# Tension in the coupling between the locomotive and the first car

A 97 250kg train pulls two freight cars. The first car has a mass of 51 355 kg and the second has a mass of 18 400 kg. The whole train accelerates at .01750m/s^2. What is the tension in the coupling between the locomotive and the first car?
A. I used Ft=ma and took (51355)(.1750)= 8987.1 N

What is the tension in the coupling between the first and the second freight cars?
a. I took (m1-m2)(.01750)=Ft so (51 355-18400)(.01750)= 576.7 N

I don't know if that is how I was supposed to do it.

also, I wanted to ask if the force of friction is the same on all inclined surfaces (going up), even if a person is trying to go up the hill instead of going down. If a person was going up a 7 degree slope and a force was needed to have constant velocity. Would I still use Ff=umgcosO to find the force of friction and then set it equal to the needed force?

OlderDan
Homework Helper
ownedbyphysics said:
A 97 250kg train pulls two freight cars. The first car has a mass of 51 355 kg and the second has a mass of 18 400 kg. The whole train accelerates at .01750m/s^2. What is the tension in the coupling between the locomotive and the first car?
A. I used Ft=ma and took (51355)(.1750)= 8987.1 N

What is the tension in the coupling between the first and the second freight cars?
a. I took (m1-m2)(.01750)=Ft so (51 355-18400)(.01750)= 576.7 N

I don't know if that is how I was supposed to do it.

also, I wanted to ask if the force of friction is the same on all inclined surfaces (going up), even if a person is trying to go up the hill instead of going down. If a person was going up a 7 degree slope and a force was needed to have constant velocity. Would I still use Ff=umgcosO to find the force of friction and then set it equal to the needed force?
Your answers are not correct. There are a couple of ways to approach this problem. The best is to draw a free body diagram for each car and consider all the forces acting. If you do that correctly, in the end you will realize that another point of view will work for this problem. Each coupling is providing the force to accelerate a certain amount of mass. If you recognize how much mass a cououpling is pulling, you can use your ma calculations to find the tension.

I'm not sure I fully inderstand your last question, but kinetic friction is always opposite the direction of motion. Static friction is in whichever direction parallel to the surface it needs to be to make ths sum of the forces zero.

would the train be pulling itself and the coupling? and the coupling pulling itself and the other coupling?

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OlderDan
Homework Helper
ownedbyphysics said:
would the train be pulling itself and the coupling? and the coupling pulling itself and the other coupling?
The train is providing the force that pulls everything. The first coupling pulls both cars. The second coupling pulls only one car.

This is exactly like the problem of 3 blocks connected by two strings. It really is worth your time to set up the free body diagrams (FBD) and equations for each car or block and then show that by adding equations you arrive at the conclusions I just stated. Some problems are more difficult than this one, and learning how to deal with FBDs will pay off down the road.

My teacher told us that "the force of friction on an inclined plane is always going uphill and is always UmgcosO," but he didn't say if that was just in the case of an object going downhill. When an object is going uphill, is the force of friction against the object or still uphill. If it is against the object, would the formula be Ff=umgsinO?

OlderDan