# Forces, Trains and Friction

## Homework Statement

A locomotive accelerates a 25 car train along a level track. Every car has a mass of 6.8 * 104 kg and is subject to a friction force f = 250v where the speed v is in meters per second and the force f is in newtons. At the instant when the speed of the train is 42 km/h, the magnitude of the acceleration is 0.20 m/s2.

(a) What is the tension in the coupling between the first car and the locomotive?
I got 412875 N
(b) If the tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at 42 km/h?
This is the part I need help with

f = 250*v
v = 11.66 m/s

## The Attempt at a Solution

Ok, for the first part,
Net Force = F - f(riction)
ma=F-f
ma+f=F
because there are 25 cars, each with its own mass and friction...
25(m)a + 25(250v) = F
25(6.8 E4)(0.2) + 25(250*11.66) = F
412875 = F

For the second part, I drew the following diagram, which should show up as an attachment.
I forgot to draw in the theta, but oh well.

Net Force = F - mgsin∅ - f - Ncos ∅ = 0
If the net force is zero, then the velocity is constant, I think...
Going on that assumption...
F-f-mgsin∅-mgcos∅ = 0
F-f = mg(sin∅ + cos∅)
(F-f)/(mg) = sin∅ + cos∅
...and that's how far I got before confusion got me.

Any help is appreciated

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You have counted friction twice... what is the $N\cos(\phi)$ part for?