A locomotive accelerates a 23 -car train along a level track. Each car has a mass of \(48.6\) metric tons and is subject to a drag force \(f=243 v\), where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and the force \(f\) is in N. At the instant when the speed of the train is \(34.5 \mathrm{~km} / \mathrm{h}\), the acceleration is \(0.182 \mathrm{~m} / \mathrm{s}^{2} .(a)\) Calculate the force exerted by the locomotive on the first car. ( \(b\) ) Suppose that the force found in part \((a)\) is the greatest force the locomotive can exert on the train. What, then, is the steepest grade up which the locomotive can pull the train at \(34.5 \mathrm{~km} / \mathrm{h} ?(1\) metric ton \(=1000 \mathrm{~kg} .\) )

Step by step solution

01

Converting Values

In the exercise, the speed is given in terms of km/h. First, it should be converted into m/s. The speed of the train in m/s is \(34.5 km/h * (1000 m / 1 km)* (1hr / 3600s) \approx 9.6 m/s \)

02

Calculating the Total Drag Force.

The total drag force on the train is given by \( f = 243v \). Multiplying this by the number of cars, hence, the total drag force would be \( f_{total} = 23 * 243* v \approx 53808.0 Newtons \).

03

Identification of Total Force on First Car

The force, F,\ exerted by the locomotive according to Newton's second law is the sum of the net force on the train and the total drag force. The net force on the train is given by \( F_{net} = m_{train} * a \) where \( m_{train} \) is the total mass of the train and \( a \) is the acceleration. The mass of the train in kg is \( m_{train} = 23cars * 48.6 MetricTons/car * 1000 kg/Metric Ton \approx 1118000 \ kg \). So the net force is \( F_{net} \approx 1118000 * 0.182 \approx 203476 newtons \). The force exerted by the locomotive is hence \( F = F_{net} + f_{total} \approx 257284.0 Newtons. \)

04

Calculating Maximum Steepest grade

To find the steepest grade, we setup a similar equation as in step 3, except this time there is an additional force, the force due to the grade \( f_{grade} \). It is given by the expression \( f_{grade} = m_{train} * g * sin(θ) \) where \( θ \) is the angle of the grade and \( g \) is the acceleration due to gravity (approximately 9.8). Since the train is not accelerating up the grade, the sum of forces is zero. Therefore, we can set the force equation equal to zero and solve for \( θ \). This gives \( θ = arcsin ((F - f_{total} ) / (m_{train} * g)) \) which comes out to be approximately 0.013 or 0.75 degrees.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!