let us study archimedes principal. i hope
you might have studied archimedes principal
at a basic level in your earlier classes.
now we are going to define it first, and then
we’ll study about some detail applications
of archimedes principal. and this principle
states, that, a body. whether partially. or
completely submerged. in a fluid. is buoyed
upward. by a force. due to fluid pressure.
that is equal to. the weight of. displaced
liquid. this is the archimedes principle . this
is the statement which you can always keep
in mind . and, a body which is buoyed upward
by force which is due to the fluid pressure
that we can calculate also, say for an example
we are given with a container . in which liquid
is filled up to a given height and inside
it say, we submerged a cylindrical body . which
is of height h. and its base area is s. it
is submerged in such a way that its top level,
is at a depth l below the free surface of
liquid. in this situation we can simply state
due to, the fluid pressure on the top face
of cylinder top circular face force f one
will act in downward direction. if the bottom
face of fluid pressure will exert, the fluid
pressure will exert a force f-two in upward
direction. and on the lateral surface as we
already discussed , the force due to fluid
pressure is normal to the surface , all these
forces will be acting in horizontally inward
direction . so all these will get cancelled
out. so here we can simply state , that. buoyant
force . on object is, here this buoyant force
which we can write as f-b this must be f-two
minus f-one . as we know net upward force
will be f-two minus f-one, and in this situation
f-one f-two can be easily obtained like the
location where f-one is applied if we, consider
point x. then pressure at point x due to,
fluid weight can be simply written as l ro
g. or net pressure can be taken as p atmospheric
plus l ro g. and if there is a point y which
is just below the bottom face of the cylindrical
object p-y can be written as, p-atmospheric
plus , the total depth of the point y, below
the free surface of liquid is l plus h. so
it’ll be p-atmospheric plus, l plus h ro
g. now in this situation if we wish to calculate
the value of f-one which is the force acting
, on the object due to fluid pressure will
be p-x multiplied by the area s so it can
be written as . p atmospheric plus l ro g
multiplied by s . similarly f-two which is
acting in upward direction will be p-y into
s which can be written as p-atmospheric plus.
l plus h ro g . multiplied by s. and if we
find out the buoyant force, on the object
this f-b we can see . if we subtract f-two
, minus f-one if you calculate this will give
you, h s ro g. an here h s will be the total
volume of the object which can be written
as v ro g. and ro we have taken here the density
of liquid , so the product of volume of object
and . the density of liquid gives us, the
mass of liquid which is displace by the volume
of this body. so this v ro g can be written
as , weight of liquid displaced. by the object.
so this is what, we have written earlier in
archimedes principle . that the total upward
force of the buoyant force acting on object
is given as weight of liquid displaced by
the object or it can be written as v ro g.
where v is the volume of liquid displace ro
is the density of, liquid and g is the gravity.
so this is the way how we calculate the buoyant
force acting on any object which is submerged.
if the object is partially submerged in the
liquid then, in this expression v would be
the submerged volume of the liquid because
, the liquid is displace only by the submerged
part of the body. lets have more applications
on archimedes principle. in continuation lets
discuss law of floatation. say if we consider
a body which is floating in a liquid . say
this is the body which is floating in a liquid.
then we can simply state. when body is floating
it’ll be partially submerged in the liquid
and. say if v is the volume of, body which
is submerged in the liquid, so we can simply
state the buoyant force acting on the body
can be written as v b , f-b is equal to v
ro g. and its weight will act in downward
direction mg if we draw the free body diagram
we can simply state . for iquilibrium of.
floating object. we can directly write in
this situation v ro g must be equal to . m
g. or we can simply state . the net buoyant
force acting on body has to balance the total
weight of the object. and you can see here
also buoyant force we are using the expression
v ro g . where v is the volume of liquid displace
in this situation it is volume of the object
which is submerged in, floating situation.
this is the way how law of floatation is define
.
