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Derive expressions for the lifting torques on the top and bottom of a slab descending into the mantle with speed \(U\) at a dip angle of \(60^{\circ}\).

Short Answer

Expert verified
The torque on the slab is \(\tau = \rho g AL \cos(60^{\circ})\).

Step by step solution

01

Understanding the Problem and Definitions

We need to derive expressions for the torques generated by the descending slab in the Earth's mantle. The slab is descending at a constant angle of \(60^{\circ}\), and its velocity is \(U\). Torque is determined by the force exerted and the distance from the pivot point. We'll assume the slab is influenced by gravitational forces and possibly viscous drag in the mantle.
02

Identify the Forces Acting on the Slab

The primary forces acting on the slab include gravitational force and possibly viscous force due to the mantle. The gravitational force can be expressed as \(F_g = \rho g A\), where \(\rho\) is the density of the slab, \(g\) is the acceleration due to gravity, and \(A\) is the cross-sectional area of the slab perpendicular to the velocity direction.
03

Calculate Gravitational Torque

The torque due to gravity on the slab can be calculated as \(\tau_g = F_g \times d\), where \(d\) is the perpendicular distance from the axis of rotation (pivot point) to the line of action of the force. Assuming this pivot is at the edge of the descending slab, and knowing the angle is \(60^{\circ}\), we can express \(\tau_g = \rho g A L \cos(60^{\circ})\), where \(L\) is the length of the slab.
04

Include the Effect of Mantle Viscosity

If we include viscous force, it can be represented as \(F_v = \eta \frac{dU}{dy}\), where \(\eta\) is the dynamic viscosity of the mantle and \(\frac{dU}{dy}\) is the velocity gradient perpendicular to the slab. The torque due to this viscous drag can be similarly calculated, but for simplicity, we are focusing mainly on gravitational force as the question implies.
05

Final Expression for Lifting Torque

The lifting torques on both the top and bottom of the slab due to gravitational force will be similar in expression but occur at different levels due to geometry. The torque is given by \(\tau = \rho g AL \cos(60^{\circ})\) on both the top and bottom, assuming symmetry regarding force distribution along the thickness.

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Key Concepts

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

Lifting Torques
When a slab descends into the Earth's mantle, it experiences torques, which are essentially rotational forces that make it want to turn. These forces are crucial because they determine how easily the slab can move and change direction as it sinks.
In this scenario, lifting torques are calculated at both the top and the bottom of the slab as it descends. The calculation involves understanding the forces acting on the slab due to gravity and possibly the resistance from the sticky, syrup-like mantle.
  • The lifting torque comes into play at the slab's edges since this is where the most prominent turning effect is felt.
  • The formula isn't too complicated; it's derived using the knowledge of forces and the geometry of the situation, such as the slab's length and the angle at which it descends.
Understanding lifting torques helps in analyzing plate tectonics and is fundamental for geophysics students exploring how slabs interact with their surroundings in the mantle.
Mantle Dynamics
Mantle dynamics is all about how the material beneath the Earth's crust moves. The mantle, located between the crust and the core, is a super thick, slow-moving layer of rock. It's not entirely solid or liquid, but a kind of in-between state that flows very slowly. The way the mantle moves hugely affects the movement of the Earth's tectonic plates.
  • The movement of these plates can lead to numerous geological phenomena like earthquakes, volcanic activity, and mountain formation.
  • These dynamics are driven by heat coming from the Earth's interior, causing the mantle to move in a convective cycle - hot material rises while cooler material sinks.
The slab's descent into the mantle, as described in geodynamic exercises, is part of these broader mantle dynamics processes, leading to the generation of torques that need to be calculated.
Gravitational Force
Gravitational force plays a critical role in the descent of slabs into the mantle. It is the force that attracts two bodies towards each other, in this case, the slab and Earth attract each other due to gravity. This force influences how the slab descends into the mantle, adding to the complexity of the movement.
  • In the calculation of torques, gravitational force is expressed using the equation: \( F_g = \rho g A \).
  • Here, \( \rho \) is the density of the slab, \( g \) is the acceleration due to gravity, and \( A \) is the slab's cross-sectional area perpendicular to the velocity.
Understanding how gravitational force works in these equations is crucial to correctly calculating the torques on descending slabs, contributing to the knowledge of plate movements.
Viscosity
Viscosity describes how thick and sticky a liquid is, which determines how easily it flows. When considering the Earth's mantle, viscosity describes how resistant the mantle material is to flow. This resistance can significantly affect how a descending slab moves through the mantle.
  • In geodynamics, viscosity is often factored into equations as the mantle provides resistance similar to how syrup would slow down an object falling into it.
  • It can be calculated using the formula: \( F_v = \eta \frac{dU}{dy} \), where \( \eta \) is the dynamic viscosity, \( U \) is the velocity, and \( y \) is the direction perpendicular to the slab's movement.
By understanding viscosity, students can grasp how the mantle's resistance impacts plate movements, which is vital for predicting geological activities and understanding tectonic processes.

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Most popular questions from this chapter

Suppose that the \(660-\mathrm{km}\) density discontinuity in the mantle corresponds to a compositional change with lighter rocks lying above more dense ones. Estimate the minimum decay time for a disturbance to this boundary. Assume \(\rho=\) \(4000 \mathrm{~kg} \mathrm{~m}^{-3}, \Delta \rho=100 \mathrm{~kg} \mathrm{~m}^{-3},\) and \(\mu=10^{21} \mathrm{~Pa} \mathrm{~s}\).

Suppose that convection extends through the entire mantle and that \(10 \%\) of the mean surface heat flow originates in the core. If the surface thermal boundary layer and the boundary layer at the core-mantle interface have equal thicknesses, how does the temperature rise across the lower mantle boundary layer compare with the temperature increase across the surface thermal boundary layer?

In the examples of folding just considered we assumed that the competent rock adhered to the incompetent rock. If the layers are free to slip, show that the wavelength of the most rapidly growing disturbance in an elastic layer of rock contained between two semi-infinite viscous fluids is given by $$\lambda=\pi h\left[E / \sigma\left(1-v^{2}\right)\right]^{1 / 2}$$ The free slip condition is equivalent to a zero shear stress condition at the boundaries of the elastic layer.

Show that the constant of integration \(A\) in the above postglacial rebound solution is given by $$A=-\left(\frac{\lambda}{2 \pi}\right)^{2} \frac{\rho g W_{m 0}}{2 \mu} e^{-t / \tau_{r}} . \quad(6-106)$$ Quantitative information on the rate of postglacial rebound can be obtained from elevated beach terraces. Wave action over a period of time erodes a beach to sea level. If sea level drops or if the land surface is elevated, a fossil beach terrace is created, as shown in Figure \(6-15 .\) The age of a fossil beach can be obtained by radioactive dating using carbon 14 in shells and driftwood. The elevations of a series of dated beach ter- races at the mouth of the Angerman River in Sweden are given in Figure \(6-16 .\) The elevations of these beach terraces are attributed to the postglacial rebound of Scandinavia since the melting of the ice sheet. The elevations have been corrected for changes in sea level. The uplift of the beach terraces is compared with the exponential time dependence given in Equation \((6-104)\). We assume that uplift began 10,000 years ago so that \(t\) is measured forward from that time to the present.

For an asthenosphere with a viscosity \(\mu=4 \times 10^{19} \mathrm{~Pa} \mathrm{~s}\) and a thickness \(h=200 \mathrm{~km},\) what is the shear stress on the base of the lithosphere if there is no counterflow \((\partial p / \partial x=0) ?\) Assume \(u_{0}=\) \(50 \mathrm{~mm} \mathrm{yr}^{-1}\) and that the base of the asthenosphere has zero velocity.

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