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Chapter 7: Problem 5

A \(0.05 \mathrm{~m}\) cube has its upper face displaced by \(0.2 \mathrm{~cm}\) by a tangential force of \(8 \mathrm{~N}\). Calculate the modulus of rigidity (in \(\left.\times 10^{4} \mathrm{Nm}^{-2}\right)\) of the material of the cube.

Short Answer

Expert verified
The modulus of rigidity is \(8 \times 10^4 \text{ N/m}^2\).

Step by step solution

01

Understand the Formula for Modulus of Rigidity

The modulus of rigidity (or shear modulus) is given by the formula: \( G = \frac{F/A}{\Delta x/h} \), where \( F \) is the force applied, \( A \) is the area over which the force is applied, \( \Delta x \) is the displacement, and \( h \) is the height or thickness of the material.
02

Identify Given Values

The problem provides the following values: the side of the cube \( h = 0.05 \text{ m} \), displacement \( \Delta x = 0.2 \text{ cm} = 0.002 \text{ m} \), and force \( F = 8 \text{ N} \).
03

Calculate the Area

The area \( A \) is the upper face of the cube, which is a square. Calculate it using the side \( h \): \( A = h^2 = (0.05 \text{ m})^2 = 0.0025 \text{ m}^2 \).
04

Compute the Shear Stress

The shear stress \( \tau \) is \( \frac{F}{A} = \frac{8 \text{ N}}{0.0025 \text{ m}^2} = 3200 \text{ N/m}^2 \).
05

Compute the Shear Strain

The shear strain \( \gamma \) is \( \frac{\Delta x}{h} = \frac{0.002 \text{ m}}{0.05 \text{ m}} = 0.04 \).
06

Calculate the Modulus of Rigidity

The modulus of rigidity \( G \) is the ratio of shear stress to shear strain: \( G = \frac{\tau}{\gamma} = \frac{3200 \text{ N/m}^2}{0.04} = 80000 \text{ N/m}^2 \).
07

Express in the Required Format

Convert this result into the desired units: \( 80000 \text{ N/m}^2 = 8 \times 10^4 \text{ N/m}^2 \).

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Key Concepts

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

Understanding Shear Stress
Shear stress is a concept that helps us understand how a material will react when a force is applied parallel to its surface. Think of it as the force trying to "shear" a material apart or make it slide over itself, just like scissors do when cutting paper.

This stress is calculated using the formula: \( \tau = \frac{F}{A} \). Here, \( F \) is the force applied, and \( A \) is the area over which that force is acting.
  • In the exercise, \( F = 8 \; \text{N} \) is the tangential force applied.
  • The face of the cube is affected, so the area \( A = 0.0025 \; \text{m}^2 \).
By plugging in these values, we find that the shear stress \( \tau = 3200 \; \text{N/m}^2 \). This number tells us how much force is applied per unit area of the material.
Exploring Shear Strain
When force is applied, materials deform, causing a change in shape or size. Shear strain quantifies how much a material has been distorted due to shear stress.

Mathematically, it is represented as \( \gamma = \frac{\Delta x}{h} \), where \( \Delta x \) is the displacement of the material, and \( h \) is the original height or thickness.
  • In the problem, \( \Delta x = 0.002 \; \text{m} \) and \( h = 0.05 \; \text{m} \).
Using these values, we calculate \( \gamma = 0.04 \).

This value shows the deformation extent of the cube, with a higher number indicating larger deformation per unit height.
Cube Deformation Measurement
Measuring how a cube deforms under a shear force gives insight into its mechanical properties and modulus of rigidity. In this exercise, a force displaces the cube's top face, translating into shear stress and strain.

Interpret the deformation as a fraction of the cube's dimensions. The side of the cube is \( 0.05 \; \text{m} \), and the displacement \( \Delta x \) is calculated to be small, at \( 0.002 \; \text{m} \).
  • A detailed understanding requires measuring these changes carefully to ensure accurate calculation of material properties like the modulus of rigidity.
By applying the concepts of shear stress and strain, we determine how the material resists deformation, offering insight into its ability to withstand applied forces without altering its structural integrity significantly.

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Most popular questions from this chapter

A metal cube of side \(10 \mathrm{~cm}\) is subjected to a shearing stress of \(10^{6} \mathrm{~N} / \mathrm{m}^{2}\). Calculate the modulus of rigidity (in \(\left._{0} \times 10^{8} \mathrm{Nm}^{-2}\right)\) if the top of the cube is displaced by \(0.05 \mathrm{~cm}\) with respect to its bottom.

A rod is made of uniform material and has non-uniform cross section. It is fixed at both the ends as shown and heated at mid-section. Which of the following statements are not correct? (1) Force of compression in the rod will be maximum at mid-section. (2) Compressive stress in the rod will be maximum at left end. (3) Since rod in fixed at both the ends, its length will remain unchanged. Hence, no strain will be induced in it. (4) None of the above.

The length of a wire is increased by \(1 \mathrm{~mm}\) on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is (1) \(0.25 \mathrm{~mm}\) (2) \(0.5 \mathrm{~mm}\) (3) \(2 \mathrm{~mm}\) (4) \(4 \mathrm{~mm}\)

According to Hooke's law, within the elastic limit stress/strain = constant. This constant depends on the type of strain or the type of force acting. Tensile stress might result in compressional or elongative strain; however, a tangential stress can only cause a shearing strain. After crossing the elastic limit, the material undergoes elongation and beyond a stage breaks. All modulus of elasticity are basically constants for the materials under stress. Just on crossing the yield region, the material will have (1) reduced stress (2) increased stress (3) breaking stress (4) constant stress

The breaking stress for a metal is \(7.8 \times 10^{9} \mathrm{Nm}^{-2}\). The density of the metal is \(7800 \mathrm{~kg} \mathrm{~m}^{-3}\). If \(g=10 \mathrm{~N} \mathrm{~kg}^{-1}\), Find the maximum length of the wire made of this metal which may be suspended without breaking.
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