Pressure

1. Overview

Pressure is a measure of how concentrated a force is on a surface. Understanding pressure allows us to explain why heavy vehicles need wide tires to avoid sinking and how deep-sea submarines withstand the immense weight of the ocean above them.

Key Definitions

• Pressure: The force acting per unit area on a surface.
• Pascal (Pa): The SI unit of pressure, equivalent to one Newton per square metre ($1\text{ N/m}^2$).
• Density ($\rho$): The mass per unit volume of a substance.

Core Content

The Pressure Equation

Pressure depends on two factors: the size of the force applied and the area over which that force is spread.

• Large Area: Spreads the force, resulting in lower pressure.
• Small Area: Concentrates the force, resulting in higher pressure.

Everyday Examples

• High Pressure: A sharp knife has a very small surface area at the edge, creating enough pressure to cut through materials with little force.
• Low Pressure: Skis or snowshoes have a large surface area to spread the wearer's weight, preventing them from sinking into soft snow.

Pressure in Liquids (Qualitative)

The pressure in a liquid behaves differently than in solids:

• Depth: Pressure increases as depth increases. This is because there is a greater weight of liquid acting downwards on the layers below.
• Density: Pressure increases if the density of the liquid increases, as a denser liquid is heavier for the same volume.
• Direction: Pressure in a fluid acts equally in all directions.

Worked Example (Core)

A box weighs $200\text{ N}$ and has a base area of $0.5\text{ m}^2$. Calculate the pressure exerted by the box on the floor.

1. State the formula: $P = F / A$
2. Substitute values: $P = 200 / 0.5$
3. Answer: $400\text{ Pa}$

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Extended Content (Extended Only)

Calculating Pressure in a Liquid

To calculate the change in pressure beneath the surface of a liquid, we use the density of the liquid and the depth.

Equation: $\Delta p = \rho g \Delta h$

Where $g$ is the acceleration due to gravity (usually $9.8\text{ m/s}^2$ or $10\text{ m/s}^2$ in IGCSE).

Worked Example (Extended)

Calculate the pressure exerted by water at the bottom of a swimming pool $3\text{ metres}$ deep. (Density of water = $1000\text{ kg/m}^3$; $g = 9.8\text{ m/s}^2$).

1. State the formula: $p = \rho g h$
2. Substitute values: $p = 1000 \times 9.8 \times 3$
3. Answer: $29,400\text{ Pa}$ (or $29.4\text{ kPa}$)

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Key Equations

Equation	Symbols	Units
$p = \frac{F}{A}$	$p$ = Pressure, $F$ = Force, $A$ = Area	$p$ (Pa), $F$ (N), $A$ ($\text{m}^2$)
$\Delta p = \rho g \Delta h$	$\rho$ = Density, $g$ = Gravitational field strength, $h$ = Depth	$\rho$ ($\text{kg/m}^3$), $g$ (N/kg), $h$ (m)

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Common Mistakes to Avoid

• ❌ Wrong: Calculating pressure using $\text{cm}^2$ but giving the answer in Pascals (Pa).
• ✅ Right: Always convert $\text{cm}^2$ to $\text{m}^2$ before calculating Pascals. (Remember: $1\text{ m}^2 = 10,000\text{ cm}^2$).
• ❌ Wrong: Thinking that if the weight stays the same, the pressure stays the same.
• ✅ Right: If the weight (force) is constant, but you reduce the contact area (e.g., standing on one foot instead of two), the pressure increases.
• ❌ Wrong: Multiplying force by area ($F \times A$).
• ✅ Right: Pressure is force divided by area ($F / A$).
• ❌ Wrong: Assuming that halving the area and doubling the force cancels out.
• ✅ Right: This actually makes the pressure four times greater ($2F / 0.5A = 4P$).

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Exam Tips

1. Unit Conversion is Key: Examiners love to give the area in $\text{cm}^2$. To convert $\text{cm}^2$ to $\text{m}^2$, divide by $10,000$ (or $100^2$). If you just divide by $100$, your answer will be off by a factor of $100$.
2. Total Pressure: In "Extended" liquid pressure questions, if the question asks for the total pressure at a certain depth, you must add the atmospheric pressure (usually $1 \times 10^5\text{ Pa}$) to the liquid pressure you calculated ($p = \text{atmospheric pressure} + \rho gh$).
3. Rearranging the Formula: Practice using the formula triangle for $P=F/A$ so you can easily calculate Force ($F = P \times A$) or Area ($A = F / P$).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) Define pressure. [2]

(b) A rectangular block of concrete has dimensions 0.5 m x 0.2 m x 0.1 m and weighs 250 N. Calculate the maximum pressure the block can exert on a horizontal surface. [3]

Worked Solution:

(a)

1. Pressure is defined as force acting normally per unit area. This is the fundamental definition.

How to earn full marks:

• State that pressure is force per area.
• State that the force acts perpendicular to the surface or normally.

(b)

1. Identify the smallest area: 0.2 m x 0.1 m = 0.02 m$^2$ The smallest area will result in the greatest pressure for a given force.

2. Calculate the pressure: $P = \frac{F}{A} = \frac{250 \text{ N}}{0.02 \text{ m}^2} = 12500 \text{ Pa}$ Applying the pressure equation.

How to earn full marks:

• Correctly calculate the smallest area: 0.02 m$^2$
• Use the pressure equation with correct substitutions.
• Correct final answer with unit: $\boxed{12500 \text{ Pa}}$

Common Pitfall: Many students struggle with identifying the correct area to use. Remember that maximum pressure occurs when the force is applied over the smallest possible area. Also, be careful with unit conversions; make sure your area is in m$^2$ before calculating pressure.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) State two ways in which the pressure at a point in a liquid increases. [2]

(b) A diver is 15 m below the surface of the sea. The density of seawater is 1030 kg/m$^3$. Calculate the pressure due to the seawater at this depth. (Assume the atmospheric pressure is negligible.) [4]

Worked Solution:

(a)

1. The pressure increases with depth. Pressure increases as you go deeper into the liquid.
2. The pressure increases with the density of the liquid. Denser liquid means more weight above.

How to earn full marks:

• Mention that increasing the depth increases the pressure.
• Mention that increasing the density of the liquid increases the pressure.

(b)

1. Recall the pressure equation: $P = \rho g h$ State the relevant equation.

2. Substitute the values: $P = (1030 \text{ kg/m}^3) \times (9.8 \text{ N/kg}) \times (15 \text{ m})$ Correct substitution of values.

3. Calculate the pressure: $P = 151410 \text{ Pa}$ Calculation of the pressure.

How to earn full marks:

• Correctly state the pressure equation $P = \rho g h$.
• Substitute all values with correct units.
• Correct final answer with unit: $\boxed{151410 \text{ Pa}}$

Common Pitfall: A common mistake is forgetting to include the gravitational field strength, g, in the pressure calculation. Also, remember that the equation $P = \rho g h$ gives the pressure due to the liquid only. If the question asks for total pressure, you might need to add atmospheric pressure.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A rectangular water tank has a base area of 2.0 m$^2$ and is filled with water to a depth of 1.5 m. The density of water is 1000 kg/m$^3$ and the gravitational field strength is 9.8 N/kg.

(a) Calculate the weight of the water in the tank. [4]

(b) Calculate the pressure exerted by the water on the base of the tank. [2]

(c) A small hole is made in the side of the tank at a depth of 0.5 m below the water surface. Describe what happens to the water as it emerges from the hole and explain why this happens. [2]

Worked Solution:

(a)

1. Calculate the volume of water: $V = A \times h = 2.0 \text{ m}^2 \times 1.5 \text{ m} = 3.0 \text{ m}^3$ Finding the volume using area and height.

2. Calculate the mass of water: $m = \rho \times V = 1000 \text{ kg/m}^3 \times 3.0 \text{ m}^3 = 3000 \text{ kg}$ Using density to find the mass.

3. Calculate the weight of water: $W = m \times g = 3000 \text{ kg} \times 9.8 \text{ N/kg} = 29400 \text{ N}$ Weight is mass times gravitational field strength.

How to earn full marks:

• Correctly calculate the volume: 3.0 m$^3$
• Correctly calculate the mass: 3000 kg
• Use the correct equation to find the weight.
• Correct final answer with unit: $\boxed{29400 \text{ N}}$

(b)

1. Calculate the pressure: $P = \rho g h = 1000 \text{ kg/m}^3 \times 9.8 \text{ N/kg} \times 1.5 \text{ m} = 14700 \text{ Pa}$ Applying the pressure equation.

How to earn full marks:

• Use the correct equation to find the pressure.
• Use correct values for density, g, and depth.
• Correct final answer with unit: $\boxed{14700 \text{ Pa}}$

(c)

1. The water emerges horizontally from the hole. The water flows out horizontally.

2. The pressure at 0.5 m depth forces the water out. The pressure is the same in all directions at that depth. The pressure at that depth causes the water to flow.

How to earn full marks:

• State that the water comes out horizontally.
• Explain that the pressure at that depth causes the water to flow out.

Common Pitfall: In part (a), be sure to calculate weight (in Newtons), not just mass (in kilograms). Many students forget the final step of multiplying by g. In part (c), remember that the water emerges horizontally due to the pressure at that specific depth.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A hydraulic lift is used in a car repair shop. A force of 200 N is applied to a small piston with a cross-sectional area of 0.01 m$^2$. This pressure is transmitted through the hydraulic fluid to a large piston with a cross-sectional area of 0.5 m$^2$.

(a) Calculate the pressure exerted on the small piston. [2]

(b) Calculate the force exerted by the large piston. [3]

(c) State one property of the hydraulic fluid that makes it suitable for use in a hydraulic lift. Explain why this property is important. [2]

(d) Explain why the hydraulic lift system does not violate the principle of conservation of energy. [2]

Worked Solution:

(a)

1. Calculate the pressure: $P = \frac{F}{A} = \frac{200 \text{ N}}{0.01 \text{ m}^2} = 20000 \text{ Pa}$ Applying the pressure equation.

How to earn full marks:

• Use the correct equation to find the pressure.
• Correct final answer with unit: $\boxed{20000 \text{ Pa}}$

(b)

1. Pressure is the same throughout the fluid: $P_{small} = P_{large}$ Stating that the pressure is the same.

2. Calculate the force on the large piston: $F = P \times A = 20000 \text{ Pa} \times 0.5 \text{ m}^2 = 10000 \text{ N}$ Rearranging and applying the pressure equation.

How to earn full marks:

• State that the pressure is the same in both pistons.
• Use the correct pressure value from part (a).
• Correct final answer with unit: $\boxed{10000 \text{ N}}$

(c)

1. The fluid is incompressible. Incompressibility is a key property.

2. If the fluid were compressible, some of the force would be used to compress the fluid rather than lift the car. So the incompressibility of fluid makes sure pressure is uniformly transmitted to the other end. Explanation of why incompressibility is important.

How to earn full marks:

• State that the fluid is incompressible.
• Explain why incompressibility is important for efficient force transmission.

(d)

1. The small piston must move a greater distance than the large piston. This is because the volume displaced is the same.

2. The work done on the small piston is equal to the work done by the large piston (assuming no energy losses due to friction). $W = F \times d$ so a smaller force over a larger distance is the same work as a larger force over a shorter distance. This explains the conservation of energy.

How to earn full marks:

• State that the small piston moves a greater distance.
• Explain that the work done by both pistons is the same, conserving energy.

Common Pitfall: Many students only focus on the force multiplication in hydraulic lifts and forget about the distances moved by the pistons. Remember that while the large piston exerts a greater force, the small piston must move a proportionally greater distance to achieve this, ensuring that the work done (and therefore energy) is conserved. Also, be sure to state incompressibility as the key property of the fluid.