How to Calculate Forces in a Train System?

[faiziqb12]

Homework Statement

the question exactly states that
a 8000 kg pulls a train of 5 wagons each of 2000 kg along a horizontal path
now if the train exerts a force of 40000 N , and the track offers a friction of then calculate
a_NET FORCE
b_ACCELERATION OF TRAIN
c_FORCE OF WAGON 1 ON WAGON 2

2. The attempt at a solution
a_i can easily calculate the net force by subtracting the frictional force from the total force provided by engine
b_now here arises my confusion
the question considered that the engine is the first wagon but i didnt
so when i filtered my answers on that
the trains acceleration was thought as the acceleration of all the wagons except ... but that doesn't make sense to me... please help?
c_now this question is doing something perfectly alien to me , it is taking the acceleration of all the wagons multiplied with the mass of the engine wagon...please help?

3.here is a link to the question and another for verification

http://www.meritnation.com/cbse/cla...engine-pulls-a-train-of-5-wagons-each-of-2000

http://schools.aglasem.com/1773 {Q.NO 15}

 

[BvU]

OK, reading is an art and rendering an exercise appears to be pretty difficult too. You missed the word locomotive and the friction force value of 5000 N .

The train's acceleration is force / mass and for the mass they add up the whole lot that gets accelerated: five wagons and a locomotive; M total = 18000 kg.
Note that the train's acceleration = the acceleration of everything that is hooked together: the locomotive has to be accelerated too !
I do agree with you that the wording of the given solution is wrong and confusing: "net accelerating force on the wagons = 35000 N" is nonsense.

Now in c answer THEY go off the track completely: the 3.5 m/s² comes out of the blue (perhaps last year's answer ?).

In this calculation, the 8000 kg is not the mass of the locomotive but the mass of four wagons.
They are accelerated to the tune of 1.944 m/s², so the force pulling on wagon 2 is 15500 N.

 

[faiziqb12]

You missed the word locomotive and the friction force value of 5000 N

RATHER IN THE QUESTION THERE IS NOT LOCOMOTIVE THERE IS ENGINE
AND I THINK THAT ITS ENGINE IS IMPLANTED IN THE FIRST WAGON.....
MY INTUTION HERE IS THAT THE ENGINE JUST ACCELERATED THE WAGON AND ITSELF WHILE THE 2ND WAGON GETS ACCELERATED BY THE FIRST WAGON(ENGINE AND WAGON)
HOPE THATS RIGHT

 

[BvU]

Nope. Engine means locomotive. See meaning number 2.

 

[faiziqb12]

Nope. Engine means locomotive. See meaning number 2.

ok...
but am i right in the other thing i stated

 

[RUber]

It should not matter if the engine is in the wagon (total mass 10000kg) or if it is in front of the first wagon (total mass 10000kg). The question is asking to find the force on the linkage between the front 10000kg and rear 8000kg of the system, given that the driving force is in the front but the friction force is distributed by mass.

 

[RUber]

In the posted solution to (c), the problem is restated to imply that the force of 40000N is being applied only to the wagons, not the total force output from the engine. The problem says that the engine "exerts a force of 40000N". Which I assume it is using to accelerate its own weight plus that of the wagons.

If you look at the system as one weighing 18000kg, the rear four cars account for 8/18 of the mass.
40000N*8/18 - 5000N*8/18 is the portion of the force acting on the rear cars. This is in line with what BvU posted above.

In the posted solution, the answer is found by only considering the force applied on the cars, so 4 cars are 8/10 the mass. Which gives the answer 40000N*8/10 - 5000N*8/10 = 28000N.

 

[faiziqb12]

so is the question incorrect

 

[RUber]

No, the posted answer is incorrect. You can't get answer (b) and answer (c) using the same assumptions about where the force is applied.
If the force is applied directly to the cars, you get acceleration of 3.5m/sec^2, otherwise if the force pulls the whole system you get 1.944m/sec^2.
For part c, it really should be as simple as F= ma, and you know the acceleration, since it is the same regardless of where you are on the train.