Torque Lab

Purpose §

To find the mass of a meter stick using the new concept of torque we just learned.

Process §

1. Set up the meter stick so that the fulcrum is directly underneath the 60 cm marking, splitting it into two segments of 60 cm and 40 cm.
2. Find the spot where you can place the 50 gram weight on the 40 cm side so that the meter stick is fully at equilibrium and balanced (not in contact with the table, but parallel).
3. Record that point.

Variables §

Given §

• $g=9.8{\text{m/s}}^2$ (acceleration due to gravity)
• $m_{\text{weight}}=50\text{grams}=0.05\text{kg}$ (mass of the additional weight)
• $r_{\text{left}}=60\text{cm}=0.600\text{m}$ (length left of the fulcrum)
• $r_{\text{right}}=40\text{ cm}=0.400\text{ m}$ (length right of the fulcrum)
• $r_{\text{weight}}=20.6\text{ cm}=0.206\text{ m}$ (distance from the fulcrum to the weight, on the right side)

Reasoned §

The meter stick is balanced on the fulcrum, so it is at equilibrium and the total torque is equal to zero $\sum \tau =0\text{Nm}$

Measured (after, by Mr. Rich) §

• $m_{stick}=103\text{ grams}$

Calculations §

Let $x$ represent the mass of the meter stick. The mass of the meter stick is distributed along its length. For the purposes of calculating torque, we’ll consider:

• 60% of the mass of the ruler on the left side $m_{\text{left}}=0.6x$
• 40% of the mass of the ruler on the right side $m_{\text{right}}=0.4x$

Torques that are counterclockwise are positive and clockwise are negative. We will consider $r_{left}$ as clockwise and $r_{right}$ as counterclockwise. You have to divide $r_{left}$ and $r_{right}$ by two because the mass isn’t all concentrated at the end— it averages out to being distributed in the middle of each segment.

Torque is equal to $r\times F$, and the force of the mass on the meter stick is equal to its weight, which is $m\times g$. So, the torques acting on the meter stick are:

• Due to the 50g weight: ${\tau }_{\text{weight}}=r_{\text{weight}}\cdot F_{\text{weight}}=0.206\text{m}\cdot 0.05\text{kg}\cdot 9.8{\text{m/s}}^2$
• Due to the mass on the left of the fulcrum: ${\tau }_{\text{left}}=\frac{r_{\text{left}}}{2}\cdot F_{\text{left}}=-0.300\text{m}\cdot 0.6x\cdot 9.8{\text{m/s}}^2$
• Due to the mass on the right of the fulcrum: ${\tau }_{\text{right}}=\frac{r_{\text{right}}}{2}\cdot F_{\text{right}}=0.200\text{m}\cdot 0.4x\cdot 9.8{\text{m/s}}^2$

If you set that equal to the zero total torque we get…

$$\begin{array}{rl}0& ={\tau }_{\text{weight}}+{\tau }_{\text{left}}+{\tau }_{\text{right}}\\ & =[0.206\times 0.05\times 9.8]-[0.300\times 0.6x\times 9.8]+[0.400\times 0.4x\times 9.8]\end{array}$$

If you solve for $x$ you then get the mass.

$$\begin{array}{rl}x& =0.103\text{ kg}\\ & =103\text{ g}\end{array}$$

Accuracy §

It turns out that our calculations perfectly match the mass measured, so we did a pretty good job and got 100% percent accuracy.