lets discuss the pressure distribution in
an accelerated frame.
as we already studied that, in a static fluid.
free surface.
of fluid is always.
normal to.
effective acceleration.
acting on it.
this we already studied and.
lets have a look on the physical situation
we can see here.
here in this figure ay we can see, that we
are having a container which is placed in
gravity . and say g is acting in downward
direction then the free surface of liquid.
is normal to g. and at any point say we consider
a point ay which is at a depth h below.
the free surface of liquid layer pressure
at point ay, can be directly written as, p-atmosphere,
plus, h ro g. this we discuss several times,
now let us consider a situation in which the
container will start moving towards right
with an acceleration ay. if the container
will start moving towards right with an acceleration
ay, the whole fluid body will experience pseudo
acceleration in backward direction or a pseudo
force m ay in backward direction.
or we can say the body will be, experiencing,
an effective acceleration in backward direction
g in downward direction, due to which the
free surface of liquid now no longer be horizontal.
and the resulting situation here we can see.
here in figure p the resulting situation is
shown.
and you can see if the container starts moving
toward right with an acceleration ay. every
medium particle of the fluid will experience
a pseudo acceleration or effective acceleration
ay in opposite direction and g it is experiencing
in downward direction.
so the resulting acceleration acting on it
will be g-effective and it is in this direction.
if we just draw the vector diagram here again
. here g is acting in downward direction effective
acceleration on the fluid is in leftward direction
. so resulting acceleration of the fluid will
be g-effective it’ll act. in this direction
and automatically the free surface of liquid
is adjusted . in such a way that it is normal
to the effective acceleration acting on it.
in this situation this effective acceleration
can be written as, as the two accelerations
are normal to each other it can be written
as root of ay-square plus g-square. and the
angle if we wish to find-out, between effective
acceleration in vertical is theta . the same
is the angle between, the free surface of
liquid with the horizontal this theta can
be, easily given as ten inverse of ay by g.
you should also keep both of these things
in your mind and you can see.
whenever a container is accelerated with an
acceleration ay we can directly state, that
the free surface of, liquid within the container
will become inclined . in such a way that
its angle with the horizontal is tan inverse
of ay by g if ay is the horizontal acceleration
of the container.
in the next slide lets discuss about the pressure
distribution, in this accelerated container.
let us discuss about pressure distribution
in the same accelerated container.
in which the free surface of liquid will become
inclined at an angle theta where theta is
given as tan inverse of ay by g which we already
studied.
now in this situation, if we wish to find
out the pressure at a point ay. with in the
fluid body.
then we need to take, its.
depth below the free surface of liquid along
the effective acceleration say this depth
is h. then directly pressure at point ay can
be given as, the pressure acting on the free
surface by atmosphere is p-atmospheric, so
we can simply write is p-atmospheric plus,
h ro g-effective which can be written as root
of g-square plus ay-square that is pressure
at point ay . this one method to find out
the pressure at point ay, and there is one
more way by which we can calculate the pressure
at point ay, by using, the pressure variation
along the direction of gravity as well as
along the direction of, effective horizontal
acceleration acting on the fluid, due to the
motion or due to the acceleration of container
let see that also.
now the alternative way to find out the pressure
at point ay is by using pressure variation
along the gravity or along the pseudo acceleration
acting on the.
fluid which is in the container which is accelerating
towards right.
now say, point ay is at a depth . h one, vertically
below the free surface of the liquid.
here we can say, along the vertical length
or vertical height h-one they wont be any
pressure variation due to this acceleration
as this will be normal to the acceleration,
and at this point pressure acting by atmosphere
is p-atmospheric, so at point ay pressure
can also be given as p-atmospheric, plus h-one
ro g here we wont take g-effective like we
have taken in this expression . because in
vertical direction pressure is only varying
due to gravity.
similarly if this point ay is taken a horizontal
distance, l. from the free surface.
again we can say at the free surface pressure
acting is p-atmospheric , so pressure at point
ay can be given as p-atmospheric plus, l ro
ay. because along the length of l, the pressure
is only varying due to the effective acceleration
ay as, due to gravity there wont be any variation
along the length l as it is horizontal, and
due to gravity no variation of pressure exist
here.
so there are three ways indirectly we can
find out the pressure at point ay, one is
along the effective gravity which we have
just now taken . other is along the vertical
and third one is along the length l. so you
just need to take care of all these cases
whenever required we can use any of these.
