Exam revision: Connected objects in motion.

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Above: What do you do if a gigantic kitten attacks the train you're on? That'll be on the exam papers one day. When I'm running the world. Don't laugh, I'm allowed to dream! Image courtesy of imagur.

This is one that keeps coming up over the years, so it seems to be a good topic for a post. It’s a favourite of exam questions and usually goes something like this (but let me be clear; this is an example I've made up, not an actual exam question): 

Question 1: 
We have a train, consisting of an engine and five coaches. The engine has a mass of 5000kg. Each coach has a mass of 3000 kg......

Like this one, although….

.  

…you’re better off to draw yourself a nice simple diagram like this to make working out the question easier.

  
a) If the train accelerates at 1.5 m/s/s then what force does the engine need to generate? 

b) What is the tension between the engine and the first car? 

c) What is the tension between the last car and the preceding car? 

d)
     i) At a certain speed each coach and the engine is feeling 450 N of wind resistance. What force does the engine now have to generate to keep the same acceleration? 

    ii) How does this affect the tension between the engine and the first car? On the face of it this kind of question is straight forwards, but there are a few mistakes that people keep making because of subtle and easily made misunderstandings (coff coff I get them wrong as well coff coff). So let’s work through it one section at a time: 

Answer: 
a) 
This is a nice and simple start: We can use the equation F = MA. This is one of Newton’s laws (if you need to revise them follow this link), and in words it means: Force needed to accelerate the train = (Mass of the train) x (Acceleration) The total mass of the train = 5000 + 5 x (3000) =20,000 kg So the force needed to accelerate the whole train = Acceleration) x (Mass of the train) = 1.5 x 20,000 Force needed = 30,000 N. 

Above: The tension in any link between coaches is equal to the mass of all the coaches behind that link, multiplied by the acceleration. This means the tension in each link gets smaller towards the rear of the train

b) 
This is a point that people fall down on, and it’s because we confuse tension in the link between the coaches with the force needed to accelerate a coach. Many people reason: The tension in the connection between the first car and the engine is what is pulling the engine forwards – therefore the tension is the F (accelerating force) in the F = M x A calculation for that coach. 
So surely we just plug the numbers for that coach into F = M x A? That would give us: F = 3000 x 1.5 F = 4500 N That’s the force needed to accelerate one coach. But the engine is pulling five of them, and enough force has to reach each coach to accelerate it – so enough force to accelerate all five coaches must be passing through the first link (otherwise how can the force get to the other coaches?). Each coach takes up a portion of the tension in order to accelerate, and the rest is passed through it and back to the other coaches. So the tension in each link is the acceleration x the mass of the coaches behind that link.  

Another way to think of this is that each coach is like a sponge for the tension from the engine and soaks up a share of it. The rest of the tension runs through it and back to the next coach. So the 1^st coach, which only needs 4500 N of force to accelerate, takes up a 4500 N portion of the 22500 N of tension in the first link, and the other 18,000 N of tension is passed to the second coach. The 2nd coach takes up its 4500 N portion and passes the rest back to the third, and so on. So to work out the tension in the link between first car and engine we need to use the mass of all 5 coaches behind that link : Mass of all 5 coaches = 3000kg x 5 = 15000 kg We can now use F =M x A to get the tension in that link: Tension in the link = (mass of the five cars) x ( acceleration) = 15,000 x 1.5 =22,500 N Remember to always include the units! 

c) 
This needs the same kind of thinking as (b), but not the same numbers: The force to accelerate the last coach comes from the tension in the link from the second to last coach. But in this case only the last car is behind the link we are considering, so only that mass is being accelerated by the tension in that link. As a result we now really only need to do the F = M x A equation for one car. Force to accelerate last car = 3000 x 1.5 =4500 N. 

d) i) 

This is a simpler question again, because we can think of the train as one big object instead of individual carriages. We already know how much force the engine needs to generate in order to accelerate the train, but now it has an extra 450 N of air resistance from each carriage to overcome as well. If the acceleration is to remain the same the engine needs to produce enough extra force to overcome the air resistance: Total air resistance = 6 x -450 N =-2700N (negative because the air resistance acts opposite to the direction of motion) So the engine must now cancel this out by producing + 2700N of extra force. So total force produced = 30,000 N + 2700N = 32700 N  

 

d ii) 
The last part of the question is also slightly tricky: It might be tempting to think that because the engine cancels out the air resistance fore, there is no extra tension. But the engine cancels the negative air resistance force by adding more force in the positive direction. This extra force needs to be transmitted to each carriage that is producing air resistance, via the tension along the train. So the tension in the train links has to go up as more force is being carried.