lets discuss about pascle’s principle of
pressure distribution.
this principle is about how pressure applied
at one point in fluid is distributed through
out its volume.
lets write down the statement of pascle’s
principle.
it states that . the pressure applied . at
one point.
in an enclosed fluid body. is transmitted.
or we can also write distributed . uniformly.
to every part of the fluid body.
or to every part of the fluid.
this is the statement we use. and here we
are talking about enclosed fluid that means
we are mainly talking about the static fluid.
in the reference frame of consideration so
we can say at any one point if some pressure
is applied it is transmitted or distributed
uniformly to every part of the fluid.
lets discuss a situation to understand it
better.
in this situation we are given with a container.
completely filled with liquid and, it is enclosed
. with the help of cork at its mouth.
now in this situation that is not opened to
atmosphere anywhere so if we consider a point
(a), in the fluid body . which is located
at a depth h below.
the top layer of the liquid . in this situation
we can simply write pressure at point ay is
given as h ro g. because this pressure is
due to the weight of the liquid which exist
above the . point ay. in this situation the
pressure is h ro g. now we’ll see what will
happen , if the cork of , this container at
its mouth is open lets see what will happen
in this situation.
now you can see, as the cork is opened, now
the free surface of liquid , is in contact
with the atmosphere and atmosphere will exert
a pressure p-atmospheric at this point.
now we can imagine the situation if spherical
liquid molecules here will experience a pressure
from atmosphere.
these will, transfer the pressure to the other
contact particles, or the other particles
in contact with these liquid particles due
to the contact reaction, and continuously
the pressure will be transmitted to every
point of the fluid body.
now in this situation again the point ay is
considered which we have seen here . which
is at a depth h below the top layer of the
liquid, so in this situation pressure at point
ay which was h ro g. due to the weight of
liquid above the point ay. will be increase
by an amount, p-atmospheric because p-atmospheric
is applied at a point . so according to pascle’s
law now the pressure at point ay can be written
as h rho g plus, p-atmospheric . if at any
one point in an enclosed fluid , an external
pressure is applied , at every point in the
fluid body we can say the pressure will be
increase by the same amount so now the new
pressure will be h ro g plus p-atmospheric.
and we are going to see many applications
of pascle’s law this is just a basic example
i have taken.
to make you understand, how the pressure is
distributed through out the body it is not
necessary that the whole , top layer of, liquid
should be exposed to atmosphere to increase
the pressure even at one point it is applied
it is distributed through out the volume.
let see, few more applications of pressure
distribution by using pascle’s law.
let us discuss application of hydraulic forces
by using pascle’s law.
here we can see, a tube, on one side of the
tube we are having a piston which is fitted
into it having cross sectional area s-one,
and the other side of tube is having a cross
sectional area s-two.
which is also fitted with another piston,
and a fluid is filled between the two piston.
now say, at this piston say it is piston ay
and this piston b. we apply a force f-one.
then, in this situation we can state, in this
case the pressure applied on the fluid from
lefthand-side will be . p-one and that can
be written as f-one by s-one, and on the other
side . obviously due to this pressure, the
fluid will be having a tendency to move toward
right, so to keep at rest we need to apply
a force f-two here . such that the same pressure
will be applied, on the fluid to keep it at
rest.
so you can say here pressure p-two can be
written as f-two by s-two.
now in this situation if either of the force
is removes like for example . we remove f-one
, then this piston will start moving toward
left and the same force will be exerted by
piston , on to the left hand side agent who
is supporting the piston.
so in this situation we can directly write,
the value of pressure p-one must be equal
to p-two because inside the fluid as, it is
horizontal there wont be any change in the
pressure at left hand or at right hand.
so in this situation if we substitute the
values it’ll be f-one by s-one is equal
to f-two by, s-two.
this implies if we calculate the value of
f-one . this f-one can be written as , f-two
multiplied by s-one by s-two.
so in this situation, if , s-one is more then
s-two which we can see here.
this implies the , value of f-one will also
be more then value of f-two.
so if a force f-two is applied, on one end
of the tube we can simply state that piston
will exert a force f-one which is higher then
the applied force onto the other hand.
or if at one end , force f-one is applied,
in a narrow cross sectional area the fluid
will exert lesser force.
and this concept we call as force amplification,
using which small amount of force can be magnified
to a large extend as the value of force which
is changing is in the ray-she o of area.
of cross section, of the tube in which the
, fluid is filled.
this concept is very useful wherever hydraulic
forces are used. and just for varying the
area of cross section of the tube in which
the fluid is filled, we can change the magnitude
of forces.
applied by a fluid or piston so on to the
external agencies or different type of mechanism
driven by the forces.
