EQUILIBRIUM OF BODIES IN LIQUID
EQUILIBRIUM OF BODIES IN LIQUID
When a heavy object
is immersed in a liquid the object appears to become lighter. Consider the
experience when a bucket of water is being drawn out of a well. While still
under the water the bucket appears very light and becomes very heavy as soon as
it is out of water.
Archimedes explains
this showing that when the bucket still in water, the body is acted upon by an
upward force U called upthrust. While still under water, the equilibrium of the
bucket is as : T = W – U.
Where T =
tension in the rope.
W
= weight.
U = upthrust.
U = W –
T. It is also known as loss in weight.
ARCHIMEDES PRINCIPLE
It states that
when an object is wholly or partially immersed in a fluid, it experiences a
loss in weight or an upthrust which is equal to the weight of the fluid
displaced by the object.
FLOATING
If a body is
denser than the liquid, it will completely sink in it, but if the density of
the body is less than that of the liquid, the body sinks until the weight of
liquid displaced is just equal to the weight of the body. The body then floats
and is said to be in equilibrium. This is the condition necessary for a body to
float.
PRINCIPLE OF FLOATING
It states that a
floating body displaces its own weight of the liquid in which it floats or a
body floats when the upthrust exerted upon it by the fluid is equal to the
weight of the body.
IMPORTANCE OF FLOATING PRINCIPLE
Ships float in
water even though they are made of steel which is denser than water. This is
because they are hollow objects containing large amount of air and so able to
displace a large amount of water, given an upthrust large enough to support the
weight of the ship. Likewise, the weight of a balloon and its content is equal
to the upthrust of air on the balloon.
EXAMPLE
An object weighs
0.08N in air and 0.01 N in a liquid of density 700kgm-3. Calculate
the upthrust of the liquid on the solid.
SOLUTION
Since upthrust = loss in weight
Upthrust is given
as (0.08 – 0.01) N = 0.07N.
DENSITY AND UPTHRUST
The density of a
substance is its mass per unit volume.
Density = mass
volume
ρ = m
v
where ρ(rho) is a
Greek letter representing density. Its S.I. unit is kgm-3.
EXAMPLE
An object weighs
0.08N in air and 0.01 N in a liquid of density 700kgm-3. Calculate
the volume of the solid.
SOLUTION
ρ = m
v
v = m
ρ
For density,
700kgm-3=
700 = 0.7gcm-3
1000
For mass,
0.007kg = 0.007 x 1000 = 7g
v = 7 = 10cm3.
0.7
RELATIVE DENSITY (R.D)
Since water is the
most common substance and its density is 1000kgm-3 or 1gcm-3, it is convenient
to use it as a standard for comparing the densities of other substances.
The relative density (R.D) of a substance is
defined as follows:
R.D = density of
the substance or
= mass of substance or
mass of equal volume of water
= weight of substance
weight of equal volume of water.
Relative Density
has no unit.
DETERMINATION OF R.D
(i)
Regular body
For a regular body, the volume can
easily be found and the mass can be obtained by direct weighing. Hence density
can be calculated.
(ii)
Irregular body
The mass can be obtained by direct weighing.
The volume is found by immersing the body in water provided it will not
dissolve in water and it will sink. The volume of water displaced will equal to
the volume of the solid. Hence, the density can be calculated.
RELATIVE DENSITY OF LIQUID BY DIRECT METHOD
A relative bottle can be used to find
the density of a liquid directly.
The relative bottle is weighed empty,
and has a mass of m. It is then filled with liquid and weighed and has mass m1.
Finally it is emptied, dried and refilled with water and weighed again, when it
has mass m2.
Recall that R.D = mass of liquid
Mass of equal
volume of water
= m1 – m
m2 – m.
Relative density of a solid in the form
of a powder (sand or steel ball bearing)
The density bottle is weighed empty (m1)
and again weighed when about one – third full of sand (m2). The
bottle with sand is filled up with water and reweighed (m3).
Finally, the bottle is emptied of its content, filled full with water, and
weighed (m4).
The relative density of the sand is
obtained as follows:
R.D = mass of any volume of sand
mass of equal volume of water
= m2 - m1
(m4 - m1) – (m3- m2)
Instrument used for measuring density.
(i) A hydrometer
is an instrument for measuring the density of liquids. Its construction and
operation are based on the fact that the depth to which a tube sinks depends on
the density of the liquid in which it is floating.
(ii) A practical
hydrometer is an instrument which gives a direct density reading of the liquid
in which it floats. It works from the principle that less the density of the
liquid, further the hydrometer sinks into it.
This shows that l
α 1/ρ.
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