Question

for a given object the buoyant force in liquids of different is

Answer 1

(P_{gauge}=\rho gh)

(P

gauge

=ρgh) increases as you go deeper in a fluid, the force from pressure exerted downward on the top of the can of beans will be less than the force from pressure exerted upward on the bottom of the can.

Essentially it's that simple. The reason there's a buoyant force is because of the rather unavoidable fact that the bottom of an object is always deeper in a fluid than the top of the object. This means the upward force from water has to be greater than the downward force from water.

OK, so it doesn't completely follow. After all, what if we considered an object where the area of the bottom was smaller than the area of the top (like a cone). Since

F=PA

F=PA, could the greater area on the top of the cone compensate for the smaller pressure on the top? Would this make the object experience a net downward buoyant force? The answer is no. It turns out that no matter what shape you make your object, the net force from water pressure will always point upward.

We can start with the fact that the water on the top of the can is pushing down

F_{down}

F

down

, and the water on the bottom of the can is pushing up

F_{up}

F

up

. We can find the total upward force on the can exerted by water pressure (which we call the buoyant force

F_{buoyant}

F

buoyant) by simply taking the difference between the magnitudes of the upward force

F_{up}

F

up

and downward force

F_{down}

F

down

.

F_{buoyant} =F_{up} - F_{down}

F

buoyant

=F

up

−F

down

We can relate these forces to the pressure by using the definition of pressure

P=\dfrac{F}{A}

P=

A

F

which can be solved for force to get

F=PA

F=PA . So the force exerted upward on the bottom of the can will be

F_{up}=P_{bottom}A

F

up

=P

bottom

A and the force exerted downward on the top of the can will be

F_{down}=P_{top}A

F

down

=P

top

A. Substituting these expressions in for each

F

F respectively in the previous equation we get,

F_{buoyant} =P_{bottom} A - P_{top}A

F

buoyant

=P

bottom

A−P

top

A

We can use the formula for hydrostatic gauge pressure

P_{gauge}=\rho gh

P

gauge

=ρgh to find expressions for the upward and downward directed pressures. The force from pressure directed upward on the bottom of the can is

P_{bottom}=\rho gh_{bottom}

P

bottom

=ρgh

bottom

and the force from pressure directed downward on the top of the can is

P_{top}=\rho gh_{top}

P

top

=ρgh

top

. We can substitute these into the previous equation for each pressure respectively to get,

F_{buoyant} =(\rho gh_{bottom}) A - (\rho gh_{top})A

F

buoyant

=(ρgh

bottom

)A−(ρgh

top

)A

Notice that each term in this equation contains the expression

\rho g A

ρgA. So we can simplify this formula by pulling out a common factor of

\rho g A

ρgA to get,

F_{buoyant} =\rho gA(h_{bottom} -h_{top})

F

buoyant

=ρgA(h

bottom

−h

top

)

Now this term

h_{bottom} -h_{top}

h

bottom

−h

top

h_{bottom}

h

bottom

and the depth of the top of the can

h_{top}

h

top

is just equal to the height of the can. (see the diagram below)

So we can replace

(h_{bottom} -h_{top})

(h

bottom

−h

top

) in the previous formula with the height of the can

h_{can}

h

can

to get,

F_{buoyant} =\rho gAh_{can}

F

buoyant

=ρgAh

can

Here's the interesting part. Since

A \times h

A×h is equal to the volume of a cylinder, we can replace the term

Ah_{can}

Ah

can

with a volume

V
V_{fluid}

V

fluid

h_{bottom}

h

bottom

\Large F_{buoyant} =\rho gA(h_{bottom} -0)

F

buoyant

=ρgA(h

bottom

−0)

But the term

Ah_{bottom}

Ah

bottom

V_f

V

f

.

\Large F_{buoyant} =\rho gV_{fluid}

F

buoyant

=ρgV

fluid

V_{can}

V

can
\Large F_{buoyant} =\rho gV_{can}

F

buoyant

=ρgV

can

\Large F_{buoyant} =\rho gV_{f}

F

buoyant

=ρgV

f That pretty much does it.

\rho

ρ in which the object is submerged, the acceleration due to gravity

g

g, and the volume of the displaced fluid

V_f

V

f

F_{net} = F_b-W

F

net

=F

b

−W

We can use the formula we derived for buoyant force to rewrite

F_b

F

b

as

\rho_f V_f g

ρ

f

V

f

g where

\rho_f

ρ

f

is the density of the fluid,

F_{net} = \rho_fV_fg - mg

F

net

=ρ

f

V

f

g−mg

We can make that second term in the formula look a whole lot

m

m in terms of the density of the object

\rho_o

ρ

o

and the volume of the object submerged

V_o

V

o

m=\rho_oV_o

m=ρ

o

V

o

F_{net} = \rho_fV_fg - \rho_oV_og

F

net

=ρ

f

V

f

g−ρ

o

V

o

g

If the object is fully submerged the two volumes

V

V are the same and we can pull out a common factor of

Vg

Vg to get,

F_{net} = Vg(\rho_f- \rho_o)

F

net

=Vg(ρ

This is your answer of for a given object the buoyant force in liquids of different