we need to know the displacement of the meterstick, formula for the $m_{2}$:

$$m _ {2} = 1 2 g \cdot \frac {8 0 c m - 2 x}{1 0 0 c m + 2 x}.$$