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

This guide covers 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. H556 is the full A level. Dynamics, free-body diagrams, drag and terminal velocity (3.2.1 and 3.2.2) are not repeated here.

See the OCR A Level Physics hub and the printable checklist. Pair this page with the revision notes and the practice questions. Try the free diagnostics to check where you stand. Resolving forces uses Foundations of Physics.

Where no value of g is given, use 9.81 N kg⁻¹.

What you must be able to do

Statement	Content
3.2.3(a)-(b)	Moment = Fx; couple and torque = Fd
3.2.3(c)-(e)	Principle of moments; centre of gravity; equilibrium
3.2.3(f)	Three coplanar forces; triangle of forces
3.2.4(a)-(b)	Density ρ = m/V; pressure p = F/A
3.2.4(c)	p = hρg; upthrust; Archimedes’ principle

3.2.3(a) Moment of a force

A force can turn an object about a point or axis called the pivot. A force’s moment about a point is the force times x, the shortest (perpendicular) distance between that point and the line along which the force acts:

moment = Fx

The unit is the newton metre (N m). If a force acts at an angle θ to a bar, only the component perpendicular to the bar turns it, so the moment is F sin θ multiplied by the distance along the bar. A force acting through the pivot has zero moment.

Worked example. A spanner is 0.24 m long. A force of 85 N is applied at its end, at 30° to the spanner. Find the moment about the nut.

Perpendicular component of the force = 85 sin 30° = 42.5 N. Moment = 42.5 × 0.24 = 10.2 N m (3 s.f.).

3.2.3(b) Couples and torque

A couple is a pair of forces that are equal in size, opposite in direction and parallel, but whose lines of action do not coincide. The resultant force is zero, so a couple produces rotation only.

The torque of a couple is one of the forces times d, the perpendicular separation of the two force lines:

torque = Fd

The torque is the same about every point, so no pivot needs choosing.

Worked example. A finger and thumb apply forces of 2.4 N each, in opposite directions, to opposite sides of a tap head. The lines of action are 0.058 m apart.

Torque = 2.4 × 0.058 = 0.139 N m (3 s.f.).

3.2.3(c) The principle of moments

For an object in equilibrium, take any one point: the total clockwise moment about it is equal to the total anticlockwise moment about it.

Method for a beam:

1. Mark every force. A uniform object’s weight acts at its midpoint.
2. Choose a pivot where an unknown force acts, so that force has zero moment.
3. Equate clockwise and anticlockwise moments.
4. Find the remaining force by balancing forces.

Worked example. A uniform scaffold board is 4.0 m long and weighs 220 N. It rests on two supports: A is 0.5 m from the left end and B is 0.8 m from the right end. A worker of weight 650 N stands 1.0 m from the left end. Find the support forces R_A and R_B.

B is 3.2 m from the left end, so A and B are 2.7 m apart. The board’s weight is 1.5 m from A and the worker is 0.5 m from A. Take moments about A: R_B × 2.7 = (220 × 1.5) + (650 × 0.5) = 330 + 325 = 655 R_B = 242.6 N, so R_B = 243 N (3 s.f.).

Balance forces vertically: R_A + R_B = 220 + 650 = 870 N. R_A = 870 − 242.6 = 627.4 N, so R_A = 627 N (3 s.f.).

3.2.3(d) Centre of mass and centre of gravity

The centre of mass is the point where the mass of an object can be considered to be concentrated. The centre of gravity is the point where the whole weight appears to act. In a uniform gravitational field they coincide. A uniform rod has its centre of gravity at its midpoint; for a ring it lies outside the material.

Finding a centre of gravity by experiment

For an irregular flat card (lamina), use a plumb line:

1. Make three small holes near the edge, well apart.
2. Hang the card freely from a pin in one hole, with a weighted string from the same pin.
3. When still, mark where the string crosses the card and draw the line. Repeat for the other holes.
4. The centre of gravity is the point where the drawn lines intersect.

At rest the centre of gravity lies directly below the pivot, otherwise the weight would have a moment about the pin.

For a rod or metre rule, the balance point on a knife edge is the centre of gravity.

Worked example. A uniform metre rule has its centre of gravity at the 50.0 cm mark. A mass of 0.100 kg is hung from the 10.0 cm mark of a different, uniform rule, and the system balances on a pivot at 42.0 cm. Find the mass M of the rule.

The rule’s weight is 8.0 cm right of the pivot (clockwise); the hanging mass is 32.0 cm left (anticlockwise). M g × 8.0 = 0.100 g × 32.0 M = 0.100 × 32.0 / 8.0 = 0.400 kg.

3.2.3(e) Equilibrium under forces and torques

An object is in equilibrium when two conditions hold:

1. The forces on it sum to zero.
2. The moments about any point sum to zero.

A hinge force has unknown size and direction, so find its components separately.

Worked example. A uniform horizontal bar, 1.2 m long and weighing 40 N, is hinged to a wall. A cable joins the free end of the bar to the wall above the hinge, making 35° with the bar. A 60 N sign hangs 0.90 m from the wall. Find the tension T in the cable and the force at the hinge.

Take moments about the hinge. The vertical component of the tension is T sin 35°, at 1.2 m.

T sin 35° × 1.2 = (40 × 0.60) + (60 × 0.90) = 24 + 54 = 78 T = 78 / (1.2 × sin 35°) = 113 N (3 s.f.).

Horizontal: the cable pulls toward the wall with T cos 35° = 92.8 N, so the hinge pushes away from the wall with 92.8 N. Vertical: T sin 35° = 65.0 N upward. The downward load is 40 + 60 = 100 N, so the hinge supplies 35.0 N upward. Hinge force = √(92.8² + 35.0²) = 99.2 N.

3.2.3(f) Three coplanar forces and the triangle of forces

Coplanar forces act in the same plane. For three non-parallel coplanar forces in equilibrium:

• their lines of action pass through one point, and
• the resultant force is zero, so the three force vectors drawn head to tail form a closed triangle (the triangle of forces).

You can draw the triangle to scale or solve it with the sine rule, cosine rule or Pythagoras.

Worked example. A lamp of weight 150 N hangs from two cables. The left cable makes 30° with the horizontal, and the right cable makes 50° with the horizontal. Find both tensions.

The angles between the force arrows are 100° (cable to cable), 120° (weight to left cable) and 140° (weight to right cable), with sines sin 100°, sin 60° and sin 40°. Each force is proportional to the sine of the angle between the other two: T_left / sin 40° = 150 / sin 100°, so T_left = 97.9 N. T_right / sin 60° = 150 / sin 100°, so T_right = 132 N.

Check: horizontal components 97.9 cos 30° and 132 cos 50° both equal 84.8 N.

3.2.4(a) Density

Density is mass per unit volume:

ρ = m/V

The SI unit is kg m⁻³. Remember 1 g cm⁻³ = 1000 kg m⁻³.

Worked example. A cylinder has diameter 2.4 cm, length 5.0 cm and mass 0.0612 kg. Find its density.

V = π × (0.012)² × 0.050 = 2.26 × 10⁻⁵ m³. ρ = 0.0612 / 2.26 × 10⁻⁵ = 2710 kg m⁻³ (3 s.f.).

For an irregular solid, find the volume from the rise in water level in a measuring cylinder.

3.2.4(b) Pressure

Pressure is the normal force per unit area:

p = F/A

The unit is the pascal (Pa), where 1 Pa = 1 N m⁻². This applies to solids, liquids and gases. For a gas, F is the force the gas particles exert on a wall.

Worked example. A paving stone of mass 38 kg measures 0.60 m × 0.40 m × 0.050 m. Find the largest and smallest pressures it exerts on level ground.

Weight = 38 × 9.81 = 373 N. Largest face (0.24 m²): p = 373 / 0.24 = 1550 Pa. Smallest face (0.020 m²): p = 373 / 0.020 = 18 600 Pa.

3.2.4(c) Pressure in a liquid, upthrust and Archimedes’ principle

Pressure at depth: p = hρg

A column of liquid of height h and cross-section A has volume hA, mass hAρ and weight hAρg. The pressure at its base is weight / area:

p = hρg

where ρ is the density of the liquid. It depends on depth and density, not container shape. Total pressure also includes the atmosphere above.

Worked example. Find the pressure due to the sea at a depth of 35 m, where ρ = 1030 kg m⁻³.

p = 35 × 1030 × 9.81 = 3.54 × 10⁵ Pa. Adding atmospheric pressure, about 1.01 × 10⁵ Pa, gives 4.55 × 10⁵ Pa.

Upthrust

Pressure increases with depth, so a submerged object feels more pressure underneath than on top. The resulting upward force is the upthrust.

For a cuboid with top and bottom faces of area A, height h, in a fluid of density ρ:

upthrust = (p_bottom − p_top) × A = hρgA = ρ g V

since V = hA is the volume of the cuboid.

Archimedes’ principle

When an object is wholly or partly immersed in a fluid, the upthrust equals the weight of fluid displaced. A fully submerged object displaces its own volume of fluid. It applies to gases too. A floating object has upthrust equal to its weight, so it displaces its own weight of fluid.

Worked example 1. A metal object weighs 7.8 N in air. It is lowered fully into water and a force meter reads 6.8 N. Find the upthrust, the volume of the object and its density. (Density of water: 1000 kg m⁻³.)

Upthrust = 7.8 − 6.8 = 1.0 N. V = U / (ρg) = 1.0 / (1000 × 9.81) = 1.02 × 10⁻⁴ m³. Mass = 7.8 / 9.81 = 0.795 kg, so ρ = 0.795 / 1.02 × 10⁻⁴ = 7800 kg m⁻³.

Worked example 2. A wooden block of density 650 kg m⁻³ floats upright in water. What fraction of its height is below the surface?

Weight = upthrust, so ρ_block V g = ρ_water V_sub g. V_sub / V = 650 / 1000 = 0.65, so 65% of the height is submerged.

Common errors

• Using the length of a bar, not the perpendicular distance.
• Mixing pivots: forces and distances must be measured about the same point.
• Using distance from a pivot for the torque of a couple. It uses the separation of the two lines of action.
• Forgetting to convert cm² to m², or g cm⁻³ to kg m⁻³.
• Saying upthrust depends on the object’s weight. It depends on the weight of fluid displaced.

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.