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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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. Use g = 9.81 N kg⁻¹ and 1000 kg m⁻³ for the density of fresh water. Revise first with the study guide or revision notes.

Questions

1. (a) State what is meant by the moment of a force. [1] (b) A force of 36 N pushes at right angles to a gate, 0.45 m from its hinge. Calculate the moment about the hinge. [2]

2. A tap is turned by a finger and thumb exerting forces of 7.5 N in opposite directions on opposite sides of the head. The lines of action are 0.12 m apart.

(a) Calculate the torque of this couple. [2] (b) Explain why a couple produces rotation but does not accelerate the centre of mass. [2]

3. A uniform beam is 2.4 m long and weighs 90 N. It is pivoted 0.20 m to the right of its centre. A crate of weight 300 N rests on the beam 0.90 m to the right of the pivot. A counterweight W is placed 1.0 m to the left of the pivot and the beam balances. Calculate W. [4]

4. Outline an experiment to locate the centre of gravity of an irregular piece of thin plywood, and explain why it works. [4]

5. A uniform metre rule of mass 0.150 kg is pivoted at the 25.0 cm mark. A mass of 0.400 kg hangs from the 5.0 cm mark. A mass m hangs from the 80.0 cm mark and the rule balances horizontally. Calculate m. [4]

6. A uniform ladder of length 5.0 m and weight 240 N leans against a smooth vertical wall, at 70° to the horizontal ground. A person of weight 700 N stands 3.0 m along the ladder from its foot.

(a) Explain why the friction force at the ground equals the force from the wall. [2] (b) Calculate the force from the wall. [4] (c) Calculate the size of the resultant force that the ground exerts on the ladder. [2]

7. A sphere of weight 36 N hangs from a string fixed to a smooth vertical wall. The string makes 20° with the wall and the sphere touches the wall.

(a) State the direction of the force from the wall and explain why it takes this direction. [2] (b) Use a triangle of forces to calculate the tension in the string and the force from the wall. [3]

8. A rectangular tank measures 2.5 m × 1.2 m × 0.80 m (length, width, depth) and is full of oil of density 870 kg m⁻³.

(a) Calculate the mass of oil. [2] (b) Calculate the pressure of the oil on the base of the tank in two ways. [2]

9. A tracked vehicle of mass 4200 kg has two tracks, each in contact with the ground over 2.8 m × 0.45 m.

(a) Calculate the pressure it exerts on level ground. [3] (b) A person of mass 55 kg stands on both feet, with a total contact area of 0.036 m². Compare the person’s pressure with the vehicle’s. [2]

10. A submersible has a circular window of diameter 0.30 m. It is 120 m below the surface of sea water of density 1030 kg m⁻³. The air pressure inside is atmospheric.

(a) Calculate the pressure due to the water at this depth. [2] (b) Calculate the resultant force on the window due to the pressure difference across it. [3]

11. A uniform block has a square base of side 0.15 m and a height of 0.20 m. It floats upright in water with 0.12 m of its height below the surface.

(a) Calculate the difference in pressure between the bottom and top faces. [2] (b) Calculate the upthrust and explain, using Archimedes’ principle, how it relates to the weight of the block. [2] (c) Calculate the density of the block. [2] (d) Calculate the mass that must be placed on the block to make it just fully submerged. [2]

12. An object of volume 7.0 × 10⁻⁵ m³ weighs 4.60 N in air. When fully immersed in a liquid, a force meter reads 3.85 N. Calculate the density of the liquid. [3]

Answers

1. (a) The force times the shortest distance between the pivot and the force’s line [1]. [1] (b) Moment = Fx = 36 × 0.45 [1] = 16.2 N m [1] [2] Examiner insight: “Distance from the pivot” alone is too loose; say the distance is perpendicular to the force’s line.

2. (a) Torque = Fd = 7.5 × 0.12 [1] = 0.90 N m [1] [2] (b) The forces are equal and opposite, so the resultant force is zero [1]. So the centre of mass does not accelerate, but the separated lines of action give a torque [1]. [2] Examiner insight: Name both facts: zero resultant force, and a torque from separated lines of action.

3. Take moments about the pivot. The beam’s weight acts at its centre, 0.20 m to the left of the pivot, giving 90 × 0.20 = 18 N m anticlockwise [1]. Crate: 300 × 0.90 = 270 N m clockwise [1]. Clockwise = anticlockwise: 270 = 18 + W × 1.0 [1]. W = 252 N [1] [4] Examiner insight: Label each moment clockwise or anticlockwise first; the beam’s weight is easy to put on the wrong side.

4. Make small holes near the edge and hang the plywood freely from a pin through one hole [1]. Hang a weighted string from the same pin and mark where it lies on the plywood [1]. Repeat with at least two other holes [1]. The lines cross at the centre of gravity, because at rest the weight has no moment about the pin, so the centre of gravity must be vertically below it [1]. [4] Examiner insight: “Hang it up and draw a line” is thin; say it hangs freely, the string is on the same pin, and you repeat.

5. Anticlockwise about the pivot: 0.400 g × 0.200 m = 0.0800 g [1]. The rule’s weight acts at 50.0 cm, 0.250 m to the right: 0.150 g × 0.250 = 0.0375 g [1]. The mass m is 0.550 m to the right: 0.0800 g = 0.0375 g + 0.550 m g [1]. m = 0.0425 / 0.550 = 0.0773 kg [1] [4] Examiner insight: g cancels, so the equation can be written in masses. Do not forget the rule’s own weight.

6. (a) The wall is smooth, so the only horizontal forces are the wall force and ground friction [1]. The ladder is in equilibrium, so the horizontal forces are equal and opposite [1]. [2] (b) Moments about the foot of the ladder. Ladder weight: 240 × 2.5 cos 70° = 205.2 N m [1]. Person: 700 × 3.0 cos 70° = 718.2 N m [1]. Wall force N has moment N × 5.0 sin 70° [1]. N = 923.4 / 4.698 = 197 N [1] [4] (c) The vertical force from the ground equals 240 + 700 = 940 N [1]. Resultant = √(940² + 197²) = 960 N [1] [2] Examiner insight: Taking moments about the foot removes two unknown forces and keeps the working short.

7. (a) Horizontal, away from the wall [1]. The wall is smooth, so there is no friction and the force acts at right angles to the wall [1]. [2] (b) Draw the closed triangle: weight 36 N vertical, wall force horizontal, tension along the string [1]. The angle between the tension and the vertical is 20°, so T = 36 / cos 20° = 38.3 N [1]. R = 36 tan 20° = 13.1 N [1] [3] Examiner insight: Draw the three forces head to tail so the arrows follow each other round a closed triangle.

8. (a) V = 2.5 × 1.2 × 0.80 = 2.4 m³ [1]. m = ρV = 870 × 2.4 = 2088 kg, about 2.1 × 10³ kg [1] [2] (b) Using p = F/A: weight = 2088 × 9.81 = 20 480 N, area = 2.5 × 1.2 = 3.0 m², so p = 6830 Pa [1]. Using p = hρg: 0.80 × 870 × 9.81 = 6830 Pa, the same [1]. [2] Examiner insight: Show both substitutions; two routes agreeing is a strong check.

9. (a) Weight = 4200 × 9.81 = 41 200 N [1]. Area = 2 × 2.8 × 0.45 = 2.52 m² [1]. p = 41 200 / 2.52 = 1.63 × 10⁴ Pa [1] [3] (b) Person: p = 55 × 9.81 / 0.036 = 1.50 × 10⁴ Pa [1]. The pressures are similar (the vehicle’s is about 9% larger) because its much greater weight acts over a much larger area [1]. [2] Examiner insight: A comparison needs both numbers and a statement linking them to force and area.

10. (a) p = hρg = 120 × 1030 × 9.81 [1] = 1.21 × 10⁶ Pa [1] [2] (b) Area = π × 0.15² = 0.0707 m² [1]. The atmospheric pressure acts on both sides, so the pressure difference is the pressure due to the water [1]. F = 1.2125 × 10⁶ × 0.0707 = 8.57 × 10⁴ N [1] [3] Examiner insight: Use the radius, not the diameter. Atmospheric pressure cancels in a pressure difference.

11. (a) Δp = hρg = 0.12 × 1000 × 9.81 [1] = 1180 Pa [1] [2] (b) Upthrust = ΔpA = 1177 × 0.15² [1] = 26.5 N. This equals the weight of the water displaced, and the block floats at rest, so it also equals the block’s weight [1]. [2] (c) Mass = 26.5 / 9.81 = 2.70 kg [1]. Volume = 0.15² × 0.20 = 4.5 × 10⁻³ m³, so ρ = 2.70 / 4.5 × 10⁻³ = 600 kg m⁻³ [1] [2] (d) Fully submerged upthrust = 1000 × 9.81 × 4.5 × 10⁻³ = 44.1 N, so the extra weight is 44.1 − 26.5 = 17.7 N [1]. Mass = 17.7 / 9.81 = 1.80 kg [1] [2] Examiner insight: Link part (b) to Archimedes in words: floating means the displaced water weighs the same as the block.

12. Upthrust = 4.60 − 3.85 = 0.75 N [1]. U = ρ g V, so ρ = 0.75 / (9.81 × 7.0 × 10⁻⁵) [1] = 1.1 × 10³ kg m⁻³ (1092 kg m⁻³) [1] [3] Examiner insight: The upthrust is the loss in apparent weight, not the weight in air.

Where marks are usually lost

• Not stating that the distance in a moment is perpendicular to the line of action.
• Missing the weight of a beam, plank or ladder.
• Taking moments about different points in one equation.
• Using the diameter instead of the radius for a circular area.
• Leaving atmospheric pressure in when asked for a pressure difference, or out when asked for total pressure.
• Describing the centre of gravity experiment without saying the card hangs freely.
• Quoting N for a torque or moment, or leaving the unit out.

Next steps

• Read the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

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.