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Two parallel and opposite forces, each of magnitude \(4000 \mathrm{~N}\), are applied tangentially to the upper and lower faces of a cubical metal block \(25 \mathrm{~cm}\) on a side. Find the displacement of the upper surface relative to the lower surface (in \(\times 10^{-5} \mathrm{~cm}\) ). The shear modulus for the metal is \(80 \mathrm{GPa}\).

Short Answer

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Step by step solution

01

Understand the Problem Dimensions

The cube has sides of 25 cm, and forces are applied on the upper and lower faces. We need to find the shear displacement due to these forces.
02

Convert Units

The shear modulus is given in GPa. Convert this to N/m²:\[80 \text{ GPa} = 80 \times 10^9 \text{ N/m}^2\]Similarly, convert the cube's side from 25 cm to meters:\[25 \text{ cm} = 0.25 \text{ m}\]
03

Use the Shear Formula

The displacement due to shear can be determined using the formula:\[\text{Shear Displacement} = \frac{F}{A \cdot G}\]where:- \(F\) is the force applied \(= 4000 \text{ N}\)- \(A\) is the area of the face \(= 0.25 \times 0.25 \text{ m}^2\)- \(G\) is the shear modulus \(= 80 \times 10^9 \text{ N/m}^2\)
04

Calculate the Area of the Cube Face

The area \(A\) of one face of the cube is:\[A = (0.25 \text{ m}) \times (0.25 \text{ m}) = 0.0625 \text{ m}^2\]
05

Calculate the Displacement

Substituting the values into the shear formula:\[\text{Shear Displacement} = \frac{4000 \text{ N}}{0.0625 \text{ m}^2 \times 80 \times 10^9 \text{ N/m}^2}\]Calculating further, we get:\[\text{Shear Displacement} = 8 \times 10^{-9} \text{ m}\]
06

Convert Displacement to Required Units

Convert the displacement from meters to centimeters, given that:\[1 \text{ m} = 100 \text{ cm}\]Thus:\[8 \times 10^{-9} \text{ m} = 8 \times 10^{-7} \text{ cm}\]
07

Scale the Result

The final answer should be in units of \(\times 10^{-5} \text{ cm}\). Therefore, we divide by \(10^{-5} \):\[\frac{8 \times 10^{-7} \text{ cm}}{10^{-5}} = 0.08\]

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Key Concepts

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

Understanding Shear Stress
In the realm of physics, when forces are applied parallel to a surface, a phenomenon known as shear stress arises. Imagine sliding a deck of cards on a table where the top cards move slightly out of alignment with the ones below. This demonstrates shear stress. In mechanical terms, shear stress is defined as the force per unit area acting parallel to the plane of interest.

When calculating shear stress, equations such as \( \tau = \frac{F}{A} \) are employed, where \( \tau \) represents shear stress, \( F \) is the force applied, and \( A \) is the area over which the force acts.
Bulleted reminders when dealing with shear stress:
  • Measure force in Newtons (N).
  • Calculate the area in square meters (\( m^2 \)).
  • Ensure forces act parallel to the material surface.
Understanding shear stress helps predict how materials behave under such forces, like how metals bend or twist when subjected to tangential forces.
Exploring Mechanics and Its Principles
Mechanics is a branch of physics focusing on the behavior of physical bodies when subjected to forces. It plays a crucial role in analyzing motion and predicting how forces affect structures. In our given problem, we see mechanics at work as it involves calculating movement of a metal cube under the application of forces.

Fundamental principles of mechanics include Newton's laws of motion, energy conservation, and material deformation under loads. Key aspects:
  • Newton's Laws: Dictate how forces cause motion.
  • Energy Principles: Describe potential and kinetic energies in systems.
  • Material Deformation: Examines how bodies change shape in response to external loads.
Understanding these principles enables us to solve complex real-world engineering problems, like ensuring buildings withstand earthquakes.
Properties of Metals under Stress
Metals, known for their durability and flexibility, exhibit unique properties when subjected to stress. These properties determine their suitability for various applications in construction and manufacturing.

When forces act on metals, several key properties come into play:
  • Elasticity: Metals can undergo temporary shape change but return to original form when forces are removed.
  • Plasticity: Beyond a certain point, metals deform permanently and do not return to the original shape.
  • Ductility: This refers to the metal's ability to be stretched into a wire. Metals like copper and aluminum are highly ductile.
  • Toughness: Indicates the amount of energy a metal can absorb before breaking. Stainless steel, for example, is incredibly tough.
In the exercise, the shear modulus relates to a metal's rigidity or stiffness when a tangential force is applied. Recognizing these properties helps engineers to design safe, efficient structures that utilize metal's strengths effectively.

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Most popular questions from this chapter

A wire of length \(L\) and radius \(r\) is fixed at one end. When a stretching force \(F\) is applied at free end, the elongation in the wire is \(l\). When another wire of same material but of length \(2 L\) and radius \(2 r\), also fixed at one end is stretched by a force \(2 F\) applied at free end, then elongation in the second wire will be (1) \(l / 2\) (2) \(l\) (3) \(2 l\) (4) \(l / 4\)

Two wires of the same length and same material but radil in the ratio of \(1: 2\) are stretched by unequal forces to produce equal elongation. The ratio of the two forces is (1) \(1: 1\) (2) \(1: 2\) (3) \(1: 3\) (4) \(1: 4\)

Find the elastic potential energy per unit volume of water (in \(\left.\times 10^{3} \mathrm{Jm}^{-3}\right)\) at a depth of \(1 \mathrm{~km}\). Given, compressibility of water \(=5 \times 10^{10} \mathrm{Sl}\) units and density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

A ring of radius \(r\) (made of wire of density \(\rho\) ) placed on a smooth horizontal surface is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring as shown in figure. Determine the angular velocity in \(\mathrm{rad} / \mathrm{s}\) at which the ring breaks. The wire breaks at a tensile stress \(\sigma .\) (given \(\sigma / \rho=9 \mathrm{~N}-\mathrm{m} / \mathrm{kg}\) and \(r=1 \mathrm{~m}\) )

Match the following: | Column I | Column II | | :--- | :--- | | i. longitudinal stress | a. Volume changes | | ii. Shear stress | b. Shape changes | | iii. Volume changes | c. Volume does not change | | iv. Tensile stress | d. Shape does not change |

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