# Tension connecting the cable between the passenger and the engine

## Homework Statement

A train consists of a caboose (1000kg), a passenger car (2000kg), and an engine (2000 kg). If the train has an acceleration of 5.5m/s^2, what is the tension connecting the cable between the passenger and the engine.

a) 27.5kN b) 16.5kN c) 11kN d) 5.5kN e)none of the above

I already know the answer is B, as this was said in class, however I am having trouble reaching the solution of 16.5kN

## Homework Equations

At first I tried using Fnet=ma, and with T1+T2-w=ma,however I did not reach the correct answer.

## The Attempt at a Solution

T1+T2-w=ma
T1-19600=4000(5.5)
T1=41600 N, 41.6 kN

i would try setting up free body diagrams for all of the parts of the train. Then you can set up equations for the sum of the x and y force components for each train part and can combine them to solve for the 2 tensions.

the way i see it.
the total mass of the system, is (the sum of the 3 train parts)
so if these were being pulled by, say, a large group of men with a rope, then
f = ma
force would equal tension.
mass, would be the total mass of the system,
acceleration would be the aceleration 0f 5.5

but, as it is a train, and it is the ENGINE, pulling the other two cariges, the tension in the string between the engine and the following two cariges, is the force that is acting on them to make them accelerate.
using f=ma
we know alread what speed they are accelerating at, and we know their mass (the sum of both these two carriges, behind the engine)

Since there is no resisting force(air resistance etc.), the tension in the cable/joint between the engine and the carriage will be the net force.
this is responsible for accelerating both the carriage and the 'caboose'
Fnet = T = ma.
'm' is the sum of the masses of the 'caboose' and the carriage.