This set practises five skills: calculating a moment, balancing clockwise and anticlockwise moments, finding a centre of mass, explaining stability, and choosing the perpendicular distance. The questions get harder as you go. All numbers and situations are invented for practice.
Answer on paper first.
Then open the answer and compare your reasoning, not only your result. Write units every time. The mistake log and retest queue is useful for recording each slip.
Part A: Moments
Q1. A force of 12 N acts at right angles to a wrench, 0.25 m from the nut. Find the moment about the nut.
Show answer
Moment = 12 × 0.25 = 3.0 N m.
Check: 3.0 ÷ 0.25 = 12 N.
Q2. A force of 15 N acts at right angles to a lever, 40 cm from the pivot. Find the moment in N m.
Show answer
Convert: 40 cm = 0.40 m. Moment = 15 × 0.40 = 6.0 N m.
Check: 15 × 40 = 600 N cm, and 600 ÷ 100 = 6.0 N m.
Q3. A nut needs a turning effect of 24 N m. You use a wrench and apply the force at right angles, 0.80 m from the nut. What force is needed?
Show answer
Force = moment ÷ distance = 24 ÷ 0.80 = 30 N.
Check: 30 × 0.80 = 24 N m.
Part B: Balance
Q4. A light see-saw is pivoted at its middle. A person of weight 250 N sits 1.8 m from the pivot. Another person sits 1.5 m from the pivot on the other side and balances it. Find the second person’s weight.
Show answer
Moment of the first person = 250 × 1.8 = 450 N m. For the second: W × 1.5 = 450, so W = 450 ÷ 1.5 = 300 N.
Check: 300 × 1.5 = 450 N m. The heavier person sits closer to the pivot, as expected.
Q5. A light beam is pivoted at its middle. On the left, a 6.0 N weight hangs 0.50 m from the pivot. On the right, a 4.0 N weight hangs 0.30 m from the pivot and a 5.0 N weight hangs 0.60 m from the pivot. (a) Does the beam balance? (b) A weight is to hang 0.40 m from the pivot on the left to balance it. What must it be?
Show answer
(a) Anticlockwise: 6.0 × 0.50 = 3.0 N m. Clockwise: 4.0 × 0.30 = 1.2 N m and 5.0 × 0.60 = 3.0 N m, so the total is 4.2 N m. The moments are not equal, so the beam does not balance. It turns clockwise.
(b) The extra anticlockwise moment needed is 4.2 − 3.0 = 1.2 N m. Weight = 1.2 ÷ 0.40 = 3.0 N.
Check: anticlockwise total becomes 3.0 + 3.0 × 0.40 = 4.2 N m, which equals the clockwise total.
Q6. A uniform metre rule of weight 1.2 N is pivoted at the 30 cm mark. A weight W hangs at the 5 cm mark and the rule balances. Find W and the upward force from the pivot.
Show answer
The rule’s weight acts at its centre, the 50 cm mark, which is 20 cm = 0.20 m to the right of the pivot. Clockwise moment = 1.2 × 0.20 = 0.24 N m.
W hangs 30 − 5 = 25 cm = 0.25 m to the left. W × 0.25 = 0.24, so W = 0.24 ÷ 0.25 = 0.96 N.
The pivot’s upward force balances the downward forces: 1.2 + 0.96 = 2.16 N.
Part C: Centre of mass and stability
Q7. A light rod is 1.0 m long. An 8 N weight is fixed at one end and a 2 N weight at the other. How far from the 8 N end is the balance point?
Show answer
Let the distance be x. Then 8 × x = 2 × (1.0 − x), so 8x = 2 − 2x, so 10x = 2 and x = 0.20 m.
Check: 8 × 0.20 = 1.6 N m and 2 × 0.80 = 1.6 N m.
Q8. Describe how you would find the centre of mass of an irregular piece of card, and explain why the method works.
Show answer
Make small holes near the edge. Hang the card freely from a pin through one hole, hang a plumb line from the same pin, and draw the line it marks. Repeat from a second hole. The centre of mass is where the lines cross. A third hole gives a check line through the same point.
It works because the card at rest hangs with its centre of mass directly below the pin, so the centre of mass lies on the vertical line under the pin.
Q9. A uniform block is 0.30 m wide and 0.80 m tall. It stands on its 0.30 m wide base. (a) Where is its centre of mass? (b) By about what angle must it tilt, about a bottom edge, before it topples? (c) Explain why a block 0.30 m wide and 0.30 m tall is more stable.
Show answer
(a) At the centre of the block: 0.40 m above the base and 0.15 m from each side.
(b) It is on the point of toppling when the centre of mass is directly above the pivot edge. tan θ = 0.15 ÷ 0.40 = 0.375, so θ is about 21°.
(c) The 0.30 m block has its centre of mass 0.15 m high and 0.15 m from the edge, so tan θ = 1 and θ = 45°. Its lower centre of mass relative to its base means it tilts much further before the vertical line from the centre of mass leaves the base. The angle calculation is an illustration; in an exam you may only need the words.
Part D: Perpendicular distance
Q10. A 50 N force acts on a bar pivoted at one end. The force acts at a point 1.0 m from the pivot, but not at right angles to the bar. The perpendicular distance from the pivot to the line of the force is 0.60 m. Find the moment and explain why 50 N m is wrong.
Show answer
Use the perpendicular distance: 50 × 0.60 = 30 N m.
50 × 1.0 = 50 N m is wrong because 1.0 m is the length along the bar, not the shortest distance from the pivot to the force’s line of action. A slanted force has less turning effect than a right-angled one of the same size at the same point.
Check: a right triangle with hypotenuse 1.0 m and one side 0.60 m has another side of 0.80 m, since 0.60² + 0.80² = 0.36 + 0.64 = 1.00.
Q11. A spanner of length 0.40 m is pushed at its end with 30 N, in a direction at 30° to the spanner’s length. Find the moment about the nut. (sin 30° = 0.5)
Show answer
Perpendicular distance = 0.40 × sin 30° = 0.40 × 0.5 = 0.20 m. Moment = 30 × 0.20 = 6.0 N m.
Check: a right-angled push of 30 N would give 30 × 0.40 = 12 N m. The slanted push is exactly half, as sin 30° = 0.5.
If you got these wrong
| What went wrong | Go to |
|---|---|
| Wrong units, or centimetres used with newtons | Calculate a moment about a pivot |
| Forces added instead of moments, or a force missed | Use clockwise and anticlockwise balance |
| Balance point placed in the middle by habit | Locate a centre of mass conceptually |
| Explanation based on mass, not base and centre of mass | Explain stability using a base and centre of mass |
| Bar length used as the distance | Check perpendicular distance in a turning problem |
Return to the moments and stability module for the study route. Students who want a teacher to go through their own working can look at online one-to-one Physics tuition.