In this lesson, we will derive Archimedes’
principle, which allows us to calculate the
buoyant force on an object.
On the left we have a large lake.
Many objects will float on the lake’s surface
including a ship made of steel and an iceberg.
If we placed a balloon filled with air underwater,
it would rise to the surface unless anchored
to the lake bed.
Some types of rocks, such as pumice, would
float, but most rocks would sink to the bottom
of the lake.
For those rocks that do not float, if you
tried to lift the rock underwater you would
notice that it would be easier to lift than
if the rock were on dry land.
Some force must be present that is keeping
the steel ship and iceberg afloat, pushing
the air-filled balloon to the surface, and
making it easier to lift the rock.
If we drain the lake, the rock, ship, and
iceberg would sit on the bottom of the lake
bed.
The balloon may rise of may sink depending
on the gas contained inside.
We want to determine the force that is causing
the objects to rise and float.
We start by examining an object of arbitrary
shape that is at rest in a fluid with specific
weight gamma f.
The fluid pressure exerts a force on the entire
surface of the object.
This force acts normal to the surface and
increases with depth.
Let’s examine the pressure force acting
on a very thin horizontal sliver of the object.
The sliver has a width w and height dz.
The force on the left side of the sliver is
equal to the pressure on the left side, Pleft,
times the small area on the left side, dAleft.
The force on the right side of the sliver
is equal to the pressure on the right side,
Pright, times the small area on the right
side, dAright.
The force vectors are orientated an angle
theta relative to the horizontal.
In general, the orientation angles may be
different for each side, which means the area
on the left side may be different from the
area on the right side.
We will call the orientation angle on the
left theta-left, and the orientation angle
on the right we will call theta-right.
Multiplying the pressure force by cos(theta)
gives the component of the pressure force
in the y-direction.
In order to find the resultant pressure force
in the y-direction, we add up the net pressure
force in the y-direction acting on each sliver
by integrating over the entire body’s surface.
The net pressure force in the y-direction
acting on each sliver is the pressure force
on the left side minus the pressure force
from the right side.
Notice that both terms contain the quantity
dA times cos(theta), which is the area projected
along the z-axis.
Since the vertical height of both sides is
the same, the projected area is the same as
well
That is, dA cos(theta) on the left side is
the same as dA cos(theta) on the right side.
For convenience, we will call this area dAproj.
Let’s redraw the sliver with an equivalent
picture containing the projected area.
dFnet becomes the pressure on the left side
times the projected area, minus the pressure
on the right side times the projected area.
We can factor out dAproj, then plug in the
expression for dFnet into the integral.
The pressure on the left side is equal to
the pressure on right side because they are
at the same depth.
This means the net pressure force in the horizontal
direction is zero.
In other words, fluid pressure forces do not
push an object sideways in a static fluid.
Now we examine the fluid pressure force acting
on a very thin vertical sliver of the object.
The sliver has a width dy and height L.
The pressure force on the top of the sliver
is equal to the pressure on the top, Ptop,
times the small area on the top, dAtop.
The pressure force on the bottom side of the
sliver is equal to the pressure on the bottom,
Pbot, times the small area on the bottom,
dAbot.
The force vectors act normal to the surface
and are orientated at an angle theta relative
to the horizontal.
In general the orientation angles may be different
for each side, which means the area on the
top may be different from the area on the
bottom.
We will call the orientation angle on the
top theta-top, and the orientation angle on
the bottom will be called theta-bottom.
Multiplying by sin(theta) gives the component
of the pressure force in the z-direction.
In order to find the resultant pressure force
in the z-direction, we add up the net pressure
force in the z-direction at each sliver by
integrating over the entire body’s surface.
The net pressure force in the z-direction
acting on each sliver is the pressure force
from the bottom minus the pressure force from
the top.
Notice that both terms contain the quantity
dA times sin(theta), which is the area projected
along the y-axis.
Since the width of each side is the same,
the projected area of both sides is the same
as well.
That is, dA sin(theta) on the top is the same
as dA sin(theta) on the bottom.
For convenience, we will call this area dAproj.
Let’s redraw the sliver with an equivalent
picture containing the projected area.
dFnet becomes the pressure at the bottom times
dAproj, minus the pressure at the top times
dAproj.
We can factor out dAproj, then plug in the
expression for dFnet into the integral.
The pressure at the top of the sliver is the
specific weight of the fluid times the depth
htop.
The pressure at the bottom of the sliver is
the specific weight of the fluid times the
depth hbot.
Plug in the expression for the pressures,
and pull the specific weight of the fluid
out of the integral because it is constant.
The difference in depth between the top and
bottom of the sliver is the length l, and
l dAproj is the volume of the small sliver,
dV.
Integrating dV over the entire body gives
the volume of the object, which we will call
VO.
We now have an expression for the resultant
force.
This force, which is caused by the fluid pressure
gradient, is called the buoyant force and
is denoted as FB.
The buoyant force points upward.
The magnitude of this force is the specific
weight of the fluid gamma-f times the volume
of object immersed in the fluid VO.
VO also could be thought of as the volume
of fluid that is displaced by the object’s
presence.
That means the buoyant force is also equal
to the weight of the displaced fluid.
This result is often called Archimedes’
principle.
The buoyant force acts through the center
of buoyancy, which is the centroid of the
displaced volume of fluid.
On the right there is a rectangular block
that is completely submerged in a liquid of
specific weight gamma2.
A different fluid with a lower specific weight
gamma1 floats on top of the liquid.
This light fluid on top could be a gas or
a different liquid.
The buoyant force acts at the center of the
block and has a magnitude of gamma2 times
the total volume of the block.
Many objects can float at the surface of a
liquid.
On the left we have a different block that
is resting at the interface of fluids 1 and
2.
That means the block is partially submerged
in fluid1 and partially submerged in fluid2.
When an object floats at the interface of
two fluids, both fluids contribute the buoyant
force.
A buoyant force acts at the centroid of the
bottom section of the block, which is submerged
in fluid2.
This force is equal to gamma2 times the volume
of the bottom section of the block.
A buoyant force also acts at the centroid
of the top section of the block, which is
submerged in fluid1.
This force is equal to gamma1 times the volume
of the top section of the block.
If the specific weight of fluid1 is much less
than the specific weight of fluid2 – for
example, if fluid1 were air and fluid2 were
water – then we can usually neglect the
buoyant force on the top section of the object.
