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Mastering Pressure: Solids and Fluids

CSEC Physics: Pressure Fundamentals

Essential Understanding: Pressure is force acting per unit area. It explains why camels walk on sand without sinking, why knives cut, and how heavy cars can be lifted using small hydraulic jacks. This guide covers the physics of solids and fluids, featuring interactive simulations.

🔑 Key Skill: Unit Conversion (\(cm^2\) to \(m^2\))
📈 Exam Focus: \(P = \rho g h\) calculations
🎯 Problem Solving: Hydraulic Press principles

Types of Pressure

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Pressure in Solids

Concept: Pressure acts perpendicular to the surface. It depends on the magnitude of the force and the area over which it is distributed.

Formula: \[ P = \frac{F}{A} \]

Examples:

  • High heels exert high pressure (small area).
  • Snowshoes exert low pressure (large area).
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Pressure in Fluids

Concept: Fluids (liquids and gases) exert pressure in all directions. Liquid pressure increases with depth due to the weight of the fluid above.

Formula: \[ P = \rho g h \]

Key Variables:

  • \(\rho\) (rho): Density (kg/m³)
  • \(h\): Depth (m)
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Atmospheric Pressure

Concept: The weight of the air column above us exerting pressure on the Earth’s surface.

Standard Value: \[ 1 \, \text{atm} \approx 1.01 \times 10^5 \, \text{Pa} \]

Measurement: Measured using a barometer (e.g., mercury barometer).

Change: Decreases as altitude increases.

Fundamental Formulas

Solid Pressure

\[ P = \frac{F}{A} \]

Unit: Pascals (Pa) or N/m²

Liquid Pressure

\[ P = \rho g h \]

Density is critical! Water = 1000 kg/m³

Hydraulic Principle

\[ P_1 = P_2 \Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

Pascal’s Principle application

Interactive Hydraulic Lab

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Pascal’s Principle Simulator

Objective: Use the principle that pressure is transmitted equally throughout a fluid. Apply a small force to the small piston to lift a heavy car on the large piston.

Input (Small Piston)

Force: 0 N

Area: 1 m²

Pressure: 0 Pa

Output (Large Piston)

Area: 5 m²

Multiplied Force: 0 N

Distance is sacrificed for Force

Graphing Pressure in Fluids

The deeper you go into a fluid, the greater the weight of the fluid above you, and thus the greater the pressure. This relationship is linear.

Analysis: This graph shows the pressure at different depths in water (\(\rho = 1000 \, \text{kg/m}^3\)). Note that the gradient represents \(\rho g\). Pressure is zero at the surface (open to atmosphere), but gauge pressure is shown here (relative to surface).

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Worked Example: Swimming Pool

Calculate the water pressure at the bottom of a swimming pool that is 2m deep. (Density of water = 1000 kg/m³, \( g = 10 \, \text{N/kg} \)).

1
Identify Variables: \( h = 2 \, \text{m} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{N/kg} \).
2
Select Formula: For a fluid at depth, use \( P = \rho g h \).
3
Calculate: \( P = 1000 \times 10 \times 2 \).
4
Final Answer: \( P = 20,000 \, \text{Pa} \) (or 20 kPa).

Key Examination Insights

Common Mistakes

  • Forgetting to convert area from \(cm^2\) to \(m^2\) (divide by 10,000).
  • Using the wrong density (e.g., using 1 for water instead of 1000).
  • Confusing “Pressure at depth” with “Total Force on the dam wall” (Force requires Average Pressure × Area).

Success Strategies

  • Always check that your final pressure unit is Pascals (N/m²).
  • For manometer questions, remember \( P = h \rho g \) for the liquid column height difference.
  • In Hydraulic Press problems, verify that \( F_1 \times d_1 = F_2 \times d_2 \) (Work input = Work output).

CSEC Practice Arena

Test Your Understanding

1
Which of the following units is equivalent to the Pascal?
N/cm²
N/m²
N/m
kg/m²
Explanation: Pressure is Force divided by Area. Force is in Newtons (N), Area is in square meters (m²). Therefore, the unit is N/m², which is named the Pascal (Pa).
2
A hydraulic press has a small piston area of 0.1 m² and a large piston area of 2.0 m². If a force of 50N is applied to the small piston, what is the force on the large piston?
50 N
2.5 N
1000 N
500 N
Solution: Using Pascal’s Principle: \( F_1/A_1 = F_2/A_2 \). Rearranging: \( F_2 = F_1 \times (A_2/A_1) = 50 \times (2.0 / 0.1) = 50 \times 20 = 1000 \, \text{N} \).
3
At what depth in water is the pressure 100,000 Pa? (Density of water = 1000 kg/m³, \( g = 10 \, \text{N/kg} \))
5 m
10 m
1 m
100 m
Solution: \( P = \rho g h \Rightarrow h = P / (\rho g) = 100,000 / (1000 \times 10) = 10 \, \text{m} \).
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CSEC Examination Mastery Tip

The “Dam Wall” Question: A common CSEC question involves calculating the total force on a dam wall.

  • The pressure is not constant; it is zero at the top and max at the bottom.
  • Therefore, you must use the Average Pressure: \( P_{\text{avg}} = (\text{Pressure at top} + \text{Pressure at bottom}) / 2 \). Since pressure at top is 0 (relative), \( P_{\text{avg}} = P_{\text{bottom}} / 2 \).
  • Then calculate Force: \( F = P_{\text{avg}} \times \text{Area of wall} \).
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