Recall and use the equation for the change in pressure beneath the surface of a liquid Δp = ρ g Δh
1.8 Pressure
Objective
Recall and use the equation for the change in pressure beneath the surface of a liquid: Δp = ρ g Δh 💧
What is Pressure?
Pressure is the force applied per unit area. Think of it like the push you feel when you press your palm against a wall. The deeper you go under water, the more “push” you feel because the water above is heavier. 📏
Key Equation
The change in pressure with depth is given by:
$$\Delta p = \rho g \Delta h$$
- ρ (rho) – Density of the liquid (kg m⁻³)
- g – Acceleration due to gravity (≈9.8 m s⁻²)
- Δh – Change in depth (m)
Units & Symbols
| Symbol | Unit | Example |
|---|---|---|
| ρ | kg m⁻³ | 1000 (water) |
| g | m s⁻² | 9.8 |
| Δh | m | 5 |
| Δp | Pa (N m⁻²) | 49000 |
Analogy: The Slinky in a Bottle
Imagine a slinky inside a bottle. When you push the top of the slinky down, the bottom feels a stronger push because the slinky above is heavier. Similarly, the deeper you go in a liquid, the more weight of liquid sits above you, increasing pressure. 🎈
Example Problem
- Water depth: Δh = 5 m
- Density of water: ρ = 1000 kg m⁻³
- Gravity: g = 9.8 m s⁻²
- Calculate Δp.
Using the formula:
$$\Delta p = 1000 \times 9.8 \times 5 = 49\,000 \text{ Pa}$$
So at 5 m depth, the pressure is 49 kPa higher than at the surface. 🚤
Quick Quiz
- What happens to Δp if you double the depth Δh? 🤔
- How would the pressure change if you used a liquid with half the density of water? 💧
- Calculate the pressure increase at 10 m depth in seawater (ρ ≈ 1025 kg m⁻³). 📐
Revision
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