Determine the tension in the coupling

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To determine the tension in the coupling of a train with 50 cars, each weighing 6.6e3 kg and accelerating at 1.5e-1 m/s², the formula F=ma is applied. For the tension between the 30th and 31st cars, the calculation involves using (51-n) where n represents the car number, leading to a total force of 19800 N. There is confusion regarding the correct application of the formula for the tension between the 49th and 50th cars, with some suggesting that n should equal 2. Clarification indicates that the correct approach is to use 50-n instead of 51-n for the last car. The discussion emphasizes the importance of correctly identifying the car numbers in the calculations.
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A train consists of 50 cars, each of which has a mass of 6.6 e3 kg. The train has an acceleration of 1.5 e-1 m/s2. Ignore friction and determine the tension in the coupling at the following places. (a) between the 30th and 31st cars

(b) between the 49th and 50th cars



F=ma



I got the first part by using the equation: (51-n)*ma = 20*6.6e3*1.5e-1 = 19800. But I can't seem to get part (b). Wouldn't n be 2?
 
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Why 2 ?
The train is 50 cars, the 50th is the last one.
It's really obvious.

Also this is not correct, why 51-n ? 50-n.
(51-n)*ma
 


Quinzio said:
Why 2 ?
The train is 50 cars, the 50th is the last one.
It's really obvious.

Also this is not correct, why 51-n ? 50-n.
(51-n)*ma

so when I had to find the acceleration of the 31st car, I did 51-31= 20, and 20*ma was correct. So if I need the acceleration of the 50th car, would I do 51-50?
 
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