Archimedes' principle indicates that the upward
buoyant force that is exerted on a body immersed in a fluid, whether fully or
partially submerged, is equal to the weight of the fluid that the body
displaces.
The Concept of Upthrust
Explain the concept of upthrust
If a heavy object is
lifted while immersed in water, it may be raised more easily than when outside
the water. This is due to the reason that when anything is placed in a liquid, it
receives an upward force called upthrust. A body appears to have lost some
weight when immersed in water due to the upthrust exerted on the body by the
water.
By
definition, upthrust is referred to as an upward force exerted by the body when
it’s partially or totally immersed in water.
Consider
the experiment below to verify the concept of upthrust.
From
this experiment, it will be observed that W1>W2. This
is because:
When
a body is partially or totally immersed in any liquid, the liquid exerts an
upward force.
A weight recorded on the spring balance of a body that is
totally or partially immersed in any liquid is called apparent weight. E.g. W2 and
the force, which temporarily reduces the weight of the body, are called upthrust
(u).
Verification of Archimedes' Principle
Verify the
Archimedes principle
Archimedes’ principle
describes two important things.
A body immersed in a fluid experiences forces, one of which is upthrust
The weight of the fluid displaced is equal to
the up-thrust.
The forces acting on a body in air and when immersed in a
liquid can be shown in the following diagrams;
When a body is immersed in a liquid, it experiences an upward force from
the liquid. This force is known as up-thrust. Up-thrust opposes the weight of
the body.
Since the
weight of the body is opposed by the up-thrust, it follows that the weight of
the body when in the liquid (W1) is
decreased and becomes smaller than the weight of the body in air (W).
The
difference between the weight of the body in air and the weight of the body in the
liquid (Apparent weight) is known as Apparent loss in weight (W2).
This apparent loss in weight (W2) is equal to the weight of the liquid which is displaced after immersing the body.
Therefore,
Hence, it can be shown that,
Activity 1
Aim: The Aim of this activity is to verify Archimedes’ principle.
Apparatus;
Spring balance
Digital /beam balance
1 beaker
Eureka can
water
1 piece of a small stone
A cotton thread or a string
Procedure;
Measure the weight of the empty beaker (Wb)
using a beam/ digital balance.
Fill the Eureka can with water until water starts to fall out of its beak.
Attach the stone to the spring balance to
measure its weight by using a cotton thread /string.
Record the weight of the stone in air as W.
Place the empty beaker under the beak of the
Eureka can.
Slowly immerse the stone in the water and
observe the changes in the readings of the spring balance.
Water will be displaced and collected by the
beaker.
Record the apparent weight of the stone
(weight of the stone in water) as W1.
Put the beaker with displaced water on the
digital balance to measure the weight of the water in the beaker =Wbw.
Discussion;
Weight of the water displaced (Ww)
can be obtained by taking (weight of beaker + weight of displaced water)
–weight of empty beaker=WbwWb. Ww=Wbw-Wb
Apparent loss in weight (W2) of
the store can be obtained by taking the weight of the store in air –Apparent weight
=W-W1
Questions;
1. Are the values Ww and W2 equal?
2.
What is the Upthrust experienced by the stone?
Example 11
Weight of a body in air =
10.0N
Weight of a body when
immersed in water = 9.2N find the upthrust.
Solution:
Data given
Weight of a body in air (W1)=
10.0N
Weight of a body when in
water (W2) = 9.2N
Formula
Upthrust = Loss of weight
in water = W1-W2
= 10.0N – 9.2N
= 0.8N
The upthrust is 0.8
Example 12
The weight of a body when totally immersed in liquid is
4.2N. if the weight of the liquid displaced is 2.5N, find the weight of the
body in the air.
Solution;
Data;
Apparent weight (W2)
= 4.2N
Weight of liquid displaced
(u) = 2.5N
Weight of body in air=?
Formula;
U = W1 –W2
W1= 4.2 +2.5 =
6.7N
The weight of the body in air is
6.7N
The Archimedes' Principle in Determining
Relative Density
Apply the Archimedes principle to
determine relative density
Relative density (R.D) of a substance can be
defined as the ratio of the mass of a substance to the mass of an equal volume
of water.
Relative
density can also be defined as the ratio of the weight of the substance to the
weight of an equal volume of water.
According to Archimedes’
principle mass of equal volume of water is the mass of the water displaced when
the substance is immersed in water.
Note that the volume of water displaced after immersing a
substance in water is equal to the volume of the substance.
That means,
Or,
Example 8
A
piece of glass weigh 1.2N in air and 0.7N when completely immersed in water.
Calculate its:
1. Relative density
2. Density of a glass
Given that density of water
= 1000kg/cm3
And
acceleration due to gravity = 10N/kg
Solution
Weight of the glass in air
(W1) = 1.2N
Weight of the glass in
water (W2) = 0.7N
R.D = W1/W1 – W2
1.2N/(1.2 – 0.7)
1.2/0.5
R.D = 2.4
R.D = Density of
glass/Density of water
Density of a glass = R.D x
Density of water
= 2.4 x 1000kg/cm
= 2400kg/m3
NB:
Relative density has no SI unit
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