Pressure
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 exerts when resting on a horizontal
table.
Solution
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 Nm2.
Pressure in liquids.
The following formula is used to
determine pressure in liquids.
Pressure = h ρ g
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 -1, determine
the pressure due to the water on the diver. (Take g = 10 Nkg)
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 × 104 kg/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 share the same bottom. Consider the following diagrams;
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.
Measuring pressure
1. Simple mercury barometer–
it is constructed using a thick-walled glass tube of length 1 m and is closed at
one end. Mercury is added into the tube then inverted and dipped into a dish containing
more mercury. The space above the mercury column is called Torricellian vacuum.
The height ‘h’ (if it is at sea
level) would be found to be 760 mm.
Atmospheric pressure can be
calculated as,
P = ρ g h =>where
ρ (mercury)- 1.36 × 104 kg/m3, g- 9.81 N/kg, h- 0.76 m.
Then P = (1.36 × 104)
× 9.81 × 0.76 = 1.014 × 105 Pa.
NOTE- this is the standard atmospheric pressure,
sometimes called one atmosphere. It is approximately one bar.
2. Fortin barometer– This is a more accurate mercury barometer. The adjusting screw is adjusted first to touch the mercury level in the leather bag.
3. Aneroid barometer– Increase in pressure causes the box to contract, the movements are magnified by the system of levers and is transmitted to the pointer by the fine chain and this causes the pointer to move. The scale is suitably calibrated to read pressure. Since pressure falls or rises as altitude falls or rises, the pointer can also be calibrated to read altitude.
4. Bourdon gauge– it is also called gauge pressure and is used in gas cylinders. When air is blown into the rubber tube, the curved metal tube tries to straighten out and this causes movement which is transmitted by levers and gears attached to a pointer. This gauge can measure both gas and liquid pressure.
Examples
Solution
Let the pressure due to water be
h1 ρ1 g1 and that of water be h ρ g. Then h1 ρ1 g1 = h ρ g.
Hence h1 = (6.7 × 10-1) × (1.36 × 104) / 1.0 × 103 = 911.2 cm or 9.11 m.
Application of pressure in
gases and liquids.
1. Rubber sucker– this is
a shallow rubber cap. Before use it is moistened to get a good seal then pressed
firmly on a smooth surface so that the air inside is pushed out. The atmospheric
pressure will then hold it firmly against the surface as shown below. They are
used by printing machines to lift papers, lifting glass panes, heavy metal
sheets etc.
2. Drinking straw– when a
liquid is drawn using a straw air is sucked through the straw to the lungs.
This leaves the space in the straw partially evacuated. The atmospheric
pressure pushing down the liquid in the container becomes greater than the
pressure inside the straw and this forces the liquid into your mouth.
3. The syringe– they work in the principle as the straw. They are used by the doctors in hospitals for giving injections.
4. Bicycle pump– it uses
two valves, one in the pump (greasy leather) and the other in the tire. When
the handle is pushed in, the pressure inside the barrel becomes greater than
the one in the tire and this pushes air inside. The valve in the tire is made
such that air is locked inside once pumped.
5. The siphon– it is used
to empty tanks which may not be easy to empty by pouring their contents out.
The tubing must be lowered below the base of the tank. The liquid flows out due
to pressure difference caused by the difference in height (h ρ g).
6. Lift pump.
7. Force pump.
Transmission of pressure in
liquids and gases.
It was first recognized by a
French mathematician and physicist called Blaise Pascal in the 17th century.
Pressure is equally distributed in a fluid and equally transmitted as shown in
the following,
a) Hydraulic brake system–
the master cylinder transmits pressure to the four slave cylinders on each
wheel. The cylinders contain brake fluid. Fluid is used because liquids are
almost incompressible. When force is applied in the pedal the resulting
pressure in the master cylinder is transmitted to the slave cylinders. This
forces the piston to open the brake shoes which then pushes the brake lining
against the drum. This force the rotation of the wheel to slow down. It is
important to note that pressure is equally distributed in all wheels so that
the car doesn’t pull or veer to one side.
b) Hydraulic press– it
consists of two pistons with different cross-sectional areas. Since pressure is
transmitted equally in fluids, when force is applied in one piston it is
transmitted to the other piston. The smaller piston is called the force while
the bigger piston is called the load. They are used to lift heavy loads in
industries, bending metals and sheets etc.
Examples
1. The area of the smaller
piston of a hydraulic press is 0.01 m2 and that of the bigger piston is 0.5 m2.
If the force applied to the smaller piston is 2 N, what force is transmitted to
the larger piston?
Solution
Pressure = force / area – hence P
= 2 / 0.01 = 200 Pa.
Force = Pressure × Area = 200 ×
0.5 = 100 N.
2. The master cylinder piston
in a car braking system has a diameter of 2.0 cm. The effective area of the
brake pads on each of the four wheels is 30 cm2. The driver exerts a
force of 500 n on the brake pedal. Calculate a) The pressure in the master
cylinder b) The total braking force in the car.
Solution
a) Area of the master
cylinder – π r2 = 3.14 cm2
Pressure = force /area = 500 /
3.14 × 10-4 = 1.59 × 106 N/m2
b) Area of brake pads =
(30 × 4) cm2. Since pressure in the wheel cylinder is the same as in
the
master cylinder)
F = Pressure × Area = (1.59 × 106) × (120 × 10-4) = 1.91 × 104 N.
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