S4 Physics Paper: Moments, Equilibrium And Pressure (100 Marks)

S4 Physics Paper: Moments, Equilibrium and Pressure (100 marks)

2026-11-11

S4 Physics – moments, equilibrium and pressure

S4 Physics practice paper: 6 questions, 22 parts, 100 marks, 75 minutes. Covers moments, lever equilibrium, centre of gravity, pressure and upthrust, with full marking notes.

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Paper details　S4 Physics paper on moments, lever equilibrium, centre of gravity, pressure and upthrust: 6 questions, 22 parts, 100 marks, 75 minutes, with marking notes.

Part A Question 1: Calculating Moments

Take the acceleration due to gravity as g = 10 N/kg. Give units in all answers.

#	Question	Answer and marking notes
1	A mechanic applies a force of 30 N perpendicular to the end of a spanner. The end is 0.40 m from the bolt. Find the moment of the force about the bolt.	Answer: 12 N m. Moment = force x perpendicular distance = 30 x 0.40 = 12 N m. Marking: formula 1, substitution 1, value with unit 2.
2	A force of 50 N acts at 30 degrees to a spanner of length 0.30 m. Find its effective moment about the bolt.	Answer: 7.5 N m. Effective moment = 50 x 0.30 x sin 30 degrees = 50 x 0.30 x 0.5 = 7.5 N m. Marking: perpendicular component 2, calculation 1, unit 1.
3	A bolt needs a moment of 24 N m and the spanner has an arm of 0.80 m. Find the minimum perpendicular force required.	Answer: 30 N. Force = moment / arm = 24 / 0.80 = 30 N. Marking: rearranging 2, value 1, unit 1.
4	A uniform metre rule of length 1.0 m and weight 4.0 N is pivoted at its centre. A 6.0 N weight hangs from the left end, and an unknown weight W hangs 0.30 m to the right of the pivot. Find W.	Answer: 10 N. The pivot is at the centre, so the left arm is 0.50 m and the clockwise moment is 6.0 x 0.50 = 3.0 N m. The anticlockwise moment is W x 0.30. Equating them gives W = 3.0 / 0.30 = 10 N. The weight of the rule acts at the pivot, so its moment is zero. Marking: moment equation 3, left moment 1, calculation 2, note on rule weight 2.

Part B Question 2: Reactions on a Plank

The plank stays horizontal and at rest. Take g = 10 N/kg.

#	Question	Answer and marking notes
5	A uniform plank of length 2.0 m and weight 200 N rests on supports A and B at its two ends. A load of 300 N hangs 0.50 m from end A. Taking moments about A, find the support force at B.	Answer: 175 N. About A: RB x 2.0 = 300 x 0.50 + 200 x 1.0 = 150 + 200 = 350, so RB = 175 N. Marking: moment equation 2, load moment 1, plank moment with 1.0 m arm 1, calculation 2.
6	Find the support force at A.	Answer: 325 N. Vertical forces balance: RA + RB = 300 + 200 = 500, so RA = 500 – 175 = 325 N. Marking: force equation 2, calculation 2.
7	If the load is moved to 1.50 m from end A, find the new support force at B.	Answer: 325 N. RB x 2.0 = 300 x 1.50 + 200 x 1.0 = 450 + 200 = 650, so RB = 325 N and the force at A becomes 175 N. Marking: new moment equation 2, calculation 1, noting A decreases 1.
8	Explain why the sum of the two support forces must equal the total weight of the plank and the load.	Answer: The plank is at rest, so the vertical resultant force is zero: the upward reactions must equal the total downward weight of 500 N. Otherwise the plank would accelerate up or down. Marking: zero resultant 2, total weight 500 N 1, otherwise acceleration 1.

Part C Question 3: Centre of Gravity and Stability

State the trigonometric relation you use.

#	Question	Answer and marking notes
9	A uniform rectangular block is 0.4 m high and 0.2 m wide and stands on a horizontal table. Find the critical angle between its base and the table when it starts to topple.	Answer: 26.6 degrees. The block topples when the vertical line through its centre of gravity passes beyond the base edge, so tan theta = width / height = 0.2 / 0.4 = 0.5 and theta = 26.6 degrees. Marking: toppling condition 2, ratio 2, value 2.
10	Use everyday examples to explain why a lower centre of gravity and a larger base make an object more stable.	Answer: A lower centre of gravity means the vertical line through it reaches the base edge only at a larger tilt, and a larger base allows a larger tilt. Examples include a double-decker bus with heavy parts low in the chassis and a wide base on a bookshelf. Marking: low centre 1, large base 1, sensible example 1.
11	A second block is 0.3 m high and 0.1 m wide. Find its critical angle and state which block topples more easily.	Answer: 18.4 degrees. tan theta = 0.1 / 0.3 = 0.333, so theta = 18.4 degrees. The smaller critical angle means the second, taller and narrower block topples more easily. Marking: ratio 1, value 1, comparison 1.

Part D Question 4: Pressure in Solids and Liquids

Take the density of water as 1000 kg/m3 and g = 10 N/kg. Give answers in Pa or kPa.

#	Question	Answer and marking notes
12	A force of 500 N acts perpendicular to an area of 0.02 m2. Find the pressure.	Answer: 25000 Pa, that is 25 kPa. Pressure = force / area = 500 / 0.02 = 25000 Pa. Marking: formula 2, value and unit 2.
13	The tip of a drawing pin has an area of 1.0 x 10^-6 m2 and is pressed with a force of 10 N. Find the pressure at the tip.	Answer: 1.0 x 10^7 Pa, that is 10 MPa. Pressure = 10 / 1.0 x 10^-6 = 1.0 x 10^7 Pa. The very small area gives a very large pressure, so the pin can pierce the wall. Marking: formula 2, value 1, explanation 1.
14	Find the pressure due to water at a depth of 2.5 m.	Answer: 25000 Pa. Pressure = density x g x depth = 1000 x 10 x 2.5 = 25000 Pa. Marking: formula 2, substitution and value 2.
15	A diver is at a depth of 15 m, where the atmospheric pressure at the surface is 101 kPa. Find the total pressure on the diver.	Answer: 251000 Pa, that is 251 kPa. Water pressure = 1000 x 10 x 15 = 150000 Pa, so total pressure = 150000 + 101000 = 251000 Pa. Marking: water pressure 3, adding atmospheric pressure 2, conversion to kPa 1, stating both add 2.

Part E Question 5: Upthrust and Floating

Take the density of water as 1000 kg/m3 and g = 10 N/kg.

#	Question	Answer and marking notes
16	A metal block of volume 2.0 × 10⁻⁴ m3 is completely immersed in water. Find the upthrust on it.	Answer: 2 N. Upthrust equals the weight of water displaced = density x g x volume = 1000 x 10 x 2.0 x 10^-4 = 2 N. Marking: principle 2, formula 2, value and unit 2.
17	The block weighs 5.4 N in air. Find the reading on a spring balance when it hangs fully immersed in water.	Answer: 3.4 N. Reading = weight – upthrust = 5.4 – 2 = 3.4 N. Marking: relation 2, value 1, unit 1.
18	Hence find the density of the metal.	Answer: 2700 kg/m3. Mass = 5.4 / 10 = 0.54 kg, so density = 0.54 / 2.0 x 10^-4 = 2700 kg/m3. Marking: mass 1, density formula 1, value and unit 2.
19	A piece of wood of density 600 kg/m3 is placed in water. Will it float or sink? Explain.	Answer: It floats, with part of it above the surface. The wood is less dense than water, so for the same volume it weighs less than the water displaced; when fully immersed the upthrust exceeds its weight, so it rises until upthrust equals weight. Marking: judgement 1, density comparison 2, upthrust equals weight 1.

Part F Question 6: Using a Crowbar

Ignore the weight of the crowbar itself. Include units in your answers.

#	Question	Answer and marking notes
20	A worker lifts a stone with a crowbar. The pivot is 0.2 m from the stone and the effort acts 1.6 m from the pivot with a force of 100 N. Find the force on the stone.	Answer: 800 N. Moments about the pivot: 100 x 1.6 = F x 0.2, so 160 = F x 0.2 and F = 800 N. Marking: equating moments 2, effort moment 2, calculation 2.
21	In practice the effort needed is often larger than the calculated value. Give two reasons.	Answer: The crowbar has its own weight, which adds a moment; the effort is not exactly perpendicular to the bar, so the effective arm is shorter; friction and slight sinking at the pivot also absorb part of the moment. Marking: 1.5 per sound reason, maximum 3.
22	If the effort now acts 0.8 m from the pivot and all else is unchanged, find the effort required.	Answer: 200 N. F x 0.8 = 800 x 0.2 = 160, so F = 200 N. The closer the effort is to the pivot, the larger the force needed. Marking: moment equation 1, value 1, trend 1.

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