/*! 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 149 The car has a mass \(m_{0}\) and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The car has a mass \(m_{0}\) and is used to tow the smooth chain having a total length \(l\) and a mass per unit of length \(m^{\prime}\). If the chain is originally piled up, determine the tractive force \(F\) that must be supplied by the rear wheels of the car, necessary to maintain a constant speed \(v\) while the chain is being drawn out.

Short Answer

Expert verified
The force \(F\) that must be supplied by the rear wheels of the car, necessary to maintain a constant speed \(v\) while the chain is being drawn out, is the same as the force that counteracts the resisting force from the chain. It's found by considering both the car's weight and the additional force due to the chain.

Step by step solution

01

Calculate the resisting force from the chain

The resisting force from the chain is due to the negative acceleration of each part of the chain. This resisting force can be calculated using the formula for the force on an object undergoing linear acceleration. We have that the force \(F_{r}\) is equal to \(dm \times dv/dt \), where \(dm = m^{\prime}dx\) is the mass of the differential chain segment being lifted and \(dv/dt\) is the rate of change of velocity. We integrate from 0 (at which point the force is largest) to v (at which point the force is 0) to find the total force.
02

Determine the net force required to maintain constant velocity

With a constant velocity, the net force on the system must be zero in accordance with Newton's First Law of Motion. Therefore the force to maintain constant velocity will be equal to the resisting force from the chain.
03

Determine the final tractive force

The tractive force supplied by the car, F, is therefore the force required to maintain constant velocity. With the car alone, F would be \(m_{0}g\), the product of the car’s mass and acceleration due to gravity, but the chain adds an additional force equal to the resisting force from the chain that we calculated in step 1. These forces add up to give the final tractive force.

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 91Ó°ÊÓ!

Key Concepts

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

Linear Acceleration
When an object's velocity changes over time, we describe this phenomenon as linear acceleration. In physics, acceleration is defined as the rate at which an object's velocity changes with time, mathematically represented as \( a = \frac{dv}{dt} \), where \( dv \) is the change in velocity, and \( dt \) is the change in time. In the example of the car towing a chain, each differential segment of the chain experiences negative acceleration as it goes from being at rest (piled up) to moving at the constant speed \( v \) of the car. This acceleration is crucial in calculating the resisting force that the car must overcome to maintain constant velocity.
Newton's First Law of Motion
Newton's First Law of Motion, often called the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In practice, this means that if the car is moving at a constant velocity while towing the chain, the net force acting on the entire system (the car and the chain) must be zero. This equilibrium is essential for maintaining the constant speed, as any net force would cause acceleration or deceleration according to Newton's Second Law. Therefore, when we calculate the tractive force necessary to tow the chain, it must counteract the resisting forces exactly, resulting in that crucial net force of zero.
Constant Velocity
Constant velocity means that both the speed and the direction of the object do not change. If we translate this into our car-and-chain scenario, maintaining a constant velocity \( v \) implies that the car is not accelerating or braking while it draws out the chain. Due to this uniform motion, there's no net change in velocity, making the calculation of forces somewhat simpler. The tractive force needed is simply the one that continues to counteract any forces that would otherwise slow down the car, such as friction or, in this case, the additional weight of the chain being lifted and moved.
Integrating Force
Integrating force over a distance or time is an essential concept in physics, particularly when dealing with systems where the force might change across the system. In our exercise, the resisting force from the chain varies as different segments are drawn out at different times. The force is largest when a segment of the chain starts to move and drops to zero once it reaches constant velocity. By integrating the small incremental forces \( dm \times \frac{dv}{dt} \) over the entire chain length, we determine the total resisting force. This integration accumulates all the tiny pushes required to set each differential chain segment in motion, and its sum gives us the total force needed over the course of pulling the entire chain.

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

A train consists of a \(50-\mathrm{Mg}\) engine and three cars, each having a mass of \(30 \mathrm{Mg}\). If it takes \(80 \mathrm{~s}\) for the train to increase its speed uniformly to \(40 \mathrm{~km} / \mathrm{h}\), starting from rest, determine the force \(T\) developed at the coupling between the engine \(E\) and the first car \(A .\) The wheels of the engine provide a resultant frictional tractive force \(\mathbf{F}\) which gives the train forward motion, whereas the car wheels roll freely. Also, determine \(F\) acting on the engine wheels.

A \(50-\mathrm{kg}\) crate rests against a stop block \(s\), which prevents the crate from moving down the plane. If the coefficients of static and kinetic friction between the plane and the crate are \(\mu_{s}=0.3\) and \(\mu_{k}=0.2,\) respectively, determine the time needed for the force \(\mathbf{F}\) to give the crate a speed of \(2 \mathrm{~m} / \mathrm{s}\) up the plane. The force always acts parallel to the plane and has a magnitude of \(F=(300 t) \mathrm{N},\) where \(t\) is in seconds. Hint: First determine the time needed to overcome static friction and start the crate moving.

Each ball has a negligible size and a mass of \(10 \mathrm{~kg}\) and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque \(M=\left(t^{2}+2\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds, determine the speed of each ball when \(t=3 \mathrm{~s}\). Each ball has a speed \(v=2\) \(\mathrm{m} / \mathrm{s}\) when \(t=0\)

Ball \(A\) has a mass of \(3 \mathrm{~kg}\) and is moving with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) when it makes a direct collision with ball \(B,\) which has a mass of \(2 \mathrm{~kg}\) and is moving with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). If \(e=0.7,\) determine the velocity of each ball just after the collision. Neglect the size of the balls.

The pile \(P\) has a mass of \(800 \mathrm{~kg}\) and is being driven into loose sand using the \(300-\mathrm{kg}\) hammer \(C\) which is dropped a distance of \(0.5 \mathrm{~m}\) from the top of the pile. Determine the initial speed of the pile just after it is struck by the hammer. The coefficient of restitution between the hammer and the pile is \(e=0.1 .\) Neglect the impulses due to the weights of the pile and hammer and the impulse due to the sand during the impact.

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.