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# What is the force in the coupling between the 37th and 38th cars in Newtons?

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A freight train consists of two 8.8 • 105-kg engines and 45 cars with average masses of 4.9 • 105 kg .
m1 = 8.8 • 105-kg     m2 = 4.9 • 105-kg          a = 4.8 4.8 • 10-2 m/s2         f = 7.9 • 105-kg

Nov 21st, 2017

I previously answered the question of how much force is required to pull the entire train. Now imagine a diagram of the coupling between the trains (usually it is just an iron bar). In the diagram, cars 1 - 37th are on the left side and cars 38th - locomotives are on the right side, and the bar in the middle connects cars 37 and 38. Now just looking at cars 37 and 38, the coupling bar between them has two forces pulling on it like a tug of war - the force of the locomotive and cars 38 - 45 and the inertia holding back cars 1 - 37. These two forces must be equal, otherwise the bar would elongate or shrink. Therefore they can be called a single force called tension, even though they have to be mathematically considered as two separate forces. Following the previous example I showed you, you can pretend cars 38-45 and the locomotive are moving as one unit, so they are a sort of locomotive together. Cars 1 - 37 are the object being pulled by the "locomotive" which is actually cars 38 - 45 and the locomotive together. Calculate the weight of cars 1 - 37 and you get 4.9x10^5kg * 37cars = 1.813x10^7 kg total. Since F = ma, we multiply by acceleration to get 1.813x10^7kg * 4.8x10^-2 m/s^2 = 8.7024x10^5 N, or 8.7024x10^5 kg*m^2/s^2.

The final answer is rounded to the same number of significant digits as the input, so we will use two significant digits to round the number to 8.7 x 10^5 Newtons.

Jul 17th, 2014

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Nov 21st, 2017
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Nov 21st, 2017
Nov 22nd, 2017
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