Question

A freight train consists of 4 engines at the front, followed by 27 rail cars. Both the engines and rail cars have the same mass of 690,888 kilograms each. Each engine exerts the same force backward on the track to push the train forward. The total force of friction on the entire train is 513,241 Newtons and is evenly distributed among all of the cars and engines. Even with this friction present, the train engines are able to accelerate the train at 0.058 meters/second2. What is the magnitude of the force in Newtons in the coupling between the engine and the first rail car?

Answer 1

Number of engines in front = 4

Number of rail cars = 27

Total number of engines and rail cars = 4 + 27 = 31

Mass of each each engine or rail car = 690888 kg

Total mass of the freight train (M) = 31 X 690888 = 21417528 kg

Total force of friction on the entire train (F​​​​​​s) = 513241 N

Acceleration of the entire train (a) = 0.058 m/s​​​​​​2​​​

Force exerted by the engines, F = Ma + F​​​​​​s = 2 1417528 X 0.058 + 513241 = 1755457.624 N

Force needed to move each engine or rail car (f) = F/31 = 56627.665 N

Force at the coupling of engines and first rail car = f X 27 = 1528946.962 N