FORM ONE PHYSICS STUDY NOTES TOPIC 7-9

TOPIC 7: PRESSURE

Concept of Pressure

Pressure is defined as the force per unit area, or the force acting normally (perpendicularly) per unit surface area.

It is calculated by the formula:

Pressure = Force (F) / Area (A)

P = F / A

Where:

• P – Pressure
• F – Force
• A – Area

The S.I Unit of Pressure

The SI unit of pressure is Newton per square metre (N/m²), commonly referred to as the Pascal (Pa).

1 Pa = 1 N/m²

Other units of pressure include atmosphere, torr, bar, and mmHg.

• 1 atmosphere = 780 mmHg
• 1 atmosphere = 1 × 10⁵ N/m² = 1 bar (used by meteorologists)

Note: For a given amount of force, the smaller the area of application, the greater the pressure exerted.

When a man lifts a bucket of water by its handle made of thin metal, he experiences discomfort. If the handle is thicker, the discomfort is much less or absent because the area over which the force is applied is larger.

Pressure due to Solids

Dependence of Pressure on Surface of Contact

Pressure in solids depends on the surface area of contact. A force (F) applied onto a small area exerts higher pressure compared to when applied onto a large surface.

Pressure in solid = Force applied / Area of contact.

Example 1

A block of wood weighing 30 N measures 5 m by 10 m by 4 m. If placed on a table with the largest possible area (5 m × 10 m) in contact, it exerts less pressure than when placed with its smallest possible area (5 m × 4 m) in contact.

Solution

Data:

• Force = 30 N
• Largest base area = ?

A_L = 10 m × 5 m = 50 m²

P = F / A = 30 N / 50 m² = 0.6 N/m²

Pressure = 0.6 N/m²

Force (F) = 30 N.

Small base area = ?

A_S = 5 m × 4 m = 20 m²

P = F / A = 30 N / 20 m² = 1.5 N/m²

Pressure = 1.5 N/m²

Example 2

A tip of a needle has a cross-sectional area of 1 × 10^-6 m². If a doctor applies a force of 20 N to a syringe connected to the needle, what pressure is exerted at the tip?

Solution

Data:

• Area (A) = 1 × 10^-6 m²
• Force (F) = 20 N
• Pressure = ?

P = F / A

P = 20 / (1 × 10^-6) = 2.0 × 10⁷ N/m²

The pressure exerted by the needle tip is 2.0 × 10⁷ N/m².

Example 3

A rectangular metal block with sides 1.05 m by 1.0 m by 1.2 m rests on a horizontal surface. If the density of the metal is 7000 kg/m³, calculate the maximum and minimum pressure the block can exert on the surface.

(Take the weight of 1 kg mass as 10 N)

Solution

Data:

• Dimensions = 1.2 m, 1.05 m, 1.0 m
• Density = 7000 kg/m³
• Maximum pressure (P_max) = ?
• Minimum pressure (P_min) = ?

Volume = 1.2 × 1.05 × 1.0 = 1.26 m³

Density = mass / Volume

7000 kg/m³ = mass / 1.26 m³

Mass = 7000 × 1.26 = 8820 kg

Weight (Force) = 8820 × 10 = 88200 N

Force of metal = 8.82 × 10⁴ N

Areas:

• A₁ = 1.2 × 1.05 = 1.26 m²
• A₂ = 1.2 × 1.0 = 1.2 m²
• A₃ = 1.05 × 1.0 = 1.05 m²

Maximum pressure is exerted on the smallest area:

P_max = F / A_min = 8.82 × 10⁴ N / 1.05 m² = 8.4 × 10⁴ N/m²

Minimum pressure is exerted on the largest area:

P_min = F / A_max = 8.82 × 10⁴ N / 1.26 m² = 7.0 × 10⁴ N/m²

The Applications of Pressure due to Solids

Applications of pressure due to solids include:

• Making objects like screws, nails, pins, spears, and arrows with sharp points to increase penetrating power.
• Helping some living organisms for self-defense, such as fish using sharp fins.
• Walking on shoes with sharp pointed heels exerts greater pressure on the ground than flat shoes.
• Construction of railways uses wide wooden or concrete sleepers below tracks to provide a larger surface area for the train’s weight, ensuring safety.
• Buildings are constructed with wide foundations to distribute the building’s weight over a larger area.

Pressure in Liquids

A liquid exerts pressure on an immersed object and on the container walls. This pressure is due to the weight of the liquid. Increasing the water level increases the pressure.

The Characteristics of Pressure in Liquids

Pressure in liquids = Force / Area

Force = h × A × ρ × g, so pressure = h × ρ × g

Hence, pressure in liquids is given as p = hρg.

Where:

• H = Height of the liquid column
• A = Area of the base

The pressure at any point in a liquid at rest depends on:

1. Depth
2. Density of the liquid

Note: It does not depend on the area.

Pressure in liquids is characterized by:

1. Pressure increases with depth.
2. Pressure acts equally in all directions.
3. Pressure increases with the density of the liquid.

Mercury exerts more pressure than an equal volume of water because mercury is denser.

The Variation of Pressure with Depth in Liquids

Demonstration of water spurting from holes at different heights:

Pressure at point A is low due to the small height of water above it; at point B, the height and distance of water increase; at point C, the height is greatest, so the water spurts the farthest.

Demonstration of pressure in a communicating vessel:

The shapes of vessels at points A, B, C, and D differ, but the pressure is the same due to the equal height (L) of the liquid above these points.

Problems Involving Pressure in Liquids

Example 4

A cube of side 2 cm is completely submerged in water so that the bottom is at a depth of 10 cm. Use g = 10 m/s² and ρ = 1000 kg/m³.

1. What is the difference between the pressure on the bottom and the top of the cube?
2. Determine the difference in force on the top and bottom.
3. What is the weight of the water displaced by the cube?

Solution

Data:

• Depth = 10 cm = 0.1 m
• ρ = 1000 kg/m³
• g = 10 m/s²
• Height of cube (h) = 2 cm = 0.02 m

Formula:

p = ρgh

Pressure at bottom = 1000 × 10 × 0.1 = 1000 N/m²

Pressure at top = 1000 × 10 × 0.08 = 800 N/m²

Change in pressure = 1000 – 800 = 200 N/m²

The difference in pressure between bottom and top is 200 N/m².

Pressure = Force / Area = F / A

Area = (0.02)² = 0.0004 m²

Difference in force = 200 N/m² × 0.0004 m² = 0.08 N

This is the upthrust acting on the cube.

Weight of water displaced = volume × density × g

Volume = (2 cm)³ = 8 cm³

For water, 1 cm³ = 1 g, so 8 cm³ = 8 g

Weight of displaced water = 8 g = 0.08 N

Example 5

Calculate the pressure at the bottom of a tank of water 15 m deep due to the water above it (ρ = 1000 kg/m³).

Solution

Data:

• Height = 15 m
• ρ = 1000 kg/m³
• g = 10 m/s²

Formula:

Pressure in liquid = ρgh

Pressure = 1000 × 10 × 15 = 150000 N/m²

The Principle of Hydraulic Pressure

It states that “Any external pressure applied to the surface of an enclosed liquid will be transmitted equally throughout the liquid.”

Illustration of Pascal’s Principle:

Note: All sides of the vessel experience equal pressure.

The hydraulic press works on Pascal’s principle. A small force can be converted into a large force and vice versa.

Note: The distance moved (d) is inversely proportional to the cross-sectional area.

Example 6

The pistons of a hydraulic press have areas of 3.0 × 10^-4 m² and 2 × 10^-2 m² respectively. If the smaller piston is pushed down with a force of 12 N, what force is required to push the larger piston?

Solution

Data:

• F₁ = 12 N
• A₁ = 3 × 10^-4 m²
• A₂ = 2 × 10^-2 m²

Application of Hydraulic Press

• Measuring pressure of a liquid.
• Used in industries to compress bulk items.
• Hydraulic brake system: Pressure applied to the brake pedal pushes the piston in the master cylinder, creating pressure in the brake fluid. This pressure is transferred to slave cylinders, multiplying the force and pushing brake shoes against the brake drum attached to the vehicle’s wheel.
• Used in metal forming industries.
• Used for lifting heavy loads, acting like a lift.
• Measuring pressure of liquids using a manometer, a device commonly used for gas pressure measurement.

Atmospheric Pressure

The Existence of Atmospheric Pressure

Atmospheric pressure results from the weight of the layer of gases surrounding the Earth. The atmosphere is a mixture of gases.

Note: Atmospheric pressure on the Earth’s surface is approximately 1.01 × 10⁵ N/m².

Experiment Demonstrating Atmospheric Pressure

Umbrella Experiment

When a glass tumbler filled with water is turned upside down gently, water does not pour out due to atmospheric pressure.

Plunger

Pulling the plunger is difficult because air is squeezed out, and high surrounding pressure causes the plunger to stick.

Crushed Bottle

When hot water is poured into a bottle and then cooled under cold water, steam condenses leaving a partial vacuum inside. The greater atmospheric pressure outside crushes the bottle inward.

Applications of Atmospheric Pressure

Applications include:

A Siphon

• Used in toilet flushing cisterns (chain and ball tank) operated by a handle that lifts water via a diaphragm-like piston pump.
• Used in siphon rain gauges to automatically drain excess water.
• A siphon cup is a reservoir attached to a gum.
• Used in drainage systems to drain water to one point.

The Lift Pump

A lift pump raises water from underground sources by lifting the liquid rather than forcing it up.

A syringe is a simple piston pump that lifts liquid by pulling and pushing a plunger inside a cylindrical tube, allowing fluid intake or expulsion through a nozzle.

Uses of Syringe:

1. Fitted with hypodermic needles for injections.
2. Measuring liquids and gases in laboratories.
3. Applying compounds such as glue or lubricant.

Bicycle Pump

A force pump consisting of a hollow metal cylinder and a movable piston.

Measuring Atmospheric Pressure

Atmospheric pressure is measured using a barometer.

Types of Barometer:

1. Simple barometer
2. Fortin barometer
3. Aneroid barometer

A simple barometer is the most fundamental type, using mercury as the barometric liquid.

It consists of a hard glass tube closed at one end.

Fortin Barometer

A Fortin barometer is a modified simple barometer. It consists of an inverted tube closed at the upper end, with the lower open end immersed in a mercury reservoir. Atmospheric pressure is measured by the height of the mercury column.

Disadvantages of Fortin Barometer:

1. Mercury is expensive and toxic; thus, an aneroid barometer is usually preferred.
2. It is not portable due to its size and liquid content.
3. Must be mounted vertically.