Physics
Physics
Chapter 2
The Force exerted
by fluids on objects
Competency
Investigates the force exerted by fluids on objects
Competency level
32
Subject Content
2' 1 Uses the pressure
exerted by solids,
liquids and gases in
daily activities.
² Concept of pressure
² Units of measuring pressure
² Pressure exerted by solids and liquids
² Atmospheric pressure
² Using pressure in daily activities
2' 2' Investigates the
forces exerted on
objects immersed
the liquids.
² Archimede's Principle
² Floatation
² Hydrometers
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unit 2
2.1
The force exerted by fluids on objects
Pressure exerted by solids, liquids and gases
• Pressure exerted by solids
When a force acts on a surface, its effect spreads
over a certain area. The average or the mean force
that acts, at right angles on a unit area is called
the pressure.
The pressure can be calculated by dividing the force
or thrust, by the area over which it acts.
Fig : 2.1
Gent's and ladies shoes
Force (F)
Contact area (A)
Pressure (P) }
When been trampled by a gent's shoe
or a ladies shoe which will cause
more pain ?
In symbols :
P
=
F
A
Example ( 1. A force of 300 N acts on an
area of 0.02 m2. Find the pressure
that acts on the surface.
Fig : 2.2
Car Jack
A broad plank put underneath when
using a car jack. Can you think why ?
300 N
0.02 m2
= 30000
2
Pressure (P) =
= 15000 N m−2
When the force is measured in N and area in m2.
The pressure is expressed in N m-2 or Pascal (Pa)
1 N m−2 = 1 Pa
Then the pressure in the above example can be
given as 15000 Pa.
When the same force acts on different extents of
Fig : 2.3
Blaise Pascal (1623-1662) surface values of pressure will vary. If the force
spreads out over a small area the pressure is high
The unit Pascal is named after Blaise and if the force spreads out over a larger area the
Pascal who did much research on
pressure is low.
fluid pressure.
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33
Physics
Fig : 2.4
Bread Knife
Fig : 2.5
Thumb stack
² A ladies stiletto heel which has a smaller area will cause
more damage and pain than a gent’s heel which has a
larger area. (Fig 2.1)
² When standing, a person will sink more into sand than
when he lies flat stretched on sand. The body weight
acts on a smaller area, in standing position, than when
the person lies stretched over a larger area.
² Usually a broad plank is kept under a jack, when it is
used to raise a vehicle. This arrangement spreads the
weight of the vehicle over a wider area, thus
reducing the pressure on the ground and preventing the
jack sinking into the ground.(Fig 2.2)
² Workers prefer to carry heavy cement bags on their
backs in reclining position, rather than on their heads. A
load spread over the back with a wider area of a worker,
exerts a lesser pressure than by the same load when
carried on his head, which gives the load only a smaller
area of contact.
² Sharpened cutting edges of knives have small areas.
When such edges are used to slice bread, the
increased pressure on bread will help to give a sharp,
fine cut (Fig 2.4).
² A force applied by the thumb on a thumb-stack will
spread over tiny area touching its pointed end. The
increased pressure at the pointed end will make it easy
to penetrate into the wood (Fig 2.5).
Solved example
The mass of a man is 60 kg. In his standing position, his feet cover an area of
100 cm2. What is the pressure on the ground due to his weight? (g = 10 m s−2)
Mass of the man
= 60 kg
Weight of the man
= 60 x 10 = 600 N
Area of contact
= 100 cm2
100
2
= 10000 m
= 0.01 m2
Pressure on the ground
= Force (weight)
Area
= 60000
1
= 600 N
0.01 m2
= 60000 N m−2 } 60000 Pa
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unit 2
Activity 1
² Make several pin - holes on
a polythene bag
²
Fill the bag with water and
hold it tight
²
Observe the water - spouts
coming out of holes.
Fig : 2.6
What does the observations
tell you?
The force exerted by fluids on objects
Pressure exerted by liquids
Liquids as well as solids, exert pressure. Pressure
due to liquids acts in all directions.
Make a few holes, one below the other on
somewhat a tall cylindrical vessel, and fill it with
water. Observe the jets of water spouts ejected.
Can you explain, what you observed ?
Fig : 2.7
² Water is ejected in all directions.
² Water spouts out faster from the
pin-holes positioned lower than
others.
² Pin holes of the same level ejects
water to equal distances
² The spout is fastest at the lowest hole
and the speed of water from the holes
Fig : 2.8 demonstrates that liquids find the same
decreased gradually one by one upwards.
level, irrespective of the shape of the container
The pressure inside a liquid increases with the depth
The following worked example shows how to calculate the pressure in a liquid.
The horizontal area of the bottom of a vessel is
0.004 m2 and it contains mercury filled to a height of
20 cm. Find the pressure due to mercury at the bottom
of the vessel.
20 cm
mercury
0.004
m2
Fig : 2.9 Pressure at the bottom
of a vessel containing mercury
Density of mercury (ρ)
= 13600 kg m-3
Gravitational acceleration (g) = 10 m s-2
Area at the bottom (A)
= 0.004 m2
20
Height of mercury column (h) = 20 cm =
m
100
Volume of mercury (A«h)
= 0.004 «
20 m3
100
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35
Physics
Density of mercury (ρ)
mass of mercury (A« h« ρ)
= 13600 kg m−3
= [0.004 « 20 « 13600] kg
100
= (0.004 « 20 « 13 600 « 10 N
(
100
20
« 13 600 « 10 N (
(0.004 «100
0.004 m2
weight of mercury (Ahρg)
Pressure due to mercury at
the bottom
(Ahρg)
A
= 27 200Nm-2 = 27 200 Pa
The mass per unit volume of any substance is called its density
mass (m)
Density (ρ) =
volume (v)
ρ
=
m
v
Units of measuring density: kg m-3 or g cm-3
20
100
P
=
13600
«
«
ρ
h
10
g
height of « density of « gravitational
liquid
liquid
acceleration
pressure liquid column
kg m−3
m
m s−2
The general formula for pressure at a point in a liquid is P = hρg
All points at the same level in a liquid are at the same depth from the surface. Hence
the liquid pressure at any point in the same level is equal.
Example 1
1) Calculate the pressure at the bottom of a mercury column 76 cm deep.
76 m
depth of the mercury (h)
=
100
density of mercury (ρ)
=
gravitational acceleration (g) =
liquid pressure
^P&
=
=
=
36
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13600 kg m-3
10 ms-2
hρg
76
«13600 « 10 = 103360 Nm-2
100
103360 Pa
unit 2
The force exerted by fluids on objects
2) Find the pressure exerted by a 10 m deep water column .
depth of the water column (h)
= 10 m
density of water (ρ)
= 1000 kg m-3
gravitational acceleration (g)
= 10 ms-2
water pressure at the bottom ^P& = hρg
= 10«1000«10 = 100000 N m-2
= 100000 Pa
Pressure exerted by gases
As well as solids and liquids, gases too exert pressure. In the laboratory, a convincing
demonstration of gas pressure can be done, as follows:
air
x
y
x
y
Water
I
Fig : 2.10
Water is first added to a U-tube
as shown in fig 2.10 (1). The
water levels on both sides are
equal.
II
Demonstration of gas pressure
Connect an inflated balloon to one arm of the
U-tube. The water level in the arm connected to
the balloon, goes down and the water level in the arm
open to air, rises up Fig : 2.10 (II)
At the same level in water pressure at X equals the
pressure at some point inY. But the pressure at some
point in Y is equal to the atmospheric pressure plus the
pressure of the water column above that point.
Now think
Can calculate the air pressure inside the balloon ?
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Physics
Atmospheric pressure
Vacuum
Hg
76 cm
The earth surface is surrounded by a thick layer of
air called the atmosphere. The atmosphere extends
to a height of about several hundred kilometeres
and its weight compresses everything around and
its effect is known as atmospheric pressure.
At sea level, the atmospheric pressure can support
a mercury column of about 76 cm.
Take a clean, dry, thick walled glass tube, closed at
one end and about 1 m in length. Use mercury to
fill the tube to the brim, taking care to exclude all
air bubbles trapped inside.
Fig : 2.11
Mercury barometer
Invert it and place the open end under mercury in
a trough as shown in Fig : 2.11 Mercury column
in the tube drops to a height of about 76 cm.
Since air was completely excluded, the space now
at the top of the tube is a vacuum; it is often called
a torricellian vacuum.
It can now be seen that the atmospheric pressure
is numerically equal to the pressure at the base
of a column of mercury of height about 76 cm.
This is the verticle height above the outside free
level of the mercury surface in the trough.
Fig : 2.12
Aneroid barometer
In its making a metre rule should be placed
vertically a little away from the glass tube so that
the zero end just touches the free mercury
surface.
The higher one goes from sea level the less dense is the air and the less weight of air
to compress. The atmospheric pressure reduces gradually.
Instruments which measure atmospheric pressure are called barometers. A simple
baro meter is shown in Fig : 2.11. With no air at the sealed end there is nothing to stop
the outside pressure of the atmosphere pushing mercury up the tube.
Aneroid Barometer
The aneroid barometer is a form of domestic barometer which contains no liquid.
They are more portable and cheaper and also easy to get the readings (Fig : 2.12).
38
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unit 2
The force exerted by fluids on objects
Application of atmospheric pressure
Fig : 2.13, 2.14 and 2.15 show three
instance which atmospheric pressure is used.
Drinking Straw
When you drink through a straw you do not really
suck up the liquid (Fig : 2.13). Instead you
remove air out of the straw. and lower the
pressure in the straw. Then the atmospheric
pressure on the surface of the liquid outside
the straw pushes the liquid up to your month.
Atmospheric
pressure
Straw
Liquid
Fig : 2.13
Drinking through a straw
Siphon
A siphon is a simple arrangement that can be used
Atmoshpheric Presure
for emptying a liquid from fixed vessels such as petrol
Rubber tube tanks or fish tanks. It consists of an open tube filled with
liquid. The tube is then placed with one end below the
liquid in the vessel as shown (Fig : 2.14). The liquid in the
Liquid
vessel is then observed to run out at A.
The pressure of the liquid at the open end A is greater
A
than the atmospheric pressure. As the pressure of air
at the end equals atmospheric pressure, it cannot support
the liquid column. Thus, allows the liquid to run out.
Fig : 2.14
function of a siphon
The Rubber sucker
When a rubber sucker is pushed against a piece of
glass, most of the air is squeezed out from under the sucker.
If you try to pull the sucker away from the glass, high
pressure air outside it and low pressure air undernealth it,
the rubber sucker is held firmly against the glass.
Fig : 2.15
2.2
Rubber plunger
Forces on objects immersed in liquids
The objects seem to weigh less when they are immersed in water. This effect can be
produced by any liquid or gas and it is the reason why some objects float.
When any object is immersed in water, the water exerts an upward force upon the
object. The upward force then causes and apparent loss in weight in the object.
If a piece of wood is held below the surface of water, an upward push can be felt by
the hand. This upward force is called upthrust.
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Physics
Activity 2
A cord is used to hang a solid piece of metal by a spring balance; which then
registers 5.5 N. It is now carefully lowered into a 'displacement can' which has been
filled up to the level of the spout. As the metal is immersed the displaced water flows
out through the spout and is collected in a beaker which rests on another balance that
can measure the weight of water collected.
First reading
When the piece of metal is partially immersed the spring balance registered 5.1 N and
the weight of water collected is 0.4 N
Second reading
When the metal is fully immersed the spring balance registers 5.0 N and the weight of
water collected is 0.5 N
5.1 N
5.5 N
1
Fig : 2.16
2
5N
3
How the apparent weight of an object is reduced when immersed in water.
Weight of the piece of metal in air
First instance
The apparent weight of the metal when it is partially
immersed
Apparent loss of weight of the metal
weight of water displaced
(Upthrust in this instance)
Second instance
The apparent weight of the metal when it is fully immersed
Apparent loss of weight of the metal
Weight of water displaced (Upthrust in this instance)
= 5.5 N
= 5.1 N
= (5.5 - 5.1) N
= 0.4 N
= 0.4 N
= 5.0 N
= (5.5 - 5.0)N
= 0.5 N
= 0.5 N
This activity clearly explains, that the upthrust upon an object, due to the liquid is equal to
the weight of liquid displaced by the object. More than 2000 years ago, Archimedes
discovered the relationship between upthrust by a liquid upon an object and the weight of
the liquid displaced.
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unit 2
The force exerted by fluids on objects
His statement about the observations is well known as Archimedes Principle.
Archimedes Principle
When an object is partially or totally immersed in a fluid, (liquid or gas) the
upthrust on the object is equal to the weight of the fluid displaced by the object
Floating
Activity 3
Weigh a wooden block with the help of a spring
balance. Gradually lower it to the water in a trough.
When the wooden block starts floating, you can notice
the spring balance reading is equals to zero. This explains
an object will float in a fluid if the upthrust can support
its weight or the weight of the object is equal to the
upthrust.
0
1
2
3
4
5
6
7
8
9
10
wooden block
Fig : 2.17 An object floats when
the up thrust is equal to the weight
of the object
If we do this activity using a 'displacement can' instead of a water trough,we can
collect and weigh the amount of water displaced.Thus you may observe that the weight
of displaced water equals to the weight of the wooden block.This means that the
weight of fluid displaced is same as the weight of the floating object.This is known
as the Law of floatation.
A floating object displaces its own weight of the fluid in which it floats.
Observe Fig : 2.18 and try to understand the forces act on a boat which is floating on
water.
Upthrust 500 N
Upthrust
Weight of displaced
water 500 N
Upthrust upon Weight of water
= displaced
the boat
Fig : 2.18
Weight
Two balancing forces upon a floating boat
When an object floats in a fluid;
weight of the object = upthrust upon the object = weight of the fluid displaced
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Physics
Activity 4
Float a test tube (with some sand)
vertically; first in water, and then
in a salt solution. Mark the two
levels to which the test tube
floated in two fluids. We can see
that the depth to which the test
tube sinks in water is a little more
than in salt solution.
Density of the salt solution is greater
than that of water.
The upthrust of fluids increases
with density and this is the reason
why the test tube sinks less in salt
solution.
1000 kg m−3
800 kg m−3
water
Fig : 2.20
alcohol
hydrometer in practice
Sand
Water
Salt solution
Fig : 2.19
Denser fluids need less to be
displaced to float objects
The same object sinks to different depths in
fluids of different densities.
Hydrometer is a small floating device
constructed following the principles of floatation.
It is used to measure densities of liquids.
Fig : 2.20 shows two hydrometers immersed in
water and in alcohol. They are calibrated to
measure densities of liquids.
There are various types of hydrometer designed
to measure densities of different liquids;
For example.
1. Metrolac : This is an instrument used to
determine the dry rubber content (DRC)
in latex which is related to the density.
2. A special hydrometer is used to test the
charge of car batteries by checking the
density of ‘sulphuric acid solution. (Fig : 2.22)
Fig : 2.22
The special
hydrometer used to find
the density of sulphuric
acid in car batteries
Fig : 2.21
This man reads a
book while floating in the dead sea
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High density of the brine in the dead sea
can keep the man floating high, without
allowing him to sink.
unit 2
The force exerted by fluids on objects
Summary
² The average or mean force acting on a unit area is called pressure.
² The formula for pressure =
Force
Average area
² The unit of pressure is newtons per square meter (Nm-2).
² Newtons per square metre is equal to one pascal (Pa).
1 N m−2 = 1 Pa
² The same force acting on different extents of area produce different amounts
of pressure. A force spread over a small area will exert a high pressure. A
force spread over a large area, will exert a low pressure.
² Pressure due to liquids acts in all directions.
² Pressure due to a liquid column = height of the liquid column x density of liquid x
gravitational acceleration.
P = hρg
² In a liquid the pressure increases and decreases with the depth.
² The pressure exerted by the atmosphere around earth is called atmospheric
pressure.
² Standard atmospheric pressure at sea level is 76 cm Hg.
² Barometer is used to measure atmospheric pressure.
² When an object is immersed in a fluid the apparent loss of weight of the object is
caused by the upthrust of the fluid.
² Until an object is immersed completely in a fluid, its apparent weight will decrease
gradually.
² Weight of the displaced liquid is equal to the upthrust by the liquid upon the immersed object.
² An object floats in a fluid, when the weight of the object equals upthrust;
² Archimedes principle
When an object is partially or totally immersed is a fluid (liquid or gas) the
upthrust on the object is equal to the weight of the fluid displaced by the object.
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Physics
Exercises
01.
The skin in pressed by the thumb with a force of 5 N. If contact area of the
thumb is 1 cm3 how much pressure is exerted on the skin ?
02.
A girl of mass 50 kg, wears a pair of slippers of area 80 cm2. What pres
sure is exerted by the slippers on the ground.
03.
04.
Find the pressure due to a water column 5 m high.
Find the height of the column of water that can be balanced by atmospheric
pressure. (density of water = 1000 kg m-3, g = 10 m s-2 )
How does the atmospheric pressure have an affect on the function of a
syringe.
It is not possible to drink soft drinks using a straw while in space.
Explain the statement.
How much is the atmospheric pressure at the sea- level in pascals (Pa).?
i
What do you mean by pressure?
ii. What is the SI unit of pressure?
iii. A force of 400 N is acting on an area of 0.08 m2. Find the pressure.
iv. It sinks more when standing, than lying flat on sand. Why?
v. Explain an activity to show that a force on a small area exerts a
higher pressure?
i. Calculate the pressure due to a column of water 6 m high.
(Density of water 1000 kg m-3, g = 10 ms-2)
ii. Draw a labeled diagram of a mercury barometer.
iii. At sea level, atmospheric pressure is 76 cm Hg. Express this pressure
in pascals (Pa)
iv. Briefly explain a simple experiment to show that there is Atmospheric
pressure.
v. Why the dams of lakes broder at the base than at the top? Explain
your answer.
i. What is ; a. Archimedes principle. b. Law of floation
ii. Why do objects immersed in water weigh less ?
iii. A block of wood hanging from a spring balance when lowered started
floating in water? What was the reading of the spring balance? Give
your reasons for your answer?
iv. An iron nail sinks in water. But a ship made of iron floats. Explain the
reason for this observation.
v. Draw a labelled diagram of a Hydrometer
vi. Name two types of hydrometers used in practical life.
05.
06.
07.
08.
09.
10.
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