O-Level Physics · Secondary 3 and 4
Pressure
Pressure tells you how concentrated a force is. In this upper secondary topic you apply p = F/A to solids, calculate the pressure in a liquid using p = hρg, see how hydraulic machines multiply forces, and learn how a mercury barometer and a manometer measure pressure. The same ideas explain why knives cut, why dams are thicker at the bottom and why your ears feel the pressure at the bottom of a swimming pool. Gas pressure is explained further in thermal physics.
What the syllabus expects
- Define pressure as force per unit area and use p = F/A
- Explain everyday examples of high and low pressure, such as knives, nails and wide tyres
- Describe how pressure is transmitted in a hydraulic system, such as a hydraulic press or car brakes
- Recall and use p = hρg to calculate the pressure due to a liquid column
- Describe how a mercury barometer measures atmospheric pressure
- Describe how a manometer measures the pressure of a gas supply
Pressure on solids
Pressure is the force acting per unit area, p = F/A, where F is the force acting at right angles to the surface. The unit is the pascal: 1 Pa = 1 N/m². A block of weight 60 N resting on a face of area 0.020 m² exerts a pressure of 60 ÷ 0.020 = 3000 Pa. The same force over a smaller area gives a larger pressure, which is why knives are sharpened, nails have points and stiletto heels dent soft floors. A larger area reduces pressure, so tractors have wide tyres, snowshoes stop you sinking and bags have wide straps. Take care with areas given in cm². Because 1 m = 100 cm, 1 m² = 100 × 100 = 10 000 cm², so to convert cm² to m² you divide by 10 000, not by 100.
Pressure in liquids
A liquid exerts pressure because of the weight of liquid above any point. The pressure due to a liquid column of depth h and density ρ is p = hρg, with h in m, ρ in kg/m³ and g = 10 N/kg, giving p in Pa. Liquid pressure increases with depth and with the density of the liquid. At a given depth it acts equally in all directions, and it does not depend on the shape or width of the container. At 5.0 m below the surface of fresh water, of density 1000 kg/m³, the pressure due to the water is 5.0 × 1000 × 10 = 50 000 Pa. The total pressure there also includes the atmospheric pressure pressing on the surface. Dams are built thicker at the base because the water pressure is greatest at the bottom.
Hydraulic systems
Liquids are almost incompressible, so a pressure applied to an enclosed liquid is transmitted equally to every part of it. A hydraulic press uses this to multiply force. A small force on a small piston creates a pressure, and the same pressure acting on a much larger piston produces a larger force. If 50 N is applied to a small piston of area 0.0010 m², the pressure is 50 ÷ 0.0010 = 50 000 Pa. On a large piston of area 0.040 m² this gives a force of 50 000 × 0.040 = 2000 N, forty times larger. There is no free energy, though. The small piston has to move forty times further than the large one. Car brakes work in the same way, transmitting the pressure from the brake pedal through brake fluid to the pistons at each wheel.
Atmospheric pressure, barometers and manometers
The air above us exerts atmospheric pressure, about 1.0 × 10⁵ Pa at sea level. A simple mercury barometer is a long glass tube filled with mercury and turned upside down in a dish of mercury. The mercury falls until the pressure of its column balances the atmospheric pressure, leaving a vacuum at the top. The vertical height of the column, about 760 mm at sea level, measures the atmospheric pressure. Tilting the tube or using a wider tube does not change this vertical height. A manometer is a U-tube containing a liquid. One arm is connected to a gas supply and the other is open to the air. If the gas pressure is higher than atmospheric, the liquid is pushed down on the gas side and up on the open side. The difference in levels h gives the extra pressure, hρg.
Worked example
A U-tube manometer contains water of density 1000 kg/m³ and is connected to a gas supply. The water level in the open arm is 25 cm higher than the level in the arm connected to the gas. Atmospheric pressure is 1.00 × 10⁵ Pa. Taking g = 10 N/kg, calculate the pressure of the gas.
- Convert the height difference to metres: h = 25 cm = 0.25 m.
- Pressure difference = hρg = 0.25 × 1000 × 10 = 2500 Pa.
- The water is higher in the open arm, so the gas is pushing harder than the atmosphere and its pressure is greater than atmospheric.
- Gas pressure = atmospheric pressure + hρg = 100 000 + 2500 = 102 500 Pa.
Answer: 1.025 × 10⁵ Pa (102 500 Pa)
Common mistakes
- Converting cm² to m² by dividing by 100. One square metre is 10 000 cm², so divide by 10 000.
- Thinking a wider container gives a greater liquid pressure at the bottom. Liquid pressure depends only on depth, density and g, not on the shape or cross-section of the container.
- Measuring depth from the bottom of the container. In p = hρg, h is the vertical depth below the liquid surface.
- Subtracting hρg in a manometer question when the gas pressure is higher than atmospheric. If the liquid is pushed down on the gas side, add hρg to atmospheric pressure.
- Forgetting atmospheric pressure when a question asks for the total pressure at a depth. The liquid pressure hρg is only the extra pressure due to the liquid.
- Claiming the barometer reading changes when the tube is tilted. The length of mercury in the tube changes, but the vertical height stays the same.
Exam tips
- Write every area in m² on its own line before substituting into p = F/A. Area conversions are where most marks are lost in this topic.
- In manometer questions, sketch the U-tube and mark which side is higher before deciding whether to add or subtract hρg from atmospheric pressure.
- When describing a barometer, say the height is the vertical distance from the mercury surface in the dish to the top of the mercury column, and mention the vacuum above it.
Practice questions
1. Easy: A crate weighing 1200 N rests on a rectangular base measuring 0.60 m by 0.50 m. Calculate the pressure it exerts on the floor.
Show answer
4000 Pa. Area = 0.60 × 0.50 = 0.30 m². Pressure = F ÷ A = 1200 ÷ 0.30 = 4000 Pa.
2. Medium: A diver is 12 m below the surface of seawater of density 1030 kg/m³. Atmospheric pressure is 1.0 × 10⁵ Pa and g = 10 N/kg. Calculate the total pressure on the diver.
Show answer
About 2.2 × 10⁵ Pa. Pressure due to the water = hρg = 12 × 1030 × 10 = 123 600 Pa. Total pressure = 100 000 + 123 600 = 223 600 Pa, which is about 2.2 × 10⁵ Pa.
3. Harder: In a hydraulic jack the small piston has an area of 2.0 cm² and the large piston has an area of 80 cm². (a) What force on the small piston will support a car of weight 12 000 N on the large piston? (b) What is the pressure in the fluid? (c) If the car rises by 1.0 cm, how far does the small piston move?
Show answer
(a) The large area is 80 ÷ 2.0 = 40 times the small area, so the force needed is 12 000 ÷ 40 = 300 N. (b) Large area = 80 ÷ 10 000 = 0.0080 m², so p = 12 000 ÷ 0.0080 = 1.5 × 10⁶ Pa (check: 300 ÷ 0.00020 gives the same). (c) The volume of fluid moved is the same on both sides, so the small piston moves 40 × 1.0 = 40 cm.
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