In this illustration, we'll analyze, the case
of pushing a cylinder in a liquid. here we
are given that a solid cylinder of density
ro is floating in a liquid of density sigma
which is greater then ro. and the cylinder
is having a cross sectional area s and length
l. and here a boy pushes the cylinder down
so that it will completely submerged in the
liquid. we are required to calculate the work
done by the buoyant pushing the cylinder.
so, here, in the initial situation when cylinder
would be, floating in a liquid. we can draw
the situation and we can see. in the floating
states say, the cylinder is submerged to a
depth h in the liquid. then here we can write.
in state of. floating, of cylinder. here we
use. the buoyant force acting, on the cylinder
is, v, ro of, liquid, g, should be equal to
m, g, where m is the mass of this cylinder.
now in this situation we are given that the
cylinder length. is l. and if we consider
it is submerged to a depth h. and its cross
sectional area is also given to us as s. so
here the volume of submerged part we can write
as h, s. multiplied by density of liquid is
given to us is sigma multiplied by g is equal
to, the mass we can calculate as l s total
volume multiplied by ro density of solid multiplied
by g. here s and g gets cancelled out and
this will give us the submerged part, of cylinder
within liquid, and this h is equal to l, ro
by, sigma. now in this situation if we further
push it down by an external force, we can
write, if, cylinder is, pushed down. slowly.
by a distance, y. then, external force, on
it is given as, here we can write if it is
further pushed down by a distance y. and say
if this is the state when, cylinder is submerged
to, a depth. h plus y. here i am considering
it is further pushed down by a distance y.
from the state of floating. then in this situation
this force would be balancing the buoyant
force minus the weight of. this cylinder so
we can write. in equilibrium. we use, here
f plus m g should be equal to h plus y multiplied
by s. sigma, g. so in this situation this
m g and this h is sigma g gets cancelled out.
and in this situation the value of force we'll
get is, s sigma g, y. so this is the force
acting on the block. cylindrical block and
slowly, pushing it down when it is submerged,
to an axis depth y. beyond the, equilibrium
depth when the cylinder was floating. so we
can now calculate the work done in. pushing
this cylinder. completely is. this work done
we can write integration of f d y, and we
integrate the value of y from zero to, l minus
h because initially y was zero and, the total
depth which is to be submerged in the liquid
is, l minus h. so in this situation we write
it s sigma g. integration of y d y from zero
to l minus h. and this will be s sigma g.
multiplied by this y square by 2 that is l
minus h whole square by 2. that is the result
of this problem.
