This note covers pressure: how force spread over an area gives pressure, why sharp knives cut and wide skis stop you sinking, and how pressure in a liquid increases with depth and density. These ideas explain dams, diving, submarines and many everyday tools.
Key ideas
Pressure on a surface
Pressure is the force acting on each unit of area, at right angles to the surface. The same force gives a high pressure on a small area and a low pressure on a large area.
- To increase pressure: increase the force, or reduce the area. A knife has a thin edge, nails and drawing pins have sharp points, and high-heeled shoes press hard on the floor.
- To decrease pressure: reduce the force, or increase the area. Skis, snowshoes and tractor tyres are wide so that the weight is spread out. Wide foundations spread the weight of a building.
The SI unit is the pascal (Pa), where 1 Pa = 1 N/m2. Pressure is also often given in N/cm2 or kPa (1 kPa = 1000 Pa). When area is in cm2, convert to m2 using 1 m2 = 10 000 cm2.
The force in the equation is often the weight of an object resting on a surface, found from W = m × g.
Pressure in liquids
A liquid exerts a pressure because its weight pushes on whatever is beneath it and its particles collide with the container walls. The pressure in a liquid:
- increases with depth, because there is more liquid above;
- increases with the density of the liquid;
- acts in all directions at a given depth, and at right angles to any surface it touches;
- does not depend on the shape or width of the container, only on the depth and density.
This is why a dam wall is thicker at the bottom, and why water spurts out of a leaking bottle faster from a hole near the bottom. Divers feel the pressure rise as they descend, and submarines need strong hulls.
(Extended) The pressure due to a liquid at depth h is found from p = ρ × g × h, where ρ is the density of the liquid, g = 9.8 N/kg and h is the vertical depth below the surface. The same equation works for the difference in pressure between two points at different depths, using the difference in depth. Note that this is only the pressure caused by the liquid. In an open container, the air above also pushes on the surface, so the total pressure is the liquid pressure plus atmospheric pressure.
The equation comes from combining the others: the weight of a column of liquid of area A and height h is ρ × A × h × g. Dividing by the area A gives pressure = ρ × g × h.
Force from a pressure
Rearranging p = F ÷ A gives F = p × A. If the pressure at a place is known, the total force on a surface there is the pressure times the area. For a surface that is deep and tall, the pressure varies with depth, so this works for a small area at a known depth.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| p = F ÷ A | Pressure (Pa) = force (N) ÷ area (m2), with the force at right angles to the area. |
| F = p × A | Force from a known pressure and area. |
| Pascal (Pa) | 1 Pa = 1 N/m2. |
| p = ρ g h (Extended) | Pressure (Pa) due to a liquid of density ρ (kg/m3) at depth h (m), with g = 9.8 N/kg. |
| Δp = ρ g Δh (Extended) | Difference in liquid pressure between two depths. |
| Density of water | 1000 kg/m3; sea water is about 1030 kg/m3. |
Worked examples
Example 1 A person of mass 65 kg stands on the ground. The area of both shoe soles in contact with the ground is 0.040 m2. Find the pressure on the ground. Then find the pressure if the person stands on skis with a total area of 0.30 m2 (g = 9.8 N/kg).
Step 1: force = weight = m × g = 65 × 9.8 = 637 N.
Step 2: pressure in shoes = F ÷ A = 637 ÷ 0.040 = 15 925 Pa, about 15 900 Pa.
Step 3: pressure on skis = 637 ÷ 0.30 = 2123 Pa, about 2120 Pa.
Answer: about 15 900 Pa in shoes and about 2120 Pa on skis. The larger area gives a much smaller pressure.
Example 2 (Extended) A diver is 15 m below the surface of the sea. The density of sea water is 1030 kg/m3. Find the pressure due to the water (g = 9.8 N/kg).
Step 1: p = ρ × g × h.
Step 2: p = 1030 × 9.8 × 15 = 151 410 Pa.
Answer: about 1.51 × 105 Pa (151 kPa), not including the atmosphere.
Example 3 (Extended) A rectangular gate of area 2.0 m2 is fixed in a dam wall. The water pressure over the gate is taken as the pressure at a depth of 8.0 m (water density 1000 kg/m3, g = 9.8 N/kg). Find the pressure and the force on the gate due to the water.
Step 1: p = ρ × g × h = 1000 × 9.8 × 8.0 = 78 400 Pa.
Step 2: F = p × A = 78 400 × 2.0 = 156 800 N.
Answer: pressure 78 400 Pa, force about 157 000 N (157 kN).
Common errors
- Using the area in cm2 with force in N and calling the answer Pa. Convert to m2 first.
- Using the length of a sloping tube or the width of the container instead of the vertical depth h.
- Thinking that a wider container gives a greater liquid pressure at the same depth. Only depth and density matter.
- Saying pressure acts only downwards in a liquid. It acts in all directions.
- Explaining skis or sharp knives in terms of force alone. The force may be the same; it is the area that changes.
Exam tip: When a question asks for an explanation of a pressure effect, state the equation idea clearly: "the same force acts over a smaller area, so the pressure is greater".
Check yourself
- A force of 240 N acts on an area of 0.0060 m2. Find the pressure.
Answer
p = 240 ÷ 0.0060 = 40 000 Pa. - A block weighs 50 N and rests on its largest face of area 0.020 m2. Then it is turned onto a face of area 0.0050 m2. Find the pressure on the table each time.
Answer
Large face: 50 ÷ 0.020 = 2500 Pa. Small face: 50 ÷ 0.0050 = 10 000 Pa. - (Extended) Find the pressure due to water at a depth of 5.0 m (ρ = 1000 kg/m3, g = 9.8 N/kg).
Answer
p = 1000 × 9.8 × 5.0 = 49 000 Pa. - Oil has a lower density than water. At the same depth, is the pressure in oil greater or less than in water? Explain.
Answer
It is less. Pressure = ρ g h, so with the same g and h, the liquid with the lower density gives the smaller pressure.