Have you ever wondered why divers need to be careful when going deep underwater, or when resurfacing? It is because the deeper you go, the higher the pressure the water exerts on the diver's body. To avoid the bends, divers stop at different depths to let the pressure of the water equalize with the pressure in their body. Read on for an introduction to the basics of hydrostatic pressure.
Explore our app and discover over 50 million learning materials for free.
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Jetzt kostenlos anmeldenNie wieder prokastinieren mit unseren Lernerinnerungen.
Jetzt kostenlos anmeldenHave you ever wondered why divers need to be careful when going deep underwater, or when resurfacing? It is because the deeper you go, the higher the pressure the water exerts on the diver's body. To avoid the bends, divers stop at different depths to let the pressure of the water equalize with the pressure in their body. Read on for an introduction to the basics of hydrostatic pressure.
So what is hydrostatic pressure?
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity.
Hydrostatic pressure varies depending on the fluid and the depth.
You will need some variables going forward:
\(P\) is the hydrostatic pressure in \(\text{N}/\text{m}^2\);
\(\rho\) is the density of the fluid in \(\text{kg}/\text{m}^3\);
\(g\) is the acceleration due to gravity in \(\text{m}/\text{s}^2\); and
\(h\) is the height of the water column in \(\text{m}\).
Then the formula for hydrostatic pressure is
\[P = \rho gh.\]
Notice that pressure increases as the height of the water column (also called the depth) increases. That fact is illustrated in the image below.
The density of water is \(1000\, \text{kg}/\text{m}^3\), and for gravity on Earth you can use \(9.81 \, \text{m}/\text{s}^2 \).
You are probably wondering what the difference between hydrostatic force and pressure is.
Hydrostatic force is the force caused by the pressure of a fluid acting on a submerged surface.
If you think about the example of the diver at the start of the article, the hydrostatic force on the diver is the force of the water on their submerged body. As you have seen, pressure increases as the height of the water column increases. In other words, the deeper the diver goes, the higher the pressure is, which in turn means the hydrostatic force acting on their body increases as well.
The hydrostatic force formula is
\[F = PA,\]
where:
\(F\) is the hydrostatic force in \(\text{N}\);
\(P\) is they hydrostatic pressure in \(\text{N}/\text{m}^2\); and
\(A\) is the area of the submerged surface in \(\text{m}^2\).
Notice that the formula works for very complicated surfaces, like that of a diver's body. However figuring out the exact hydrostatic force on a diver is a bit of a big problem to start with, so instead let's look at submerging a flat plate in a fluid instead.
In the picture above, there is a flat hexagonal plate submerged \(d\) units under the water. The rectangle shown has height \(\Delta y\) and width \(x_i\), so the area of the rectangle is \(x_i\Delta y\). That means the hydrostatic force on the rectangle is
\[ \begin{align} F_{\text{rectangle}} &= PA \\ &= (\rho gh)(x_i\Delta y ) \\ &= (1000\cdot 9.81 d_i)(x_i\Delta y) \\ &= 9810d_ix_i\Delta y \, \text{N}. \end{align}\]
An approximation of the hydrostatic force on the entire plate would be the sum of the hydrostatic forces on all of the rectangles, so if there are \(n\) rectangles,
\[ F_{\text{plate}} \approx \sum\limits_{i=i}^n 9810d_ix_i\Delta y .\]
This should look familiar to you since finding areas this way is one of the applications of integrals!
Let's continue on with the plate example started in the previous section. The first step is to make a properly labeled diagram with the \(x\) and \(y\)-axis included. This plate is vertically symmetric, so it makes sense to center it on the \(y\)-axis with the bottom of the plate on the \(x\) axis.
The top of the plate is \(6\) meters below the surface of the water. The plate can be thought of in two sections: the upper section where \(y \ge 2\); and the lower section where \(y \le 2\).
Then the total area of the plate is the sum of the upper section area and the lower section area. Therefore when integrating you can split the integral at \(y=2\) to make calculations easier. In fact, since the area is symmetric across the \(y\) axis, you can find the area of the right half of the plate and then multiply by \(2\) to get the total area of the plate.
For the upper part of the plate, the depth below the surface is \(d_i\). The equation of the line describing the edge of the plate is \(y = -0.8x + 8\).
The summation for the hydrostatic force on the upper part of the plate is
\[ F_{\text{upper}} \approx \sum\limits_{i=i}^n 9810d_ix_i\Delta y ,\]
which has a \(\Delta y\) in it. This means the integration will be done with respect to \(y\) and not with respect to \(x\). Therefore solving the equation of the line for \(x\) in terms of \(y\) gives
\[x = -1.25y+10.\]
You will need to be careful since writing it this way only gives you the right half of the plate! Since the plate is symmetric, to find the area from the left side of the plate to the right side of the plate, you can multiply the summation by \(2\). Substituting the line equation into the summation and multiplying by \(2\) gives
\[ F_{\text{upper}} \approx 2\sum\limits_{i=i}^n 9810d_i(-1.25y_i+10)\Delta y .\]
That just leaves writing \(d_i\) in terms of \(y_i\). As you can see from the picture above,
\[y_i = 12-d_i,\]
and now the summation becomes
\[ F_{\text{upper}} \approx 2\sum\limits_{i=i}^n 9810(12-y_i)(-1.25y_i+10)\Delta y .\]
Then the total force on the upper half of the plate is
\[\begin{align} F_{\text{upper}} &= 2\lim\limits_{n\to \infty} \sum\limits_{i=1}^n 9810(12-y_i)(-1.25y_i+10) \Delta y \\ &= 2\int_2^6 9810(12-y)(-1.25y+10) \, \mathrm{d}y \\ &= \left. 2(9810)\left(\frac{5y^3-150y^2+1440y}{12}\right)\right|_{y=2}^{y=6} \\ &= 2(9810)\left(\frac{500}{3}\right) \\ &= 3270000 \, \text{N}.\end{align} \]
Let's look at the lower plate. The diagram below shows the equation of the line for that part of the plate.
The equation of the line is \(y=0.4x-1\). As in the calculations for the upper half of the plate, you need to write this equation in terms of \(x\) so it can be substituted in. Rewriting the equation of the line in terms of \(x\) you get \(x = 2.5y+2.5\). Since \(d_i = 12-y_i\), the hydrostatic force on a horizontal rectangular strip running from the \(y\)-axis to the line is
\[ \begin{align} F_i &= P_iA_i \\ &= 9810d_i(2.5y_i+2.5)\Delta y \\ &= 9810(12-y_i)(2.5y_i+2.5 ) \Delta y. \end{align}\]
The total force on the upper half of the plate is then
\[\begin{align} F_{\text{lower}} &= 2\lim\limits_{n\to \infty} \sum\limits_{i=1}^n 9810(12-y_i)(2.5y_i+2.5 ) \Delta y \\ &= 2(2.5)(9810)\int_0^2 (12-y)(y+1 ) \, \mathrm{d}y \\ &= \left. (49050)\frac{-2y^3+33y^2+72y}{6}\right|_{y=0}^{y=2} \\ &=(49050)\left(\frac{130}{3}\right) \\ &= 2125500\, \text{N}.\end{align} \]
As you would expect, the hydrostatic force on the lower part of the plate is smaller than that on the upper part of the plate because the area on the lower plate is smaller than the area on the upper part of the plate.
Then to get the total hydrostatic force on the plate, you just need to add the hydrostatic force on the upper part of the plate to the hydrostatic force on the lower part of the plate:
\[ \begin{align} F_{\text{total}} &= F_{\text{upper}} + F_{\text{lower}} \\ &= 3270000 + 2125500 \\ &= 5395500 \, \text{N}.\end{align} \]
As you would expect, the hydrostatic force on the lower part of the plate is smaller than that on the upper part of the plate because the area on the lower plate is smaller than the area on the upper part of the plate.
\[P = \rho gh.\]where:
\(P\) is the hydrostatic pressure in \(\text{N}/\text{m}^2\);
\(\rho\) is the density of the fluid in \(\text{kg}/\text{m}^3\);
\(g\) is the acceleration due to gravity in \(\text{m}/\text{s}^2\); and
\(h\) is the height of the water column in \(\text{m}\).
Hydrostatic force is the force caused by the pressure of a fluid acting on a submerged surface.
The hydrostatic force formula is
\[F = PA,\]
where:
\(F\) is the hydrostatic force in \(\text{N}\);
\(P\) is they hydrostatic pressure in \(\text{N}/\text{m}^2\); and
\(A\) is the area of the submerged surface in \(\text{m}^2\).
Hydrostatic pressure is the pressure exerted by a column of fluid.
Gravity.
It depends on the depth in the fluid the object is submerged and the shape of the object. But for any point in a water column, to calculate the pressure we use:
P=ρgh
Here h is the depth, g is the gravity and is the density ρ of the fluid.
P=ρgd, where P is pressure, ρ is the density of the fluid, g is gravitational acceleration, and d or h is the depth below the surface.
The hydrostatic force that acts on an area is F=PA, where F is force, P is hydrostatic pressure, and A is the area of the object.
Usually with a gauge submerged in the fluid which returns the force per unit area.
What is the definition of hydrostatic force?
Hydrostatic force is the force caused by the pressure of a fluid acting on a submerged surface.
What is the definition of hydrostatic pressure?
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity.
Which of the following would increase the hydrostatic force?
Increasing the area of the submerged object.
Which of the following would increase the hydrostatic pressure?
Having a higher gravity.
In the formula for hydrostatic pressure \(P=\rho gh\) what do the variables stand for?
\(P\) is the hydrostatic pressure in \(\text{N}/\text{m}^2\);
\(\rho\) is the density of the fluid in \(\text{kg}/\text{m}^3\);
\(g\) is the acceleration due to gravity in \(\text{m}/\text{s}^2\); and
\(h\) is the height of the water column in \(\text{m}\).
In the hydrostatic force formula \(F=PA\) what do the variables stand for?
\(F\) is the hydrostatic force in \(\text{N}\);
\(P\) is they hydrostatic pressure in \(\text{N}/\text{m}^2\); and
\(A\) is the area of the submerged surface in \(\text{m}^2\).
Already have an account? Log in
Open in AppThe first learning app that truly has everything you need to ace your exams in one place
Sign up to highlight and take notes. It’s 100% free.
Save explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of StudySmarter.
Already have an account? Log in
Already have an account? Log in
The first learning app that truly has everything you need to ace your exams in one place
Already have an account? Log in