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A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.?
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A local train uses motor and trailer coaches in the ratio of 1:2. The ...
Problem Statement: A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.?

Solution:
Step 1: Calculation of Weight of Motor and Trailer Coach
Given, ratio of weight of motor and trailer coach = 1:2
Let the weight of motor coach be x tonnes
Therefore, weight of trailer coach = 2x tonnes
Total weight of train = x + 2x + 2x = 5x tonnes
Given, weight of trailer coach = 35 tonnes
Therefore, 2x = 35
x = 17.5 tonnes
Therefore, weight of motor coach = 17.5 tonnes and weight of trailer coach = 35 tonnes

Step 2: Calculation of Tractive Effort
Tractive effort (TE) = coefficient of adhesion x weight on driving wheels
Weight on driving wheels = weight of motor coach + weight of trailer coach = 17.5 + 35 = 52.5 tonnes
TE = 0.2 x 52.5 = 10.5 tonnes

Step 3: Calculation of Train Resistance
Train resistance (R) = rolling resistance + gradient resistance + curve resistance + acceleration resistance
Given, in resistance = 30 N/tonne
Effective rotating mass is 10% of the dead weight.
Rolling resistance (RR) = in resistance x weight of train x effective rotating mass
RR = 30 x 5x x 0.1 = 15x N
Gradient resistance (GR) = weight of train x gradient
As the track is level, gradient resistance is zero.
Curve resistance (CR) = k x weight of train
As the radius of curvature is not given, curve resistance cannot be calculated.
Acceleration resistance (AR) = TE - R
AR = 10.5 - (15x/1000) = 10.5 - 0.15x tonnes

Step 4: Calculation of Maximum Train Acceleration on a Level Track
Maximum train acceleration (a) = AR / (weight of train)
a = (10.5 - 0.15x) / 5x

Substituting the value of x, we get
a = (10.5 - 0.15 x 17.5) / (5 x 17.5) = 0.07 m/s^2

Therefore, maximum train acceleration on a level track is 0.07 m/s^2.

Step 5: Calculation of Maximum Acceleration if Motor and Trailer Coaches are Used in the Ratio of 1:1
If motor and trailer coaches are used in the ratio of 1:1, then weight of motor coach = weight of trailer coach = (52.5 / 2) = 26
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A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.?
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A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A local train uses motor and trailer coaches in the ratio of 1:2. The weight of a motor coach is connes and that of trailer 35 tonnes. All the wheels in a motor coach are driving wheels. The in resistance is 30 N/tonne. Effective rotating mass is 10% of the dead weight. If the coefficient adhesion is 0.2, calculate (a) The maximum train acceleration on a level track. (b) What will be the maximum acceleration if the motor and trailer coaches are used in the ratio of 1:1.?.
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