OCR A Level Physics A: Moments, equilibrium, density and pressure (H556) – Revision Notes

These revision notes condense sections 3.2.3 (Equilibrium) and 3.2.4 (Density and pressure) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Read the study guide, which explains each point in full, then test yourself with the practice questions.

Course pages: OCR A Level Physics hub | checklist | free diagnostics. Use g = 9.81 N kg⁻¹ unless a question says otherwise.

Definitions to recall

Term	Statement
Moment of a force (3.2.3(a))	Force times the shortest distance between the pivot and the line along which the force acts
Couple (3.2.3(b))	Two forces equal in size, opposite in direction, parallel, with different lines of action
Torque of a couple (3.2.3(b))	One of the forces times the perpendicular separation of the two force lines
Principle of moments (3.2.3(c))	In equilibrium, for any chosen point, total clockwise moment = total anticlockwise moment
Centre of mass (3.2.3(d))	The point where the mass of the object can be considered to act
Centre of gravity (3.2.3(d))	The point where the weight of the object can be considered to act
Equilibrium (3.2.3(e))	Resultant force zero and resultant moment zero
Density (3.2.4(a))	Mass per unit volume
Pressure (3.2.4(b))	Normal force per unit area
Upthrust (3.2.4(c))	Upward force on an object in a fluid, from higher pressure on its lower surface than its upper surface
Archimedes’ principle (3.2.4(c))	The upthrust on an object in a fluid equals the weight of fluid it displaces

Equations and units

Quantity	Equation	Unit
Moment	moment = Fx	N m
Torque of a couple	torque = Fd	N m
Density	ρ = m/V	kg m⁻³
Pressure	p = F/A	Pa (N m⁻²)
Pressure due to a liquid	p = hρg	Pa
Upthrust	U = ρ g V (V = volume of fluid displaced)	N

Conversions: 1 g cm⁻³ = 1000 kg m⁻³. 1 cm² = 1 × 10⁻⁴ m². 1 litre = 1 × 10⁻³ m³.

Moments and couples (3.2.3(a)-(b))

• x or d is always perpendicular to the force. If the force is at angle θ to the bar, use F sin θ.
• A force through the pivot has zero moment.
• A couple has zero resultant force, so it rotates an object but does not accelerate its centre of mass.
• The torque of a couple is the same about every point.

Reminder. A force of 120 N acts 0.55 m from a pivot along a lever, at 40° to the lever. Moment = 120 × sin 40° × 0.55 = 42.4 N m.

Method: principle of moments (3.2.3(c))

1. Sketch the object and label every force, including weight at the centre of gravity.
2. Pick the pivot where the most unknown forces act.
3. Write: sum of clockwise moments = sum of anticlockwise moments.
4. Solve for one unknown.
5. Apply the force condition (upward = downward) to find the others.
6. Check the answer is sensible: the support nearer the load carries more.

Centre of gravity (3.2.3(d))

• A regular uniform object has its centre of gravity at its geometric centre.
• Irregular lamina: hang it freely from at least two holes in turn. Each time hang a weighted string from the same point and draw the vertical. The centre of gravity is at the intersection. A third hole is a check.
• Rod or rule: it balances on a pivot at its centre of gravity.
• The centre of gravity lies directly below a point of suspension when the object is at rest.
• Hanging an object from a point and finding the position of the plumb line gives a line, not a point, which is why a second suspension point is needed.

Equilibrium (3.2.3(e)) and three forces (3.2.3(f))

Two conditions, both required:

1. Resultant force = 0 (resolve horizontally and vertically).
2. Resultant moment = 0 about any point.

Three coplanar forces in equilibrium:

• Their lines of action meet at a point.
• Their force vectors form a closed triangle when drawn head to tail.
• Solve with a scale drawing, Pythagoras, trigonometry or the sine rule: each force divided by the sine of the angle between the other two is constant.

Reminder. A 60 N weight hangs from a rope and is pulled sideways by a horizontal force until the rope makes 40° with the vertical. Tension = 60 / cos 40° = 78.3 N. Horizontal force = 60 tan 40° = 50.3 N.

Density, pressure and upthrust (3.2.4)

• ρ = m/V. Convert volumes to m³ before substituting.
• p = F/A. Shrinking the area raises the pressure for a given force.
• p = hρg gives the pressure due to the liquid only. Add the pressure of the atmosphere for the total.
• Pressure depends on vertical depth, not on the shape or width of the container.
• Upthrust comes from p = hρg: pressure on the lower face is greater than on the upper face. For a cuboid, U = hρgA = ρgV.
• Archimedes: U = weight of displaced fluid = ρ_fluid g V_displaced.
• Apparent weight = actual weight − upthrust.
• Floating: upthrust = weight, so V_displaced / V_object = ρ_object / ρ_fluid.

Reminder. Ice of density 920 kg m⁻³ floats in sea water of density 1030 kg m⁻³. The fraction submerged is 920 / 1030 = 0.893, so about 89% is below the surface.

Must-know distinctions

Pair	Difference
Moment and couple	A moment is about a chosen pivot. A couple is two forces whose resultant is zero and whose torque is the same about every point.
Centre of mass and centre of gravity	Mass versus weight. They coincide in a uniform gravitational field.
Pressure and force	Pressure is force per unit area. A large force can give a small pressure.
Pressure due to liquid and total pressure	Total = hρg + atmospheric pressure.
Upthrust and weight	Upthrust depends on the fluid displaced. Weight depends on the object.

Quick self-test

1. A force of 250 N acts at right angles to a spanner at a distance of 0.30 m from the nut. Calculate the moment.
2. A couple has a torque of 3.6 N m. The forces are 0.15 m apart. Calculate the size of each force.
3. A 400 N child sits 1.5 m from the pivot of a light seesaw. How far from the pivot must a 600 N adult sit on the other side to balance it?
4. A mass of 0.52 kg of oil fills a container of volume 6.0 × 10⁻⁴ m³. Calculate the density of the oil.
5. A force of 90 N acts normally on an area of 3.0 × 10⁻³ m². Calculate the pressure.
6. Calculate the pressure due to the water at a depth of 12 m in a lake. Density of water: 1000 kg m⁻³.
7. A fully submerged object has volume 5.0 × 10⁻⁴ m³. Calculate the upthrust in water.
8. State the two conditions for an object to be in equilibrium.
9. Describe how to find the centre of gravity of an irregular flat piece of card.
10. A uniform metre rule is pivoted at the 40.0 cm mark. A 0.30 kg mass hangs at the 10.0 cm mark and the rule balances. Calculate the mass of the rule.

Answers

1. 250 × 0.30 = 75 N m.
2. F = torque / d = 3.6 / 0.15 = 24 N.
3. 400 × 1.5 = 600 × x, so x = 1.0 m.
4. ρ = 0.52 / 6.0 × 10⁻⁴ = 870 kg m⁻³ (867 to 3 s.f.).
5. p = 90 / 3.0 × 10⁻³ = 3.0 × 10⁴ Pa.
6. p = 12 × 1000 × 9.81 = 1.2 × 10⁵ Pa (117 720 Pa).
7. U = 1000 × 9.81 × 5.0 × 10⁻⁴ = 4.9 N.
8. The forces sum to zero and the moments about any point sum to zero.
9. Make three holes. Hang the card freely from one hole with a weighted string from the same pin and draw the vertical. Repeat for the other holes. The intersection is the centre of gravity.
10. The rule’s weight acts at 50.0 cm, 10.0 cm from the pivot. M × 10.0 = 0.30 × 30.0, so M = 0.90 kg.

Where marks are usually lost

• Writing “force × distance” without saying the distance is perpendicular.
• Taking moments about one point but using distances measured from another.
• Leaving out the weight of a beam, or putting it at the end instead of the centre of gravity.
• Not stating both conditions for equilibrium when asked to explain why an object is stationary.
• Using the length of a diagonal rope instead of the shortest distance from the pivot to the rope’s line.
• Using the wrong area, or forgetting to convert cm² to m².
• Using p = hρg with a slanting length instead of the vertical depth.
• Quoting a final answer without a unit, or with the wrong one (N m for torque, not N).
• Describing the centre of gravity method without saying the card must hang freely and the plumb line must be hung from the same pin.
• Calling the displaced fluid’s weight “the object’s weight” in an Archimedes explanation.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 3.2.3 Equilibrium and 3.2.4 Density and pressure.