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Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle

Physics

Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle

مقدمة الدرس

مقدمة الدرس

  1. This lesson covers density
  2. the pressure at a point and inside a liquid
  3. atmospheric pressure
  4. the manometer
  5. and Pascal’s principle with its use in the hydraulic press.
  • Define density and calculate it from mass and volume.
  • Calculate the pressure at a point inside a liquid, including the atmospheric pressure acting on the free surface.
  • Find the relative density of a liquid such as oil from the heights of two immiscible liquids in a U-shaped tube.
  • Explain atmospheric pressure, standard atmospheric pressure and its units.
  • Explain how a manometer measures the pressure of a confined gas.
  • State Pascal’s principle and apply it to the hydraulic press.
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 1 / 6
المفاهيم والمصطلحات الأساسية

المفاهيم والمصطلحات الأساسية

Comparison aspectMain ideaKeywords
Density (ρ)
  • The mass per unit volume of a material
  • ρ = m/V. Its SI unit is kg/m3
massunitvolumematerial
Relative density
  • The ratio of the density of a material to the density of water at 4 °C
  • which is 1000 kg/m3. It has no units
ratiomaterialwater1000
Pressure at a point
  • The average force acting perpendicularly on a unit area surrounding that point
  • P = F/A. Its unit is the newton per square meter (N/m2)
averageforceactingperpendicularly
Pascal (Pa)
  • The SI unit of pressure
  • where 1 Pascal = 1 N/m2
unitpressure
Atmospheric pressure
  • The pressure caused by the weight of the column of air above the Earth’s surface
  • acting on a unit area
causedweightcolumnabove
Standard atmospheric pressure, The air pressure measured at sea level at 0 °C. It equals 1 atm = 0.76 m Hg = 1.013 × 105 N/m2.measuredlevelequals
Comparison aspectMain ideaKeywords
Manometer
  • A U-shaped tube containing a liquid of known density
  • used to measure the difference between the pressure of a confined gas and atmospheric pressure
shapedtubecontainingliquid
Pascal’s principle
  • When pressure is applied to a confined fluid
  • the pressure is transmitted in full to all parts of the fluid and to the walls of the container
pressureappliedconfinedfluid
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 2 / 6
الشرح والتفصيل العلمي الشامل

الشرح والتفصيل العلمي الشامل

Comparison aspectMain ideaKeywords
Density.The density of a material depends on the atomic or molecular mass of its particles and on the spacing between them. An object with lower density floats on a liquid with higher densitymaterialdependsatomicmolecular
Pressure in a liquid.The pressure of the liquid alone at depth h is P = hρg, where g is the acceleration due to gravity. If the free surface is exposed to the atmosphere, the total pressure is P = Pa + hρg. The pressure increases with depth and with the density of the liquid.alonedepthaccelerationgravity
  • All points on the same horizontal level inside one liquid have the same pressure.
  • In connected vessels, the liquid rises to the same level in every branch, regardless of the shape of the vessels, provided their bases are at the same horizontal level.
  • Dams are thicker at the base, because the pressure increases with depth.
  • Immiscible liquids. In a U-shaped tube, the pressure at the separating level is the same on both sides, so ρoho = ρwhw. By measuring the two heights, the relative density of oil can be found, and then its density, if the density of water is known.
  • Atmospheric pressure. Torricelli’s mercury barometer shows that atmospheric pressure equals the pressure of a mercury column about 0.76 m high. The height of this column stays the same when the tube is tilted. Atmospheric pressure is therefore Pa = ρgh, where ρ is the density of mercury and h is the height of the column.
Comparison aspectMain ideaKeywords
Manometer.When the gas pressure P is greater than atmospheric pressure, P = Pa + ρgh. When it is less, P = Pa − ρgh. The pressure difference is ΔP = P − Pa = ρgh, where h is the difference in height between the two liquid surfaces.pressuregreateratmosphericless
Pascal’s principle.A liquid is nearly incompressible, so an extra pressure ΔP on a piston is transmitted in full to every point of the liquid. The pressure at depth h then becomes P = P1 + hρg + ΔP, where P1 is the pressure at the liquid surface under the piston. If the pressure rises above a certain limit, the container may break.liquidnearlyincompressibleextra
Hydraulic press.A small force f on a small piston of area a balances a large force F on a large piston of area A when f/a = F/A. The mechanical advantage is η = F/f = A/a. The small piston moves a greater distance than the large piston, since f·y1 = F·y2.smallforcepistonarea
Applications.→Car brakes, the hydraulic lift that raises cars in gas stations, and the hydraulic excavator all depend on Pascal’s principle. In a brake, a small force on the pedal, acting over a relatively long distance, produces a large force on the piston of the master cylinder. This pressure is transmitted through the brake fluid to the wheel cylinders, which push the brake shoes or pads against the drum or disc.
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 3 / 6
المقارنات وخريطة المفاهيم الجوهرية

المقارنات وخريطة المفاهيم الجوهرية

Quantity Relation Unit
Density ρ = m/V kg/m3
Relative density Density of material / density of water at 4 °C No units
Pressure at a point P = F/A N/m2
Pressure of a liquid alone at depth h P = hρg N/m2
Total pressure in a liquid with an open surface P = Pa + hρg N/m2
Manometer, pressure difference ΔP = ρgh N/m2
Hydraulic press, mechanical advantage η = F/f = A/a No units
  • Pressure in a liquid depends on depth and density. The base area of the container does not appear in P = Pa + hρg.
  • Points at the same depth in one liquid have the same pressure.
  • Pascal’s principle explains how a small force on a small piston can produce a large force on a large piston.
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 4 / 6
أمثلة ومسائل محلولة بالخطوات

أمثلة ومسائل محلولة بالخطوات

  • Example 1 Total pressure and force on the bottom of a tank
  • A tank contains salt water of density 1030 kg/m3.
  • Its base area is 1000 cm2, the water height is 1 m, the water surface is exposed to the air, g = 10 m/s2, and the atmospheric pressure is 1.013 × 105 N/m2.
  • Find the total pressure and the total force on the bottom.
  • Answer
  • Total pressure P = Pa + ρgh = 1.013 × 105 + 1030 × 10 × 1 = 1.116 × 105 N/m2.
  • Base area 1000 cm2 = 0.1 m2.
  • Total force F = P × A = 1.116 × 105 × 0.1 = 1.116 × 104 N.
  • Example 2 Hydraulic press
  • The small piston of a hydraulic press has an area of 10 cm2, and a force of 100 N acts on it.
  • The large piston has an area of 800 cm2.
  • Find the force on the large piston, the mechanical advantage, and the distance the small piston moves when the large piston rises 1 cm.
  • Answer
Force on the large pistonF = f × A / a = 100 × 800 / 10 = 8 × 103 N.
Mechanical advantageη = A / a = 800 / 10 = 80.
Distancef·y1 = F·y2, so y1 = 8000 × 1 / 100 = 80 cm.
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 5 / 6
تريكات الامتحان وكبسولة التفوق الوزاري

تريكات الامتحان وكبسولة التفوق الوزاري

  • Convert every quantity to SI before substituting density in kg/m3, area in m2 (1 cm2 = 10-4 m2), force in N, and pressure in N/m2.
  • Use P = hρg for the liquid alone, and add Pa when the liquid surface is open to the air.
  • The pressure at a point depends on depth and density, not on the base area of the container.
  • Relative density has no units, because it is a ratio of two densities.
  • In a hydraulic press, the large piston gives a larger force but moves through a smaller distance than the small piston.
  • Standard atmospheric pressure 1 atm = 0.76 m Hg = 760 mm Hg = 1.013 × 105 N/m2 = 1.013 bar.
Lesson 1 Summary: Density, Liquid Pressure and Pascal Principle · 6 / 6