A train of n identical wagons, each of mass m , is pulled by…

A train of $n$ identical wagons, each of mass $m$, is pulled by a force $F$. Neglecting friction, find the tension in the couplings between the wagons.

Explanation of Method

Coupling tensions in a train.

Step 1: Acceleration

\[ a = \frac{F}{nm}. \]

Step 2: Coupling behind the $k$-th wagon

Number the wagons from the front. The coupling between wagons $k$ and $k + 1$ pulls the $n - k$ wagons behind it: $T_k = (n - k)\,m\,a$.

\[T_k = \frac{n - k}{n}\,F\]

$F$ $T_1 = \tfrac{3}{4}F$ $T_2 = \tfrac{2}{4}F$ $T_3 = \tfrac{1}{4}F$ $1$ $2$ $3$ $4$ $T_k = \frac{n - k}{n}\,F$

Example with $n = 4$ wagons, numbered from the front. Each coupling pulls only the wagons behind it, so the tension drops by $F/n$ from one coupling to the next.

Result

\[\boxed{\,T_k = \frac{n - k}{n}\,F\,}\]

Remark

Check. The original result is correct. The first coupling carries almost the whole force, $(n-1)F/n$; the last one only $F/n$.