/*! 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 85 The end of a pile of loose-link ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 85

The end of a pile of loose-link chain of mass \(\rho\) per unit length is being pulled horizontally along the surface by a constant force \(P .\) If the coefficient of kinetic friction between the chain and the surface is \(\mu_{k},\) determine the acceleration \(a\) of the chain in terms of \(x\) and \(\dot{x}\).

Short Answer

Expert verified
The acceleration is \( a = \frac{P}{\rho x} - \mu_k g \).

Step by step solution

01

Understand the Forces Involved

When a force \( P \) pulls the chain horizontally, friction opposes this motion. The frictional force \( F_f \) is given by \( F_f = \mu_k F_N \), where \( \mu_k \) is the coefficient of kinetic friction and \( F_N \) is the normal force. Since the chain slides on a horizontal surface, \( F_N \) is equal to the gravitational force acting on the section of the chain in contact with the surface.
02

Define the System鈥檚 Dynamics

To define the chain's movement, we need to relate the external force \( P \), the frictional force, and the chain's mass and acceleration. The net force \( F_{net} \) acting on the mass \( m \) of the chain segment is \( F_{net} = P - \mu_k m g \), where \( g \) is the acceleration due to gravity.
03

Express the Mass in Terms of Length

The mass \( m \) of the chain segment on the surface is \( m = \rho x \), where \( \rho \) is the mass per unit length and \( x \) is the length of the chain in contact with the surface. Substitute \( m \) in the frictional force and the net force equations.
04

Calculate the Net Force

Substitute \( m = \rho x \) into the expression for the net force: \[ F_{net} = P - \mu_k (\rho x) g \]. This relates the net force to the length of the chain on the surface.
05

Apply Newton鈥檚 Second Law

According to Newton's second law, the net force on an object equals its mass times its acceleration: \( F_{net} = m a \). Substitute \( m = \rho x \) into this formula to find the acceleration: \[ \rho x a = P - \mu_k \rho x g \].
06

Solve for the Acceleration

Solve for the acceleration \( a \) by dividing both sides of the equation by \( \rho x \): \[ a = \frac{P}{\rho x} - \mu_k g \]. This final equation gives the acceleration of the chain in terms of \( P \), \( \rho \), \( \mu_k \), \( g \), and \( x \).

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.

Kinetic Friction
Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. It is an important concept in dynamics, as it affects the movement of objects. Understanding kinetic friction helps us predict how an object will behave when forces are applied to it.

For the pile of loose-link chain in the exercise, kinetic friction acts against the constant pulling force. It is calculated using the equation:
  • \( F_f = \mu_k F_N \)
where \( F_f \) is the frictional force, \( \mu_k \) is the coefficient of kinetic friction, and \( F_N \) is the normal force, which equals the weight of the chain segment on the ground.The coefficient of kinetic friction, \( \mu_k \), is a dimensionless value that quantifies the friction between the chain and the surface. The greater the \( \mu_k \), the larger the frictional force opposing the movement.
Newton's Second Law
Newton's second law of motion is a cornerstone in the understanding of dynamics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be mathematically expressed as:

  • \( F_{net} = m a \)
where \( F_{net} \) is the total force acting on the object, \( m \) is the mass, and \( a \) is the acceleration.In the context of the exercise, Newton's second law allows us to formulate the relationship between the forces acting on the chain and its resulting acceleration. By substituting the expression for mass in terms of length, we can accurately determine how the chain accelerates under a given force, accounting for frictional forces.
Force Analysis
Force analysis involves evaluating all the forces acting on a system to understand its movement. For the loose-link chain, we consider both the pulling force \( P \) and the opposing force of kinetic friction.

The net force \( F_{net} \) is calculated by subtracting the frictional force \( F_f \) from the pulling force \( P \):
  • \( F_{net} = P - \mu_k m g \)
where \( m \) is the mass of the chain segment, \( g \) is acceleration due to gravity.By understanding the interactions between these forces, we can predict the behavior of the chain as it is pulled. This analysis helps in designing systems or solving problems efficiently, ensuring all contributing forces are considered.
Mass Per Unit Length
Mass per unit length, often denoted by \( \rho \), is a measure of how much mass is contained in a unit length of an object. It is particularly useful when dealing with objects like cables, ropes, or chains, where mass distribution influences physical behaviors.

In the exercise, the chain's mass \( m \) in contact with the surface is expressed in terms of its length \( x \) and mass per unit length \( \rho \):
  • \( m = \rho x \)
This relationship is vital in calculating the net force and, subsequently, the acceleration. By expressing mass this way, we simplify the calculations needed to analyze the dynamics of systems involving elongated objects under various forces.

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

Salt water flows through the fixed 12 -in.-insidediameter pipe at a speed \(v_{0}=4 \mathrm{ft} / \mathrm{sec}\) and enters the \(150^{\circ}\) bend with inside radius of 24 in. The water exits to the atmosphere through the 6 -in.-diameter nozzle \(C\). Determine the shear force \(V\), axial force \(P\), and bending moment \(M\) at flanges \(A\) and \(B\) which result from the flow of the salt water. The gage pressure in the pipe at flange \(A\) is 250 lb/in. \(^{2}\) and the pressure drop between \(A\) and \(B\) due to head loss may be neglected.

The experimental ground-effect machine has a total weight of 4200 lb. It hovers 1 or \(2 \mathrm{ft}\) off the ground by pumping air at atmospheric pressure through the circular intake duct at \(B\) and discharging it horizontally under the periphery of the skirt \(C\). For an intake velocity \(v\) of \(150 \mathrm{ft} / \mathrm{sec},\) calculate the aver age air pressure \(p\) under the 18 -ft-diameter machine at ground level. The specific weight of the air is \(0.076 \mathrm{lb} / \mathrm{ft}^{3}\).

A tank truck for washing down streets has a total weight of 20,000 lb when its tank is full. With the spray turned on, 80 lb of water per second issue from the nozzle with a velocity of 60 ft/sec relative to the truck at the \(30^{\circ}\) angle shown. If the truck is to accelerate at the rate of \(2 \mathrm{ft} / \mathrm{sec}^{2}\) when starting on a level road, determine the required tractive force \(P\) between the tires and the road when \((a)\) the spray is turned on and ( \(b\) ) the spray is turned off.

Air is pumped through the stationary duct \(A\) with a velocity of 50 ft/sec and exhausted through an experimental nozzle section \(B C\). The average static pressure across section \(B\) is 150 lb/in. \(^{2}\) gage, and the specific weight of air at this pressure and at the temperature prevailing is \(0.840 \mathrm{lb} / \mathrm{ft}^{3} .\) The average static pressure across the exit section \(C\) is measured to be \(2 \mathrm{lb} / \mathrm{in} .^{2}\) gage, and the corresponding specific weight of air is \(0.0760 \mathrm{lb} / \mathrm{ft}^{3}\). Calculate the force \(T\) exerted on the nozzle flange at \(B\) by the bolts and the gasket to hold the nozzle in place.

The space shuttle, together with its central fuel \(\operatorname{tank}\) and two booster rockets, has a total mass of \(2.04\left(10^{6}\right) \mathrm{kg}\) at liftoff. Each of the two booster rockets produces a thrust of \(11.80\left(10^{6}\right) \mathrm{N},\) and each of the three main engines of the shuttle produces a thrust of \(2.00\left(10^{8}\right) \mathrm{N}\). The specific impulse (ratio of exhaust velocity to gravitational acceleration) for each of the three main engines of the shuttle is 455 s. Calculate the initial vertical acceleration a of the assembly with all five engines operating and find the rate at which fuel is being consumed by each of the shuttle's three engines.
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Radiation

Read Explanation

Modern Physics

Read Explanation

Solid State Physics

Read Explanation

Waves Physics

Read Explanation

Rotational Dynamics

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.