/*! 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 135 The \(6-\mathrm{kg}\) cylinder i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 135

The \(6-\mathrm{kg}\) cylinder is released from rest in the position shown and falls on the spring, which has been initially precompressed \(50 \mathrm{mm}\) by the light strap and restraining wires. If the stiffness of the spring is \(4 \mathrm{kN} / \mathrm{m},\) compute the additional deflection \(\delta\) of the spring produced by the falling cylinder before it rebounds.

Short Answer

Expert verified
Additional deflection \(\delta\) is approximately 29 mm.

Step by step solution

01

Identify Initial Conditions and Forces

The cylinder has a mass of 6 kg and is initially at rest before falling on the spring. The spring is initially precompressed by 50 mm. The force due to gravity acting on the cylinder can be calculated using the formula: \( F = mg \), where \( g = 9.81 \text{ m/s}^2 \).
02

Equate Potential Energy and Spring Energy

When the cylinder falls on the spring, its potential energy is converted into the energy stored in the spring. The potential energy just before hitting the spring can be expressed as \( E_p = mgh \) where \( h \) is the height equivalent to the additional compression caused by the fall. The spring energy can be expressed as \( E_s = \frac{1}{2}k(x_0 + \delta)^2 \), where \( x_0 = 0.05 \text{ m} \) is the initial compression and \( \delta \) is the additional deflection.
03

Solve for Additional Deflection \(\delta\)

Assume the entire potential energy converts to spring energy: \( mgh = \frac{1}{2}k(x_0 + \delta)^2 \). Plug in the values: \( 6 \times 9.81 \times (x_0 + \delta) = \frac{1}{2} \times 4000 \times (0.05 + \delta)^2 \). This equation can get simplified and can be solved for \( \delta \).
04

Calculation

Expand and simplify the equation from Step 3:\( 58.86 \times (0.05 + \delta) = 2000 \times (0.05^2 + 2 \times 0.05 \times \delta + \delta^2)\). Solve this quadratic equation to find \( \delta \). On solving, you get \( \delta \approx 0.029 \text{ m} \) or 29 mm.

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.

Potential Energy
Potential energy is the energy an object possesses due to its position or height above the ground. In this exercise, the 6-kg cylinder has potential energy because it is elevated above the spring. The formula for potential energy is given by \( E_p = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (9.81 m/s虏), and \( h \) is the height of the object.

When the cylinder falls, this energy is converted into other forms, primarily into the energy stored in the spring. Understanding this conversion process is crucial in analyzing systems involving heights and potential energy.
Spring Force
The spring force is the force exerted by a spring when it is compressed or stretched. This force is governed by Hooke鈥檚 Law, which states that the force exerted by a spring is proportional to its displacement \( x \) from its rest position. This is expressed mathematically as \( F = -kx \), where \( k \) is the stiffness or spring constant, and \( x \) is the displacement from the equilibrium position.

In this problem, the spring is initially compressed by 50 mm, and we need to find the additional compression caused by the falling cylinder. With a spring constant \( k = 4000 \, \text{N/m} \), the spring's ability to resist and store energy when compressed is significant, affecting the dynamics of the entire system.
Mass-Spring System
A mass-spring system consists of a mass attached to a spring. When the mass moves, the spring compresses or extends, storing mechanical energy as potential energy, and later releasing it as kinetic energy.

In this exercise, the cylinder represents the mass in the mass-spring system. When it falls onto the spring, the entire potential energy of the cylinder (due to its height) transfers into the spring, causing it to compress further. Understanding how the mass-spring system functions helps in predicting the behavior of the spring under the weight of the mass, and is a fundamental concept in mechanical dynamics.
Energy Conversion
Energy conversion in a mechanical context often involves transforming potential energy to kinetic energy, or into energy stored in a spring. In this exercise, when the cylinder falls, potential energy due to its height gets converted completely into the stored energy of the spring as the spring compresses.

This transformation is described by the equation \( mgh = \frac{1}{2}k(x_0 + \delta)^2 \), which equates the potential energy of the cylinder to the spring energy. This equality is the basis for calculating how much the spring will compress under the weight of the falling cylinder, illustrating the principles of energy conservation and conversion.
Deflection Calculation
Deflection calculation involves determining how much a spring compresses or deflects when a force is applied. In this exercise, we aim to find the additional compression, \( \delta \), beyond the initial 50 mm precompression.

Using the equation from energy conversion, the potential energy converts into stored spring energy, leading to a quadratic equation: \[58.86 \times (0.05 + \delta) = 2000 \times (0.05^2 + 2 \times 0.05 \times \delta + \delta^2)\]Solving this equation gives the additional deflection \( \delta \), which is approximately 29 mm. Mastering deflection calculations is vital in understanding and designing systems that involve springs and other elastic materials.

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

When a particle is dropped from rest relative to the surface of the earth at a latitude \(\gamma\), the initial apparent acceleration is the relative acceleration due to gravity \(g_{\mathrm{rel}} .\) The absolute acceleration due to gravity \(g\) is directed toward the center of the earth. Derive an expression for \(g_{\text {rel }}\) in terms of \(g, R\) \(\omega,\) and \(\gamma,\) where \(R\) is the radius of the earth treated as a sphere and \(\omega\) is the constant angular velocity of the earth about the polar axis considered fixed. Although axes \(x-y-z\) are attached to the earth and hence rotate, we may use Eq. \(3 / 50\) as long as the particle has no velocity relative to \(x-y-z\). (Hint: Use the first two terms of the binomial expansion for the approximation.)

The hydraulic braking system for the truck and trailer is set to produce equal braking forces for the two units. If the brakes are applied uniformly for 5 seconds to bring the rig to a stop from a speed of \(20 \mathrm{mi} / \mathrm{hr}\) down the 10 -percent grade, determine the force \(P\) in the coupling between the trailer and the truck. The truck weighs 20,000 lb and the trailer weighs \(15,000 \mathrm{lb}\).

A spacecraft \(m\) is heading toward the center of the moon with a velocity of \(2000 \mathrm{mi} / \mathrm{hr}\) at a distance from the moon's surface equal to the radius \(R\) of the moon. Compute the impact velocity \(v\) with the surface of the moon if the spacecraft is unable to fire its retro-rockets. Consider the moon fixed in space. The radius \(R\) of the moon is \(1080 \mathrm{mi}\), and the acceleration due to gravity at its surface is \(5.32 \mathrm{ft} / \mathrm{sec}^{2}\).

The retarding forces which act on the race car are the drag force \(F_{D}\) and a nonaerodynamic force \(F_{R}\) The drag force is \(F_{D}=C_{D}\left(\frac{1}{2} \rho v^{2}\right) S,\) where \(C_{D}\) is the drag coefficient, \(\rho\) is the air density, \(v\) is the car speed, and \(S=30 \mathrm{ft}^{2}\) is the projected frontal area of the car. The nonaerodynamic force \(F_{R}\) is constant at 200 lb. With its sheet metal in good condition, the race car has a drag coefficient \(C_{D}=0.3\) and it has a corresponding top speed \(v=200 \mathrm{mi} / \mathrm{hr}\) After a minor collision, the damaged front-end sheet metal causes the drag coefficient to be \(C_{D}^{\prime}=\) \(0.4 .\) What is the corresponding top speed \(v^{\prime}\) of the race car?

The motor unit \(A\) is used to elevate the 300 -kg cylinder at a constant rate of \(2 \mathrm{m} / \mathrm{s}\). If the power meter \(B\) registers an electrical input of \(2.20 \mathrm{kW}\) calculate the combined electrical and mechanical efficiency \(e\) of the system.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Electricity and Magnetism

Read Explanation

Famous Physicists

Read Explanation

Geometrical and Physical Optics

Read Explanation

Engineering Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
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.