Find the tension between two train carts as the train brakes.

[ichivictus]

This was from my last exam. I got this problem wrong. I feel like I should have known how to do it but I blanked out or something. Just coming here to figure out how it's supposed to be done.

(This is from what I remember, numbers may be off from what was on the test)

Q: A train with three carts, each with a 9800 kg mass, is traveling at 14 m/s and the last cart of the train applies brakes with 40,000 N. Find the tension of "A." Treat "A" as a frictionless and extensible rope.

40,000 N <-- (Train1)-(Train2)-(Train3) --> 14 m/s

A is the tension between Train2 and Train3.

I know I got it wrong, but taking another look I think I have to take the net force and subtract all forces till I'm left with the tension. Not entirely sure though.

I was at a blank and tried to equate both sides of "A" and solve for something I think. It looked silly.

Relevant equations would be F=ma and possibly kinematics since all other problems on the test were kinematics. However, I'm convinced that this is just something dealing with subtracting forces. I just can't figure it out.

 

[Simon Bridge]

What you do is draw a free-body diagram for each cart.
Then apply F=ma for each one separately.
Then solve the simultaneous equations.