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Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

Short Answer

Expert verified
The ratio of shear strain for the larger object to the smaller one is \( \frac{1}{9} \).

Step by step solution

01

Understanding Shear Strain and Hooke's Law

Shear strain is defined as the change in shape due to applied forces and is given by the ratio of the deformation (displacement) to the original dimension perpendicular to the force direction. According to Hooke's Law for shear, shear strain is proportional to shear stress (force per unit area), given by \( \text{shear strain} = \frac{F}{A} \times \frac{1}{G} \), where \( A \) is the area and \( G \) is the shear modulus.
02

Determine the Area for Each Solid

The original rectangular solid has dimensions \( a \), \( b \), and \( c \), resulting in an area exposed to shear \( A = a \times b \). The larger solid has dimensions \( 3a \), \( 3b \), and \( 3c \), giving an area \( A' = 3a \times 3b = 9ab \).
03

Shear Strain Calculation for the Smaller Object

For the smaller object, with area \( A = ab \), the shear strain \( \gamma_1 \) is calculated as \( \gamma_1 = \frac{F}{ab} \times \frac{1}{G} \).
04

Shear Strain Calculation for the Larger Object

For the larger object, with area \( A' = 9ab \), the shear strain \( \gamma_2 \) is calculated as \( \gamma_2 = \frac{F}{9ab} \times \frac{1}{G} \).
05

Calculate the Ratio of Shear Strain

The ratio of the shear strain for the larger object to that of the smaller object is \( \frac{\gamma_2}{\gamma_1} = \frac{\frac{F}{9ab} \times \frac{1}{G}}{\frac{F}{ab} \times \frac{1}{G}} = \frac{1}{9} \).
06

Conclusion

The shear strain ratio for the larger object to the smaller object is \( \frac{1}{9} \), meaning the larger object has significantly less shear strain under the same applied forces.

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Key Concepts

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

Hooke's Law
Hooke's Law is fundamental in understanding the relationship between the deformation of materials and applied forces. It's a principle that states that the deformation of a material (strain) is proportional to the applied force (stress), as long as the deformation is within the elastic limit of the material. This means that when you apply a small enough force to a material, it will return to its original shape once the force is removed.

In the context of shear stress, Hooke's Law can be expressed through the equation:
  • \( ext{shear strain} = \frac{F}{A} \times \frac{1}{G} \)
where:
  • \( F \) is the force applied,
  • \( A \) is the area over which the force is applied,
  • \( G \) is the shear modulus of the material.
The shear modulus is a measure of the material's rigidity, indicating how easily it can be deformed when a shear force is applied. This law helps predict how materials will respond under various stress conditions.
Shear Stress
Shear stress is a key concept when discussing how materials respond to forces that are applied parallel to a surface. Unlike normal stress, which acts perpendicular to the surface, shear stress acts horizontally and tends to cause layers of the material to slide past one another.

The formula for calculating shear stress is:
  • \( ext{Shear stress} = \frac{F}{A} \)
where:
  • \( F \) is the force applied,
  • \( A \) is the area of the surface facing the force.
Understanding shear stress is crucial in determining how a material will behave when forces are applied in a manner that could potentially lead to deformation or failure. This concept is widely used in engineering, especially in materials science and mechanical engineering.
Shear Modulus
The shear modulus, often denoted as \( G \), is an important property of a material that measures its ability to withstand shear stress without deforming. It is one of the material constants that describe the elastic properties, alongside Young's modulus and Poisson's ratio.

The shear modulus can be defined mathematically as:
  • \( G = \frac{ ext{Shear stress}}{ ext{Shear strain}} \)
This means that a material with a high shear modulus is very rigid and doesn't deform easily when subjected to shear forces. Conversely, a low shear modulus implies that a material deforms more easily. Knowing the shear modulus helps engineers choose the right materials for structures that will endure varying shear forces.

Choosing materials with the appropriate shear modulus is crucial in applications where durability and resistance to deformation under shear stress are essential. It ensures reliability and safety in construction and design.
Rectangular Solid
A rectangular solid, also known as a rectangular prism, is a three-dimensional figure with six rectangular faces. Understanding the geometry of a rectangular solid is important in physics and engineering, especially when analyzing stresses and strains under various forces.

When shear forces are applied to a rectangular solid:
  • The area exposed to shear is the face of the rectangle over which the force acts.
  • Forces may cause a change in shape, or shear strain, as one face moves relative to the opposite face.
In our exercise, shear forces are applied to two rectangular solids of the same material but different sizes. This highlights an important aspect of scaling in physics:
  • When dimensions are increased, both the surface area and volume change, affecting how forces impact the object.
  • The exercise demonstrates how shear strain is affected by changing dimensions and helps students understand the practical implications of material properties and geometry in real-world applications.
Practicing these concepts on various shapes and sizes helps in gaining a deeper understanding of how forces interact with different structures.

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Most popular questions from this chapter

Mountain Climbing. Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. 11.33 . Suppose that an 82.0 -kg climber, who is 1.90 \(\mathrm{m}\) tall and has a center of gravity 1.1 \(\mathrm{m}\) from his feet, rapels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1,40 \mathrm{m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the forctical compo- face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

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A metal rod that is 4.00 \(\mathrm{m}\) long and 0.50 \(\mathrm{cm}^{2}\) in cross- sectional area is found to stretch 0.20 \(\mathrm{cm}\) under a tension of 5000 \(\mathrm{N}\) . What is Young's modulus for this metal?

A nonuniform fire escape ladder is 6.0 \(\mathrm{m}\) long when extended to the icy alley below. It is held at the top by a frictionless pivot, and there is neghigible frictional force from the icy surface at the bottom. The ladder weighs 250 \(\mathrm{N}\) , and its center of gravity is 2.0 \(\mathrm{m}\) along the ladder from its bottom. A mother and child of total weight 750 \(\mathrm{N}\) are on the ladder 1.5 \(\mathrm{m}\) from the pivot. The ladder makes an angle \(\theta\) with the horizontal. Find the magnitude and direction of \((a)\) the force exerted by the icy alley on the ladder and (b) the force exerted by the ladder on the pivot. (c) Do your answers in parts (a) and (b) depend on the angle \(\theta ?\)

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 \(\mathrm{kg}\) from a vertical steel wire 0.50 \(\mathrm{m}\) long and \(2.5 \times 10^{-3} \mathrm{cm}^{2}\) in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 \(\mathrm{kg}\) . For each wire, compute (a) the tensile strain and \((\mathrm{b})\) the elongation.

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