where:

• $\rho$ (rho) = density (kg/m³)
• $m$ = mass (kg)
• $V$ = volume (m³)

Common Densities:

Specific Gravity

Specific Gravity (SG) is the ratio of a substance's density to water's density:

$SG = \frac{\rho_{substance}}{\rho_{water}}$

• Dimensionless (no units)
• SG < 1: floats in water
• SG > 1: sinks in water

Pressure

Pressure is force per unit area:

$P = \frac{F}{A}$

where:

• $P$ = pressure (Pa or N/m²)
• $F$ = force perpendicular to surface (N)
• $A$ = area (m²)

Common Units:

• Pascal (Pa): 1 Pa = 1 N/m²
• Atmosphere (atm): 1 atm = 101,325 Pa ≈ 101 kPa
• Bar: 1 bar = 100,000 Pa
• mmHg: 760 mmHg = 1 atm
• psi: 14.7 psi = 1 atm

Pressure in Fluids

Unlike solids, fluids cannot sustain shear stress - they flow. This creates unique pressure properties:

1. Pressure acts perpendicular to surfaces - fluid pressure always pushes normal to any surface

2. Pressure is isotropic - equal in all directions at a point

3. Pressure increases with depth

Pressure vs. Depth

For a fluid at rest, pressure increases linearly with depth:

$P = P_0 + \rho g h$

where:

• $P$ = pressure at depth $h$
• $P_0$ = pressure at surface (usually atmospheric)
• $\rho$ = fluid density (kg/m³)
• $g$ = 9.8 m/s²
• $h$ = depth below surface (m)

💡 Key Insight: Pressure depends only on vertical depth, not container shape!

Gauge vs. Absolute Pressure

• Absolute pressure: Total pressure including atmosphere $P_{absolute} = P_{atm} + P_{gauge}$

• Gauge pressure: Pressure relative to atmosphere $P_{gauge} = \rho g h$

Most pressure gauges (like tire gauges) read gauge pressure.

Pascal's Principle

Pascal's Principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid.

Hydraulic Systems

This principle enables hydraulic lifts:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

Force multiplication: $F_2 = F_1 \frac{A_2}{A_1}$

Trade-off: Volume conservation means: $d_1 A_1 = d_2 A_2$

Small piston moves far, large piston moves little distance.

Atmospheric Pressure

At sea level:

$P_{atm} = 101,325 \text{ Pa} = 101.3 \text{ kPa} = 1 \text{ atm}$

This is equivalent to:

• 10.3 m column of water
• 760 mm column of mercury

Atmospheric pressure decreases with altitude.

Problem-Solving Strategy

1. Identify pressure type needed (gauge or absolute)
2. Choose reference point (usually surface)
3. Apply $P = P_0 + \rho g h$
4. Remember $h$ is vertical depth only
5. Check units (convert to SI)

Common Mistakes

❌ Forgetting atmospheric pressure in absolute calculations ❌ Using horizontal distance for $h$ ❌ Mixing gauge and absolute pressure ❌ Unit confusion (1 atm ≠ 1 Pa) ❌ Thinking pressure depends on container shape

Find: Mass $m$

Solution:

Step 1: Calculate volume of cube. $V = L^3 = (0.050)^3 = 1.25 \times 10^{-4} \text{ m}^3$

Step 2: Use density formula. $m = \rho V = (2700)(1.25 \times 10^{-4}) = 0.338 \text{ kg}$

Answer: 338 g