The density of seawater is equal to 1.025 g/cm3, Find:
a) The pressure exerted only by the water column at a point 50 m deep;
b) The pressure at this point, taking into account atmospheric pressure, which at sea level is
1.013×105 Pa.
Problem data:
- Density of seawater: ρ = 1.025 g/cm3;
- Acceleration due to gravity: g = 9.8 m/s2.
Problem diagram:
Solution:
First, we must convert the water density given in grams per cubic centimeter (g/cm3) to kilograms per
cubic meter (kg/m3), used in the International System of Units (SI)
\[
\begin{align}
\rho= & 1.025\;\mathrm{\frac{\cancel g}{cm^3}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel g}}\times\frac{(100\;\mathrm{cm})^3}{(1\;\mathrm m\;)^{3}}=\\
= & 1.025\;\mathrm{\frac{1}{\cancel{cm^3}}}\times\frac{\mathrm{kg}}{1\cancel{000}}\times\frac{1 000 \cancel{000}\;\mathrm{\cancel{cm^3}}}{\mathrm m^3}=\\
= & 1025\;\mathrm{kg/m^3}
\end{align}
\]
a) The pressure of the liquid column, pc, is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{p_{c}=\rho gh}
\end{gather}
\]
\[
\begin{gather}
p_c=\left(1025\;\mathrm{\frac{kg}{m^{\cancel 3}}}\right)\times\left(9.8\;\mathrm{\frac{\cancel m}{s^2}}\right)\times\left(50\;\mathrm{\cancel m}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{p_c=5.023\times10^5\;\mathrm{Pa}}
\end{gather}
\]
b) The total pressure is given by
\[
\begin{gather}
p=p_0+p_c
\end{gather}
\]
where p0 is atmospheric pressure at sea level
\[
\begin{gather}
p=1.013\times 10^5\;\mathrm{Pa}+5.023\times 10^5\;\mathrm{Pa}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{p=6.036\times 10^5\;\mathrm{Pa}}
\end{gather}
\]
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