/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 A square plate is \(1.0 \times 1... [FREE SOLUTION] | 91影视

91影视

A square plate is \(1.0 \times 10^{-2} \mathrm{~m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{~m}\) on a side, and has a mass of \(7.2 \times 10^{-2} \mathrm{~kg}\). The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\). One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is \(0.90 .\) A force is applied to the top of the plate, as in Figure \(10-32 a\). Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10-32 b\) ) that can be created by the applied force just before the plate begins to move.

Short Answer

Expert verified
(a) Maximum shear stress: 705.56 N/m虏. (b) Maximum shear strain: 3.53 x 10鈦烩伕. (c) Shear deformation \( \Delta X \): 3.53 x 10鈦宦光伆 m.

Step by step solution

01

Identify Shear Stress Formula

The shear stress \( \tau \) is calculated as the force \( F \) divided by the area \( A \). The area \( A \) of the square face on a side is \((3.0 \times 10^{-2} \text{ m})^2\). The maximum force that can be applied without sliding is impacted by static friction: \( F_{\text{max}} = \mu \cdot \text{normal force (mass \( \times \) gravity)} \). Substitute to find \( \tau = \frac{F_{\text{max}}}{A} \).
02

Calculate Area

The area of the square face is: \[A = (3.0 \times 10^{-2} \text{ m})^2 = 9.0 \times 10^{-4} \text{ m}^2.\]
03

Calculate Maximum Force due to Friction

The maximum force from static friction \( F_{\text{max}} \) is: \[F_{\text{max}} = \mu \cdot m g = 0.90 \cdot 7.2 \times 10^{-2} \text{ kg} \times 9.8 \text{ m/s}^2.\]Calculating gives \( F_{\text{max}} \approx 0.635 \text{ N} \).
04

Compute Maximum Shear Stress

Substitute the values into the shear stress formula:\[\tau = \frac{F_{\text{max}}}{A} = \frac{0.635 \text{ N}}{9.0 \times 10^{-4} \text{ m}^2} = 705.56 \text{ N/m}^2.\]Thus, the maximum shear stress is \( 705.56 \text{ N/m}^2 \).
05

Determine Shear Strain using Shear Modulus

Shear strain \( \gamma \) is related to shear stress \( \tau \) and shear modulus \( G \) by:\[\gamma = \frac{\tau}{G} = \frac{705.56 \text{ N/m}^2}{2.0 \times 10^{10} \text{ N/m}^2}.\]Calculating, \( \gamma \approx 3.53 \times 10^{-8} \).
06

Calculate Shear Deformation \( \Delta X \)

Shear deformation \( \Delta X \) can be found using:\[\Delta X = \gamma \times \text{thickness} = 3.53 \times 10^{-8} \times 1.0 \times 10^{-2} \text{ m } = 3.53 \times 10^{-10} \text{ m} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Shear Strain
Shear strain, denoted as \( \gamma \), is a measure of how much a material deforms under shear stress. It describes the relative displacement between parallel planes within a material. Think of shear strain as observing how much a square distorts into a parallelogram when a force is applied parallel to its surfaces.

In the given exercise, we calculate it using the formula \( \gamma = \frac{\tau}{G} \). Here, \( \tau \) represents the shear stress and \( G \) is the shear modulus of the material, which quantifies how easily the material deforms under shear stress. For example, by applying a calculated shear stress of \( 705.56 \text{ N/m}^2 \) to our material with a shear modulus of \( 2.0 \times 10^{10} \text{ N/m}^2 \), we calculate the shear strain to be approximately \( 3.53 \times 10^{-8} \).

This tiny amount of strain indicates that the material is quite resistant to shear deformation, showing only minimal bending under the applied force. Understanding shear strain is crucial for materials used in structures like bridges or buildings, ensuring they withstand forces without bending too much.
Shear Deformation
Shear deformation refers to the change in shape of a material due to shear stress. While shear strain measures the angle change, shear deformation provides a physical measurement of distortion. It particularly looks at the horizontal shift between layers within the structure.

The formula \( \Delta X = \gamma \times \text{thickness} \), helps calculate this transformation, where \( \Delta X \) indicates the amount of shear deformation and "thickness" is the initial dimension perpendicular to the force. From the exercise, we found the shear strain to be \( 3.53 \times 10^{-8} \) and when multiplied by the thickness \((1.0 \times 10^{-2} \text{ m})\), we determine the shear deformation to be \( 3.53 \times 10^{-10} \text{ m} \).

Considering its minute measure, this shows how resilient the material is against external forces, underscoring why the choice of material and its properties are key in design and engineering.
Frictional Force
Frictional force is the force resisting the relative motion of solid surfaces. In the context of the exercise, it is crucial for preventing the square plate from sliding. The maximum frictional force, \( F_{\text{max}} \), acts before the plate starts to move and is influenced by the static friction coefficient and the normal force.

We calculated it using the formula: \( F_{\text{max}} = \mu \cdot m \cdot g \), where \( \mu = 0.90 \) is the coefficient of static friction, \( m = 7.2 \times 10^{-2} \text{ kg} \) is the mass of the plate, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. This calculation provided \( F_{\text{max}} \approx 0.635 \text{ N} \).

Understanding the role of frictional force is vital as it ensures stability and prevents unwanted motion in systems, making it an integral part of safe and reliable design.
Shear Modulus
The shear modulus, represented by \( G \), is a fundamental property that describes how a material responds to shear stress. It links the amount of shear stress a material can withstand to its resultant shear strain.

A high shear modulus signifies that the material requires a lot of force to change shape, making it strong and rigid under shear forces. For the embodiment in our exercise, \( G \) is given as \( 2.0 \times 10^{10} \text{ N/m}^2 \), indicating a robust material that deforms minimally under stress.

Such properties are significant when choosing materials for construction or manufacturing, as the shear modulus helps predict how a material will behave when an external force is applied parallel to its surface. Understanding this property assists engineers in ensuring that materials chosen for specific applications have the necessary strength and resilience.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The front spring of a car's suspension system has a spring constant of \(1.50 \times 10^{6} \mathrm{~N} / \mathrm{m}\) and supports a mass of \(215 \mathrm{~kg}\). The wheel has a radius of \(0.400 \mathrm{~m}\). The car is traveling on a bumpy road, on which the distance between the bumps is equal to the circumference of the wheel. Due to resonance, the wheel starts to vibrate strongly when the car is traveling at a certain minimum linear speed. What is this speed?

A simple pendulum is made from a \(0.65\) -m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

In a room that is \(2.44 \mathrm{~m}\) high, a spring (unstrained length \(=0.30 \mathrm{~m}\) ) hangs from the ceiling. A board whose length is \(1.98 \mathrm{~m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board ( 104 N) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

A \(1.0 \times 10^{-3}-\mathrm{kg}\) house spider is hanging vertically by a thread that has a Young's modulus of \(4.5 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) and a radius of \(13 \times 10^{-6} \mathrm{~m}\). Suppose that a \(95-\mathrm{kg}\) person is hanging vertically on an aluminum wire. What is the radius of the wire that would exhibit the same strain as the spider's thread, when the thread is stressed by the full weight of the spider?

A \(1.00 \times 10^{-2} \mathrm{~kg}\) bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless, horizontal spring \((k=845 \mathrm{~N} / \mathrm{m})\). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of \(0.200 \mathrm{~m}\). What is the speed of the bullet?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.