This problem works to familiarize us with the
weight of
objects underwater. The concept required to solve this problem includes the
force density formula and the
Buoyant force. Buoyancy, more commonly known as
upthrust, is an
upward force exercised by a
fluid that resists the
weight of a
partly or
fully submerged object.
In simple words, it is the
ability of an
object to float in a
fluid either wholly or
partly immersed. The
formula to calculate
buoyancy force is:
\[ F_b = – \rho g V \]
Where $F_b$ is the
buoyancy force,
$\rho$ is the
fluid density,
$g$ is the
acceleration due to
gravity, and
$V$ is the
volume of the fluid.
Expert Answer
We will start by
calculating the
force that the
iceberg exerts on the fluid in the form of its
weight. Using the
buoyancy force formula:
\[ F_w = S_{g_{ice}}\times \rho_{water}\times V \]
Where $F_w$ is the
force due to the
weight of the
iceberg, $S_{g_{ice}}$ is the
specific gravity of the iceberg = $0.917$, $\rho_{water}$ is the
density of
water = $62.4 lb/ft^3$, and $V$ is the
volume of the
water.
Substituting the values in the above
equation gives us:
\[ F_w = 0.917\times 62.4\times V \]
\[ F_w =57.22V\space lb \]
This is the
force that the
iceberg exerts on the water. Now to find the
buoyancy force that
acts upwards and cancels out the
weight of the iceberg:
\[ F_b = S_{g_{sea water}}\times \rho_{water}\times V’ \]
Here, $S_{g_{sea water}}$ is the
specific gravity of the
seawater= $1.025$, $V’$ is the
volume of the
immersed portion of the iceberg and everything else
remains the same.
Substituting the values in the above equation gives us:
\[ F_b = 1.025\times 62.4\times V’ \]
\[ F_b = 63.96V’\space lb \]
Now, we are
left with $2$ unknowns here, $V$ and $V’$, but we are to find the
percent volume of the iceberg that is
immersed in water. To find the
volume, we will equate the two
forces to relate the
formula for the
percentage:
\[F_b=F_w \]
\[57.22V\space lb=63.96V’\space lb\]
\[\dfrac{V’}{V} = \dfrac{57.22}{63.96}\]
\[= 0.895 \]
Since the
ratio is $0.895$, the percentage of the
immersed volume to the total volume is:
\[\bigtriangleup V’ (\%) = 89.5\% \]
Numerical Result
$89.5\%$ of the
volume of the
iceberg is
underwater.
Example
An iceberg is
sailing somewhat immersed in seawater. The
density of seawater is $1.03 g/cm^3$ and that of
solid ice is $0.92 g/cm^3$. Find the
percentage of the total
volume of the iceberg
above the seawater.
Let’s assume $V$ to be the
volume of the
iceberg, and $x$ to be the
volume of the
immersed portion of the
iceberg.
Then with the help of
Archimedes’ principle, we can have,
\[\text{iceberg weight}=\text{weight of the water displaced (weight of x volume of water)}\]
\[\implies V\times 0.92\times g=x\times 1.03\times g\]
\[\dfrac{x}{V} = \dfrac{92}{103}\]
The
percent volume of the
iceberg in
seawater is =$\dfrac{92}{103}\times 100=89.32\%$.
Hence, the
percentage of the
total volume of the iceberg above the
seawater is:
\[=100−89.32=10.68\%\approx 11\% \]
