lets discuss about the concept of calculation
of force on side walls of a random shaped
vessel. let us first discuss about the typical
vessel. which we usually use , say we are
given with a conical bucket . in which we
fill up the water. and in this situation we
know well that this water due to its weight
, will push the bottom in downward direction
and will push the side walls in the direction
normal to the surface. obviously due to its
reaction the bottom will, apply a normal reaction
on the fluid body say n-one in upward direction
. and n-two in the direction normal to the
side walls of the vessel. so in this situation
we can say overall weight of this water , is
balanced by one is n-one other is the vertical
component of the , force exerted by side walls.
as the horizontal component of the force exerted
by side walls will always cancel out as the
fluid is in iquilibrium . so in this situation
say we are required to find this n. which
is the force exerted by side walls. then for
this n horizontal component will get cancel
out and only component which will be exerting
force on liquid is n v, that is the , vertical
component of the force, applied by the side
walls. on the fluid, so for its calculation
we can directly state if we calculate, pressure.
at bottom. if h is the height up to which
the liquid is filled this pressure at bottom
can be simply written as h ro g which is due
to the weight of liquid. and if we find out
say we are also given with area of this bottom
which is say ay, so we can find out the force
acting. on the bottom of container . is, this
force can be directly written as h ro g multiplied
by the area. and this is basically the normal
reaction n-one which the bottom is exerting
on fluid. and in this situation. we can directly
state . the weight of this liquid is more
then this normal reaction as it is balanced
by the side walls also . sow we can directly
state in this situation if we draw the free
body diagram of fluid body. then it can be
written that for this fluid body its weight
m-g is acting in downward direction . if m
is the mass of fluid taken, in the bucket.
and in upward direction it’ll be experiencing
one is the n-one the normal reaction due to
the bottom. and other is n-v which is the
. vertical component of the force exerted
by side walls as the horizontal component
is already balanced. so in this situation
we can directly write for, iquilibrium of
. fluid body. here we can write n-one + n-v
is balancing the weight of, this whole liquid
. so this n-v which is actually the force
exerted by side walls on. the fluid body can
be written as m g minus n-one, of which the
value we already calculated as h ro g into
ay. which is the force exerted by fluid pressure
on bottom. so this is the way how we calculate
the force acting on the side wall . here the
sides walls are exerting force on the liquid
in upward direction. but actually it depends
on the shape of surface lets see, few more
situation in which we take , different kind
of vessels which are exerting, force by the
side walls in different directions also. let
us discuss few more situation like, say we
are given with a vessel , of such a shape
in which the top is circular . and the side
walls are given like this. and the bottom
is having a cross sectional area ay. in this
situation if it is filled with liquid. we
can see in this situation the liquid is, pushing
the side walls normal to the surface. and
side walls are exerting a force in the liquid
in opposite direction. so below the middle
height you can say the side walls will exert
, a force on the liquid in a direction, normal
to, the surface, and . always we can say the
horizontal components will be balanced out,
and vertical components will be left over.
but you can simply state , the side walls
which are above the middle height will be
pushing the fluid in downward direction. due
to the shape. and these are having vertical
component in downward direction . so in this
situation we can simply state if we draw the
free body diagram of the fluid. the fluid
weight m-g will act in downward direction
. due to the normal reaction . n-one. which
is acting at the bottom n-one will act in
upward direction. now in this situation in
the container, if weight is, more then the
normal reaction n-one then obviously in this
situation this n-v will act in upward direction
. but in some situation , will also take some
example . in some situation, if n-one is more
then the weight of the fluid which is, contain
in the vessel. in that case n-v will act in
downward direction for the iquilibrium. so
we can simply state, for iquilibrium. of fluid
will always right the similar situation n-one
plus n-v is equal to m-g. so n-v can be written
as m-g minus n-one and n-one can be calculated
by multiplying fluid pressure at the bottom
. with, the total area of the bottom. say
if we consider a situation in which we are
. given with a vessel like this. in which
a very small . opening is there at the top
of the vessel and the bottom is closed. by
a circular base. and say if in this vessel
a liquid is filled. here in this situation
we can say if the base area is ay. here we
can see the bottom will be exerting a normal
reaction in upward direction and, all the
side walls will be exerting the force on fluid
in inward direction like this which i am drawing.
and again the horizontal component of all
these forces will cancel and vertical component
will exist only in downward direction. so
if we draw the free body diagram of this fluid
body , m-g will act in downward direction
here n-one will act in upward direction , and
obviously here n-v will be acting in downward
direction because in this situation , the
normal reaction applied by the bottom on the
fluid body is more then the wait that’s
why for its iquilibrium in this situation
. n-one must be equal to m-g plus n-v. so
here, the force exerted by side walls on the
fluid which always in vertical direction and
here it is in vertically , downward direction
can be given as n-one minus m-g. and as usual
if, h is the height up to which the liquid
is filled n-one can be written as , h ro g
multiplied by ay. this is the way how we calculate
the force exerted by the, fluid on side walls
or force exerted by side walls . on the fluid
, it is vise versa in opposite direction.
