Specification: AQA A-level Physics (7408) 3.4.1.2 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam
๐ All the equations in one place: the AQA A-level Every Key Equation PowerPoint has 127 equations, 381 practice questions and 15 derivations, for ยฃ12.
๐ KEY DEFINITIONS โ learn them word for word
Moment of a force: the force multiplied by the perpendicular distance from the pivot to the line of action of the force.
Principle of moments: for an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
Couple: a pair of equal and opposite parallel forces whose lines of action do not coincide. Its torque = one of the forces ร the perpendicular distance between the lines of action of the forces. A couple produces rotation only, with no resultant force.
Centre of mass: the single point through which the weight of an object can be taken to act. For a uniform object it is at the geometric centre.
The equation
moment = force ร perpendicular distance from the pivot to the line of action of the force moment = F d
| Quantity | Symbol | Unit |
|---|---|---|
| moment (or torque) | M | newton metres (N m) |
| force | F | newtons (N) |
| perpendicular distance | d | metres (m) |
Rearranged: F = moment รท d | d = moment รท F | Torque of a couple = F ร d, where d is the perpendicular distance between the two forces. If the force is at an angle ฮธ to the lever, use only its perpendicular component, F sin ฮธ.
Use FIFA for every answer
- F โ Formula: write the equation as it appears on the sheet.
- I โ Insert: put in the numbers (convert to standard units first).
- F โ Fix: rearrange to make the unknown the subject.
- A โ Answer: calculate and always give the unit.
Core: Q1โ4. Exam standard: Q5โ8. Stretch: Q9โ10.
Questions
Q1 (Core) A mechanic pushes on the end of a spanner with a force of 25 N at right angles to the spanner, 0.20 m from the centre of the nut. Calculate the moment about the nut.
Show solution
F: moment = F d
I: moment = 25 ร 0.20
F: moment is already the subject
A: 5.0 N m (2 s.f.)
Q2 (Core) A child of weight 320 N sits on a see-saw 1.5 m from the pivot. Calculate the moment of her weight about the pivot.
Show solution
F: moment = F d
I: moment = 320 ร 1.5
F: moment is already the subject
A: 480 N m (2 s.f.)
Q3 (Core) A moment of 12.0 N m is needed to open a stiff door. The handle is 78.0 cm from the hinge. Calculate the smallest force on the handle that will open the door, to 3 s.f.
Show solution
Convert first: 78.0 cm = 0.780 m
F: moment = F d
I: 12.0 = F ร 0.780
F: F = 12.0 รท 0.780
A: 15.4 N (3 s.f.), applied at right angles to the door
Watch out: the force is smallest when it is perpendicular to the door, because then d is the full 0.780 m.
Q4 (Core) A driver turns a steering wheel of diameter 380 mm by pushing up on one side and down on the other, with equal and opposite forces of 15 N acting tangentially to the rim. Calculate the torque of the couple.
Show solution
Convert first: 380 mm = 0.380 m (the perpendicular distance between the two forces is the full diameter)
F: torque = F ร d
I: torque = 15 ร 0.380
F: torque is already the subject
A: 5.7 N m (2 s.f.)
Watch out: for a couple, use one force ร the distance between the forces. Do not double it and do not use the radius.
Q5 (Exam) A 28 kg child sits on a see-saw 1.5 m from the pivot. Her 72 kg father sits on the other side so that the see-saw balances horizontally. Calculate how far from the pivot he sits, to 2 s.f. (Assume the see-saw’s own weight acts through the pivot.)
Show solution
F: sum of clockwise moments = sum of anticlockwise moments
I: 72 ร g ร d = 28 ร g ร 1.5
F: g cancels, so d = (28 ร 1.5) รท 72 = 0.5833โฆ m
A: 0.58 m (2 s.f.)
Q6 (Exam) A person holds a 5.0 kg mass in their hand with the forearm horizontal. The mass is 0.35 m from the elbow joint and the biceps muscle pulls vertically upwards on the forearm 4.0 cm from the elbow. Ignoring the weight of the forearm, calculate the force in the biceps, to 2 s.f. (g = 9.81 N kgโปยน)
Show solution
Convert first: 4.0 cm = 0.040 m; weight = m g = 5.0 ร 9.81 = 49.05 N
F: sum of clockwise moments = sum of anticlockwise moments (about the elbow)
I: 49.05 ร 0.35 = F ร 0.040, so 17.1675 = F ร 0.040
F: F = 17.1675 รท 0.040 = 429.1875 N
A: 430 N (2 s.f.)
Watch out: take moments about the elbow so the unknown joint force has no moment. The muscle force is almost nine times the load because its distance is so small.
Q7 (Exam) A uniform metre rule of mass 0.120 kg is pivoted at the 35.0 cm mark. A mass hung from the 5.0 cm mark makes the rule balance horizontally. Calculate the hung mass in grams, to 3 s.f.
Show solution
Step 1 โ centre of mass: the rule is uniform, so its weight acts at the 50.0 cm mark, 50.0 โ 35.0 = 15.0 cm = 0.150 m from the pivot. The hung mass is 35.0 โ 5.0 = 30.0 cm = 0.300 m from the pivot.
Step 2 โ principle of moments
F: sum of clockwise moments = sum of anticlockwise moments
I: 0.120 ร g ร 0.150 = m ร g ร 0.300
F: g cancels, so m = (0.120 ร 0.150) รท 0.300 = 0.0600 kg
A: 60.0 g (3 s.f.)
Watch out: measure distances from the pivot, not from the zero end of the rule.
Q8 (Exam) A wheelbarrow has its wheel axle as the pivot. The load has a weight of 650 N acting 0.45 m horizontally from the axle, and the barrow itself has a weight of 150 N acting 0.60 m from the axle. The gardener lifts the handles 1.35 m from the axle. Calculate the vertical force needed to hold the barrow steady, to 3 s.f.
Show solution
F: sum of clockwise moments = sum of anticlockwise moments (about the axle)
I: F ร 1.35 = (650 ร 0.45) + (150 ร 0.60) = 292.5 + 90 = 382.5 N m
F: F = 382.5 รท 1.35 = 283.33โฆ N
A: 283 N (3 s.f.)
Watch out: add every moment acting in the same direction before you rearrange.
Q9 (Stretch) A uniform plank of mass 40 kg and length 4.0 m rests horizontally on two supports, A and B, one at each end. A 70 kg decorator stands on the plank 1.0 m from A. Calculate the reaction force at each support, to 3 s.f. (g = 9.81 N kgโปยน)
Show solution
Step 1 โ weights: plank = 40 ร 9.81 = 392.4 N acting at the centre, 2.0 m from A; decorator = 70 ร 9.81 = 686.7 N acting 1.0 m from A
Step 2 โ moments about A (this removes the unknown force at A)
F: sum of clockwise moments = sum of anticlockwise moments
I: (392.4 ร 2.0) + (686.7 ร 1.0) = RB ร 4.0, so 784.8 + 686.7 = 1471.5 = RB ร 4.0
F: RB = 1471.5 รท 4.0 = 367.875 N
Step 3 โ vertical equilibrium: RA + RB = 392.4 + 686.7 = 1079.1 N, so RA = 1079.1 โ 367.875 = 711.225 N
A: RA = 711 N and RB = 368 N (3 s.f.)
Watch out: the support nearer the decorator takes the bigger force. Check your answer makes sense this way round.
Q10 (Stretch) A uniform trapdoor of mass 12 kg and width 0.90 m is hinged along one edge. It is held horizontal by a rope that runs from its free edge to a point on the wall above the hinge. The rope makes an angle of 40ยฐ with the trapdoor. Calculate (a) the tension in the rope and (b) the horizontal component of the force from the hinge on the trapdoor. Give both to 3 s.f. (g = 9.81 N kgโปยน)
Show solution
Step 1 โ weight: W = 12 ร 9.81 = 117.72 N, acting at the centre, 0.45 m from the hinge
Step 2 โ perpendicular component of the tension: only T sin 40ยฐ acts at right angles to the trapdoor, at a distance of 0.90 m from the hinge
F: sum of clockwise moments = sum of anticlockwise moments (about the hinge)
I: 117.72 ร 0.45 = T sin 40ยฐ ร 0.90, so 52.974 = T ร 0.57850โฆ
F: T = 52.974 รท 0.57850โฆ = 91.569โฆ N
Step 3 โ horizontal equilibrium: the hinge must balance the horizontal component of the tension, T cos 40ยฐ = 91.569โฆ ร cos 40ยฐ = 70.146โฆ N
A: (a) T = 91.6 N (b) 70.1 N (3 s.f.), directed away from the wall
Watch out: using T ร 0.90 without the sin 40ยฐ is the most common error. Only the perpendicular component of a force produces a moment.
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