A vertical off-shore structure is built to withstand a maximum stress of $10^{9}\ \text{Pa}$. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly $3\ \text{km}$, and ignore ocean currents.

> [!TLDR] > The key idea is to compare the hydrostatic pressure at the ocean floor (due to the water column above) with the structure’s maximum stress tolerance. The pressure at 3 km depth is about $3.03 \times 10^{7}\ \text{Pa}$, which is far less than $10^{9}\ \text{Pa}$. So the structure is **suitable**. --- ### 1. What is the structure actually experiencing? When you place a vertical structure on the ocean floor, the main force it must withstand is the **hydrostatic pressure** of the water column above it. This pressure acts equally from all sides at a given depth (Pascal’s principle), pushing inward on the structure. The structure’s material has a maximum stress it can bear — here given as $10^{9}\ \text{Pa}$ — which is essentially the pressure it can tolerate before failing. The question is: does the pressure at 3 km depth exceed this limit? If yes, the structure would be crushed. If no, it’s safe. > [!WARNING] > A common mistake is to think the structure must support the *weight* of the water above it like a column. But hydrostatic pressure is not the same as weight — it’s the force per unit area exerted by the fluid, and it acts omnidirectionally. The structure’s stress tolerance is directly comparable to this pressure. --- ### 2. Calculate the hydrostatic pressure at depth The pressure due to a fluid column of height $h$ is given by: $$ P = \rho g h $$ where: - $\rho$ = density of seawater ≈ $1030\ \text{kg/m}^3$ (slightly denser than fresh water) - $g$ = acceleration due to gravity ≈ $9.8\ \text{m/s}^2$ - $h$ = depth = $3\ \text{km} = 3000\ \text{m}$ Plug in the numbers: $$ P = (1030)(9.8)(3000) $$ First, $1030 \times 9.8 = 10094$. Then $10094 \times 3000 = 3.0282 \times 10^{7}\ \text{Pa}$. So: $$ P \approx 3.03 \times 10^{7}\ \text{Pa} $$ > [!TIP] > You can remember the rule of thumb: every 10 m of water adds roughly 1 atm of pressure (≈ $10^{5}\ \text{Pa}$). At 3000 m, that gives about $300 \times 10^{5} = 3 \times 10^{7}\ \text{Pa}$ — a

A vertical off-shore structure is built to withstand a maximum stress of 10⁹Pa. Is the structure suitable for

Exercises · 9.7

10^{9}\ \text{Pa}

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Concept understanding — Hydrostatic Pressure Balance

Hydrostatic Pressure Balance

Imagine you're standing at the bottom of a swimming pool. You feel pressure on your ears — and the deeper you go, the more intense that pressure becomes. Now think about a column of water above you: every kilogram of that water is being pulled down by gravity. That weight has to be supported by the water below it. The deeper you go, the more water is stacked above you, so the greater the weight pressing down.

That's the core intuition: pressure in a fluid at rest increases with depth because the fluid above has to be supported by the fluid below.

The Precise Statement

Hydrostatic pressure balance is the condition that holds for any fluid at rest in a uniform gravitational field. It says:

P_{\text{below}} = P_{\text{above}} + \rho g h

where:

$P_{\text{below}}$ is the pressure at a lower point,
$P_{\text{above}}$ is the pressure at a higher point,
$\rho$ is the density of the fluid (assumed constant),
$g$ is the acceleration due to gravity,
$h$ is the vertical depth between the two points.

Equivalently, the pressure gradient in the vertical direction is:

\frac{dP}{dz} = -\rho g

where $z$ increases upward. The minus sign tells you pressure decreases as you go up.

Why This Makes Sense

Take a thin horizontal slab of fluid of area $A$ , thickness $dz$ , at some depth. Its weight is $dW = \rho g A \, dz$ . For the slab to be in equilibrium (not accelerating), the net upward force from pressure must exactly balance this weight.

The upward force on the slab's bottom face is $P(z) A$ , and the downward force on its top face is $P(z+dz) A$ . The net upward force is:

P(z) A - P(z+dz) A = - \frac{dP}{dz} A \, dz

Setting this equal to the weight $\rho g A \, dz$ gives:

-\frac{dP}{dz} = \rho g

which is exactly the differential form above.

Note

This balance assumes the fluid is static — no flow, no acceleration. If the fluid moves, additional terms (like viscous forces or inertial effects) appear.

Key Implications

Pressure depends only on depth, not on the shape of the container. A tall thin tube and a wide shallow tank give the same pressure at the same depth — because only the vertical height of fluid above matters.
Pressure is the same at all points on the same horizontal level. If you move sideways at constant depth, $\rho g h$ doesn't change, so $P$ doesn't change.
Gases are compressible, so $\rho$ is not constant. For air, the density changes with pressure itself, leading to an exponential decrease — but the same principle applies locally.

A Common Mistake …

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The key idea is to compare the hydrostatic pressure at the ocean floor (due to the water column above) with the structure’s maximum stress tolerance. The pressure at 3 km depth is about $3.03 \times 10^{7}\ \text{Pa}$ , which is far less than $10^{9}\ \text{Pa}$ . So the structure is suitable.

1. What is the structure actually experiencing?

When you place a vertical structure on the ocean floor, the main force it must withstand is the hydrostatic pressure of the water column above it. This pressure acts equally from all sides at a given depth (Pascal’s principle), pushing inward on the structure. The structure’s material has a maximum stress it can bear — here given as $10^{9}\ \text{Pa}$ — which is essentially the pressure it can tolerate before failing.

The question is: does the pressure at 3 km depth exceed this limit? If yes, the structure would be crushed. If no, it’s safe.

Watch out

A common mistake is to think the structure must support the weight of the water above it like a column. But hydrostatic pressure is not the same as weight — it’s the force per unit area exerted by the fluid, and it acts omnidirectionally. The structure’s stress tolerance is directly comparable to this pressure.

2. Calculate the hydrostatic pressure at depth

The pressure due to a fluid column of height $h$ is given by:

P = \rho g h

where:

$\rho$ = density of seawater ≈ $1030\ \text{kg/m}^3$ (slightly denser than fresh water)
$g$ = acceleration due to gravity ≈ $9.8\ \text{m/s}^2$
$h$ = depth = $3\ \text{km} = 3000\ \text{m}$

Plug in the numbers:

P = (1030)(9.8)(3000)

First, $1030 \times 9.8 = 10094$ . Then $10094 \times 3000 = 3.0282 \times 10^{7}\ \text{Pa}$ .

So:

P \approx 3.03 \times 10^{7}\ \text{Pa}

Tip

You can remember the rule of thumb: every 10 m of water adds roughly 1 atm of pressure (≈ $10^{5}\ \text{Pa}$ ). At 3000 m, that gives about $300 \times 10^{5} = 3 \times 10^{7}\ \text{Pa}$ — a quick sanity check that matches our calculation.

3. Compare with the structure’s limit …

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Q.A vertical off-shore structure is built to withstand a maximum stress of 109 Pa10^{9}\ \text{Pa}109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km3\ \text{km}3 km, and ignore ocean currents.

1. What is the structure actually experiencing?

2. Calculate the hydrostatic pressure at depth

3. Compare with the structure’s limit …

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Q.A vertical off-shore structure is built to withstand a maximum stress of $10^{9}\ \text{Pa}$ . Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly $3\ \text{km}$ , and ignore ocean currents.

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