In-Depth Notes on Hydrostatics

Expected Learning Outcomes

  • Students should be able to:
    • Explain density, specific gravity, fluid pressure, and pressure of fluid inside a vessel.
    • Compute the effect of fluid on submerged bodies.

Hydrostatic Overview

  • Hydrostatics: Study of fluids at rest, typically confined within a vessel.
  • Types of fluids:
    • Liquids (e.g., bottled mineral water, blood, hydraulic fluid)
    • Gases (e.g., air)

Key Properties of Materials

Density
  • Defined as mass per unit volume.
  • Formula: p=mVp = \frac{m}{V} where:
    • pp = density in g/cm³ or kg/m³
    • mm = mass in grams or kg
    • VV = volume in cm³ or m³
Common Densities
  • Copper: p=8.9extg/cm3p = 8.9 ext{ g/cm}^3
  • Lead: p=11.3extg/cm3p = 11.3 ext{ g/cm}^3
  • Water: p=1extg/cm3p = 1 ext{ g/cm}^3 (or 1000 kg/m³)
  • Useful conversion: 1 ml = 1 cc = 1 cm³
Specific Gravity (sp.gr)
  • Defined as the ratio of the density of a substance to the density of a standard substance (usually water).
    • Formula:
      sp.gr.=Density<em>substanceDensity</em>standardsp.gr. = \frac{Density<em>{substance}}{Density</em>{standard}}
  • Unitless value approximately equals the density.

Example Calculations

Density and Specific Gravity of Glycerin
  • Given: Mass = 60.4 g, Volume = 48 ml.
    • Density Calculation:
      p=60.4extg48extml=1.26extg/cm3p = \frac{60.4 ext{ g}}{48 ext{ ml}} = 1.26 ext{ g/cm}^3
    • Specific Gravity Calculation:
      sp.gr.=1.261=1.26sp.gr. = \frac{1.26}{1} = 1.26

Applications

  • Blood density (normal: 1.04 to 1.06 g/cm³), indicating anemia with unusually low values.
  • Urine density (normally around 1.02 g/cm³), indicating salt excretion levels.

Weight Density

  • Defined as weight per unit volume.
  • Formula:
    D=WeightVolumeD = \frac{Weight}{Volume}
  • Units: N/m³, dynes/cc, lbs/ft³ for water: 62.4 lbs/ft³

Hydrostatic Pressure

  • The pressure exerted by a column of liquid in a vessel.
  • Formula: P=FAP = \frac{F}{A} where:
    • PP = pressure (N/m²)
    • FF = force (N)
    • AA = area (m²)
Depth and Pressure
  • Pressure depends on the depth of the liquid and its weight density: P=ρghP = \rho gh where:
    • ρ\rho = density of liquid
    • gg = gravitational acceleration (9.8 m/s²)
    • hh = height of liquid column
Example: Hydrostatic Pressure Exerted by Mercury
  • Given: Density of mercury = 13.6 g/cm³, height = 12 cm.
    • Pressure Calculation:
      P=(13.6extg/cm3)(980extcm/s2)(12extcm)=159,936extdyne/cm2P = (13.6 ext{ g/cm}^3)(980 ext{ cm/s}^2)(12 ext{ cm}) = 159,936 ext{ dyne/cm}^2

Pascal’s Principle

  • States that pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid and walls of the container.

Hydraulic Systems

Pascal’s Law for Hydraulic Systems
  • Relationship between force, area, and pressure:
    • P<em>1=P</em>2P<em>1 = P</em>2
    • Thus,
    • F<em>1A</em>1=F<em>2A</em>2\frac{F<em>1}{A</em>1} = \frac{F<em>2}{A</em>2}
Example: Hydraulic Press
  • Given:
    • Small piston area = 0.005 m²
    • Large piston area = 0.03 m²
    • Force applied to small piston: 500 N
    • Force on larger piston:
      F<em>2=F</em>1A<em>2A</em>1=5000.030.005=3000extNF<em>2 = F</em>1 \frac{A<em>2}{A</em>1} = 500 \frac{0.03}{0.005} = 3000 ext{ N}

Buoyancy and Archimedes’ Principle

  • States that an immersed body experiences an upward force equal to the weight of the fluid it displaces.
Buoyant Force (BF)
  • Formula:
    BF=W<em>airW</em>liquidBF = W<em>{air} - W</em>{liquid}
Conditions for Floating and Sinking
  1. Weight of the body < Buoyant Force: Floats
  2. Weight of the body > Buoyant Force: Sinks
  3. Weight of the body = Buoyant Force: Suspended
Example: Solid Weighing in Different Mediums
  • Given: Wair = 350 dynes, Wwater = 210 dynes.
    • Buoyant Force Calculation:
      BF=W<em>airW</em>water=350extdynes210extdynes=140extdynesBF = W<em>{air} - W</em>{water} = 350 ext{ dynes} - 210 ext{ dynes} = 140 ext{ dynes}
    • Specific Gravity Calculation:
      sp.gr.=Weight<em>objectWeight</em>displaced=350140=2.5sp.gr. = \frac{Weight<em>{object}}{Weight</em>{displaced}} = \frac{350}{140} = 2.5