Determine the tension in the coupling

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SUMMARY

The discussion focuses on calculating the tension in the coupling of a train consisting of 50 cars, each with a mass of 6.6e3 kg, under an acceleration of 1.5e-1 m/s². The correct formula to determine the tension between the cars is derived from Newton's second law, F=ma, where 'n' represents the position of the car in the sequence. For the tension between the 30th and 31st cars, the calculation is (51-n)*ma, resulting in a force of 19800 N. The confusion arises regarding the application of 'n' for the last car, where it is clarified that for the 50th car, the correct calculation is 51-50.

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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?
 
Last edited:

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