The EF Express, a train similar to the one shown above, has 3 locomotives (70 to

Question

The EF Express, a train similar to the one shown above, has 3 locomotives (70 tons each) and the direction the train is going. 64 coal cars 48 tons cach). The train is gong down a 1.6% incline Use positve as EXAMPLE EXAMPLE NOTES IMAGES DISCUSS UNITS STATS HELP Part Description What is the mass of the entire train? (include units with answer) Answer 3282kg Format Check A. What is the acceleration of the train if it is allowed to roll down the track (no friction, braking, or tractive force)? B. (include units with answer) Formal Check What is the magnitude of the force in the coupling between the locomotives and the cars while rolling down C. the incline? (include units with answer) Format Check The engines are started and the tractive force provided by each locomotive is 10500 lbs. What is the acceleration of D. the train as it goes down the 1.6% incline? (include units with answer) What is the magnitude of the force in the coupling between the locomotives and the cars while going down the 1.6% incline with a tractive force provided by each locomotive of 10500 lbs? (include units with answer) E. Format Check The engines are turned off so there is no tractive force. The train is moving at 19 mph down the inclinc. What is the required magnitude of the braking force to bring the Format Check train to a stop in 1900 ft? (include units with answer)

Explanation / Answer

given three locomotives, 70 ton each, 64 coal cars, 48 toms each
incline = 1.6 %

A. Mass of train, m = 70*3 + 64*48 = 3282 tonnes = 3282,000 Kg
B. now, tan(theta) = 16/1000
       where theta is the angle of the slope
       theta = 0.9165 deg

       so when the train is allowed to roll without any friciton
       acceleration = g*sin(theta) = 0.1569 m/s/s

C. while rolling down the incline, let the force between the couplings of locomotive and cars be F
       then F + 64*48*1000*gsin(theta) = 64*48*1000*gsin(theta)
       F = 0

D. Tractive force by each locomotive, F = 10500 lbs
   Net tractive force = 3F = 10500*3 lbs = 31500 lbs = 14288.16*9.81 = 140166.8496 N
   so net acceleration = gsin(theta) + 140166.8496/ 3282,000 = 0.199 m/s/s

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