Cambridge AS & A Level9702

Forces, density and pressure

Physics 9702 Chapter Notes

What this chapter covers

Forces, density and pressure - Turning effects of forcesForces, density and pressure - Equilibrium of forcesForces, density and pressure - Density and pressure
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1. Density: Mass in a Given Space

Density is a fundamental property of matter that tells us how much 'stuff' (mass) is packed into a given space (volume). A small, heavy object is very dense, while a large, light object is not. For example, a kilogram of lead takes up much less space than a kilogram of feathers because lead is much denser. Density is an intrinsic property of a substance, meaning a pure substance will have the same density regardless of how much of it you have. To find the density (ρ) of an object, you measure its mass(m) and its volume (V) and use the formula ρ = m/V. The standard unit for density is kilograms per cubic metre (kg/m³), although grams per cubic centimetre (g/cm³) is also common.

ρ = m / V

Key term

Density (ρ): The mass per unit volume of a substance.

Common pitfall

When converting between g/cm³ and kg/m³, a common mistake is to multiply by 100 instead of 1000. Remember that 1 g/cm³ = 1000 kg/m³.

Fun fact

The human body is slightly denser than water, which is why most people need to make a small effort to float.

Worked example 13 marks

A rectangular block of aluminium has dimensions 2.0 cm x 3.0 cm x 5.0 cm. It has a mass of 81 g. Calculate the density of aluminium in g/cm³ and kg/m³.

  1. 1

    Step 1: Calculate the volume of the block. Volume V = length × width × height = 2.0 cm × 3.0 cm × 5.0 cm = 30 cm³.

  2. 2

    Step 2: Calculate the density in g/cm³. Density ρ = mass / volume = 81 g / 30 cm³ = 2.7 g/cm³.

  3. 3

    Step 3: Convert the density to kg/m³. We know 1 g = 10⁻³ kg and 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³. So, ρ = 2.7 g/cm³ = 2.7 × (10⁻³ kg) / (10⁻⁶ m³) = 2.7 × 10³ kg/m³ = 2700 kg/m³.

Recap

  • Density is mass divided by volume.
  • The SI unit for density is kilograms per cubic metre (kg/m³).
  • The density of a substance is constant at a given temperature and pressure.
  • To find the volume of an irregular object, you can use a displacement can.

Quick check

  1. A substance has a mass of 500 kg and a volume of 0.5 m³. What is its density?1 mark

2. Understanding Pressure

Pressure is defined as the force acting perpendicularly on a unit area of a surface. A large force on a small area produces a high pressure, which explains why a sharp knife (small area) cuts easily, while a blunt knife (larger area) does not. The formula is p = F/A, where 'p' is pressure, 'F' is the normal force, and 'A' is the cross-sectional area over which the force acts. The term 'normal' is crucial – it means the force is at a right angle (90°) to the surface. The SI unit of pressure is the Pascal (Pa), which is defined as one newton per square metre (N/m²).

p = F / A

Key term

Pressure (p): The normal force exerted per unit cross-sectional area.

Examiner insight

Examiners reward clear statements that pressure acts 'normally' or 'perpendicularly' to a surface. Always convert areas to m² before calculating pressure in Pascals.

Fun fact

A woman in high heels can exert more pressure on the ground than an elephant, because her weight is concentrated on a much smaller area.

Worked example 14 marks

A student of mass 60 kg stands on the floor. If the area of one shoe in contact with the floor is 150 cm², calculate the pressure exerted on the floor when the student stands on both feet. (Take g = 9.81 N/kg).

  1. 1

    Step 1: Calculate the student's weight. Weight is the force, F. F = mass × g = 60 kg × 9.81 N/kg = 588.6 N.

  2. 2

    Step 2: Calculate the total area of contact. The student stands on two feet, so the total area A = 2 × 150 cm² = 300 cm².

  3. 3

    Step 3: Convert the area to SI units (m²). Since 1 m = 100 cm, 1 m² = (100 cm)² = 10000 cm². So, A = 300 / 10000 m² = 0.030 m².

  4. 4

    Step 4: Calculate the pressure. p = F / A = 588.6 N / 0.030 m² = 19620 Pa, or 19.6 kPa.

Recap

  • Pressure is the normal force per unit area.
  • The formula for pressure is p = F/A.
  • The SI unit for pressure is the Pascal (Pa), equal to 1 N/m².
  • A smaller area results in a larger pressure for the same force.

Quick check

  1. Why are the tyres on a tractor very wide?2 marks

3. Pressure and Depth in Fluids

In a fluid (a liquid or a gas), pressure increases with depth. This is because the fluid at a lower level has to support the weight of all the fluid above it. The pressure difference (Δp) between two points in a fluid is determined by the fluid's density (ρ), the acceleration of free fall (g), and the vertical height difference (Δh) between the points. This relationship is given by the equation Δp = ρgΔh. This pressure is often called 'gauge pressure' as it is the pressure relative to the surroundings. To find the absolute pressure at a certain depth, you must add the atmospheric pressure to the gauge pressure.

Δp = ρgΔh

Key term

Hydrostatic Pressure: The pressure exerted by a fluid at equilibrium at a given depth due to the force of gravity.

Examiner insight

The ability to derive the formula p = ρgh from first principles (Weight = mg, Density = m/V) is a key skill that can be tested for several marks.

Fun fact

The pressure at the bottom of the Mariana Trench is over 1000 times the atmospheric pressure at sea level, equivalent to having 50 jumbo jets stacked on top of you.

Worked example 13 marks

A submarine is at a depth of 250 m in seawater. The density of seawater is 1030 kg/m³. Calculate the gauge pressure on the hull of the submarine. (Take g = 9.81 N/kg).

  1. 1

    Step 1: Identify the given values. Depth h = 250 m, density ρ = 1030 kg/m³, g = 9.81 N/kg.

  2. 2

    Step 2: State the formula for pressure in a fluid. p = ρgh.

  3. 3

    Step 3: Substitute the values into the formula. p = 1030 kg/m³ × 9.81 N/kg × 250 m.

  4. 4

    Step 4: Calculate the result. p = 2,526,075 Pa. This is approximately 2.53 x 10⁶ Pa or 2.53 MPa.

Worked example 23 marks

Derive the formula for the pressure exerted by a fluid column of height h, density ρ, and base area A.

  1. 1

    Step 1: Consider a column of fluid of height h and base area A. The volume of the fluid is V = A × h.

  2. 2

    Step 2: The mass of the fluid is given by density × volume. So, m = ρ × V = ρ × A × h.

  3. 3

    Step 3: The weight of the fluid column is the force it exerts on the base. F = weight = m × g = (ρ × A ×h) × g.

  4. 4

    Step 4: Pressure is force per unit area. p = F / A = (ρ × A × h ×g) / A.

  5. 5

    Step 5: Cancel the area A from the numerator and denominator. This leaves p = ρgh.

Recap

  • Pressure in a fluid increases with depth.
  • The formula for pressure difference in a fluid is Δp = ρgΔh.
  • This pressure is independent of the shape of the container.
  • Absolute pressure = gauge pressure + atmospheric pressure.

Quick check

  1. If you double the depth in a liquid, what happens to the gauge pressure?1 mark

4. Upthrust and Archimedes' Principle

When an object is submerged in a fluid, it experiences an upward force called upthrust. This force arises because the pressure on the bottom surface of the object is greater than the pressure on its top surface (since pressure increases with depth). Archimedes' principle provides a simple way to calculate this force: it states that the upthrust on a submerged object is equal to the weight of the fluid that the object displaces. If the upthrust is greater than the object's weight, the object will float or rise. If the upthrust is less than the object's weight, it will sink. If they are equal, the object will float in equilibrium.

Upthrust = ρ_fluid × g × V_displaced

Key term

Upthrust: The upward buoyant force exerted on a body immersed in a fluid, arising from the pressure increase with depth.

Examiner insight

Clear explanations linking upthrust to the pressure difference on the top and bottom surfaces of an object score highly. When applying Archimedes' principle, be careful to use the density of the fluid, not the object.

Fun fact

Icebergs float with about 90% of their volume underwater because the density of ice is about 90% of the density of seawater.

Worked example 15 marks

A block of wood of volume 0.05 m³ and mass 40 kg is held fully submerged in water of density 1000 kg/m³.(a) Calculate the upthrust on the block.(b) When released, will the block sink or float? (Take g = 9.81 N/kg).

  1. 1

    Part (a): Calculate the upthrust.

  2. 2

    Step 1: The volume of water displaced is equal to the volume of the block, V = 0.05 m³.

  3. 3

    Step 2: Use the upthrust formula: Upthrust = ρ_water × g × V.

  4. 4

    Step 3: Substitute values: Upthrust = 1000 kg/m³ × 9.81 N/kg × 0.05 m³ = 490.5 N.

  5. 5

    Part (b): Determine if it sinks or floats.

  6. 6

    Step 4: Calculate the weight of the block. Weight = mass × g = 40 kg × 9.81 N/kg = 392.4 N.

  7. 7

    Step 5: Compare the upthrust with the weight. Upthrust (490.5 N) is greater than the weight (392.4 N).

  8. 8

    Step 6: Conclude that the block will experience a net upward force and will rise to the surface to float.

Recap

  • Upthrust is an upward force on an object in a fluid.
  • It is caused by the pressure difference between the top and bottom of the object.
  • Archimedes' principle states upthrust equals the weight of the displaced fluid.
  • An object floats if its weight is less than or equal to the upthrust.
  • The formula for upthrust is F = ρgV, where V is the volume of displaced fluid.

Quick check

  1. A 1 kg block of iron and a 1 kg block of wood are placed in water. Which experiences the greater upthrust?2 marks

5. Hooke's Law and Material Deformation

When forces are applied to a solid object, they can cause it to change shape, a process called deformation. A tensile force stretches an object, while a compressive force squashes it. For many materials, like springs and wires, the extension(x) is directly proportional to the applied force (F), provided a certain limit is not exceeded. This relationship is Hooke's Law, F = kx. The constant 'k' is the force constant (or spring constant), a measure of the object's stiffness. A graph of force against extension is a straight line through the origin up to the 'limit of proportionality'. If the force is removed within the 'elastic limit', the object returns to its original shape (elastic deformation). Beyond this limit, the object is permanently deformed (plastic deformation).

F = kx

Key term

Hooke's Law: The extension of an object is directly proportional to the force applied, as long as the elastic limit is not exceeded.

Examiner insight

Candidates must be able to accurately sketch and label a force-extension graph, clearly indicating the limit of proportionality, elastic limit, and regions of elastic and plastic behaviour.

Common pitfall

Confusing the 'limit of proportionality' (where the F-x graph is no longer linear) with the 'elastic limit' (where permanent deformation begins). For many materials they are very close, but they are distinct concepts.

Worked example 14 marks

A spring has a force constant of 200 N/m.(a) What force is needed to extend it by 5.0 cm?(b) What is its extension when a force of 15 N is applied?

  1. 1

    Part (a): Calculate the force.

  2. 2

    Step 1: Convert the extension to metres. x = 5.0 cm = 0.050 m.

  3. 3

    Step 2: Use Hooke's Law: F = kx.

  4. 4

    Step 3: Substitute values: F = 200 N/m × 0.050 m = 10 N.

  5. 5

    Part (b): Calculate the extension.

  6. 6

    Step 4: Rearrange Hooke's Law to find x: x = F/k.

  7. 7

    Step 5: Substitute values: x = 15 N / 200 N/m = 0.075 m.

  8. 8

    Step 6: Convert the extension to cm if required: 0.075 m = 7.5 cm.

Recap

  • Hooke's Law states that force is proportional to extension (F ∝ x).
  • The formula is F = kx, where k is the force constant in N/m.
  • Elastic deformation is temporary; the object returns to its original shape.
  • Plastic deformation is permanent.
  • The limit of proportionality is where the F-x graph stops being a straight line.

Quick check

  1. What does the gradient of a force-extension graph represent?1 mark

6. Stress, Strain, and the Young Modulus

To compare the 'stretchiness' of materials without being confused by their shape or size, we use stress and strain. Stress (σ) is a measure of the force applied per unit of cross-sectional area (σ = F/A), essentially normalising the force. Strain (ε) is the fractional change in length (ε = x/L), where x is the extension and L is the original length. Strain is a dimensionless quantity. For an elastic material, the ratio of stress to strain is a constant known as the Young Modulus (E). E = σ/ε. The Young Modulus is a measure of a material's stiffness; a higher value means the material is stiffer and harder to stretch. To measure E for a wire, one applies known forces (weights), measures the extension with a precise instrument (like a vernier scale), and measures the wire's original length and diameter (to find area).

Stress, σ = F / A

Strain, ε = x / L

Young Modulus, E = σ / ε = (F/A) / (x/L)

Key term

Young Modulus (E): A measure of a material's stiffness, defined as the ratio of tensile stress to tensile strain within the limit of proportionality.

Examiner insight

When describing the experiment to measure the Young Modulus, marks are awarded for mentioning methods to ensure accuracy, such as using a long wire for a measurable extension and using a micrometer to measure the diameter at several points.

Common pitfall

Using the diameter instead of the radius when calculating the cross-sectional area (A = πr²). Always remember to halve the diameter first.

Worked example 15 marks

A steel wire of length 2.0 m and diameter 0.80 mm is stretched by a force of 150 N. It extends by 1.9 mm. Calculate the Young Modulus of steel.

  1. 1

    Step 1: List all quantities in SI units. L = 2.0 m, F = 150 N, x = 1.9 mm = 1.9 × 10⁻³ m. Diameter = 0.80 mm, so radius r = 0.40 mm = 0.40 × 10⁻³ m.

  2. 2

    Step 2: Calculate the cross-sectional area, A. A = πr² = π × (0.40 × 10⁻³ m)² = 5.027 × 10⁻⁷ m².

  3. 3

    Step 3: Calculate the stress, σ. σ = F / A = 150 N / (5.027 × 10⁻⁷ m²) = 2.984 × 10⁸ Pa.

  4. 4

    Step 4: Calculate the strain, ε. ε = x / L = (1.9 × 10⁻³m) / 2.0 m = 9.5 × 10⁻⁴.

  5. 5

    Step 5: Calculate the Young Modulus, E. E = σ / ε = (2.984 × 10⁸ Pa) / (9.5 × 10⁻⁴) = 3.14 × 10¹¹ Pa.

Recap

  • Stress is force per unit area (σ = F/A).
  • Strain is the fractional change in length (ε = x/L).
  • The Young Modulus (E) is the ratio of stress to strain (E = σ/ε).
  • Young Modulus is a property of a material, indicating its stiffness.
  • To measure E, you need to measure F, A, x, and L accurately.

Quick check

  1. Material A has a Young Modulus of 200 GPa. Material B has a Young Modulus of 100 GPa. Which is stiffer?1 mark

7. Energy Stored in Deformed Materials

When you stretch or compress an elastic material, you do work on it. This work is stored in the material as elastic potential energy (also called strain energy). When the force is removed, this stored energy is released, often causing the object to spring back. The amount of work done (and energy stored) is equal to the area under the force-extension graph. For a material that obeys Hooke's Law, the graph is a triangle. The area of a triangle is ½ × base × height, which for the graph corresponds to ½ × extension × force. This gives the formula for elastic potential energy, E_p = ½Fx. Since F = kx, we can substitute this in to get an alternative formula, E_p = ½kx².

E_p = (1/2)Fx

E_p = (1/2)kx²

Key term

Elastic Potential Energy: The potential energy stored in an elastic object as a result of its deformation, equal to the work done to deform it.

Examiner insight

Examiners look for the understanding that the area under a force-extension graph represents the work done or energy stored. Be prepared to calculate this area for non-linear graphs by counting squares.

Common pitfall

Forgetting the '½' in the energy formulas. The energy is not simply force × extension because the force is not constant as the spring is stretched; it increases from 0 to F.

Worked example 15 marks

A spring with a force constant k = 80 N/m is compressed by 6.0 cm.(a) How much energy is stored in the spring?(b) The spring is used to launch a 0.40 kg trolley on a frictionless track. What is the maximum speed of the trolley?

  1. 1

    Part (a): Calculate the stored energy.

  2. 2

    Step 1: Convert extension to metres. x = 6.0 cm = 0.060 m.

  3. 3

    Step 2: Use the formula E_p = ½kx². E_p = ½ × 80 N/m × (0.060 m)².

  4. 4

    Step 3: Calculate the energy. E_p = 40 × 0.0036 = 0.144 J.

  5. 5

    Part (b): Calculate the trolley's speed.

  6. 6

    Step 4: Apply conservation of energy. The stored elastic potential energy is converted into kinetic energy of the trolley. E_p = E_k.

  7. 7

    Step 5: State the formula for kinetic energy: E_k = ½mv². So, 0.144 J = ½ × 0.40 kg × v².

  8. 8

    Step 6: Rearrange to find v². 0.144 = 0.20 × v². So, v² = 0.144 / 0.20 = 0.72.

  9. 9

    Step 7: Find v by taking the square root. v = √0.72 ≈ 0.85 m/s.

Recap

  • Work done in stretching a material is stored as elastic potential energy.
  • The energy stored is the area under the force-extension graph.
  • For a material obeying Hooke's Law, E_p = ½Fx.
  • An alternative formula is E_p = ½kx².
  • This stored energy can be converted into other forms, like kinetic energy.

Quick check

  1. If you double the extension of a spring, by what factor does the stored energy increase?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A crown has a mass of 1.20 kg. When fully submerged in water (density 1000 kg/m³), its apparent weight is 11.4 N. Taking g = 9.81 N/kg, calculate the density of the crown.5 marks
  2. A rectangular block of dimensions 0.5 m x 0.4 m x 0.2 m has a weight of 800 N. Calculate the maximum and minimum pressure it can exert when resting on a horizontal surface.4 marks
  3. A U-tube manometer containing mercury (density 13600 kg/m³) is connected to a gas supply. The level of mercury in the arm open to the atmosphere is 12 cm higher than the level in the arm connected to the gas supply. Atmospheric pressure is 1.01 x 10⁵ Pa. Calculate the absolute pressure of the gas supply.4 marks
  4. A spring is hung vertically. When a mass of 200 g is attached, its length increases from 15 cm to 19 cm. (a) Calculate the force constant of the spring. (b) Calculate the energy stored in the spring when it is extended by this amount.5 marks
  5. Define tensile stress, tensile strain, and the Young modulus. Explain why the Young modulus is a more useful property for a material scientist than the force constant.5 marks
  6. A nylon guitar string is 65 cm long and has a diameter of 1.0 mm. When under a tension of 90 N, it is in tune. The Young modulus for nylon is 2.0 GPa. Calculate the extension of the string.4 marks
  7. Describe an experiment to determine the Young modulus of a metal in the form of a wire. Your description should include a list of the apparatus, the measurements to be taken, and how the Young modulus is calculated from these measurements.6 marks
  8. A 1500 kg car is supported equally by four tyres. The gauge pressure in each tyre is 250 kPa. Calculate the area of contact of one tyre with the road.4 marks
  9. The graph shows the force-extension curve for a rubber band. (a) Explain why the band does not obey Hooke's Law. (b) Use the graph to determine the work done in stretching the band by 40 cm.4 marks
  10. A spherical weather balloon of diameter 3.0 m is filled with helium (density 0.18 kg/m³). The balloon material has a mass of 5.0 kg. The density of the surrounding air is 1.22 kg/m³. Calculate the net upward force on the balloon.6 marks

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