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INTRODUCTION

 Pressure is defined as the force acting normally (perpendicularly) per unit area .

The SI units for pressure is newton per metre squared (N/m²).

  One Nm^-2 is known as one Pascal (Pa).

 Pressure = normal force / area or pressure = thrust / area . Another unit for measuring pressure is the bar. 1 bar = 105 N/m².

 1 millibar = 100 N/m².

Calculating Pressure

 Examples

1. A rectangular brick of weight 10 N, measures 50 cm × 30 cm × 10 cm.

 calculate the values of the maximum and minimum pressures which the block exert when resting on a horizontal table.

•  Area of the smallest face = 0.3 × 0.1 = 0.03 m².
• Area of the largest face = 0.5 × 0.3 = 0.15 m².
• Maximum pressure = 10 N / 0.03 = 3.3 × 10² N/m².
• Minimum pressure = 10 N / 0.15 = 67 N/m².

2. A man of mass 84 kg stands upright on a floor. If the area of contact of his shoes and the floor is 420 cm2, determine the average pressure he exerts on the floor. (Take g = 10 N/Kg)

 Solution

   Pressure = force / area = 840 / 0.042 = 20,000 Nm^-2.

 Pressure in liquids.

The following formula is used to determine pressure in liquids.

• Pressure = h ρg,

where h – height of the liquid, ρ – density and g – is force of gravity.

•  Examples

1. A diver is 10 m below the surface of water in a dam. If the density of water is 1,000 kgm ^-3, determine the pressure due to the water on the diver. (Take g = 10 Nkg^-1)

 Solution

• Pressure = h ρ g = 10 × 1000 × 10
•  = 100,000 Nm^-2.

2. The density of mercury is 13,600 kgm^-3. Determine the liquid pressure at a point 76 cm below the surface of mercury. (Take g = 10 Nkg^-1)

 Solution

•  Pressure = h ρ g = 0.76 × 13,600 × 10
• = 103,360 Nm^-2

3. The height of the mercury column in a barometer is found to be 67.0 cm at a certain place. What would be the height of a water barometer at the same place? (Densities of mercury and water are 1.36 × 10⁴kg/m3 and 1.0 × 10³ kg/m³ respectively.)

 Solution

•  Let the pressure due to water be h1ρ1g1 = h ρ g,
• hence;
• h1 = h ρ / ρ1= (6.7 × 10^-1) × (1.36 × 10⁴)
• = 911.2 cm or 9.11 m.

 U-tube manometer

It is a transparent tube bent into U-shape. When a liquid is poured into a u-tube it settles at equal level since pressure depends on height and they s hare the same bottom.

•  Consider the following diagrams;

 For the levels to differ the pressure P1 must be greater than P2, hence

P1 = P2 + hρg.

If P1 is the lung pressure, P0 is the atmospheric pressure, then if the difference is ‘h’ then lung pressure can calculated as follows.

P1 = P0 + hρg.

 Example

 A man blows into one end of a U-tube containing water until the levels differ by 40.0 cm. if the atmospheric pressure is 1.01 × 10⁵ N/m² and the density of water is 1000 kg/m3, calculate his lung pressure.

 Solution

• Lung pressure = atmospheric Pressure + liquid pressure
•  P1 = P0 + hρg. Hence P1 = (1.01 × 10⁵) + (0.4 × 10 × 1000)
•  = 1.05 × 10⁵ N/m².