Torque Lab with Meter Stick and Weights Answers
Understanding how torque works is fundamental in physics, and one of the most hands-on ways to explore this concept is through a torque lab using a meter stick and weights. Worth adding: this experiment demonstrates the principles of rotational equilibrium, torque, and lever arms, providing practical insights into how forces act on objects around a pivot point. Whether you’re a student tackling this lab for the first time or an educator designing a lesson, this guide will walk you through the setup, procedure, and key answers to common questions that arise during the experiment Worth keeping that in mind..
Materials and Setup
To conduct a torque lab with a meter stick and weights, you’ll need the following materials:
- A meter stick (ruler or wooden dowel)
- A pivot point (fulcrum, such as a knife edge or support stand)
- Weights (masses like metal coins, washers, or calibrated masses)
- A ruler or measuring tape for precise distance measurements
- A scale (optional, to verify weight values)
The meter stick should be placed horizontally on the pivot point, ensuring it can rotate freely. Now, if the stick is uniform, it will naturally balance at its center of mass (50 cm mark). That said, adding weights to one side will shift the balance point, requiring adjustments to achieve equilibrium That alone is useful..
Procedure Steps
- Balance the meter stick alone: Place the pivot at the 50 cm mark and observe if the stick balances horizontally. If it doesn’t, adjust the pivot until equilibrium is reached.
- Add weights to one side: Attach a known weight (e.g., 200 g) to the stick at a specific distance from the pivot (e.g., 20 cm). Note the distance as the lever arm.
- Balance with a counterweight: Place a second weight on the opposite side of the pivot (e.g., at 80 cm) and adjust its mass until the system balances.
- Record data: Document the positions and masses of all weights, as well as the distances from the pivot.
- Repeat with varying configurations: Test different weight combinations and distances to observe how they affect rotational equilibrium.
Scientific Explanation of Torque and Equilibrium
Torque (τ) is the rotational analog of force, calculated as the product of the applied force and the lever arm (r), the perpendicular distance from the pivot to the force’s line of action:
τ = r × F
When the meter stick is in rotational equilibrium, the sum of all torques acting on it equals zero:
Στ = 0
This means the clockwise torques must equal the counterclockwise torques. To give you an idea, if a weight on the left side creates a clockwise torque, a weight on the right side must generate an equal counterclockwise torque to maintain balance Easy to understand, harder to ignore..
The lever arm is critical here. Also, a smaller mass at a greater distance can balance a larger mass at a shorter distance, illustrating the principle of torque multiplication. This concept is foundational in understanding tools like wrenches and seesaws.
Sample Calculations and Answers
Example 1: A 200 g weight is placed 20 cm to the left of the pivot. What mass is needed at 80 cm to the right to balance the system?
- Left torque: τ₁ = (20 cm) × (200 g) = 4,000 g·cm (clockwise)
- Right torque: τ₂ = (80 cm) × (m₂)
- Set τ₁ = τ₂: 4,000 = 80 × m₂ → m₂ = 50 g
Example 2: If the meter stick itself has a mass of 100 g and is uniform, where should a 300 g weight be placed 10 cm to the left of the pivot to balance a 150 g weight on the right?
- Left torque: τ₁ = (10 cm) × (300 g) + (50
Continuing the Sample Calculations
Example 2 (cont.) – The meter stick itself contributes a torque about the pivot because its centre of mass is located at the 50 cm mark. Assuming the stick is uniform, its weight (100 g) acts at the 50 cm point. If the pivot is currently positioned at the 50 cm mark, the torque due to the stick is:
- Stick torque: τₛ = (50 cm – pivot) × (100 g).
Since the pivot coincides with the centre of mass, the lever arm is zero and τₛ = 0.
That said, if the pivot is shifted to a new position d cm from the 50 cm mark, the stick’s torque becomes τₛ = d × 100 g (positive if it tends to rotate clockwise, negative otherwise).
For the purpose of this illustration, let’s keep the pivot at the 50 cm mark, so the stick’s own torque does not affect the balance. The condition for equilibrium is therefore:
[ \tau_{\text{left}} = \tau_{\text{right}} ]
where
[ \tau_{\text{left}} = (10\ \text{cm}) \times (300\ \text{g}) = 3{,}000\ \text{g·cm} ]
and [ \tau_{\text{right}} = (x\ \text{cm}) \times (150\ \text{g}) ]
Solving for the unknown distance x:
[3{,}000 = 150 \times x ;;\Longrightarrow;; x = \frac{3{,}000}{150} = 20\ \text{cm} ]
Thus, a 150 g mass must be placed 20 cm to the right of the pivot to counteract the 300 g load situated 10 cm to the left. If the pivot is fixed at the 50 cm mark, the 150 g weight should be attached at the 70 cm graduation line Small thing, real impact. That alone is useful..
Additional Worked Example
Example 3 – Suppose we now add a 50 g weight at the 30 cm mark (left side) while keeping the 300 g weight at 10 cm left of the pivot and the 150 g weight at 70 cm right of the pivot. Determine whether the system remains balanced Easy to understand, harder to ignore..
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Compute net left‑hand torque
- Torque from the 300 g weight: ( \tau_{300} = 10\ \text{cm} \times 300\ \text{g} = 3{,}000\ \text{g·cm} ) (clockwise)
- Torque from the 50 g weight: ( \tau_{50} = 30\ \text{cm} \times 50\ \text{g} = 1{,}500\ \text{g·cm} ) (clockwise)
- Total clockwise torque = ( 3{,}000 + 1{,}500 = 4{,}500\ \text{g·cm} )
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Compute right‑hand torque (unchanged)
- ( \tau_{150} = 20\ \text{cm} \times 150\ \text{g} = 3{,}000\ \text{g·cm} ) (counter‑clockwise)
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Compare
Since ( 4{,}500 \neq 3{,}000 ), the clockwise torque now exceeds the counter‑clockwise torque by 1{,}500 g·cm. The stick will rotate clockwise until a new equilibrium is reached, either by sliding the 150 g mass outward or by adding an additional counter‑weight It's one of those things that adds up..
To restore balance, we can solve for a new distance x for the 150 g mass:
[ 4{,}500 = 150 \times x ;;\Longrightarrow;; x = \frac{4{,}500}{150} = 30\ \text{cm} ]
Hence, moving the 150 g weight to the 80 cm mark (30 cm from the pivot on the right) will re‑establish equilibrium
