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An average thickness of the oceanic crust is \(6 \mathrm{~km}\). Its density is \(2900 \mathrm{~kg} \mathrm{~m}^{-3}\). This is overlain by \(5 \mathrm{~km}\) of water \(\left(\rho_{w}=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\) in a typical ocean basin. Determine the normal force per unit area on a horizontal plane at the base of the oceanic crust due to the weight of the crust and the overlying water.

Short Answer

Expert verified
The normal force per unit area at the base of the oceanic crust is \(219,940,000 \, N/m^2\).

Step by step solution

01

Calculate the Weight of the Oceanic Crust

The weight of the oceanic crust per unit area can be calculated using the formula: \( W_{crust} = ext{thickness} \times ext{density of crust} \times g \) where \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \).\[ W_{crust} = 6 \, km \times 2900 \, kg/m^3 \times 9.8 \, m/s^2 \]Note that 6 km should be converted to meters: \( 6 \, km = 6000 \, m \).\[ W_{crust} = 6000 \, m \times 2900 \, kg/m^3 \times 9.8 \, m/s^2 \]After calculating, \( W_{crust} \approx 170,940,000 \, N/m^2 \).
02

Calculate the Weight of the Overlying Water

The weight of the overlying water per unit area can be calculated using the same type of formula: \( W_{water} = ext{thickness} \times ext{density of water} \times g \).\[ W_{water} = 5 \, km \times 1000 \, kg/m^3 \times 9.8 \, m/s^2 \]Convert 5 km to meters: \( 5 \, km = 5000 \, m \).\[ W_{water} = 5000 \, m \times 1000 \, kg/m^3 \times 9.8 \, m/s^2 \]After calculating, \( W_{water} \approx 49,000,000 \, N/m^2 \).
03

Sum the Forces to Find the Total Normal Force per Unit Area

The total normal force per unit area at the base of the oceanic crust is the sum of the forces due to the oceanic crust and the water above it.\[ W_{total} = W_{crust} + W_{water} \]\[ W_{total} = 170,940,000 \, N/m^2 + 49,000,000 \, N/m^2 \]\[ W_{total} = 219,940,000 \, N/m^2 \].
04

Conclusion

The normal force per unit area on a horizontal plane at the base of the oceanic crust is \( 219,940,000 \, N/m^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oceanic Crust
The oceanic crust is the outermost layer of the Earth's lithosphere that is found underneath the ocean basins. It is distinct in composition and thickness when compared to the continental crust. Typically, the oceanic crust is about 6 kilometers thick, composed mainly of basalt, a dense volcanic rock.
Some key characteristics of oceanic crust include:
  • Composed primarily of basalt.
  • Thin but dense, compared to continental crust.
  • Younger than continental crust, often being less than 200 million years old.
  • Constantly being created and destroyed through processes like seafloor spreading and subduction.
The thickness and density make oceanic crust significant in calculations of geological forces, such as determining the pressure it exerts at its base.
Density Calculations
Density is a measure of mass per unit volume and is essential in calculations of force and pressure, especially in geophysics and oceanography. In simple terms, density determines how much matter is packed into a space. For the oceanic crust, which has a density of 2900 kg/m³, this means it is quite dense, leading to a substantial weight per unit area when combined with its thickness.
When performing density calculations:
  • Use the formula: \( \text{density} = \frac{\text{mass}}{\text{volume}} \).
  • In geophysical terms, density is often used along with the thickness of layers to calculate weights.
  • It's crucial to convert units correctly; for instance, converting kilometers to meters.
Using density, one can determine the weight of both the oceanic crust and the water above it, as seen in the problem where these values are crucial for calculating the force at the base of the crust.
Normal Force
In physics, a normal force is a force that is exerted by a surface to support the weight of an object resting on it and acts perpendicular to the surface. When analyzing geological formations like the oceanic crust, understanding the normal force helps in comprehending the stresses that are present at the base of the crust due to the above layers' weights.
The normal force calculation involves:
  • Calculating the individual weights of both the crust and the water resting atop it.
  • Summing these forces to find the total normal force acting per unit area.
  • Applying the concept of pressure, since the normal force here reflects pressure exerted by two layers.
For the oceanic crust problem, the calculated force (219,940,000 N/m²) represents the total pressure exerted on a horizontal plane at the crust’s base.
Earth's Gravity
Gravity, a fundamental force of nature, has a profound impact on all calculations involving mass and force. It is the force that attracts two bodies towards each other, with everything being pulled towards the center of the Earth by this force. In geophysics, the standard value used for Earth's gravity is approximately 9.8 m/s².
Key roles of Earth's gravity include:
  • Affecting the weight of objects (Weight = mass × gravity).
  • Influencing oceanic and atmospheric phenomena.
  • Being a constant that helps simplify calculations across different regions.
In the given problem, Earth's gravity acts on both the oceanic crust and the overlying water, contributing to the overall force exerted at the base of the crust. Calculating this force accurately involves using the constant gravitational value for determining both water and crust weights.

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Most popular questions from this chapter

Show that the principal strains are the minimum and the maximum fractional changes in length.

A mountain range has an elevation of \(5 \mathrm{~km}\). Assuming that \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \rho_{c}=2800 \mathrm{~kg}\) \(\mathrm{m}^{-3}\), and that the reference or normal continental crust has a thickness of \(35 \mathrm{~km}\), determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable.

Consider a block of rock with a height of \(1 \mathrm{~m}\) and horizontal dimensions of \(2 \mathrm{~m}\). The density of the rock is \(2750 \mathrm{~kg} \mathrm{~m}^{-3}\). If the coefficient of friction is \(0.8,\) what force is required to push the rock on a horizontal surface?

A sedimentary basin has a thickness of \(7 \mathrm{~km}\). Assuming that the crustal stretching model is applicable and that \(h_{c c}=35 \mathrm{~km}, \rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}\), \(\rho_{c c}=2700 \mathrm{~kg} \mathrm{~m}^{-3},\) and \(\rho_{s}=2450 \mathrm{~kg} \mathrm{~m}^{3},\) determine the stretching factor.

Consider a simple two-layer model of a planet consisting of a core of density \(\rho_{c}\) and radius \(b\) surrounded by a mantle of density \(\rho_{m}\) and thickness \(a-b\). Show that the gravitational acceleration as a function of radius is given by $$ \begin{aligned} g(r) &=\frac{4}{3} \pi \rho_{c} G r \quad 0 \leq r \leq b \\ &=\frac{4}{3} \pi G\left[r \rho_{m}+b^{3}\left(\rho_{c}-\rho_{m}\right) / r^{2}\right] \quad b \leq r \leq a \end{aligned} $$

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