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//Chapter 10:Traction Drives
//Example 5
clc;
//Variable Initialization
Mm=40 //weight of the motor coach in tonne
Mt=35 //weight of the trailer in tonne
u=0.2 //co-efficient of adhesion
r=30 //train resistance N/tonne
//Solution
//(a)when the motor to trailer ratio is 1:2
M=Mm+2*Mt //weight of one unit
Me=1.1*M
Md=40 //weight on the driving wheels
Fm=9810*u*Md //total tractive effort without the wheel
Ft=Fm //at maximum accelaration
alpha=(Ft-M*r)/(277.8*Me) //required accelaration since Ft=277.8*u*alpha*M*r
//(b)when the motor to trailer ratio is 1:1
M=Mm+Mt //weight of one unit
Me=1.1*M
Md=40 //weight on the driving wheels
Fm=9810*u*Md //total tractive effort wihout the wheel
Ft=Fm //at maximum accelaration
alpha1=(Ft-M*r)/(277.8*Me) //required accelaration since Ft=277.8*u*alpha*M*r
//Results
mprintf("(a)Maximum accelaration on a level track is alpha : %.4f kmphps",alpha)
mprintf("\n(b)Maximum accelaration when motor to trailer coaches ratio is 1:1 is alpha : %.3f kmphps",alpha1)
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