KENYA FORM 1

1. PRESSURE

2. (i)A person makes deeper marks while
walking on a soft ground in very narrow and
sharp-heeled shoes than when in ordinary shoes
with flat heels.
(ii)It is easier to push a sharp pin through a
cardboard than it is to push a blunt one through
the same material using the same force ,see
figure 4.1(a) and (b).

3. In (i) above,sharp pointed heels dig deeper on
soft ground because the force applied acts on
smaller area than when the blunt pin is used.
In all these cases , a force acting on a surface
produces a penetration effect.This penetration
effect is larger when the force acts on a small
area than when the same force acts on a larger
area.

4. Pressure is defined as force
acting per unit area.

5. Units of pressure
The SI unit of pressure is thus Newton per Square metre (N/m2),
which is also called the Pascal(Pa).
1N/m2=1 Pa
Other units include mmHg,cmHg and atmosphere(atm).

6. Ex1:A man of mass 84 kg stands upright on a
floor.If the area of contact of his shoes and flor
is 420 cm2,determine the average pressure he
exerts on the floor.(take g=10 N/kg)

7. Ex 2: A metallic block of mass 40 kg exerts a
pressure of 20 N/m2 on aflat surface.Determine
the area of contact between the block and the
surface.(take g=10 N/kg)

8. Ex3: A rectangular aluminium solid block of
density 2700 kg/m3 has dimensions of
40cmx12cmx6cm.The block rests on a horizontal
flat surface.Calculate
a)the minimum pressure,
b)the maximum pressure it can exert.

10. PRESSURE IN LIQUIDS
Exp 4.1: to show variation of pressure in liquid
Observation: the lower hole,A,throws water farthest,followed by B
and lastly C.
Conclusion : Pressure of water at A is greater than pressure at B
and Pressure at B is bigger than at C.
So,Pressure increases with depth.

11. Liquid levels
This show that the liquid flows to find its own
level.
Pressure in liquid is equal at all points at the
same level.

12. Liquid levels in a U-tube
When water is poured into a U-tube, it will flow
into the other arm.The water will setle in the
tube with the levels on both arms being the
same,see in fig 4.5(a)

13. When one arm of the U tube is blown into with
the mouth, the level moves downwards,while on
the arm other arm it rises, see in figure 4.5(b).
This caused by the pressure difference between
the two arms.

14. Exp :4.2(a) to investigate the variation of
liquid pressure with depth and density
1. Take a beaker filled with water. Insert the
manometer at different levels. You will notice, as
seen earlier, that greater the depth, higher is the
pressure.

15. 2. Take two beakers, one filled with water and
one filled with a salt solution. If you keep the
manometer funnel at same depths in the two
beakers, you will notice that pressure is
dependent on the density of the liquid. Higher
the density, higher the pressure.

16. 3. Take a beaker filled with water. Insert the
manometer at a certain depth. Rotate the
manometer mouth in all direction. You will notice
that the pressure at a certain depth inside a
liquid is equal in all directions.

17. 4. In a beaker filled with water, insert the
manometer funnel at a fixed depth, move the
funnel back and forth at the same depth. You
will notice that the manometer pressure does
not change. This means that the pressure inside
a liquid is same at all places at the same depth.

18. In summary :
(i) pressure in a liquid increases with depth
below its surface.
(ii) pressure in a liquid at a particular depth is
the same in all directions.
(iii) pressure in a liquid increases with the
density of the liquid.

19. Homework :Exercise 4.1

20. Fluid Pressure Formula
If A is the cross-section area of the column, h
is the height of the column and p the density of
the liquid, then;
Volume of the liquid=cross-section area x
height
=A x h

21. Mass of the liquid=volume of the liquid x
density
=Axhxp
So,weight of the liquid=mass of the liquid x
gravity
=A x h x p x g

22. From the definition of pressure,
P=h x p x g

23. P=h x p x g
From the formula , it can be seen that the
pressure due to a liquid column is directly
proportional to:
(i) height h of the column.
(ii) the density p of the liquid.

24. Example 4: A diver is 10 m below the surface
of water in a dam.If the density of water is
1000kg/m3,determine the pressure due to the
water on the diver.(take g=10 N/kg)

25. Example5: The density of mercury is 13600
kg/m3.Determine the liquid pressure at a point
76 cm below the surface of mercury.(take
g=10N/kg) .

26. ATMOSPHERIC PRESSURE
There is an envelope of air that surrounds the
earth. We call it the atmosphere. The
atmosphere is bound to the earth by the
gravitational attraction of the earth.
This atmosphere of air around us produces
pressure at ground level or other levels due to
the weight of the air above that level.

27. Simple demonstrations of Air Pressure
Atmospheric pressure acts in all directions.
(a)Take a little water in a tin and boil the
water to drive the air out of it. Cork the tin
tightly and then cool it by pouring water on it.
What happens?

28. When the air inside the tin is driven out,the
atmospheric pressure outside is not
counterbalanced by the pressure inside(when
the steam cools , a partial vacuum is created
inside). The tin collapses.
You can understand now why an open empty
tin does not get squashed with the atmosphere
pressing on the outside.

29. (b) Fill glass or lipless beaker with water to
the brim and slide a paper card over the
top.Take care not to leave any air bubble in
water. Invert the glass. What happens?
The card is held in position by the atmospheric
pressure pushing upwards on the card and the
water does not fall out. The upward atmospheric
pressure is greater than the downward force due
to water.

30. A more convenient method is to use a glass
tube sealed at one end ,as in fig 4.18(a)

31. İf mercury , which is much denser than water
is used, the column supported is found to be
much shorter,see figure 4.18(b). At sea level ,
the atmospheric pressure supports
approximately 76 cm of mercury column or
approximately 10 m of water column.

35. MEASUREMENT OF PRESSURE
U-tube Manometer
A manometer is an instrument that can
measure fluid pressure. It consists of a U-tube
filled with water or any other suitable liquid,see
figure 4.19

36. Due to pressure of the gas Pg,
The water level in the aother limb rises to, say,Y.
This difference in water levels is the difference between
gas pressure Pg and the atmospheric pressure Pa.
Since X and Z are at the same horizontal level ,pressure
at X equals pressure at Z.Pressure at X is pressure of gas
Pg.
Pressure at Z=atmospheric pressure+pressure due to
the column of water
Therefore ,Pg=Pa+hpg

37. Example:

38. Mercury Barometer
It has been shown that atmospheric pressure
supports a liquid column in a tube. When this
arrangement is used to measure pressure, it is
called a barometer.
At sea level, acolumn of mercury and water
supported by atmospheric pressure is
approximately 76 cm.

39. The height h of the column is a measure of the
atmospheric pressure. At sea level , h=76cmHg.
Since density p of mercury is 13600 kg/m3;
This is the Standard atmospheric pressure ,
and is sometimes referred to as one
atmosphere.

40. Testing the vacuum in a Barometer
If the barometer has air at the top, it will be
faulty.
If the mercury column is tilted, the mercury
from the dish flows to the tube and completely
fills it.

41. If there was air,the mercury would not fill the
tube completely. This shows that space above
the mercury in the tube is a vacuum.
The space above the mercury in the tube when
upright is called Toricellian vacuum and
contains a little mercury vapour.

42. Fortin Barometer
The simple mercury barometer cannot be used
for accurate measurements of atmospheric
pressure. An improved version called the Fortin
barometer is used where high precision is
required.

43. video

44. Aneroid Barometer
The mercury barometer is the most reliable
type of barometer, but is not readily portable.
The aneroid barometer is a portable type of
barometer consisting of asealed,corrugated
metal box,see figure 4.22.

45. Video

46. Pressure Gauges
Pressure gauges are portable anda re used
mostly for measuring gas pressure,tyre
pressure,pressure of compressed air in
compressors and steam pressure.

48. TRANSMISSION OF PRESSURE IN LIQUIDS
When the plunger is pushed in,the liquid
squirts out the holes with equal force. If the
plunger exerts a force F and the piston area is
A,then the pressure P developed is F/A. This
pressure is transmitted equally to all parts of
liquid.

49. EXP 4.3:To investigate how pressure is
transmitted in liquids(Using Identical Syringes)

50. Procedure :
Place a mass m on one of the plungers and
observe what happens.
When the first mass is placed on the plunger,
the plunger moves downwards and the second
plunger moves up.
Place an identical mass on the other plunger
and observe what happens.
When an identical mass is placed on the
second plunger, the first plunger with the mass
on it moves upwards and stops when their levels
are the same.

51. The pressure in two syringes is the same. This
is because the masses and diameters of the
syringes are the same.

52. Using different Syringes

53. At balance , the pressure due to the mass in P
is equal to the pressure due to the other mass in
Q.
Pressure applied at one part in a liquid is
transmitted equally to all other parts of the
enclosed liquid. This is called the Principle of
Transmission of Pressure in Liquids(Pascal’s
Principle).
Note: gases may transmit pressure in a similar
way when they are confined and incompressible.

54. Hydraulic Machines
Pascal’s Principle enables a small force to be
multiplied into a large force. This principle of
transmission of pressure in liquids is made use
of in hydraulic machines.

55. Hydraulic Lift
When a force is applied on piston S,the
pressure exerted by the force is transmitted
throughout the liquid to piston L,see figure 4.12.

56. The pressure P1 exerted on the liquid by the
piston S due to F1 is given by;

57. This pressure will be transmitted by the liquid
to the larger piston L.
So, the force F2 produced on the large piston is
given by;
Note:Hydraulic lifts are used for hoisting cars in garages.

58. Example : in figure 4.12 ,a force of 100 N acts
on the smaller piston of area 2 cm2. Calculate
the upward force acting on the larger piston of
area 900 cm2.

59. Example: figure 4.13 shows two masses
placed on light pistons.The pistons are held
stationary by the liquid, whose density is 0.8
g/cm3.Determine the value of the force.

60. Hydraulic Brake System
The force applied on the foot pedal exerts pressure on
the master cylinder. The pressure is transmitted by the
brake fluid to the slave cylinder. This causes the pistons of
the slave cylinder to open the brake shoe and hence the
brake lining presses the drum. The rotation of the Wheel is
thus resisted.

61. When the force on the foot pedal is withdrawn, the
return spring pulls back the brake shoe which then pushes
the slave cylinder piston back.
The advantage of this system is that the pressure
exerted in master cylinder is transmitted equally to all the
four Wheel cylinders. So the braking force obtained is
uniform.
Brake fluid should have the following properties:
• Be incompressible
• Have low freezing point and high boiling point
• Should not corrode the parts of the brake system

62. homework
Exercise : 4.2

63. APPLICATIONS OF PRESSURE IN GASES
AND LIQUIDS
THE BICYCLE PUMP
It consist of a hollow metal cylinder C with a lid
at the top and a smallhole at the bottom.
The piston rod passes through the lid.
The upper end of P has a handle while at its
lower end, a cup-shaped leather washer W. The
washer touches the sides of the cylinder tightly.

64. (i)when the psiton is withdrawn , the air in
section A of the cylinder expands. Hence the
pressure inA is less than the atmospheric
pressure .So air can pass the washer into A.

65. (ii)when the piston is lowered , the air inA is
compressed. The pressure exerted by the
compressed air presses the leather washer
tightly against the walls of the cylinder and does
not allow the air to escapeoutside. When the
pressure in A exceeds that in the tyre, the valve
in the tyre opens and more air enters the tyre.

66. The lift Pump
A lift pump is used to raise water from wells. İt
consists of a cylindrical meatal barel with a side
tube. İt has two valves ,P and Q, as shown in
figure 4.25.

67. Upstroke
When the plunger moves up during the upstroke , valve P
closes due to its weight and pressure of water above it. At the
same time , air above valve expands and its pressure reduces
below atmospheric pressure.
The atmospheric pressure on the water in the well below thus
pushes water up past valve Q into barrel, as shown fig.4.25(a)
The plunger is moved up and down until the space between P
and Q is filled with water.

68. Downstroke
During downstroke , valve Q closes due to its
weight and pressure of water above the
piston,see fig 4.25(b)

69. Limitations of this pump
The atmospheric pressure can only support a
column of water of about 10 m.
In practice , the possible height of water that
can be raised by this pump is less than 10 m
because of:
(a) low atmospheric pressure in places high
above sea level.
(b)leakages at the valves and pistons.

70. The force Pump
This pump can be used to raise water to
heights of more than 10 m.
In this case the valve V2 is not in the piston,
but in the side delivery tube which is near the
bottom of the cylinder. Valve V1 must be about
7-8 m from the surface of water. This pump can
force water to great heights as the water is
forced out due to the pressure applied at the
piston. Thus , water is raised from the well into
the pump under atmospheric pressure , but it is
forced into the tank under the pressure applied
by the piston. Fire- Brigade pumps are of this
type.

72. Siphon
Tanks are difficult to empty out simply by
pouring out their contents. To make easier , we
use a siphon.
Aflexible pipe filled with the liquid is fitted as
shown below. One end is outside (below the
surface of the tank) and the other is in the tank
(insiude the liquid). The liquid pours out of the
pipe at the lower end.

73. The water pours out due to the pressure
between A and B:pg(h2-h1). End B is at a
higher pressure than end A due to the difference
ib heights: h=h2-h1.
Note : if the pipe is initially empty, the siphon
would not work. The air needs to be sucked out
of the pipe fort he liquid tor ise and pour out.

74. HOMEWORK:
REVISION EXERCISE