Forces, Density and Pressure — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Types of contact and non-contact force, centre of mass and moments, coplanar force equilibrium rules, fluid density and pressure calculations, and Archimedes' principle for buoyancy.

You should already know: IGCSE Physics, basic algebra and trigonometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Physics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Forces, Density and Pressure?

Forces, Density and Pressure is a core AS-level mechanics topic for A-Level Physics, bridging fundamental force interactions and fluid statics. It describes how contact and non-contact forces act on solid and fluid objects, how stationary objects remain balanced, and how mass per unit volume and force per unit area govern the behaviour of static liquids and gases. This topic accounts for 5-10% of AS-level marks annually, per past examiner reports, and is tested across Paper 1 (MCQs), Paper 2 (structured questions) and Paper 3 (practical assessments).

2. Types of force — gravity, normal, friction, tension, drag, upthrust

All forces are vector quantities (with magnitude and direction) measured in Newtons (N). You will encounter 6 core force types in this topic:

1. Gravitational force (weight): Non-contact force acting on all objects with mass, calculated as $W=mg$, where $m$ is mass in kg and $g$ is gravitational field strength ($9.81m/s^2$ for A-Level exams, directed towards the Earth’s centre).
2. Normal contact force: Perpendicular reaction force between two touching surfaces, arising from compression of the surface material. For a stationary object on a horizontal surface, normal force equals the object’s weight.
3. Friction: Parallel force acting along contact surfaces that opposes relative motion. Static friction $F_s\le {\mu }_{s}N$ prevents motion, while kinetic friction $F_k={\mu }_{k}N$ acts on moving objects, where $\mu$ is the dimensionless coefficient of friction and $N$ is normal force.
4. Tension: Pulling force transmitted through a massless, inextensible string, rope or cable when stretched; tension is equal at both ends of the string for standard exam assumptions.
5. Drag: Force opposing motion of an object through a fluid (gas or liquid), which increases with object speed, cross-sectional area and fluid density. For small spheres moving at low speed, drag follows Stokes’ Law: $F=6\pi \eta rv$, where $\eta$ is fluid viscosity, $r$ is sphere radius and $v$ is speed.
6. Upthrust: Upward buoyant force on objects fully or partially submerged in fluid, caused by the pressure difference between the top and bottom of the object.

Mini worked example: A 3kg block is pulled horizontally across a rough table with a kinetic friction coefficient ${\mu }_k=0.25$. Calculate the friction force opposing motion: $$F_k={\mu }_{k}mg=0.25\times 3\times 9.81=7.36N$$

3. Centre of mass and moment of a force

Centre of mass (CoM)

The centre of mass of an object is the single point where the entire weight of the object can be considered to act. For uniform, regular-shaped objects (e.g. rectangular blocks, spheres, straight rods), the CoM is at the geometric centre. For irregular objects, the CoM can be found via the plumb line experiment (recapped from IGCSE Physics).

Moment of a force

The moment of a force is its turning effect around a fixed pivot, calculated as: $$M=F\times d_{\perp }$$ where $d_{\perp }$ is the perpendicular distance from the pivot to the line of action of the force. Moments are measured in Nm, and use the standard A-Level convention that anticlockwise moments are positive, while clockwise moments are negative.

Worked example: A 25N force is applied 40cm from a pivot at an angle of 30° to the lever arm. Calculate the moment of the force:

1. Convert distance to SI units: $40cm=0.4m$
2. Calculate perpendicular distance: $d_{\perp }=0.4\times \sin (30°)=0.2m$
3. Calculate moment: $M=25\times 0.2=5.0Nm$ anticlockwise

Only the component of force perpendicular to the lever contributes to the turning effect; the parallel component only pulls or pushes on the pivot and does not create rotation.

4. Equilibrium of coplanar forces — torque sum to zero

Coplanar forces are forces that all act in the same 2D plane. For an object to be in static equilibrium (no linear or angular acceleration), two conditions must be satisfied (you must state both for full marks in 3+ mark exam questions):

1. Translational equilibrium: The vector sum of all forces is zero, so $\sum F_x=0$ (sum of horizontal forces is zero) and $\sum F_y=0$ (sum of vertical forces is zero).
2. Rotational equilibrium: The sum of all moments (torques) around any pivot is zero, so $\sum M_{clockwise}=\sum M_{anticlockwise}$.

Exam tip: Choose a pivot point where an unknown force acts to eliminate it from your moment equation, simplifying calculations drastically.

Worked example: A uniform 1.2m long plank of mass 5kg rests on two supports: Support A at the 0cm mark, and Support B at the 90cm mark. Calculate the normal reaction force at each support:

1. Weight of plank: $W=5\times 9.81=49.05N$, acting at the CoM (60cm mark)
2. Take moments around Support A to eliminate reaction force $R_A$:

• Clockwise moment: $49.05\times 0.6=29.43Nm$
• Anticlockwise moment: $R_B\times 0.9$
• Equate: $R_B=29.43/0.9=32.7N$

3. Use translational equilibrium to find $R_A$: $R_A+R_B=49.05$ → $R_A=49.05-32.7=16.35N\approx 16.4N$

5. Density and pressure in fluids

Density

Density $\rho$ is the mass per unit volume of a substance, calculated as: $$\rho =\frac{m}{V}$$ where $m$ is mass in kg and $V$ is volume in $m^3$, giving SI units of $kg/m^3$. To convert from common $g/c{m}^3$ to $kg/m^3$, multiply by 1000 (e.g. water density = $1g/c{m}^3=1000kg/m^3$).

Pressure

Pressure $p$ is the force per unit area applied perpendicular to a surface, calculated as: $$p=\frac{F}{A}$$ where $F$ is force in N and $A$ is area in $m^2$, giving SI units of Pascals (Pa) = $1N/m^2$. For static fluids, pressure at a given depth acts equally in all directions, and is independent of container shape.

Fluid pressure at depth

The total pressure at depth $h$ below the surface of a fluid is equal to atmospheric pressure plus the pressure from the weight of fluid above the point: $$p=p_0+\rho gh$$ where $p_0$ is atmospheric pressure (~$1.01\times 10^{5}Pa$ at sea level), $\rho$ is fluid density, $g$ is gravitational field strength and $h$ is depth in m. The term $\rho gh$ is known as gauge pressure, the pressure above atmospheric pressure.

Worked example: Calculate the absolute pressure 12m below the surface of a lake, where water density is $1000kg/m^3$:

1. Gauge pressure: $\rho gh=1000\times 9.81\times 12=117720Pa\approx 1.18\times 10^{5}Pa$
2. Absolute pressure: $p=1.01\times 10^5+1.18\times 10^5=2.19\times 10^{5}Pa$

6. Archimedes' principle and buoyancy

Archimedes' principle states that when an object is fully or partially submerged in a fluid, the upthrust force acting on the object is equal to the weight of the fluid displaced by the object. The formula for upthrust is: $$F_{up}={\rho }_{fluid}{V}_{displaced}g$$ where $V_{displaced}$ is the volume of fluid displaced, equal to the volume of the submerged part of the object.

Key floating condition: If an object floats, the upward upthrust equals the downward weight of the object, so ${\rho }_{object}{V}_{object}g={\rho }_{fluid}{V}_{displaced}g$. Rearranged, the fraction of the object submerged is equal to the ratio of the object’s density to the fluid’s density: $\frac{V_{displaced}}{V_{object}}=\frac{{\rho }_{object}}{{\rho }_{fluid}}$.

Worked example: A plastic block of density $820kg/m^3$ and volume $0.003m^3$ floats in water ($\rho =1000kg/m^3$). Calculate the volume of the block that remains above water:

1. Weight of block: $W=820\times 0.003\times 9.81=24.13N$
2. Equate weight to upthrust: $24.13=1000\times V_{displaced}\times 9.81$
3. Solve for displaced volume: $V_{displaced}=0.00246m^3$
4. Volume above water: $0.003-0.00246=0.00054m^3=5.4\times 10^{-4}{m}^3$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using straight line distance instead of perpendicular distance when calculating moments. Why students do it: They forget only the force component perpendicular to the lever contributes to rotation. Correct move: Always use $M=Fd\sin \theta$ where $\theta$ is the angle between the force and the lever arm, to get the perpendicular component.
• Wrong move: Forgetting to add atmospheric pressure when asked for absolute fluid pressure. Why students do it: They mix up gauge pressure (pressure above atmospheric) and total pressure. Correct move: Check question wording explicitly: "gauge pressure" = only $\rho gh$, "absolute/total pressure" = $p_0+\rho gh$.
• Wrong move: Using the object’s density instead of fluid density when calculating upthrust. Why students do it: They confuse the weight of the object with the weight of displaced fluid. Correct move: Upthrust depends only on the fluid density and volume of displaced fluid, not the material of the submerged object.
• Wrong move: Only stating one equilibrium condition in exam questions. Why students do it: They confuse static equilibrium with just force balance. Correct move: For any 3+ mark equilibrium question, explicitly write both force balance ($\sum F_x=0,\sum F_y=0$) and moment balance ($\sum M=0$) to get full marks.
• Wrong move: Using non-SI units for density or distance in calculations. Why students do it: Questions often give density in $g/c{m}^3$ or distance in cm. Correct move: Convert all units to SI before substituting into formulas: $g/c{m}^3\times 1000=kg/m^3$, $cm/100=m$.

8. Practice Questions (A-Level Physics Style)

Question 1 (AS Paper 1 MCQ, 1 mark)

A uniform 0.8m long rod of mass 2kg is pivoted at its left end. A 4kg mass is hung from the right end of the rod. What minimum upward force applied 0.6m from the pivot is required to keep the rod horizontal? A) 58.9 N B) 65.4 N C) 71.9 N D) 78.5 N

Solution:

1. Calculate forces: Weight of rod = $2\times 9.81=19.62N$ at 0.4m mark; weight of mass = $4\times 9.81=39.24N$ at 0.8m mark.
2. Take moments around pivot: Clockwise moments = $19.62\times 0.4+39.24\times 0.8=7.848+31.392=39.24Nm$.
3. Anticlockwise moments = $F\times 0.6$. Equate to clockwise moments: $F=39.24/0.6=65.4N$. Correct answer: B.

Question 2 (AS Paper 2 Structured, 3 marks)

Calculate the upthrust acting on a solid iron cube of side length 0.05m fully submerged in oil of density $890kg/m^3$. Solution:

1. Volume of cube = displaced volume: $V=(0.05{)}^3=1.25\times 10^{-4}{m}^3$ (1 mark)
2. Substitute into upthrust formula: $F_{up}=890\times 1.25\times 10^{-4}\times 9.81$ (1 mark)
3. Final answer: $F_{up}\approx 1.09N$ (1 mark, unit required)

Question 3 (A2 Paper 4 Structured, 5 marks)

A 5m long uniform ladder of mass 12kg rests against a smooth frictionless vertical wall, making a 50° angle with the horizontal ground. The coefficient of static friction between the ladder and ground is 0.45. Calculate the maximum mass of a person that can stand at the top of the ladder before it slips.

Solution:

1. State equilibrium conditions: $\sum F_x=0,\sum F_y=0,\sum M=0$ (1 mark)
2. Force balance: Horizontal: Wall reaction $R_w=F_f=\mu R_g$; Vertical: $R_g=(12+m)g$ where $m$ is mass of person (1 mark)
3. Take moments around base of ladder: Clockwise moments = $12g\times 2.5\cos 50°+mg\times 5\cos 50°$; Anticlockwise moments = $R_w\times 5\sin 50°$ (1 mark)
4. Substitute $R_w=0.45(12+m)g$ into moment equation, cancel $g$ from all terms: $12\times 2.5\times 0.6428+m\times 5\times 0.6428=0.45(12+m)\times 5\times 0.7660$ $19.28+3.214m=0.45(12+m)\times 3.83=20.68+1.7235m$ (1 mark)
5. Rearrange to solve: $1.4905m=1.4$ → $m\approx 0.94kg$ (1 mark, accept 0.9-1.0 kg)

9. Quick Reference Cheatsheet

10. What's Next

This topic is the foundation for all subsequent mechanics content in the A-Level Physics syllabus. You will use force equilibrium rules to solve Newton’s laws of motion, work-energy, and circular motion problems later in AS level, while fluid pressure and upthrust concepts extend to ideal gas laws, viscous flow, and hydraulic system applications tested in A2 Paper 4. Examiners frequently combine this topic with kinematics (e.g. terminal velocity calculations from drag and weight balance) in 5+ mark structured questions, so mastering these fundamentals directly impacts your performance across 30% of total syllabus marks.

If you are struggling with moment calculations, buoyancy problems, or any other content in this guide, you can get instant personalised support from Ollie, our AI physics tutor, by visiting the homepage. You can also access topic-specific MCQ and structured practice tests aligned with official A-Level Physics mark schemes to test your understanding and identify gaps before your exam.

Aligned with the Cambridge International AS & A Level Physics 9702 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.