Physics · Simulation and virtual lab Free
Moments: balance a metre rule on a knife edge
Hang masses on a metre rule, slide them until it balances and discover the principle of moments.
- You will learn to
- Calculate moments in N·m.
- Verify the principle of moments with a balanced metre rule.
- Find the mass of a metre rule using moments.
- You will measure
- Masses (g) and distances from the pivot (cm); moments in N·m.
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: metre rule, knife edge, thread, slotted masses, balance.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To verify the principle of moments with a metre rule balanced on a knife edge, and to use it to find the mass of the metre rule.
What to do
- Keep the pivot at 50 cm. Set mass A = 100 g at 20 cm and mass B = 100 g. Slide B until the rule balances, then record.
- Make A = 200 g. Find the balance position of B again and record. Repeat with other masses and positions (at least 5 results).
- For each row calculate mass × distance on each side. Look at the graph: what do you notice?
- Choose Find the mass of the metre rule. The pivot is now at 40 cm and only mass A hangs on the rule.
- Slide A until the rule balances and record. Repeat with 50 g, 100 g and 200 g.
- Use M × (50 − 40) = mA × dA to calculate the mass M of the rule each time, then take the average.
Results
📏 Read the instruments carefully
| Instrument | Smallest reading | Uncertainty |
|---|
| Metre rule | 1 mm | ± 0.5 mm for each position |
| Slotted masses | 50 g | ± 1 g (check with a balance) |
| Knife edge (balance point) | — | about ± 0.5 cm because of friction |
Write every reading with its unit, and with the same number of decimal places as the instrument. Use the 🔍 Magnifier on the picture to read small scales at eye level.
⚠️ Avoid these mistakes
- Measuring distances from the end of the rule instead of from the pivot.
- Forgetting the weight of the rule when the pivot is not at its centre.
- Recording before the rule is really level: wait until it stays horizontal.
✍️ Conclusion
Write your own conclusion first: what do your results and graph show, and which law or principle do they support?
Check your conclusion
When the rule is balanced, the sum of the anticlockwise moments about the pivot equals the sum of the clockwise moments (principle of moments). The weight of a uniform rule acts at its centre, so moments can be used to find the mass of the rule.
Questions
Define the moment of a force about a point.
The moment is the force multiplied by the perpendicular distance from the pivot to the line of action of the force (unit N·m).
State the principle of moments.
When a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.
Why is the rule balanced at 50 cm first, with no masses?
So that its centre of gravity is exactly over the pivot: its own weight then has no moment and does not affect the results.
A 100 g mass hangs 30 cm from the pivot. Find the distance needed for a 150 g mass to balance it.
100 × 30 = 150 × d, so d = 20 cm on the other side.
Why do we use thread loops (not glue) to hang the masses?
The loops can be moved easily along the rule, and the force acts at a clear, single point whose position can be read.
What is a couple?
Two equal and opposite parallel forces whose lines of action do not meet; they produce turning only. Moment of a couple = one force × perpendicular distance between them.
For teachers: expected results and how the model works
g = 9.8 N/kg. The rule balances when the net moment is less than 0.0025 N·m (like friction at a real knife edge), so learners must place masses to about ±0.5 cm. The rule mass is 115 g: with the pivot at 40 cm, 50 g balances at 17.0 cm, 100 g at 28.5 cm. “Help me balance” places the mass at the exact position.
Quick check 3 questions · scored
Choose one answer for each question. You see at once if it is right, why, and your score at the end.
1. A 0.50 N force acts 0.40 m from a pivot. What is its moment?
Moment = force × distance = 0.50 × 0.40 = 0.20 N·m.
2. 100 g hangs 30 cm left of the pivot. Where must 150 g hang to balance it?
100 × 30 = 150 × d, so d = 20 cm on the right.
3. A metre rule balances at 40 cm with 50 g at the 15 cm mark. Mass of the rule?
50 × 25 = M × 10, so M = 125 g.
Your score:
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A uniform plank 4.0 m long and weighing 200 N rests on a pivot at its centre. A child of weight 300 N sits 1.5 m from the pivot. Where must a child of weight 450 N sit to balance the plank? (3 marks)
Marking scheme: Clockwise moment = anticlockwise moment (1): 300 × 1.5 = 450 × d (1); d = 1.0 m on the other side of the pivot (1).
A metre rule is balanced at the 40 cm mark when a 50 g mass hangs at the 15 cm mark. Calculate the mass of the rule. (4 marks)
Marking scheme: The weight of the rule acts at 50 cm, 10 cm from the pivot (1). 50 × (40 − 15) = M × 10 (1); 1250 = 10 M (1); M = 125 g (1).
State two conditions for a body to be in equilibrium. (2 marks)
Marking scheme: The resultant force on the body is zero (1); the sum of the moments about any point is zero (clockwise = anticlockwise) (1).
Explain why it is easier to open a door by pushing at the handle than near the hinges. (3 marks)
Marking scheme: Moment = force × perpendicular distance (1). The handle is far from the hinge (pivot), so the distance is large (1); a smaller force gives the same moment needed to turn the door (1).
Explain the difference between stable and unstable equilibrium, with one example of each. (4 marks)
Marking scheme: Stable: when slightly displaced, the body returns to its original position because its centre of gravity rises (1); e.g. a cone on its base (1). Unstable: when slightly displaced, it moves further away because its centre of gravity falls (1); e.g. a cone on its tip (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Moments: balance a metre rule on a knife edge
Aim: To verify the principle of moments with a metre rule balanced on a knife edge, and to use it to find the mass of the metre rule.
Apparatus (real laboratory): metre rule, knife edge, thread, slotted masses, balance.
What to do (with the simulation on www.donatstem.co.rw)
- Keep the pivot at 50 cm. Set mass A = 100 g at 20 cm and mass B = 100 g. Slide B until the rule balances, then record.
- Make A = 200 g. Find the balance position of B again and record. Repeat with other masses and positions (at least 5 results).
- For each row calculate mass × distance on each side. Look at the graph: what do you notice?
- Choose Find the mass of the metre rule. The pivot is now at 40 cm and only mass A hangs on the rule.
- Slide A until the rule balances and record. Repeat with 50 g, 100 g and 200 g.
- Use M × (50 − 40) = mA × dA to calculate the mass M of the rule each time, then take the average.
Results
| # | mA (g) | dA (cm) | mB (g) | dB (cm) | mA × dA | mB × dB |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
Graph
Plot the anticlockwise moment on the vertical axis against the clockwise moment. Draw the best line and find its gradient.
Questions
- Define the moment of a force about a point. [2]
- State the principle of moments. [2]
- Why is the rule balanced at 50 cm first, with no masses? [2]
- A 100 g mass hangs 30 cm from the pivot. Find the distance needed for a 150 g mass to balance it. [2]
- Why do we use thread loops (not glue) to hang the masses? [2]
- What is a couple? [2]
- A uniform plank 4.0 m long and weighing 200 N rests on a pivot at its centre. A child of weight 300 N sits 1.5 m from the pivot. Where must a child of weight 450 N sit to balance the plank? [3]
- A metre rule is balanced at the 40 cm mark when a 50 g mass hangs at the 15 cm mark. Calculate the mass of the rule. [4]
- State two conditions for a body to be in equilibrium. [2]
- Explain why it is easier to open a door by pushing at the handle than near the hinges. [3]
- Explain the difference between stable and unstable equilibrium, with one example of each. [4]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Moments: balance a metre rule on a knife edge
Expected results
g = 9.8 N/kg. The rule balances when the net moment is less than 0.0025 N·m (like friction at a real knife edge), so learners must place masses to about ±0.5 cm. The rule mass is 115 g: with the pivot at 40 cm, 50 g balances at 17.0 cm, 100 g at 28.5 cm. “Help me balance” places the mass at the exact position.
Answers
- Define the moment of a force about a point. [2]
The moment is the force multiplied by the perpendicular distance from the pivot to the line of action of the force (unit N·m).
- State the principle of moments. [2]
When a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.
- Why is the rule balanced at 50 cm first, with no masses? [2]
So that its centre of gravity is exactly over the pivot: its own weight then has no moment and does not affect the results.
- A 100 g mass hangs 30 cm from the pivot. Find the distance needed for a 150 g mass to balance it. [2]
100 × 30 = 150 × d, so d = 20 cm on the other side.
- Why do we use thread loops (not glue) to hang the masses? [2]
The loops can be moved easily along the rule, and the force acts at a clear, single point whose position can be read.
- What is a couple? [2]
Two equal and opposite parallel forces whose lines of action do not meet; they produce turning only. Moment of a couple = one force × perpendicular distance between them.
- A uniform plank 4.0 m long and weighing 200 N rests on a pivot at its centre. A child of weight 300 N sits 1.5 m from the pivot. Where must a child of weight 450 N sit to balance the plank? [3]
Clockwise moment = anticlockwise moment (1): 300 × 1.5 = 450 × d (1); d = 1.0 m on the other side of the pivot (1).
- A metre rule is balanced at the 40 cm mark when a 50 g mass hangs at the 15 cm mark. Calculate the mass of the rule. [4]
The weight of the rule acts at 50 cm, 10 cm from the pivot (1). 50 × (40 − 15) = M × 10 (1); 1250 = 10 M (1); M = 125 g (1).
- State two conditions for a body to be in equilibrium. [2]
The resultant force on the body is zero (1); the sum of the moments about any point is zero (clockwise = anticlockwise) (1).
- Explain why it is easier to open a door by pushing at the handle than near the hinges. [3]
Moment = force × perpendicular distance (1). The handle is far from the hinge (pivot), so the distance is large (1); a smaller force gives the same moment needed to turn the door (1).
- Explain the difference between stable and unstable equilibrium, with one example of each. [4]
Stable: when slightly displaced, the body returns to its original position because its centre of gravity rises (1); e.g. a cone on its base (1). Unstable: when slightly displaced, it moves further away because its centre of gravity falls (1); e.g. a cone on its tip (1).