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A tangential force equal to the weight of \(500 \mathrm{~kg}\) is applied to the opposite face of a cube of alminium of side \(10 \mathrm{~cm}\) has its one face fixed. Find the shearing strain produced Rigidity modulus of aluminium is \(2.5 \times 10^{10} \mathrm{Nm}^{-2}\).

Short Answer

Expert verified
The shearing strain produced is approximately \(1.96 \times 10^{-5}\).

Step by step solution

01

Determine the Force

The weight of an object is calculated using the formula: \[ W = m imes g \] where \( m = 500 \) kg and \( g = 9.8 \text{ m/s}^2 \). Thus, the force, \( F \), is:\[ F = 500 \times 9.8 = 4900 \text{ N} \]
02

Calculate the Area of One Face of the Cube

The side length of the cube is given as \( 10 \text{ cm} \) or \( 0.1 \text{ m} \). The area \( A \) of one face of the cube is:\[ A = ext{side}^2 = (0.1)^2 = 0.01 \text{ m}^2 \]
03

Compute the Shear Stress

Shear stress \( \tau \) is the force applied per unit area. It is given by the formula:\[ \tau = \frac{F}{A} \]where \( F = 4900 \text{ N} \) and \( A = 0.01 \text{ m}^2 \). Thus:\[ \tau = \frac{4900}{0.01} = 490000 \text{ N/m}^2 \]
04

Use the Rigidity Modulus to Find Shearing Strain

The shearing strain \( \gamma \) is related to stress \( \tau \) and the rigidity modulus \( G \) by the formula:\[ \gamma = \frac{\tau}{G} \]where \( G = 2.5 \times 10^{10} \text{ N/m}^2 \). Thus:\[ \gamma = \frac{490000}{2.5 \times 10^{10}} = 1.96 \times 10^{-5} \]

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Key Concepts

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

Tangential Force
When a tangential force is applied, it means that the force is acting along a surface, parallel to it, rather than perpendicular. In the exercise, the tangential force comes from the weight of an object. The force is equal to the weight of a 500 kg object, which is calculated using the formula \( F = m \times g \). Here, \( m \) is mass and \( g \) represents the acceleration due to gravity, approximately 9.8 m/s² on Earth.

Tangential forces are critical in real-world applications such as in wheelchair ramps, frictional forces between tires and roads, and more. In the problem, the tangential force causes a deformation that is measured by the strain it produces.
  • This force tries to move one layer of the cube over another.
  • Understanding how materials react to such forces helps in designing structures and products that are safe and effective.
Rigidity Modulus
The rigidity modulus, also known as the shear modulus, is a material property that measures its ability to resist shearing deformations. It represents the ratio of shear stress to the shearing strain. More formally, it is defined by the equation:
\[ G = \frac{\text{Shear Stress}}{\text{Shearing Strain}} \]
For rigid materials like aluminum, the rigidity modulus is a large number, indicating that a considerable amount of stress is required to produce a small strain.

In our problem, the rigidity modulus of aluminum is given as \(2.5 \times 10^{10} \text{ N/m}^2\).
  • This high value means aluminum is quite resistant to shape changes when sheared.
  • It is a crucial parameter for engineers to understand while designing anything from simple structures to complex machinery.
The higher the rigidity modulus, the harder it is to deform the material. This concept helps us predict and calculate deformations, ensuring materials will behave as expected under different forces.
Shear Stress
Shear stress refers to the force per unit area exerted parallel to a material. It typically happens when opposing forces act on the material's opposite faces. In the exercise, the force applied was 4900 N acting on an area of \(0.01 \text{ m}^2 \).

The shear stress equation is:
\[ \tau = \frac{F}{A} \]
where \(F\) is the force, and \(A\) is the area. This calculation shows how much force is concentrated over a unit area, giving us a stress value of \(490,000 \text{ N/m}^2\).
  • Understanding shear stress helps in analyzing the internal forces at play within materials.
  • It aids in predicting failure conditions where materials might undergo permanent deformation or fracture.
The concept of shear stress is crucial in fields like materials science and civil engineering. It guides the design of everyday objects, from bridges to buildings, ensuring that they are safe and sound under various forces.

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