Fluid Statics: Pressure, Buoyancy & Pascal’s Law

Why Study Fluid Statics?

Fluid statics is the study of fluids (liquids and gases) at rest. It helps explain:

• Why a dam doesn’t collapse under water pressure.
• How a submarine stays submerged or rises.
• Why blood pressure increases as you go deeper in your body.
• How hydraulic machines (like car lifts) work.

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1. Pressure in Fluids

What is Pressure?

Pressure is the force per unit area exerted by a fluid on its surroundings. Mathematically: $$P=\frac{F}{A}$$ where:

• $P$ = Pressure (N/m² or Pascal, Pa)
• $F$ = Force (N)
• $A$ = Area (m²)

Pressure in a Fluid at Rest

In a fluid at rest:

• Pressure acts perpendicular to any surface.
• Pressure increases with depth because the fluid above exerts more weight.
• Pressure is the same at all points on a horizontal plane (Pascal’s Law).

How Pressure Varies with Depth

Consider a small cube of fluid at depth $h$:

```figure
{"type":"graph","fns":[{"expr":"1000 * 9.81 * x","label":"Pressure (P) = ρgh (Pa)"}],"x":[0,10],"ylabel":"Pressure (Pa)","xlabel":"Depth (m)","caption":"Pressure increases linearly with depth in a fluid (ρ = 1000 kg/m³, g = 9.81 m/s²)."}

• The pressure at depth ( h ) is due to the weight of the fluid above it.
• For a fluid of density ( \rho ) (rho), the pressure at depth ( h ) is: $$ P = P_0 + \rho g h $$ where:
  • ( P_0 ) = Atmospheric pressure (1.01 × 10⁵ Pa at sea level)
  • ( \rho ) = Density of the fluid (kg/m³)
  • ( g ) = Acceleration due to gravity (9.8 m/s²)
  • ( h ) = Depth below the surface (m)

Solved Example 1: Calculating Pressure at Depth

A fish swims at a depth of 10 m in seawater. What is the pressure on the fish? (Density of seawater = 1030 kg/m³, ( g = 9.8 , \text{m/s}^2 ))

Solution:

1. Use the formula: $$ P = P_0 + \rho g h $$
2. Plug in the values: $$ P = 1.01 \times 10^5 + (1030)(9.8)(10) $$
3. Calculate: $$ P = 1.01 \times 10^5 + 100,940 = 2.0194 \times 10^5 , \text{Pa} $$ Answer: The pressure on the fish is 2.02 × 10⁵ Pa.

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2. Gauge Pressure vs. Absolute Pressure

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3. Buoyancy and Archimedes’ Principle

What is Buoyancy?

Buoyancy is the upward force exerted by a fluid on an immersed object. It depends on:

• The volume of the displaced fluid.
• The density of the fluid.

Archimedes’ Principle

"An object immersed in a fluid experiences an upward buoyant force equal to the weight of the displaced fluid."

Mathematically: $$ F_b = \rho_{\text{fluid}} V_{\text{displaced}} g $$ where:

• ( F_b ) = Buoyant force (N)
• ( \rho_{\text{fluid}} ) = Density of the fluid (kg/m³)
• ( V_{\text{displaced}} ) = Volume of displaced fluid (m³)
• ( g ) = Acceleration due to gravity (9.8 m/s²)

Why Does an Object Float or Sink?

• If ( F_b > \text{Weight of object} ) → Floats (e.g., wood in water).
• If ( F_b < \text{Weight of object} ) → Sinks (e.g., stone in water).
• If ( F_b = \text{Weight of object} ) → Suspended (e.g., submarine at equilibrium).

Solved Example 2: Buoyant Force on a Submerged Rock

A rock with a volume of 0.002 m³ is fully submerged in water (( \rho_{\text{water}} = 1000 , \text{kg/m}^3 )). What is the buoyant force on it?

Solution:

1. Use Archimedes’ principle: $$ F_b = \rho_{\text{water}} V g $$
2. Plug in the values: $$ F_b = (1000)(0.002)(9.8) = 19.6 , \text{N} $$ Answer: The buoyant force is 19.6 N.

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4. Pascal’s Law (Hydraulic Press Principle)

What is Pascal’s Law?

"Pressure applied to a fluid in a closed container is transmitted equally in all directions."

Mathematically: $$ P_1 = P_2 $$ where ( P_1 ) and ( P_2 ) are pressures at any two points in the fluid.

Applications of Pascal’s Law

1. Hydraulic Lifts (e.g., car lifts, construction cranes).
2. Braking Systems in vehicles.
3. Hydraulic Presses for shaping metals.

How a Hydraulic Lift Works

```mermaid
flowchart TD
    A["Small piston (P₁)"] -->|Force F₁| B["Pressure P₁ = F₁/A₁"]
    B -->|P₁ = P₂| C["Large piston (P₂)"]
    C -->|Force F₂ = P₂ × A₂| D["Lifts heavy object"]
    style A fill:#f9f,stroke:#333
    style C fill:#bbf,stroke:#333

• Small force on a small piston creates high pressure.
• This pressure is transmitted to a large piston, producing a large force.

Solved Example 3: Hydraulic Lift Calculation

A hydraulic lift has a small piston with area 0.01 m² and a large piston with area 0.5 m². If a force of 100 N is applied to the small piston, what is the force on the large piston?

Solution:

1. Since $P_1=P_2$: $$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$
2. Rearrange to find $F_2$: $$F_2=F_1\times \frac{A_2}{A_1}$$
3. Plug in the values: $$F_2=100\times \frac{0.5}{0.01}=5000\text{N}$$ Answer: The force on the large piston is 5000 N.

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5. Atmospheric Pressure

What is Atmospheric Pressure?

Atmospheric pressure is the pressure exerted by Earth’s atmosphere due to the weight of air above us.

• At sea level: 1.01 × 10⁵ Pa (1 atm).
• Decreases with altitude.

How is Atmospheric Pressure Measured?

• Barometer: Uses a column of mercury to measure pressure.

• Aneroid Barometer: Uses a flexible metal box to measure pressure changes.

Solved Example 4: Mercury Column Height

What is the height of the mercury column in a barometer at sea level? (Density of mercury = 13,600 kg/m³)

Solution:

1. Use the pressure formula: $$P=\rho gh$$
2. Rearrange to find $h$: $$h=\frac{P}{\rho g}$$
3. Plug in the values: $$h=\frac{1.01\times 10^5}{(13,600)(9.8)}=0.76\text{m}=76\text{cm}$$ Answer: The mercury column height is 76 cm.

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6. Pressure in Liquids vs. Gases

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Exam Tips for Fluid Statics

1. Memorize Key Formulas:

   • $P=\frac{F}{A}$
   • $P=P_0+\rho gh$
   • $F_b={\rho }_{\text{fluid}}Vg$
   • $P_1=P_2$ (Pascal’s Law)

2. Understand Buoyancy:

   • Always compare $F_b$ with the weight of the object.
   • Use ${\rho }_{\text{object}}$ vs. ${\rho }_{\text{fluid}}$ to predict floating/sinking.

3. Diagrams are Key:

   • Draw free-body diagrams for submerged objects.
   • Show pressure variation with depth in graphs.

4. Real-World Applications:

   • Dams: Pressure increases with depth → thicker base.
   • Submarines: Adjust buoyancy by changing water intake.
   • Hydraulic Machines: Multiplication of force using Pascal’s Law.

5. Common Mistakes to Avoid:

   • Forgetting to include $P_0$ (atmospheric pressure) in calculations.
   • Misapplying Archimedes’ principle (only displaced fluid volume counts).
   • Confusing gauge pressure with absolute pressure.

6. NEB Board-Style Questions (Practice)

   • Short Answer (2-3 marks):

     1. Explain why a dam is built thicker at the bottom.
     2. What is the difference between gauge pressure and absolute pressure?
     3. How does a hydraulic press work?

   • Long Answer (5-7 marks):

     1. A wooden block floats in water. If salt is added to the water, will the block float higher or lower? Explain.
     2. Derive the expression for buoyant force and explain its significance.
     3. A U-tube contains water and oil. If the oil is denser than water, how will the levels adjust?

   • Numerical (5 marks):

     1. Calculate the pressure at a depth of 5 m in freshwater ($\rho =1000{\text{kg/m}}^3$).
     2. A metal block of volume 0.001 m³ is suspended in oil ($\rho =800{\text{kg/m}}^3$). Find the buoyant force.
     3. In a hydraulic lift, the area ratio of the small piston to the large piston is 1:100. If a force of 50 N is applied to the small piston, what is the force on the large piston?

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Good luck with your NEB exam! 🚀