1

Up Thrust Force
• Up thrust is upward force acting on an object floating or
immersed in the fluid appears lighter that its actual
weight due to up thrust force (force of buoyancy).
• Up thrust causes some bodies to float in liquid or in air.
.

2
 Demonstration Of Upthrust
Procedure
Weigh the block in air= 𝑾𝟏
w1.
Note the reading on the top pan balance
with a beaker. 𝑾𝟎
Immerse the block in water in the overflow w2.
can as shown in the diagram below
Note the weight of the block when fully
immersed as w2.
Note the new reading on top pan balance of
water displaced 𝑾𝟑
Weight of water displaced 𝑾𝟒 =𝑾𝟑 -𝑾𝟎
Apparent loss of weight=𝑾𝟏 − 𝑾𝟐
The upthrust = 𝑾𝟏 − 𝑾𝟐 .=𝑾𝟒
It is noted that Apparent loss of weight =
weight of the fluid dis[laced upthrust on object

3
 Archimedes Principle
States principle states that when a body is fully or partially
immersed in a fluid it experiences an up thrust equal to the
weight of displaced fluid.
• Archimedes Principle says that if an object of
volume Vobj is totally submerged in a fluid of
density Pfluid
• The upward buoyant force is
B = Pfluid gVobj
• Of course, the downward gravitational force is
Fg = Mg = Pfluid gVobj

𝑼𝒑𝒕𝒉𝒓𝒖𝒔𝒕 = 𝝆𝑽𝒈
𝝆 = 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒔 𝑽 = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅 𝒈 = 𝒈𝒓𝒂𝒗𝒊𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Upthrust depends on:
• Density of fluid
• Volume of fluid displaced = equal to the volume/part of object in liquid
4
Showing Forces Acting On A Body

Upthrust

T Weight

Upthrust (pvg) = Tension (T) Weight (mg)

5
5
 Up thrust In Gases
• Gases experiences small up thrust because of their low density.
• A balloon filled with hydrogen or helium rises up because of low
density.

𝒘𝒆𝒊𝒈𝒉𝒕 < 𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕

stationary
𝒘𝒆𝒊𝒈𝒉𝒕 = 𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕

𝒘𝒆𝒊𝒈𝒉𝒕 > 𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕

6
 Showing Forces Acting On A Balloon
Density of air
= ρa

Volume of balloon = 𝑽𝑩

Density of helium = 𝝆𝑯

Maximum Load

Upward forces = Downward forces

Upthrust = weight of gas + max load
𝝆𝒂 x 𝑽𝑩 x g = ρ𝑯 𝐱 𝑽𝑩 x g + max load
7
7
Example 1
A stone weighs 2.0 N in air and 1.2 N when totally immersed in water. Calculate:
(a) The volume of the stone.
(b) The density of the stone
(Take g = 10 N 𝐤𝐠 −𝟏 )
Solution
(a) Upthrust = weight in air – apparent weight
= (2.0 – 1.2) N
= 0.8 N
Upthrust = weight of water displaced by the stone.
Weight of water displaced = 0.8 N
𝟎.𝟖
Mass of water displaced =
𝟏𝟎
= 0.08 kg
Since density of water = 1000 𝐤𝐠𝐦−𝟑
𝟎.𝟎𝟖
Volume of the water displaced = 𝟏𝟎𝟎𝟎 𝐦𝟑
= 8.0 x 𝟏𝟎−𝟓 𝐦𝟑
Volume of stone = 8.0 x 𝟏𝟎−𝟓 𝐦𝟑
𝟐.𝟎
(b) Mass of the stone = 𝟏𝟎
= 0.2 kg
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐭𝐨𝐧𝐞
Density of the stone = 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐭𝐨𝐧𝐞
𝟎.𝟐 𝐤𝐠
= 𝟖.𝟎 𝐱𝟏𝟎−𝟓𝐦𝟑
= 2500 𝐤𝐠𝐦−𝟑
8
Example 2
A meteorological balloon has a volume of 36 𝐦𝟑 and is filled with helium of
density 0.18 𝐤𝐠𝐦𝟑 . If the weight of its fabric is 120 N, calculate the
maximum load which the balloon can lift. (Take density of air as 1.3 𝐤𝐠𝐦𝟑 )
Solution
Volume of air displaced by the balloon = 36 𝐦𝟑 .
Mass of air displaced by the balloon = (36 x 1.3) kg
Weight of air displaced = 36 x 1.3 x 10
= 468 N
Weight of air displaced = upthrust
= 468 N
Mass of helium in the balloon = (36 x 1.8 x 𝟏𝟎−𝟏 ) kg
Weight of helium in the balloon = (36 x 1.8 x 𝟏𝟎−𝟏 x 10)
= 64.8 N
Weight inflated balloon = (64.8+120) N
= 184.8 N
Upthrust = maximum load + weight of inflated balloon
Maximum load = upthrust – weight of balloon
= 468 – 184.8
= 283.2 N
9
 Law of Floatation

In this case we consider the floating object and weight of the fluid displaced.
It is observed that
• the test tube sinks deeper when more sand is added.
• the weight of the test tube and its contents is equal to
weight of displaced water.

10
 Law of Flotation
Law of flotation states that, a floating object displaces its own
weight of the fluid in which it floats.
When a body is in fluid there are two forces acting on it.
1. A force equal to its weight, acting downwards.
2. Buoyant force, acting upwards.
According to the principle of floatation, the weight of a
floating body is equal to the weight of the liquid
displaced by its submerged part.
Weight of floating object = Weight of fluid displaced = buoyant force
Apparent weight of a floating body:
According to the principle of floatation, when a body floats in a liquid, the
weight of the body is equal to the weight of the liquid displaced by its
submerged part. Thus, the upthrust by the liquid is equal to the weight of the
body.
Apparent weight of body = True weight – Upthrust
=0
Thus, a floating body appears to have no weight (or its apparent weight is zero)
11
 Law of Floatation
When weight of object When weight of object When weight of object is
is greater than upthrust, is equal to upthrust a less than upthrust the
the a body completely body immerses in liquid body floats in liquid if its
sinks in liquid if density but remains floating in density is greater than
of body is greater than liquid. If its density is density of liquid. The
density of liquid. greater than density of portion of the body that
liquid. If the density of lies inside the liquid
liquid = density of solid displaces the liquid equal
the the upper surface of to its own weight.
body is at the level of
liquid.

12
Example 3
A boat of mass 1000 kg floats on fresh water. If the boat enters sea water, determine
the load that must be added to it so that it displaces the same volume of water as
before. (Take density of fresh water as 1000 𝐤𝐠𝐦𝟑 and density of sea water as 1030
𝐤𝐠𝐦𝟑 )
Weight of the boat = 1000 x 10
= 10 000 N
By the law of floatation;
Weight of fresh water displaced – weight of the boat
= 10 000 N
𝟏𝟎𝟎𝟎𝟎
Mass of fresh water displaced = 𝟏𝟎
= 1000 kg
𝟏𝟎𝟎𝟎
Volume of fresh water displaced = 𝟏𝟎𝟎𝟎
= 1 𝐦𝟑
Volume of sea water displaced on addition of load = 1 𝐦𝟑
Mass of sea water displaced = 1 x 1030
= 1030 kg
Weight of sea water displaced = (1030 x 10)
= 10300 N
Extra load required = weight of sea water displaced – weight of fresh water displaced
= 10300 - 10000
= 300 N
13
Example 4
A balloon of volume 6.0 𝐦𝟑 is filled with hydrogen of density 0.09
𝐤𝐠𝐦𝟑 and floats in air of density 1.3 𝐤𝐠𝐦−𝟑 . Calculate the weight of the
fabric of the balloon.
Solution
Mass of hydrogen in the balloon = 6.0 x 0.09
= 0.54 kg
Weight of hydrogen = 0.54 x 10
= 5.4 N
Let W be the weight of the fabric of the balloon.
Total weight of the inflated balloon = (5.4 + W) N
Volume of air displaced by the balloon = 6.0 𝐦𝟑
Mass of air displaced = 6.0 x 1.3
= 7.8 kg
Weight of the air displaced = 7.8 x 10
= 78 N
By the law of floatation, 5.4 + W = 78
W = 78 – 5.4
= 72.6 N

14
Example 5
A cube of side 4 cm weighs 1.12 N in air. Calculate:
(a) Its apparent weight when immersed in a liquid of density 0.79 𝐠𝐜𝐦−𝟑
(b) The density of the material of the cube.
Solution
(a) Volume of the cube = 4 x 4 x 4 x 𝟏𝟎−𝟒
= 6.4 x 𝟏𝟎−𝟓 𝐦𝟑
Volume of a liquid displaced by the cube = 6.4 x 𝟏𝟎−𝟓 𝐦𝟑
Mass of liquid displaced = (6.4 x 𝟏𝟎−𝟓 x 7.90 x 𝟏𝟎𝟐 )
= 5.06 x 𝟏𝟎−𝟐 kg
Weight of a liquid displaced = 5.06 x 𝟏𝟎−𝟐 x 10
= 5.06 x 𝟏𝟎−𝟏 N
But weight of liquid displaced = upthrust on cube
Apparent weight = 1.12 – 0.506
= 6.14 x 𝟏𝟎−𝟏 N
(b) Weight of the cube = 1.12 N
𝟏.𝟏𝟐
Mass of the cube = 𝟏𝟎
= 1.12 x 𝟏𝟎−𝟏 kg
Volume of the cube = 6.4 x𝟏𝟎−𝟓 𝐦𝟑
𝟏.𝟏𝟐 𝐱𝟏𝟎−𝟏
Density of the cube = 𝟔.𝟒 𝐱𝟏𝟎−𝟓
= 1.75 x 𝟏𝟎𝟐 𝐤𝐠𝐦−𝟑
15
Example 6
The wooden block in figure 3.6 floats in two liquids X and Y. given that the densities of X and
Y are 1 𝐠𝐜𝐦−𝟑 and 0.8 𝐠𝐜𝐦−𝟑 respectively, determine:
(a) The mass of the block
(b) The density of the block
Solution
(a) Volume V of liquid Y is displaced = 5 x 4 x 3 x 10−6 m3
= 6 x 10−5 m3
Upthrust= ρVg
= 800 x 6 x 10−5 x 10
= 0.48 N
Volume V of liquid X displaced = 5 x 4 x 6 x 10−6 m3
= 1.2x 10−4 m3
Upthrust = ρVg
= 1000 x 1.2 x 10−4 x 10
= 1.2 N
Total upthrust on block= 0.48 + 1.2
= 1.68 N
The block is floating, hence weight of block equals total upthrust.
1.68
Weight of block =
10
= 0.168 kg
(b) Volume of block = 5 x 4 x 12 x 10−6 m3
= 2.4 x 10−4 m3
mass 0.168𝑘𝑔 3
Density = = 3 =700kg/m
volume 0.00024 m

16
 Upthrust and Relative Density
• Relative density is the ratio of the density of a substance
to the density of water.
• To find relative density of a solid or a liquid several methods or
formulas are used.
𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒐𝒇 𝒂 𝒔𝒖𝒃𝒔𝒕𝒂𝒏𝒄𝒆
• 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓
• If equal volumes of the substance and water are considered,
𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅
• 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑴𝒂𝒔𝒔 𝒐𝒇 𝒆𝒒𝒖𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓
𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒆𝒒𝒖𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓

𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅
𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅 𝒘𝒂𝒕𝒆𝒓

𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒏𝒔𝒊𝒕𝒚
𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅
=
𝒂𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓/𝒖𝒑 𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓
17
 Relative Density Of A Floater Body
 Measure the weight of a solid in
air.
 Measure the weight of solid in
in liquid A
 Measure the weight of same
solid in liquid B
 Calculate the upthrust on solid in A B
liquid A
 Calculate the upthrust of solid in
liquid B

𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒍𝒊𝒒𝒖𝒊𝒅

𝑹. 𝑫 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅𝒔 =
𝒖𝒑𝒕𝒉𝒓𝒖𝒔𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓

18
Example 7
A solid mass 0.8 kg suspended by a string is totally immersed in water. If the tension in
the string is 4.8 N, calculate:
(a) The volume of the solid
(b) The relative density of the solid
Solution
(a) Weight of the solid = (0.8 x 10) N
= 8.0 N
Apparent weight of the solid = 4.8 N
Upthrust = (8.0 – 4.8) N
= 3.2 N
But upthrust = weight of water displaced.
Weight of the water displaced = 3.2 N
𝟎.𝟑𝟐
Mass of water displaced = 𝟏𝟎 g
= 0.32 kg
𝟎.𝟑𝟐
Volume of water displaced =
𝟏𝟎𝟎𝟎
= 3.2 x 𝟏𝟎−𝟒 𝐦𝟑
Volume of water displaced = volume of the solid
Volume of the solid = 3.2 x 𝟏𝟎−𝟒 𝐦𝟑
𝐰𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐨𝐥𝐢𝐝
(b) Relative density of the solid = 𝐮𝐩𝐭𝐡𝐫𝐮𝐬𝐭 𝐢𝐧 𝐰𝐚𝐭𝐞𝐫
𝟖.𝟎
= 𝟑.𝟐
= 2.5
19
Example 8
In an experiment with a metal cube, the following
results were obtained:
Weight of the cube in air = 0.5 N
Weight of cube when completely immersed in water = 0.44 N
Weight of the cube when completely immersed in oil = 0.46 N
Calculate the relative density of the oil.

Solution
𝐮𝐩𝐭𝐡𝐫𝐮𝐬𝐭 𝐢𝐧 𝐨𝐢𝐥
Relative density of the oil =
𝐮𝐩𝐭𝐡𝐫𝐮𝐬𝐭 𝐢𝐧 𝐰𝐚𝐭𝐞𝐫
𝟎.𝟓 −𝟎.𝟒𝟔
=
𝟎.𝟓 −𝟎.𝟒𝟒
= 0.667

20
APPLICATIONS OF ARCHIMEDES PRINCIPLE AND RELATIVE DENSITY

Hydrometer
It is an instrument used to find relative densities of density of liquids.
It applies the law of flotation in its operation.
Thin stem to increase
sensitivity

Large bulb containing air to

displace large volume of liquid
to provide sufficient upthrust to
float the hydrometer

Weighted with lead shots to make

the hydrometer float verticaly

21
21
Example 9
A hydrometer of mass 20g floats in oil of density 0.8 𝐠𝐜𝐦−𝟑 , with 5cm of its stem
above the oil. If the cross sectional area of the stem is 0.4 𝐜𝐦𝟐 , calculate:
(a) The total volume of the hydrometer
(b) Length of its stem out of water, if it floats in water (density of water = 1
𝐠𝐜𝐦−𝟑 )
𝟐𝟎
(c) Volume of oil displaced = 𝟎.𝟖
= 25 𝐜𝐦𝟑
This is the volume of hydrometer immersed in oil.
Volume of part of hydrometer above oil = 5 cm x 0.4 𝐜𝐦𝟐
= 2.0 𝐜𝐦𝟑 Volume of water displaced = 𝟐𝟎 𝐠
Total volume of hydrometer = 25 + 2.0 𝟏 𝐠𝐜𝐦−𝟑

= 27 𝐜𝐦𝟑 = 20 𝐜𝐦𝟐
(b) Mass of hydrometer = 2 x 𝟏𝟎𝟐 kg But this is volume of part of
hydrometer immersed in water.
Weight of hydrometer = 2.0 x 𝟏𝟎−𝟏 N
Volume of part of hydrometer above
Weight of water displaced = 2.0 x 𝟏𝟎−𝟏 N
water = 27 – 20
Weight of the water displaced = 3.2 N 𝟐
𝟐.𝟎 𝐱 𝟏𝟎 −𝟏 = 7 𝐜𝐦
Mass of water displaced = Length of stem out of water =
𝟏𝟎
𝐯𝐨𝐥𝐮𝐦𝐞
= 2.0 x 𝟏𝟎−𝟐 kg
𝐚𝐫𝐞𝐚
= 20 g 𝟕
= 𝟎.𝟒= 17.5 cm
22
Example 10
A glass tube of uniform diameter 2.4 cm is weighted to float
vertically in a liquid. The length of tube immersed in the liquid
is 14 cm. if the density of the liquid is 11.2 𝐠𝐜𝐦−𝟑 , find the
mass of the tube and its contents.

Solution
Area of cross section of the tube = 𝛑𝐫 𝟐
= 3.142 x (𝟏. 𝟐)𝟐
= 4.52 𝐜𝐦𝟐
Volume of liquid displaced by the tube = 4.52 x 14
= 63.3 𝐜𝐦𝟑
Mass of liquid displaced by the tube = 6𝟑. 𝟑 𝐱 𝟏. 𝟐
= 75.96 g
Mass of tube and contents = 0.076 kg

23
Special Purpose Hydrometers
Lactometer range 1.015 – 1.0045 𝑔𝑐𝑚−3 so as to
measure relative density of milk.

Brewers’ hydrometer to measure relative density of

alcohol and wine. The rubber bulb is
squeezed and
Rubber bulb released so that the
acid is drawn into
the glass tube ant
Glass tube
the density of acid
read. The acid is
hydrometer then returned to the
battery
24
 Archimedes’s Principle and RelativeDensity
b) Balloons
• Used by meteorologists where a gas which
is less dense than air like hydrogen is used.
The balloon moves upwards because up
thrust force is greater than weight of the
balloon.
(c) Ships
• They are made of steel which is denser than
water but floats because they are hollow
and their average density is less than that of
water.
(d) Sub-marine
• It can sink or float. It is fitted with ballast
tanks that can be filled with air or water
hence varying its weight .To sink, ballast
tanks are filled with water so that its
weight is greater than upthrust.
• To float compressed air is pumped into the
tank displacing water so that upthrust is
greater than weight of the submarine.
25
1.State Archimedes principle.
2. A block of metal of volume 80cm3 weighs 3.80N in air. Determine it’s
weight when fully sub merged in a liquid of density 1200kgm-3

3. State the law of floatation.

4. Explain how a submarine is made to float and sink in water.
5. A solid copper sphere will sink in water while a hollow copper sphere of the
same mass may float. Give a reason for this.
6. Give a reason why a steel rod sinks in water while a ship made of steel floats
on water.
7. A flat test tube containing lead shots is immersed in a fluid, where it floats
as shown. What is the use of the lead shots?
8. A test-tube does not float upright on water, but floats upright when loaded
with sand. Give one reason for this observation.
9. A cork and a stone are both held under water and released at the same time.
i) State the observation that would be made
ii) Explain the observation above
10. A balloon of volume 1.2x107 cm3 is filled with hydrogen gas of density 9.0
x 10-5/g/cm3. Determine the weight of the fabric of the balloon.
26
11. A mass of 120g half immersed in water displaced a volume of 20cm3.
Calculate the density of the object.
12. A solid displaced 5.5 cm3 of paraffin when floating and 20cm3. Calculate the
density of the object.
13. An object weighs 1 .04N in air, 0.64N when fully immersed in water and 0.
72N when fully immersed in a liquid. If the density of water is 1000 kgm-3, find the
density of the liquid.
14. A body weighs 40N in air, 30N when in water and 35N when in liquid X. Find
the relative density of liquid X.
15. A block of glass of mass 250g floats in mercury. What volume of the glass lies
under the surface of the mercury? (Density of mercury is 13.6 x 103).
16. A piece of marble of mass 1.4kg and relative density 2.8 is supported by a light
string from a spring balance. It is then lowered into the water fully. Determine the
upthrust.
17. A body weighs 22N in kerosene and 20N in water. If it weighs 30N in air, find
the relative density of kerosene.
18. A solid of mass 100g and density 2.5g/cm3 weighs 0.5N when totally
submerged in a liquid. Determine the density of the liquid.
19. A solid Y weighs 40N in air, 30N when in water and 35N in liquid X. Find the
density of;
(i) Solid Y
(ii) Liquid X

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