Question

a meter stick is found to balance at the 49.7 - cm mark when placed on a fulcrum. when a 43.5 - gram mass is attached at the 26.5 - cm mark, the fulcrum must be moved to the 39.2 - cm mark for balance. what is the mass of the meter stick?

Explanation:

Step1: Calculate the torque due to the attached mass

The torque $\tau$ is given by $\tau = r\times F$, where $F = mg$. The distance of the attached mass from the new fulcrum is $r_1=(39.2 - 26.5)\text{ cm}=12.7\text{ cm}$. Let $m_1 = 43.5\text{ g}$. The torque due to the attached mass is $\tau_1=m_1g\times r_1$.

Step2: Calculate the distance of the center - of - mass of the meter - stick from the new fulcrum

The center - of - mass of the meter - stick is at its original balance point. The distance of the center - of - mass of the meter - stick from the new fulcrum is $r_2=(49.7 - 39.2)\text{ cm}=10.5\text{ cm}$. Let the mass of the meter - stick be $m_2$. The torque due to the meter - stick is $\tau_2=m_2g\times r_2$.

Step3: Apply the condition for rotational equilibrium

For rotational equilibrium, $\tau_1=\tau_2$. Substituting the expressions for $\tau_1$ and $\tau_2$:

$$m_1g\times r_1=m_2g\times r_2$$

Since $g$ cancels out on both sides of the equation, we have $m_2=\frac{m_1\times r_1}{r_2}$.
Substituting $m_1 = 43.5\text{ g}$, $r_1 = 12.7\text{ cm}$, and $r_2 = 10.5\text{ cm}$:

$$m_2=\frac{43.5\times12.7}{10.5}$$

$$m_2=\frac{552.45}{10.5}$$

$$m_2 = 52.6\text{ g}$$

Answer:

$52.6$ g