Question

A freight train consists of two 8.1 x 10^4kg engines and 45 cars with average masses of 5.75 x10^4 kg each.

m=8.1x10^4kg

m2=5.75x10^4kg

f=7.25x10^5 N

Part (a): What is the magnitude of the force that each engine must exert backward on the track to accelerate the train at a rate of 5.00x10^-2m/s^2 if the force of friction is 7.6x10^5N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-eficient transportation systems.

Part(b): What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

I ALREADY ANSWERED PAT A NEED HELP WITH B

Answer 1

(a.)

Total mass of car = M = 2*Me + 45*Mc

here, Me = mass of engine

Nc = mass of car

So, M = 2*8.1*10^4 + 45*5.75*10^4 = 2749500 kg

now by newton's law,

Fnet = M*a

given, a = 5.00*10^-2 m/s^2

So, Fnet = (2749500)*5.00*10^-2

Fnet = 137475 N

by force balance on engine in horizontal direction,

Fnet = Fe - fr

here, Fe = force provide by both engine = ??

fr = friction force = 7.25*10^5 N

So, Fe = Fnet + fr = 137475 + 7.25*10^5

Fe = 862475 N

So, force exert by each engine = F = Fe/2 = 862475/2 = 431237.5

F = 4.31*10^5 N

(b.)

Mass attached to coupling in backward direction behind the 37th car is given by,

m = 8*Mc

m = 8*5.75*10^4 = 46*10^4 kg

now friction force on last eight cars is given by,

(47 since there are 45 cars and two engines)

fr ' = 8*(fr/47) = 8*(7.25*10^5)/47

fr ' = 123404.26 N

now by force balance on coupling,

Fnet = F' - fr '

here, Fnet = m*a = (46*10^4)*(5.00*10^-2)

Fnet = 23000 N

So, F' = force in coupling = Fnet + fr '

F' = 23000 + 123404.26

F' = 146404.26 N

F' = 1.46*10^5 N

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